Reeds Marine Engineering and Technology Volume 7: Advanced Electrotechnology for Marine Engineers 9781472987556, 9781408176047

Table of contents :
Cover
Copyright
Contents
PREFACE
ACKNOWLEDGEMENTS
THE S.I. SYSTEM
1 TRANSFORMER OPERATION
Brief Early History of the Transformer
Principle of Operation
The e.m.f. Equation
The Transformer on No Load
Primary phasor diagram
Secondary phasor diagram
Combined phasor diagram
Voltage and turns relationship
The Transformer on Load
On-load phasor diagram
On-load voltage diagram
Referred Values of Resistance, Reactance and Impedance
Transformer Efficiency
Efficiency
2 THE TRANSFORMER (TESTING AND EFFICIENCY)
The Combined Phasor Diagram
The Equivalent Circuit
Voltage Regulation
Internal Voltage-Drop Formula (For Approximation)
Transformer Testing
The open-circuit test
The short-circuit test
Direct-Loading Test (Sumpner’s Test)
Percentage Resistance, Reactance and Impedance
Efficiency
Conditions for maximum efficiency
Conditions for all-day efficiency
3 THE TRANSFORMER APPLICATIONS
Instrument Transformers
The current transformer (C.T.)
The voltage transformer (V.T.)
The Auto-Transformer
Fixed ratio type
Variable ratio type
Three-phase transformation
Methods of connection
The Saturable Reactor
The transductor
The magnetic amplifier
Solid-State Transformers
4 D.C. MACHINES
Testing of D.C. Machines
Output
Efficiency
Losses
Testing methods
D.C. Generators in Parallel
Parallel operation
Load sharing
Commutation and Armature Reaction
Commutation
Armature reaction
Special D.C. Machines
The rotary transformer
The rotary converter
The rotating amplifier
5 D.C. RECTIFICATION
Electromagnetic Induction
Inductance
The Direct Current LR Circuit
Growth of current
Decay of current
The field switch and discharge resistor
The Direct Current CR Circuit
Growth of current
Discharge conditions
The Alternating Currents LR and CR Circuit
Current asymmetry
Rectification
Terms
Rectifier arrangements
Battery charging by rectifier
6 THE A.C. GENERATOR
The A.C. Generator
Rotating-armature type
Rotating-field type
Excitation Arrangements
Rotary excitation systems
Static excitation systems
The Speed-Frequency Equation
E.M.F. equation
Waveform of generated e.m.f.
Stator Windings
Types of windings
The Alternator on Load
7 THE MARINE ALTERNATOR
The Rotating Magnetic Field
Armature reaction
The phasor diagram (continued)
Prediction of voltage regulation
Synchronising torque
8 THE INDUCTION MOTOR
Principle of Operation
Rotor to stator relationships
Relation between rotor loss, rotor input power and rotor output
Torque conditions
The phasor diagram
The Circle Diagram
Explanation of the circle diagram
Testing procedure
Construction of the circle diagram
9 A.C. MACHINES – OPERATION
A.C. Generators
A.C. generators in parallel
The synchroscope
Parallel operation
Load sharing
The Synchronous Motor
Operating action
Starting
Induction Motors
Starting
Speed and torque control
Speed adjustment
10 ELECTRONIC EMISSION PROCESSES AND DEVICES
Electron Emission
Thermionic Emission
The Vacuum Diode
Static characteristics
Dynamic characteristics – load line
The Vacuum Triode
Static characteristics
Valve parameters
Parameter relationships
Ionisation
The hot-cathode discharge lamp (low pressure)
The hot-cathode discharge lamp (high pressure)
The Cathode Ray Oscilloscope
Image intensifiers
Image intensifier device operation
11 SOLID-STATE ELECTRONICS: THE DIODE
Semiconductors
Basic theory
Conduction control
The P-N Junction
The junction diode
Rectifier operation
Rectifier Circuits
Half-wave
Full-wave
Filter Circuit
Voltage Doubler Circuit
Stabilised power supplies
The zener diode
Series Stabilisation
12 SOLID-STATE ELECTRONICS: THE TRANSISTOR
The Transistor
The junction transistor
Transistor characteristics
Load lines
Leakage current
The practical amplifier
The transistor as a switch
The thyristor
13 ELECTRONIC AIDS TO NAVIGATION
Radar
Maritime radar applications
Maximum Detection Range (MDR) and Radar Cross Section (RCS)
Antennas
Signal Processing
Clutter
Noise
Environmental effects
Lasers
Lasers for marine applications
Stimulated emission
Laser safety
Laser range finding and maintaining ships’ proximity in RAS
Automatic Radar Plotting Aid (ARPA)
Electronic Chart Displays
Electronic Chart Display and Information System (ECDIS)
War-fighting ECDIS
Radio Navigation Aids
Interference-based navigation-related sensors
LORAN-C
Limitations of LORAN
Global Positioning Systems
NAVSTAR
GLONASS
GALILEO
Global Maritime Distress Safety System (GMDSS)
Automatic Identification System (AIS)
Emergency Beacons
How a typical beacon operates
Activation
Hydrostatic Release Unit
GPS beacon operation
High-precision registered beacon operation
Non-GPS Doppler location
Satellites
Search and rescue response
14 DISPLAY DEVICES
Cathode Ray Tube
Production of displays
Oscilloscope CRTs
CRT advantages
CRT disadvantages
Liquid Crystal Displays
A brief history of liquid crystals
Liquid crystal phases
Switching speeds
Liquid crystal alignment
Liquid crystal technology
The Plasma Display Device
History
Plasma display advantages
Plasma display disadvantages
Screen burn-out
ElectroLuminescent Displays
History
Material fabrication technologies
Polymer Light Emitting Diodes
Phosphorescent materials
Device structure: Bottom or top emission?
Fabrication
OLED advantages
OLED disadvantages
Outdoor performance
Power consumption
Holographic Displays
SOLUTIONS TO PRACTICE EXAMPLES
SELECTION OF TYPICAL EXAMINATION QUESTIONS
SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS
INDEX
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
R
S
T
V
W
Z

Citation preview

REEDS MARINE ENGINEERING AND TECHNOLOGY

ADVANCED ELECTROTECHNOLOGY FOR MARINE ENGINEERS

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REEDS MARINE ENGINEERING AND TECHNOLOGY SERIES

Vol. 1 Mathematics for Marine Engineers Vol. 2 Applied Mechanics for Marine Engineers Vol. 3 Applied Heat for Marine Engineers Vol. 4 Naval Architecture for Marine Engineers Vol. 5 Ship Construction for Marine Students Vol. 6 Basic Electrotechnology for Marine Engineers Vol. 7 Advanced Electrotechnology for Marine Engineers Vol. 8 General Engineering Knowledge for Marine Engineers Vol. 9 Steam Engineering Knowledge for Marine Engineers Vol. 10 Instrumentation and Control Systems Vol. 11 Engineering Drawings for Marine Engineers Vol. 12 Motor Engineering Knowledge for Marine Engineers Vol. 13 Ship Stability, Powering and Resistance Vol. 14 Stealth Warship Technology Vol. 15 Electronics, Navigational Aids and Radio Theory for Electrotechnical Officers

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7 REEDS MARINE ENGINEERING AND TECHNOLOGY

ADVANCED ELECTROTECHNOLOGY FOR MARINE ENGINEERS

Revised by Christopher Lavers Edmund G R Kraal

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Published by Adlard Coles Nautical an imprint of Bloomsbury Publishing Plc 50 Bedford Square, London WC1B 3DP www.adlardcoles.com Bloomsbury is a trademark of Bloomsbury Publishing Plc Copyright © Christopher Ralph Lavers and Adlard Coles Nautical 1970, 1977, 2014 First edition published by Thomas Reed Publications 1970 Second edition 1977 Reprinted 1979, 1996, 1999, 2002 Reprinted by Adlard Coles Nautical 2006, 2007, 2008, 2010 and 2011 This third edition published by Adlard Coles Nautical 2014 Print ISBN 978-1-4081-7603-0 ePDF ISBN 978-1-4081-7604-7 ePub ISBN 978-1-4081-7136-3 All rights reserved. No part of this publication may be reproduced in any form or by any means – graphic, electronic or mechanical, including photocopying, recording, taping or information storage and retrieval systems – without the prior permission in writing of the publishers. The right of the author to be identified as the author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act, 1988. A CIP catalogue record for this book is available from the British Library. Note: while all reasonable care has been taken in the publication of this book, the publisher takes no responsibility for the use of the methods or products described in the book.

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CONTENTS PREFACE

xi

ACKNOWLEDGEMENTS

xiii

THE S.I. SYSTEM

xiv

1 TRANSFORMER OPERATION Brief Early History of the Transformer Principle of Operation The e.m.f. Equation The Transformer on No Load Primary phasor diagram Secondary phasor diagram Combined phasor diagram Voltage and turns relationship The Transformer on Load On-load phasor diagram On-load voltage diagram Referred Values of Resistance, Reactance and Impedance Transformer Efficiency Efficiency

1 2 2 5 7 7 8 9 10 11 12 15 18 20 21

2 THE TRANSFORMER (TESTING AND EFFICIENCY) The Combined Phasor Diagram The Equivalent Circuit Voltage Regulation Internal Voltage-Drop Formula (For Approximation) Transformer Testing The open-circuit test The short-circuit test Direct-Loading Test (Sumpner’s Test) Percentage Resistance, Reactance and Impedance Efficiency Conditions for maximum efficiency Conditions for all-day efficiency

25 26 29 30 35 38 38 40 42 43 44 45 46

3 THE TRANSFORMER APPLICATIONS Instrument Transformers The current transformer (C.T.) The voltage transformer (V.T.) The Auto-Transformer Fixed ratio type

50 50 51 57 58 58

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vi • Contents Variable ratio type Three-phase transformation Methods of connection The Saturable Reactor The transductor The magnetic amplifier Solid-State Transformers

61 63 65 68 70 70 78

4 D.C. MACHINES Testing of D.C. Machines Output Efficiency Losses Testing methods D.C. Generators in Parallel Parallel operation Load sharing Commutation and Armature Reaction Commutation Armature reaction Special D.C. Machines The rotary transformer The rotary converter The rotating amplifier

82 82 83 84 84 87 98 98 101 104 104 110 115 115 116 116

5 D.C. RECTIFICATION Electromagnetic Induction Inductance The Direct Current LR Circuit Growth of current Decay of current The field switch and discharge resistor The Direct Current CR Circuit Growth of current Discharge conditions The Alternating Currents LR and CR Circuit Current asymmetry Rectification Terms Rectifier arrangements Battery charging by rectifier

119 120 120 121 122 126 128 130 131 133 135 135 137 137 137 148

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Contents • vii 6 THE A.C. GENERATOR The A.C. Generator Rotating-armature type Rotating-field type Excitation Arrangements Rotary excitation systems Static excitation systems The Speed-Frequency Equation E.M.F. equation Waveform of generated e.m.f. Stator Windings Types of windings The Alternator on Load

151 152 152 153 156 157 159 162 163 169 175 176 179

7 THE MARINE ALTERNATOR The Rotating Magnetic Field Armature reaction The phasor diagram (continued) Prediction of voltage regulation Synchronising torque

185 186 193 195 199 203

8 THE INDUCTION MOTOR Principle of Operation Rotor to stator relationships Relation between rotor loss, rotor input power and rotor output Torque conditions The phasor diagram The Circle Diagram Explanation of the circle diagram Testing procedure Construction of the circle diagram

211 212 214 217 218 225 228 230 231 232

9 A.C. MACHINES – OPERATION A.C. Generators A.C. generators in parallel The synchroscope Parallel operation Load sharing The Synchronous Motor Operating action Starting

240 240 241 242 243 250 255 255 258

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viii • Contents Induction Motors Starting Speed and torque control Speed adjustment

260 260 264 268

10 ELECTRONIC EMISSION PROCESSES AND DEVICES Electron Emission Thermionic Emission The Vacuum Diode Static characteristics Dynamic characteristics – load line The Vacuum Triode Static characteristics Valve parameters Parameter relationships Ionisation The hot-cathode discharge lamp (low pressure) The hot-cathode discharge lamp (high pressure) The Cathode Ray Oscilloscope Image intensifiers Image intensifier device operation

272 273 274 274 277 280 283 284 286 287 289 290 291 292 293 294

11 SOLID-STATE ELECTRONICS: THE DIODE Semiconductors Basic theory Conduction control The P-N Junction The junction diode Rectifier operation Rectifier Circuits Half-wave Full-wave Filter Circuit Voltage Doubler Circuit Stabilised power supplies The zener diode Series Stabilisation

298 298 300 302 304 306 309 312 312 313 316 317 317 318 319

12 SOLID-STATE ELECTRONICS: THE TRANSISTOR The Transistor The junction transistor Transistor characteristics Load lines

324 324 325 330 342

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Contents • ix Leakage current The practical amplifier The transistor as a switch The thyristor

13 ELECTRONIC AIDS TO NAVIGATION Radar Maritime radar applications Maximum Detection Range (MDR) and Radar Cross Section (RCS) Antennas Signal Processing Clutter Noise Environmental effects Lasers Lasers for marine applications Stimulated emission Laser safety Laser range finding and maintaining ships’ proximity in RAS Automatic Radar Plotting Aid (ARPA) Electronic Chart Displays Electronic Chart Display and Information System (ECDIS) War-fighting ECDIS Radio Navigation Aids Interference-based navigation-related sensors LORAN-C Limitations of LORAN Global Positioning Systems NAVSTAR GLONASS GALILEO Global Maritime Distress Safety System (GMDSS) Automatic Identification System (AIS) Emergency Beacons How a typical beacon operates Activation Hydrostatic Release Unit GPS beacon operation High-precision registered beacon operation Non-GPS Doppler location Satellites Search and rescue response

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343 343 344 346

352 353 354 357 359 361 361 361 362 362 362 363 364 364 365 366 366 366 367 367 367 368 369 372 372 373 373 374 376 376 377 377 377 378 378 379 379

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x • Contents 14 DISPLAY DEVICES Cathode Ray Tube Production of displays Oscilloscope CRTs CRT advantages CRT disadvantages Liquid Crystal Displays A brief history of liquid crystals Liquid crystal phases Switching speeds Liquid crystal alignment Liquid crystal technology The Plasma Display Device History Plasma display advantages Plasma display disadvantages Screen burn-out ElectroLuminescent Displays History Material fabrication technologies Polymer Light Emitting Diodes Phosphorescent materials Device structure: Bottom or top emission? Fabrication OLED advantages OLED disadvantages Outdoor performance Power consumption Holographic Displays

382 382 384 385 386 386 387 388 391 393 395 396 401 401 402 402 403 404 405 408 408 408 409 409 410 410 410 410 411

SOLUTIONS TO PRACTICE EXAMPLES

414

SELECTION OF TYPICAL EXAMINATION QUESTIONS

493

SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS

498

INDEX

521

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PREFACE This book is companion to Volume 6 (Basic Electrotechnology for Engineers) and covers aspects of theory lying outside the scope of Volume 6. The syllabus covers fully the Department of Trade and Maritime and Coastguard Agency syllabuses’ requirements in Electrotechnology for Marine Engineers, and beyond, and provides practical applications and illustrations to Coastguard, Merchant and Naval students, whether studying for engineering qualifications and certificates, or not. In common with Volume 7 in this series, numerous fully worked problems are included in the text. In addition, test examples and typical examination questions are provided – with solutions, for the students to attempt on their own. As for Volume 6, the subject matter has been treated in the order and in the manner in which it would likely be taught in a maritime college and this book should be viewed as complementary to any lecture notes that should be taken, combined with the use of modern multimedia approaches. It is also for this reason that A.C. and D.C. theories are progressed side by side, with new technology introduced in as chronological sequence as possible. Volume 7 attempts to balance the tension between A.C. and D.C. machines – a tension having its roots in early electrical history. At the close of the 19th century, electricity had revolutionised everyday life, with Thomas Edison’s development of cheap electrical lighting accelerating this transformation. Edison’s initial monopoly of D.C. electrical generation and lighting was soon challenged by millionaire George Westinghouse, who had developed a compressed air system that ensured every railway car in a train would brake simultaneously. Westinghouse believed he could develop a system using A.C. rather than Edison’s D.C. system. However, A.C. technology was then (as now) more complex and little understood, and only progressed significantly after Westinghouse enlisted the creative genius of Nikola Tesla, who had parted company with Edison on less than amicable terms. It is worth considering that A.C. and D.C. provide suitable alternative solutions to electrical transmission and operational device problems, and that a D.C. signal is after all, only an A.C. signal whose frequency is 0 Hz! The creative genius of engineers and inventors like Nikola Tesla is not a product of accident, there is a pattern that may be traced back to the dawn of the industrial revolution. Thomas Newcomen, another such early genius, came from Dartmouth, England (home of Britannia Royal Naval College and of Samuel Lake who pioneered Britain’s first steam trawlers). Newcomen (1663/4–1729) was the first to drive a piston through the agency of steam and created the world’s first steam engine, which from 1717 pumped water from mines and so enabled what we now know as the industrial revolution to start. Newcomen was a committed non-conformist Christian

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xii • Preface and pastor whose creative ability and life rose above the narrow conventional mindset of his day, and was not well treated at this time. Unfortunately today’s climate of political correctness pervades every area of society, an effective mechanism to prevent individuals expressing genuine diversity of views from the state’s sanctioned prevailing ‘correct model’. Social orthodoxy and conformity rarely provide innovation and avenues of new thought and invention. On the other hand those of strong moral conviction, often committed Christians, provide safeguards to society against otherwise ethic-less application of science. If not for the moral and ethical integrity of Sir Michael Faraday who refused to develop chemical weapons for the then battlefields of the Crimea (1853–1856), terrible weapons of the 20th century may otherwise have made their first appearance some 50 years earlier (incidentally on the flood preventing embankment of modern Dartmouth first promoted by Samuel Lake sit several Imperial Canons seized from an otherwise ill-fated Crimea Campaign). Future wars over energy and natural resources may by the mid-21st century dominate the military and political landscape unless suitable energy sources can be found. It is to the innovative engineers and physicists that we look to provide a lead in this quest. Unless Hot or Cold Fusion, the ‘holy grail’ of energy physicists, is achieved we must consider other alternative energy sources which abound in the form of: solar power, wind, geothermal energy, and marine renewables from tidal devices. Such technologies are being explored extensively by Plymouth University, the leading Maritime University in the South West of England, which is providing a pivotal role in this endeavour with large-scale investment in enterprise and research, and as such is at the forefront of future ‘green’ marine global energy provision. You can never cross the ocean until you have the courage to lose sight of the shore. Christopher Columbus

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ACKNOWLEDGEMENTS I would like to thank my wife Anne and family for their ongoing support while completing this Volume, which has taken over 2 years to complete, sometimes subject to the phenomenon of the ‘moving goalpost’. I would also like to express my profound gratitude for those who have mentored and encouraged me in my work over the past 30 years, especially Professor Roy Sambles FRS, who as my Ph.D. supervisor and more recently as president-elect of the Institute of Physics (president from 2015), has profoundly influenced my interest in practical physics and engineering applications generally. Having worked with many physics models it is a rare event to find one that works well, but it is a rarer event indeed to find those we might ourselves wish to model or imitate in part, demonstrating that it is possible to be both a Christian and a scientist in the 21st century without sacrificing academic credibility. Thanks for the example Roy. He stilled the storm to a whisper; the waves of the sea were hushed. Psalm 107.29

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THE S.I. SYSTEM Prefixes, Symbols, Multiples and Submultiples Prefix

Symbol

Units multiplying factor

tera

T

× 1012

giga

G

× 109

mega

M

× 106

kilo

k

× 103

milli

m

× 10−3

micro

μ

× 10−6

nano

n

× 10−9

pico

p

× 10−12

Examples: 1 megawatt (MW) = 1 × 103 kilowatts (kW) = 1 × 106 watts (W) 1 kilovolt (kV)

= 1 × 103 volts (V)

1 milliampere (mA) = 1 × 10−3 ampere (A) 1 microfarad (μF)

= 1 × 10−6 farad (F)

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The S.I. System • xv

Physical Quantities (Electrical), Symbols and Units The table has been compiled from recommendations in B.S. 1991 and the List of Symbols and Abbreviations issued by the I.E.E. Quantity

Symbol

Unit

Abbreviation of unit after numerical value

Force

F

newton

Work or Energy

W

joule or newton metre

Torque

T

newton metre

Power

P

watt

W

Time

t

second

s

Angular velocity Speed

ω (omega)

radians per second

N J or Nm Nm

rad/s

N

revolutions per minute

rev/min

n

revolutions per second

rev/s

Electric charge

Q

coulomb

C

Potential difference (p.d.)

V

volt

V

Electromotive force (e.m.f.)

Ɛ

volt

V

Current

I

ampere

A

Resistance

R

ohm

Resistivity (specific resistance)

ρ (rho)

ohm-metre

Ω (omega) Ωm

Conductance

G

siemens

S

Magnetomotive force (m.m.f.)

F

ampere-turn

At

Magnetic field strength

H

ampere-turn per metre or ampere per metre

Magnetic flux

Φ (phi)

weber

Magnetic flux density

B

tesla

Reluctance

S

ampere-turn or ampere per weber

Absolute permeability of free space Absolute permeability

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At/m A/m Wb T At/Wb or A/Wb

μ0 (mu)

henry per metre

H/m

μ

henry per metre

H/m

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xvi • The S.I. System

Quantity

Symbol

Unit

Abbreviation of unit after numerical value

Relative permeability

μr

–

–

Self inductance

L

henry

H

Mutual inductance

M

henry

H

Reactance

X

ohm

Ω

Impedance

Z

ohm

Ω

Frequency

f

hertz

Hz

Capacitance

C

farad

F

ε0 (epsilon)

farad per metre

F/m

Absolute permittivity

ε

farad per metre

F/m

Relative permittivity (dielectric constant, specific inductive capacity)

εr

–

Electric field strength electric force

E

volt per metre

Absolute permittivity of free space

Electric flux

Ψ (psi)

coulomb

–

V/m C C/m2

Electric flux density, electric displacement

D

coulomb per square metre

Active power

P

watt

W

Reactive power

Q

volt ampere reactive

VAr

Apparent power

S

volt ampere

VA

Phase difference

φ (phi)

degree

°

Power factor (P.F.)

cos φ

–

–

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1

TRANSFORMER OPERATION The transformer, ‘the heart of the alternating current system’. William Stanley Jr

A brief introduction to the static transformer principle was made in Volume 6, Chapter 6, but it is vital to stress the importance of understanding its operating principle and applications as these devices are of practical use to marine engineers. This device in its various forms and its applications will be discussed in some detail in the next three chapters. A transformer is an essential piece of equipment providing electrical energy in an A.C. current form, and it is often classed as a machine because, although it has no moving parts, it is capable of converting energy. However, unlike a generator, it receives the energy in an electrical and not a mechanical form; nevertheless energy conversion takes place, in that, electrical energy is given out at a higher or lower voltage than that in which it is received. If the output voltage is greater than the input voltage it is considered to be a step-up transformer. A transformer whose output voltage is less than its input voltage is known as a step-down transformer. Not wanting to state the obvious the transformer, although using Faraday’s laws raises or lowers A.C. voltage it cannot increase power, so that if voltage is raised, the current is proportionally lowered and vice versa. Since operation does not involve rotation of any armature, field system or commutator, rotational and windage losses do not occur and its efficiency is thus high. For electrical ‘power’ purposes, i.e. transformers operating at 50 or 60 Hz, iron cores are essential and iron losses will occur. Winding copper losses are also present when current is supplied, nonetheless the transformer is the most efficient of electrical machines and a fullload efficiency of 95.5% for units of 5 kVA and 97.5% for units up to 1 MVA may be achieved.

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2 • Advanced Electrotechnology

Brief Early History of the Transformer The property of induction was discovered in the 1830s and demonstrated by early engineers such as Michael Faraday, but it was not until 1886 that William Stanley working for Westinghouse built the first mass-produced affordable commercial transformer. William Stanley, Jr (1858–1916) was an American physicist born in Brooklyn, New York, who over his long career with Westinghouse obtained 129 patents covering a variety of electric devices. The need for the transformer arose out of the problems that were being encountered at this time in terms of transmission of power. D.C. power was mainly used in the 1880s and it was difficult to achieve over any great distance because of either practical safety issues using high voltages on thin wires or low voltages on very thick ones. However, with A.C. systems you can use a high voltage to transmit the electrical current down very long wires and then use a device (the transformer) to drop or step the voltage down to more manageable voltage levels. In reality, power can be stepped down several times on power distribution systems. Stanley first demonstrated a full A.C. power distribution system using both step-up and step-down transformers in 1886 in Great Barrington, MA, followed by development of the first three-phase transformer in Germany by Russian-born engineer Mikhail Dolivo-Dobrovolsky in 1889, two essential factors of electrical distribution systems used to this day, in land, sea and air applications.

Principle of Operation We begin our discussion with the ‘traditional’ view of the transformer, introducing the ‘phasor’ approach before finishing our discussion with modern versions of the device once the basic principle of operation is understood. Since the transformer functions using the principle of mutual induction, it is beneficial to revise briefly some of the earlier theoretical aspects. When an A.C. current passes through a solenoid, an A.C. flux is produced which is in phase with the current, i.e. if the former varies sinusoidally then the flux will also vary sinusoidally. This relationship applies specifically if the solenoid is air-cored; however, in order to use the magnetising ampere-turns to maximum advantage, an iron core is needed. Provided the iron is not worked at too high a flux density, i.e. does not saturate, working conditions are similar to those for air. In basic theory, a condition of flux-linkages exists, and as the current varies

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Transformer Operation • 3 in a sinusoidal manner the flux-linkages will vary similarly and an e.m.f. will be induced through self-induction. The e.m.f. magnitude and direction will be in accordance with the laws of electromagnetic induction, which will be used later to develop a formula applicable to transformer operation. Of course the effect of mutual inductance is to cause the primary coil to take more power from the electrical supply in response to an increased load on the secondary. Consider next a second coil, isolated from the first but wound on the same iron core. The first or primary coil of insulated wire, when carrying an A.C. current, produces the accompanying flux which now links with the secondary or second coil of insulated wire. Since the flux varies, not only the flux-linkages associated with the primary, but also those associated with the secondary vary and again an e.m.f. is induced through mutual induction. As for the primary, the e.m.f. magnitude and direction can be deduced from first principles as shown in Volume 6. The transformer can be considered to be an arrangement of two or more coils, electrically separated but linked by the common thread of a magnetic circuit. If an A.C. voltage is applied to the primary causing an A.C. current to flow, by mutual induction an A.C. voltage is generated across the secondary. If the secondary circuit is completed, a current will flow and energy will transfer from the primary to the secondary. Figure 1.1 illustrates the basic arrangement of the primary and secondary windings and the iron core. Although this diagram is used for explanatory purposes, it is noted that this is an impractical arrangement. For efficient working, primary and secondary are never placed on separate limbs of the core. The ‘ideal’ transformer of course neglects losses due to resistive heating – the primary coil assumes perfect (unity) coupling to the secondary (i.e. no magnetic losses), factors which of course are not really true! Although practical transformer design and construction are best dealt with in specialist books on the subject, an illustration of a single-phase, air-cooled, marine-type transformer is given, with the main parts labelled (figure 1.2). Iron core

Primary Secondary

▲ Figure 1.1

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4 • Advanced Electrotechnology Laminated iron circuit

Secondary small gauge insulated wire or strip

Packing strips or insulating material Primary heavy gauge insulated copper wire or strip

▲ Figure 1.2 Φ V1

V1

E1

I I

Φ

π 2π

E1 1/f sec

▲ Figure 1.3

In accordance with the operating principle already described, the conditions now considered are those which occur when the primary winding is ‘energised’ by connection to the supply voltage V1. The secondary winding is neglected for present. Let I be the resulting current which sets up a flux Φ, of maximum value Φm. Figure 1.3 illustrates the waveform relationships for the conditions considered. I and Φ are shown in phase as discussed earlier and an induced e.m.f. E1 lags the flux by 90° in accordance with Faraday’s law which states that the value of induced e.m.f. is proportional to the rate of change of flux-linkages. Since the maximum rate of change of flux occurs when the wave passes through its zero value, the maximum value of induced e.m.f. will occur at this instant. According to Lenz’s law the induced e.m.f. E1 must at all times be equal yet opposite to the applied voltage V1 which produces the magnetising current I and the resulting flux Φ.

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Transformer Operation • 5 The voltage waveform phases are shown and approximate phasors drawn to produce a phasor diagram with the flux used as the reference. The diagram shows I and Φ in phase, with E1 lagging Φ by 90°. V1 is 180° out of phase with E1 or 90° ahead of Φ.

The e.m.f. Equation Let the frequency of the applied voltage be f hertz and of r.m.s. value V1 volts. Let N1 be the number of turns of the primary winding and N2 the number of turns of the secondary winding. Φ is the flux linking these windings, the maximum value of which is Φm. 1 In a quarter of a cycle or seconds, the flux falls from a value of Φm weber to zero. 4f From Faraday’s law, the average e.m.f. induced value per turn = the rate of change of Φ −0 flux-linkages = m = 4f Φ m volts . 1 4f Similarly the average value for N1 turns is Eav = 4Φm fN1 volts. As sine-wave conditions are considered it is known that the r.m.s. value is 1.11 times the average value, so the r.m.s. e.m.f. induced in N1 turns for a sine-wave voltage is: E1 = 4.44Φm fN1 volts. Again since the induced e.m.f. is equal and opposite to the supply voltage then: E1 V1 = 4.44 44 Φ m fN1 volts. The same flux is associated with the secondary winding so that by similar reasoning, it can be shown that the e.m.f. induced in this winding is given by: E2

f 2 volts. fN

An approach to some proofs using calculus was made in Volume 6. Experience has shown that some students prefer this method and the transformer equation can be derived in this alternative manner. If flux variation is assumed to be sinusoidal, it can be shown that φ = Φm sin ω t. From Faraday’s law e

N1

dφ (Reference – Volume 6, Chapter 6.) dt

The e.m.f. induced in the primary winding (Lenz’s law) is expressed as: e1

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N1

d(

m

dt

ωt)

volts

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6 • Advanced Electrotechnology The – ve sign indicates that the self-induced e.m.f. opposes the change of current and flux. After the required differentiation: e1 = – N1 ω Φm cos ωt volts or e1 = N1 ω Φm sin(ωt – 90°). Thus the induced e.m.f. takes the form of a sine wave lagging behind the flux wave by 90°. The maximum value of e occurs when cos ωt = 1 and this condition can be written as: Em1

π ffN1Φ m

Remembering that ω = 2πf and the r.m.s. value of a sine wave is 0.707 Em = or

Em 2

,

then: E1 =

2π fN1Φ m 2

= 4.44 Φ m fN1 volts as deduced.

Example 1.1. The core of a single-phase, 6600/440 V, 50 Hz transformer is of square cross-section, each side being 160 mm. If the maximum flux density in the core iron does not exceed 1.4 T, find the number of turns required for each winding. Since Flux = Flux density × Area Flux = 1.4 × 160 × 10–3 × 160 × 10–3 = 1.4 × 25 600 × 10–6 = 1.4 × 2.56 × 10–2 = 3.584 × 10–2 weber. To avoid saturation of the iron, the Φm value must not exceed 0.03584 Wb or 35.85 mWb. Thus for the primary winding at V1: 6600 = 4.44 3.584 × 10 −2 × 50 × N1 or N1 =

6600 6600 = = 829.6 (say 830 turns) 2.22 3.584 7.956

For the secondary winding at V2: N2 =

440 = 55.33 (say 55 turns) 2.22 3.584

As a check it can be deduced that: N2 V2 830 × 440 2 = or N2 = = 83 × = 55.33 (say 55 turns) N1 V1 6600 3

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Transformer Operation • 7

The Transformer on No Load The conditions of working on no load, i.e. with the secondary circuit ‘open’, can be investigated if suitable instruments are connected in the circuit. Since a temperature rise of the iron core will be noted soon after energising the primary, it is clear some electrical energy is taken from the supply and lost (converted into heat). A wattmeter should be included in the test circuit together with a voltmeter and an ammeter. Figure 1.4 shows the arrangement.

W A V1

V2

▲ Figure 1.4

Primary phasor diagram The transformer phasor diagram has been introduced and the test instruments’ readings are used to explain the diagram. The no-load current will be the ammeter reading I0 and the wattmeter reading indicates the power input P0 V1I0 cosφ 0 . From such readings it is deduced that the no-load current I0 is resolved into a power component I0 cosφ 0 (the working component), i.e. responsible for the unwanted heat, and regarded as a loss. I0 cosφ0 is denoted by Iw, the power component in the phasor diagram, which is in phase with the voltage. The no-load phasor diagram is given in figure 1.5 and is considered separately for primary and secondary windings. The primary phasor diagram also shows the other component of the no-load current I0. This component will be I0 sinφ 0 , i.e. a reactive or wattless component, which, being at right angles to the voltage, and in phase with the flux Φ, can be regarded as producing the latter. Thus I0 sinφ 0 is denoted by Im and termed the magnetising component. The following can be deduced: I0

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Iw2 + Im2 where Iw = I0

0

d Im = I0 sin i φ0 .

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8 • Advanced Electrotechnology V1

(a)

Iw

(b)

I0 Im

Φ

Φ

E2

E1

▲ Figure 1.5

We also have V1I0 cosφ 0 = V1Iw = heat loss due to the iron loss in the core + a small copper loss in the primary. It is noted that as the no-load current is so small and the resistance of the winding so low, copper loss can be neglected and the input power taken as a measure of the iron loss which remains constant over the transformer’s load range (assuming a constant V1 and ƒ at any load), which in turn ensures a constant В value of flux density.

Secondary phasor diagram The illustrations of figure 1.5 show two phasor diagrams. The first is for the primary winding, that is for E1 drawn dotted (because it is usual to omit this). The second diagram of figure 1.5 is for the secondary winding and the induced e.m.f. E2 drawn at right angles to the common core flux Φ. Unlike the primary since no supply voltage is applied, no anti-phase voltage phasor is drawn, i.e. there is no phasor corresponding to V1. When the secondary circuit is completed, the induced e.m.f. becomes an active voltage responsible for the current through the connected load and may be expressed as V2 (the secondary terminal voltage). Thus on no-load V 2 = E 2. Example 1.2. For the no-load test on a transformer, an ammeter was found to read 0.18 A and a wattmeter 12 W. The reading on the primary voltmeter was 400 V and that on the secondary voltmeter 240 V. Calculate the magnetising component of the no-load current (2 decimal places), the iron loss component (3 decimal places) and the transformation ratio.

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Transformer Operation • 9

Iron-loss component Iw =

P 12 = = 0 03 A V1 400

I02 − Iw2 = 0.182 − 0.032

Magnetising component Im

= 10 −1 1.82 − 0.32 = 10 −1 3.24 − 0.09 = 10 −1 3.15 = 10 10 −1 × 1.775 = 0.178 A Transformation ratio =

400 5 = or 1.66 : 1 240 3

Combined phasor diagram From the secondary diagram it is observed there is a 180° phase shift between the primary supply voltage V1 and the secondary induced e.m.f. E2 (the secondary terminal voltage V2). The phasor E1, though explaining the transformer’s action, is not essential to the phasor diagram and if omitted, can be replaced by E2. The primary and secondary diagrams are thus combined with flux, Φ being used as the reference phasor to show the primary and secondary phase relationships. This arrangement is better, as it will be seen later that any effects in one winding will affect the other. Figure 1.6 shows the combined phasor diagram for a transformer with a 2:1 turns ratio. The phasor for voltage V2 is thus drawn half the length of the phasor for voltage V1. V1

Iw

I0 Im

Φ

V2

▲ Figure 1.6

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10 • Advanced Electrotechnology

Voltage and turns relationship Since the induced e.m.f.s in the primary and secondary windings are in accordance with the e.m.f. equation then: E1

m

f 1 a d E2 = 4.44 Φ m fN2 fN

Since E1 = V1 and E2 can be written as V2 then: V1 4.44 Φ m fN1 V N = or 1 = 1 V2 4.44 Φ m fN2 V2 N2 Attention was drawn to this relationship earlier but is repeated here because of its similarity to a current relationship, shortly deduced. Example 1.3. A 440 V transformer has 3000 turns on the primary. If a tapping is to be made available for 400 V, find its position on the winding. If one secondary winding suitable for 200, 220 and 240 V is also to be provided, find the number of turns and the tapping positions. Since voltage is proportional to turns then: For the Primary

440 400

3000 3000 × 400 30 000 or N = = = 2727.2 N 440 11

The tapping should be 3000 – 2727 or 273 turns from the end which is remote from that made common for 440 and 400 V. Figure 1.7 shows the arrangement. For the Secondary

V2 V1

N2 220 × 3000 or N2 = = 1500 turns N1 440

440 V 400 V 2727 turns

1364 turns

273 turns

136 turns

200 V 220 V 240 V

▲ Figure 1.7

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Transformer Operation • 11 Thus 1500 turns will give a voltage of 220 V. For 200 V, the turns required will be half that is needed for 400 V =

2727 = 1364 2

Thus 1500 – 1364 = 136 turns, making a difference of 20 V. The turns required for 240 V will be 1500 + 136 = 1636, for a fully tapped secondary. The secondary consists of 1636 turns with tappings at 136 and 272 turns from the end which is remote from that made common for 240, 220 and 200 V.

The Transformer on Load This condition occurs when the secondary transformer circuit is completed and current flows. As the secondary now acts as a generator, the secondary current magnitude will be decided by the supplied load impedance. Assume a current of I2 is to be produced by the terminal voltage V2. The passage of I2 through the secondary winding turns constitutes demagnetising ampere-turns which result in the flux, shown in figure 1.8 as Φ L2 . This secondary flux links with the primary and by Lenz’s law produces a demagnetising effect which results in the core flux Φ reducing. Flux reduction through the primary results in a small reduction of the primary induced e.m.f. E1. The effect of the supply voltage V1 then produces a larger primary current, most of which counters the secondary current’s demagnetising effect. The increase in I1 due to I2 flowing, caused by the greater difference between V1 and E1, is made up of an extra primary-current component I1 which will add to the no-load current I0 which continues to flow to produce the components Im and Iw. The new primary current I1 consists of I1 and I0, and a phasor summation is shown.

Φ

I1

V1

ΦL 1

I2

ΦL 2

V2

L o a d

▲ Figure 1.8

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12 • Advanced Electrotechnology Returning to the demagnetising effect of the secondary flux Φ L2 , it is remembered that the core flux value Φ is constant, and decided by the expression: V1

ffN N1

If V1, ƒ and N1 are constant, Φm and Φ must be constant, so any secondary demagnetising effect of I2N2 must be offset by the magnetising effect of the extra primary current. The resulting primary magnetising ampere-turns I1N1 restore the main flux to its original value by producing an extra primary flux L2 . Since Φ L1 must equal Φ L2 then I1′N1 must equal I1′N2 or I1′N1 = I2 N2 . The full representation of the flux conditions is shown in figure 1.8, but an explanation of the phases of these fluxes is needed to show the complete action. Fluxes Φ L1 and Φ L2 are equal in magnitude and as they nullify the effects of each other on the non-related windings, it follows that they must be in anti-phase and oppose each other in the main iron core. They are thus forced out into the air, as shown, for most of their magnetic circuit paths and are called Leakage Fluxes, hence the suffix L1 and L2 as used for L1 and Φ L2 . Also since the magnetic circuits’ reluctances are due mainly to air paths, fluxes are proportional to the energising currents and in phase with them, as shown on the onload phasor diagram.

On-load phasor diagram To follow the development of the diagram it is essential to understand the theory of transformer working. It is worth summarising transformer action by considering its similarity to that of a D.C. motor-generator set. If the generator is loaded the motor will immediately draw more current from the supply, and as load is applied the speed falls. The motor’s back e.m.f. falls accordingly, and the supply voltage ensures more current is supplied until a condition is attained when, the power put in is sufficient to provide the output power plus the power required to overcome any losses. For the transformer, the output current causes a demagnetising effect on the core which lowers the primary’s back e.m.f. Here the supply voltage ensures that extra current is supplied until the balance condition is attained, when input power equals the output power plus the power required to overcome the losses. Although complex, the phasor in figure 1.9 represents the conditions occurring when an inductive load is supplied, which is the most common case. Assume a secondary current of I2 for a load operating at a P.F. of cos φ2. On the diagram for the secondary side, I2 is drawn lagging V2 by an angle φ2 and, as before, a transformation ratio of 2:1 is assumed. On the primary side, the original no-load current conditions are shown but in

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Transformer Operation • 13 V1 I 1′

φ1

I1 ΦL 1

φ2 Iw ΦL

Io φ2

Φ

Im

2

V2

I2

▲ Figure 1.9

addition we have I1′ , which is in anti-phase with I2, lagging V1 by an angle equal to φ2. It is drawn half the size of I2 because of the transformation ratio. It is seen that I1 is the phasor sum of I0 and I1′ but I0 is drawn larger than it would be for an actual practical transformer. Although I1′ N = I2N2, for purposes of approximation, we write I1N1 = I2N2 because the noload current I0 is small enough to be neglected when the transformer is on load. The assumption of I1 I′ I1 is acceptable for practical purposes and I0 will be omitted from most phasor diagrams from now on. The current/turns relationship deduced shortly is made on this basis. Phasor diagram shows Φ L2 drawn in phase with I2 and Φ L1 in phase with I1. Both Φ L2 and Φ L1 are small in comparison to Φ because, although they are produced by currents of load magnitudes compared to the very small value of Im, the magnetic circuits are through air, while Φ is through iron. Leakage fluxes can be drawn in phase with the energising currents but in the case of the primary, I1′ and I1 are assumed identical, I0 being neglected, no error is introduced if Φ L1 is assumed to be caused by I1 and drawn in phase with it. THE CURRENT AND TURNS RELATIONSHIP. Theory of operation shows that the demagnetising ampere-turns of the secondary are equal and cancel the magnetising ampere-turns produced by the primary. Thus I2 N2 = I1′N N1 Since I1 is assumed equal to I1′ − I0 being neglected, the usual accepted expression is: I1N1 = I2 N2 or

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I1 N2 = I2 N1

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14 • Advanced Electrotechnology

Inversion of the proportionality is noted, i.e. unlike that for voltage where

V1 N1 = V2 N2

Example 1.4. A single-phase transformer with a ratio of 440/200 V takes a no-load current of 8 A at 0.25 P.F. (lagging). If the secondary supplies a current of 220 A at 0.8 P.F. (lagging), estimate the current taken by the primary from the supply (1 decimal place). Load component of primary current I1′=

I2 × N2 I × V 220 × 200 or I1′= 2 2 = = 100 A N1 V1 440

Resolving I1′ and I0 into horizontal and vertical component the resultant is obtained with the accepted phasor summation method: Sum of vertical components I V

I1′ cos φ 2 + I0 cos φ = 100 × 0.8 + 8 × 0.25 2 = 80 + 2 82 A

Sum of horizontal components IH

I1′ sinφ 2 + I0 sin i φ0 = 100 × 0.6 8 × 0.97 = 60 + 7.76 = 67.76 A

So I

IH2 + I V 2 = 822 + 67.762 = 10 8.22 + 6 782 = 10 67.24 + 45.97 = 10 113.21 10 10 64 = 106.4 A

THE kVA RELATIONSHIP. The deductions from the voltage, current and turns relationships are combined to produce the following: From the voltage/turns relationship

V1 N1 = V2 N2

From the current/turns relationship

I2 N1 = I1 N2

By equating

V1 N1 I2 V I = = or 1 = 2 or V1I1 = V2 I2 V2 N2 I1 V2 I1

This can be written as: Primary volt-amperes = Secondary volt-amperes Dividing by 1000

Primary VA Secondary VA = or Primary kVA = Secondary kVA 1000 1000

Thus there is only one kVA transformer rating, applicable to both primary and secondary.

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Transformer Operation • 15 Example 1.5. If the transformer iron core of example 1.1 is used for 50 kVA operation, estimate the wire area used for primary and secondary windings. The wire should not work at a current density above 2.33 A mm2, to ensure a safe working temperature in mm2 (2 decimal places). Data already known about this transformer from example 1.1. Primary voltage 6600 V Turns 830 Secondary voltage 440 V Turns 55 Estimated primary current I1 =

50 000 250 = = 7.576 A 6600 33

50 000 1250 = = 113.636 A 440 11 IV or secondary current from I1V1 = I2V2 gives I2 = 1 1 V2

Estimated secondary current I2 =

whence I2 = 7.576 ×

6600 = 7.576 × 15 = 113.64 A 440

Cross-section of primary = Cross-section of secondary =

7.576 = 3 25 mm2 2 33 113.64 = 48.77 mm2 2 33

On-load voltage diagram Since the diagram involves voltage phasors it will be incomplete unless all e.m.f. or voltage drop sources are accounted for. Any voltage drops are ‘internal’ as their effect is only evident when a transformer is on load, due to the normal causes of A.C. circuit current limitation: namely, resistance and reactance. Their existence and effects on transformer operation are now examined. INTERNAL VOLTAGE DROPS: RESISTANCE. The primary and secondary windings’ ohmic resistances are independent of each other and determined by length, area and the wires’ specific resistance. Since currents passing through these windings are also dissimilar in value, the primary and secondary voltage drops should be considered separately first and then co-related if a suitable relationship can be found. A transformer is represented as an idealised unit with all winding resistances contained in external units R1 and R2 as shown. Figures 1.10 and 1.11 show the arrangement and phasor diagram discussed respectively.

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16 • Advanced Electrotechnology R1

I1

R2 V ′2

V′1

V1

I2 V2

▲ Figure 1.10 V1

I1 R1 V′1

I1

Φ

V2 V′2

I2 R2

I2

▲ Figure 1.11

Passage of a load current causes voltage drops I1R1 and I2R2 which, being due to resistance, are in phase with the current. These voltage drops, if considered with the primary and secondary terminal voltages, are like the armature voltage drops in a motor-generator introduced to understand transformer working. Thus for a generator or secondary of a transformer, as terminal voltage is less than the generated voltage by an internal voltage drop, it follows that the latter must be subtracted from the generated voltage. For the transformer a construction is involved, V2 V2′ − I2 R2 the generated or induced e.m.f. is V2 and the voltage-drop phasor I2R2 drawn in phase with the current. Note the slight phase displacement of terminal-voltage phasor V2 from the original no-load position V2. Similarly for the primary, as for a motor, the supply voltage V1 must overcome the induced e.m.f. V1 and armature internal resistance or, in this case, the primary winding.

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Transformer Operation • 17 Here V1 V1 − I1R1 and an allowance is made for phasor construction, I1R1 is drawn in phase with I1. INTERNAL VOLTAGE DROPS: REACTANCE. When the secondary circuit is completed and current flows, the demagnetising secondary ampere-turns I2N2 are countered by the magnetising primary ampere-turns I1N1. This was described earlier, where the 2 ampereturn effects or m.m.f.s set up fluxes which oppose each other and take the path shown in figure 1.11. These fluxes are termed ‘leakage fluxes’ and being mainly through air are in phase with the currents producing them. Fluxes cut the coils which produce them and being alternating, result in self-induced e.m.f.s in quadrature with the fluxes or energising currents. An A.C. current, through the medium of flux, produces a back e.m.f. or Eb IIX . Thus no error is proportional to the current. This e.m.f. is written as Eb I o introduced in crediting the winding with a current limiting factor added to the resistance and responsible for the voltage drop. This factor or quantity is the leakage reactance and, provided the phase of its voltage drop with respect to the producing current is given, the terms reactance, resistance and impedance can be used for primary and secondary. The on-load diagrams (figures 1.12 and 1.13) take into account both resistance and reactance voltage drops. Points to note are: (1) The secondary can be viewed as a generator and the internal voltage drop treated by phasors to give the terminal voltage on load. Thus V2 V2 − I2 R2 − I2 X 2 where V2 is the open-circuit induced voltage, I2R2 is the resistance voltage drop in phase with I2, I2X2 is the leakage reactance voltage drop in quadrature with I2, and V2 the resulting terminal voltage. The shaded triangle is called the secondary impedance triangle and the phasor V2V2′ is the secondary voltage drop given by I2 Z 2 where Z 2

R22 + X 22 .

(2) The primary is viewed as a motor whose supply voltage V1 must overcome all internal voltage drops which add as phasors. Thus V1 V1′+ I1 R1 + I1 X . On no-load V1 coincides with V1′ but on load there are internal voltage drops due to resistance I1R1 in phase with I1 and the leakage reactance I1X1 in quadrature with I1. These are added as phasors to the primary induced voltage V1′V1′ produced by the main flux Φ. I1 R1 V1

X2

X1 V′1

V ′2

R2

I2 V2

▲ Figure 1.12

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18 • Advanced Electrotechnology I1 X1

V1

I1 R1

I1 Z 1 V′1

I1 φ1 Φ φ2

V2 I2 Z 2

I2 R2 I2 X2

V′2

I2

▲ Figure 1.13

The shaded triangle is the primary impedance triangle and V1V1′ is the impedance voltage drop: I1Z1 , where Z1 = R12 X12 There exists a practical range of resistance for operation of a transformer to achieve maximum output power. With too large a load drawn from the secondary the primary coil can heat up considerably, sometimes denoted by an increased audible ‘hum’ from the transformer (which is of course further wasted energy).

Referred Values of Resistance, Reactance and Impedance Since the voltages at the primary and secondary are related by the turns ratio, any secondary voltage drop can be considered in terms of the primary side if the turns ratio is applied. The same reasoning may be applied to a primary voltage drop if referred to the secondary. It is convenient here to deduce the equivalent resistance and equivalent reactance for one winding in terms of the other winding. Consider the resistance voltage drop in the primary given by I1R1. Then if the turns ratio ⎛N ⎞ is applied this gives an equivalent voltage drop in the secondary namely: I1R1 ⎜ 2 ⎟ ⎝ N1 ⎠

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Transformer Operation • 19

But I1

⎛N ⎞ ⎛N ⎞ I2 ⎜ 2 ⎟ so the primary voltage drop in secondary terms is written as I2 R1 ⎜ 2 ⎟ ⎝ N1 ⎠ ⎝ N1 ⎠

2

The secondary voltage drop due to the secondary current and resistance is given by: I2R2 Thus the total resistance voltage drop of the primary referred to the secondary is: ⎛N ⎞ I2 R2 + I2 R1 ⎜ 2 ⎟ ⎝ N1 ⎠

2

2 ⎧⎪ ⎛ N ⎞ ⎫⎪ or = I2 ⎨R2 + R1 ⎜ 2 ⎟ ⎬ ⎝ N1 ⎠ ⎪⎭ ⎪⎩

The primary is regarded as having no resistance and the secondary as having resistance: 2 ⎛N ⎞ R2 R1 ⎜ 2 ⎟ . This total resistance value is called the Equivalent Resistance written R2 ⎝ N1 ⎠ ⎛ N2 ⎞ The transformer’s Equivalent Resistance, referred to the secondary is: R 2 = R2 + R1 ⎜ ⎟ ⎝ N1 ⎠ ⎛N ⎞ Similarly the Equivalent Resistance referred to the primary side is: R1 = R1 + R2 ⎜ 1 ⎟ ⎝N ⎠

2

2

2

By considering the reactance voltage drops, a similar deduction for Equivalent Reactance is produced. Thus for the primary, the reactance voltage drop is I1X1, equivalent to a ⎛N ⎞ voltage drop on the secondary side given by I1 X1 ⎜ 2 ⎟ ⎝ N1 ⎠ 2 ⎛N ⎞ I2 ⎜ 2 ⎟ . So the primary voltage drop becomes I X ⎛ N2 ⎞ if referred to the 2 1⎜ ⎝ N1 ⎠ ⎝ N1 ⎟⎠ secondary.

But I1

However, there exists a secondary reactance voltage drop I2X2 so the total reactance 2 2 ⎧⎪ ⎛ N2 ⎞ ⎛ N2 ⎞ ⎫⎪ voltage drop, referred to the secondary is I2 X 2 + I2 X1 ⎜ ⎟ or = I2 ⎨ X 2 + X1 ⎜ ⎟ ⎬ . ⎝ N1 ⎠ ⎝ N1 ⎠ ⎭⎪ ⎩⎪ As for resistance, the transformer primary can be regarded as having no reactance 2 ⎛N ⎞ and the secondary to have reactance value: X 2 X1 ⎜ 2 ⎟ . This expression gives X2, the ⎝ N1 ⎠ Equivalent Reactance, a reactance value referred to the secondary. So the Equivalent 2 ⎛N ⎞ Reactance of the transformer referred to the secondary side is: X 2 = X 2 + X1 ⎜ 2 ⎟ and, ⎝ N1 ⎠ a similar expression derived for the primary side.

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20 • Advanced Electrotechnology Equivalent Reactance referred to the primary side is ⎛N ⎞ X 1 = X1 + X 2 ⎜ 2 ⎟ ⎝N ⎠

2

1

The expression for the Equivalent Impedance is Z1

R12 + X12

and

R22 + X 22

2

Example 1.6. A marine, dry-type, 17.5 kVA, 460/115 V, single-phase, 50/60 Hz transformer has primary and secondary resistances of 0.36 and 0.02 Ω respectively and winding leakage reactances of the windings are 0.82 and 0.06 Ω respectively. Determine the voltage applied to the primary to obtain full-load current with the secondary winding short-circuited. Neglect the magnetising current (1 decimal place). Full load primary current I1 = Transformation ratio = 2

17 500 = 38 A 460

460 4 N1 4 = or = 115 1 N2 1

⎛N ⎞ ⎛ 4⎞ Then R1 = R1 + R2 ⎜ 1 ⎟ = 0.36 0.02 × = 0.36 36 ⎝ 1⎠ ⎝ N2 ⎠ 2

2

⎛N ⎞ ⎛ 4⎞ X 1 = X1 + X 2 ⎜ 1 ⎟ = 0.82 0.06 × = 0.82 82 ⎝ 1⎠ ⎝ N2 ⎠ 2

(0.02 × 16 ) − 0.36 + 0.32 = 0.68 Ω

82 (0.06 + 16 ) − 0.82

0.96 = 1.78 Ω

Z = 0.682 + 1.782 = 0.4624 6 + 3.1684 = 3.6308 = 1.905 Ω Voltage to be applied to the primary = I1Z1 Z 1 = 38 × 1.905 − 72.4 V

Transformer Efficiency Only a basic treatment is made here regarding losses and efficiency. Since a transformer is a static device, the only losses which occur are: (1) iron losses and (2) copper losses. IRON LOSSES (PFe). These consist of hysteresis and eddy-current losses in the iron core. These are affected by flux density and frequency as shown in Volume 6. At a constant applied voltage and frequency, core flux is unaltered by the load current. Iron losses PFe can thus be assumed constant and load independent.

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Transformer Operation • 21 COPPER LOSSES (PCu). It is known that this is proportional to current2 and resistance. Then Primary copper loss = I12 R1 Secondary copper loss = I22 R2 or the total copper loss PCu

⎛N ⎞ I12 R1 + I22 R2 But I1 = I2 ⎜ 2 ⎟ ⎝ N1 ⎠ 2

⎛N ⎞ so the total copper loss = I2 ⎜ 2 ⎟ R1 I22 R2 ⎝N ⎠ 2

1

2 ⎧⎪ ⎛ N2 ⎞ ⎫⎪ = I 2 ⎨R2 + R1 ⎜ ⎟ ⎬ ⎝ N1 ⎠ ⎪⎭ ⎪⎩ 2

2

⎛N ⎞ It is already known that R 2 = R2 + R1 ⎜ 2 ⎟ so total copper loss referred to the secondary ⎝ N1 ⎠ = I22 R 2 In a similar manner the total copper loss can be shown to be I12 R1. Thus PCu

I22 R 2 o I12 R1

Efficiency The expression for efficiency is developed from first principles thus: Output Output = Input Output + Losses Output = Output + Iron + Copper loss

Efficiency =

Introducing the appropriate units the above equation is expressed as: η=

kW

kW + (

+

)

where PFe is the iron loss in kW and PCu is the copper loss in kW, or η= =

or η =

kVA cos φ kVA cos φ + PFe + PCCu kVA cos φ ⎛ I 2 R2 ⎞ kVA cos φ + PFe + 2 1000 ⎟⎠ ⎝ kVA cos φ ⎛ I 2 R1 ⎞ kVA cos φ + PFe + 1 ⎟ 1000 ⎠ ⎝

The maximum efficiency of a transformer occurs when the copper loss = the iron loss.

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22 • Advanced Electrotechnology Example 1.7. The primary resistance of a 440/110 V, single-phase transformer is 0.28 Ω and the secondary resistance is 0.018 Ω. If iron loss is 160 W when the correct primary voltage is applied, find the kW loading to give maximum efficiency at unity P.F. (3 significant figures). Turns ratio = so R 2 = 0.018 0 8 + 0.28

440 4 = 110 1 ⎛ 1⎞ ⎝ 4⎠

2

=0 0.018 018 + 0 0.0175 0175 = 0.0355 Ω For maximum efficiency, copper loss equals the iron loss so: PCu Thus: I22 =

60

o I22 R2 = 160

160 12.64 giving i i I2 = = 67.4 A 0.0355 0 0355 0.188

Load rating

110 × 67.4 1000

7.43 kVA

or since cos φ = 1 and kW = kVA then ‘loading’ for maximum efficiency = 7.43 kW. Example 1.8. The core of a single-phase transformer has a cross-sectional area of 15 000 mm2 and the windings are chosen to operate the iron at a maximum flux density of 1.1 T from a 50 Hz supply. If the secondary winding consists of 66 turns, estimate the kVA output if the winding is connected to a load of 4 Ω impedance value (2 decimal places). Since area = 15 000 ×10–6 and Bm = 1.1 T Φ m = 150 × 10 −4 × 1.1 = 165 × 10 −4 Wb Substituting in the e.m.f. equation: V2

f 2 fN

4.44 × 165 × 10 −4 × 50 × 66

= 2.22 × 165 × 10 −2 × 66 = 1.11× 33 × 66 × 10 −1 = 241.758 V × 10 −1 = 241.758 V Estimated current = Output rating

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241.758 = 60.44 A 4

241.8 8 × 60 44 1000

2.42 6.04 = 14.62 kVA.

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Transformer Operation • 23

Practice Examples 1.1 A single-phase transformer is designed to operate at 2 V per turn and a turns ratio of 3:1. If the secondary winding is to supply a load of 8 kVA at 80 V, determine: (a) the primary supply voltage, (b) the number of turns on each winding and (c) the current in each winding (all 3 significant figures). 1.2 A 25 kVA, 440/110 V, 50 Hz, single-phase, marine-type, step-down, ‘engine room’ transformer is designed to work with 1.5 V per turn with a flux density not exceeding 1.35 T. Calculate: (a) the required number of turns on the primary and secondary windings respectively, (b) the cross-sectional area of the iron core (4 significant figures) and (c) the secondary current (2 decimal places). 1.3 A single-phase, marine-type, step-down transformer has the following particulars: Turns ratio 4:1; no-load current 5.0 A at 0.3 at 0.3 P.F. (lagging); secondary voltage 110 V; secondary load 10 kVA at 0.8 P.F. (lagging). Calculate: (a) the primary voltage, neglecting the internal voltage drop (3 significant figures), (b) the secondary current on load (3 significant figures), (c) the primary current (1 decimal place) and (d) the primary P.F. (3 decimal places). 1.4 A 6.6 kV, 50 Hz, single-phase transformer with a transformation ratio of 1:0.06 takes a no-load current of 0.7 A and a full-load current of 7.827 A when the secondary is loaded to 120 A at 0.8 P.F. (lagging). What is the no-load P.F. (2 decimal places)? 1.5 A 1 kVA, single-phase transformer has an iron loss = 15 W and a full-load copper loss of 30 W. Calculate its efficiency on full-load output at 0.8 P.F. (lagging) (2 decimal places). 1.6 A single-phase power transformer supplies a load of 20 kVA at a power-factor of 0.81 (lagging). The iron loss of the transformer is 200 W and the copper loss at this load is 180 W. Calculate: (a) the efficiency (1 decimal place), (b) if the load is now changed to 30 kVA at a power-factor of 0.91 (lagging) and (c) calculate the new efficiency (1 decimal place). 1.7 A 20 kVA, 2000/220 V, single-phase transformer has a primary resistance of 2.5 Ω and a secondary resistance of 0.028 Ω. Corresponding leakage reactances are 2.8 Ω and 0.028 Ω. If the secondary terminals were accidentally short-circuited, estimate the current which would flow in the primary circuit (3 significant figures). 1.8 The resistances of the primary and secondary windings of a 27.5 kVA, 450/112 V, single-phase, marine-type transformer are 0.055 and 0.00325 Ω respectively. At the rated supply voltage iron loss is 170 W. Calculate for this transformer: (a) the

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24 • Advanced Electrotechnology full-load efficiency at 0.8 P.F. (lagging), (b) the kVA output at which efficiency is a maximum at 0.8 P.F. (lagging) and (c) the value of maximum efficiency at 0.8 P.F. (lagging) (all 2 decimal places). 1.9 A three-phase, marine, dry-type transformer is used to step down the voltage of a three-phase, star-connected alternator to provide the supply for 120 V lighting. The transformer has a 4:1 phase turns ratio and is delta connected on the primary side and star connected on the secondary side. If the lighting is supplied at the line voltage of the transformer, what must be the phase voltage of the starconnected alternator (3 significant figures)? 1.10 A 200 kVA, 6600/415 V, three-phase transformer connected in delta/star supplies a 120 kW, 415 V, 50 Hz three-phase motor whose P.F. and efficiency are 0.8 (lagging) and 83% respectively. Neglecting transformer losses, calculate the current in each transformer winding (Primary 2 decimal places, Secondary 1 decimal place).

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2

THE TRANSFORMER (TESTING AND EFFICIENCY) Machine Efficiency – getting the most output for the least input. Chris Lavers

The operating principle of the transformer, with and without load, was introduced in Chapter 1 and developed until the causes of internal voltage drop were considered. Phasor diagrams were added to help visualise the vector quantities and attention given to the fact that secondary effects can be seen to be reflected in the primary side, provided they are equated, using the turns ratio. Thus a step-down transformer with a 2:1 turns ratio (Primary:Secondary) and a 3 V drop in the secondary is equivalent to a voltage drop of 6 V in the primary. It is possible to equate the secondary phasor diagram with corresponding primary conditions provided the turns ratio relationship is used now that the phasor method is understood. This technique is here explored further for a better marine engineer’s understanding of how transformer efficiency, percentage resistance, reactance and impedance are obtained. It is unlikely that most of those reading this volume will have to make all these measurements and perform the consequent calculations to find the values, but it is a part of the STCW Code that as a marine engineer criteria exists for evaluating your competence in the isolation, dismantling and re-assembly of plant and equipment in accordance with accepted practice.

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26 • Advanced Electrotechnology

The Combined Phasor Diagram Since the primary and secondary currents are drawn in anti-phase, the following procedure is possible if a primary current I1 is used as the reference phasor (drawn horizontally), and the no-load current I0 neglected. Figures 2.1a and 2.1b show the 2 stages of the operation. The secondary diagram is rotated through 180°, so current I2 lines up with the reference phasor I1. The secondary voltage-drop triangle stands upright and the overall secondary diagram (figure 2.1b) is similar to the primary (figure 2.1a) and can be matched to the primary if the secondary voltages are adjusted to the primary conditions multiplied by the turns ratio, i.e. OV2′ is scaled up to equal OV1′ if multiplied ⎛N ⎞ by ⎜ 1 ⎟ . The secondary voltages and voltage drops must be multiplied by the turns ⎝ N2 ⎠ ⎛N ⎞ ratio, and the current-turn relationship, since I2 I1 ⎜ 1 ⎟ ⎝ N2 ⎠ (a)

V1

(b)

I1X1 V′1 I1R1

V ′2 I 2 X2 V2 φ

φ1

φ2 φ I1

O

I2R2

O

I2

▲ Figure 2.1

Figures 2.2a and 2.2b now take the form shown. It is seen that the secondary resistance ⎛N ⎞ voltage drop (originally I2R2) becomes I2 R2 ⎜ 1 ⎟ referred to the primary, and thus ⎝ N2 ⎠ 2

⎛N ⎞ I1R2 ⎜ 1 ⎟ expressed completely in primary current and voltage terms. Similarly the ⎝ N2 ⎠

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The Transformer (Testing and Efficiency) • 27 (a)

(b)

V1

V ′1 = V ′2

((

I1R1

I2X2

φ2

((

– I1X1= I1

(( (( (( ((

I2R2

φ1

( ( (( 2

N1 – N2

V2

V1

I1X1

N1 N1 = I1X2 – – N2 N2

2

2

N1 X1+X2 – N2

( ( (( 2

N1 = I R N1 – 1 2 – N2 N2

N1 – I1R1= I1 R1+R2 – N2

((

N1 – N2

N1 V2 – N2 φ2

I1

φ1 I1

▲ Figure 2.2 2

⎛N ⎞ secondary reactance voltage drop I2X2 becomes I1 X 2 ⎜ 1 ⎟ referred to the primary. ⎝ N2 ⎠ The next step is to draw the combined diagram with the in-phase voltage components added, to give the total resistance and reactance voltage drops. 2 ⎧⎪ ⎛ N ⎞ ⎫⎪ The total resistance voltage drop is: I1R1 = I1 ⎨R1 + R2 ⎜ 1 ⎟ ⎬ an expression deduced ⎝ N2 ⎠ ⎭⎪ ⎩⎪ 2

⎛N ⎞ R1 + R2 ⎜ 1 ⎟ and is the ⎝ N2 ⎠ equivalent resistance referred to the primary side, i.e. the total resistance of the windings is considered to be in the primary only. in connection with the Equivalent Resistance R1. Thus: R1

2 ⎧⎪ ⎛ N1 ⎞ ⎫⎪ The diagrams show the reactance voltage drops given by I1 ⎨ X1 + X 2 ⎜ ⎟ ⎬ an expression ⎝ N2 ⎠ ⎭⎪ ⎩⎪ 2 ⎛N ⎞ deduced in Chapter 1 regarding Equivalent Reactance X1. Thus X1 X1 + X 2 ⎜ 1 ⎟ and ⎝ N2 ⎠

is the equivalent reactance referred to the primary side, i.e. the total reactance of the windings is now considered to be in the primary only. The equivalent impedance Z 1 follows the usual relation, thus: Z1

R12 + X12

A simplified combined phasor is shown in figure 2.3. Note V2′ = V2, since all voltage drops are now accounted for on the primary side.

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28 • Advanced Electrotechnology V1

– I1 X1

– I1 Z 1

– I1 R1

((

N1 V2 – N2

φ2

φ1 I1

▲ Figure 2.3

The above deductions may be made by referring the primary diagram to the secondary, in which case it may be shown that: 2

R2

⎛N ⎞ R2 + R1 ⎜ 2 ⎟ , X 2 ⎝ N1 ⎠

2

⎛N ⎞ X 2 + X1 ⎜ 2 ⎟ and Z 2 ⎝ N1 ⎠

R22 + X 22

Example 2.1. A 660/220 V, single-phase transformer has a primary resistance of 0.29 Ω and a secondary resistance of 0.025 Ω. Corresponding reactance values are 0.44 and 0.04 Ω. Estimate the primary current flowing if a short-circuit occurs across the secondary terminals (1 decimal place). Turns ratio

V1 N1 660 3 = = = V2 N2 220 1

2

Also R1

⎛N ⎞ R1 + R2 ⎜ 1 ⎟ = 0.29 29 ⎝N ⎠

(0.025 × 9 )

0.29 + 0.225 = 0.515 515 Ω

2

2

Similarly X1

⎛N ⎞ X1 + X 2 ⎜ 1 ⎟ = 0.44 44 ⎝ N2 ⎠

(0.04 × 9 )

0.44 + 0.36 0.8 Ω

And Z1

R12 + X12 = 0.515 5 52 + 0.82 = 10 0 = 10 −11 26.52 + 64 = 10

1

1

5 152

82

90.52 = 10 −1 × 9.514 = 0.95 Ω

660 = 694.7 A. This is hard to 0 95 calculate any other way except by referring the secondary to the primary! The short-circuit (S.C.) current in the primary will be

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The Transformer (Testing and Efficiency) • 29

The Equivalent Circuit In Chapter 1, the phasor diagram of a transformer (figure 1.10) was developed to illustrate the conditions. Similarly figure 2.4 should be referred to when considering the development of the equivalent circuit. The first illustration of figure 2.4 shows the primary and secondary resistances and reactance values as separate components outside the windings. The second illustration shows how these values can be transferred from one winding to the other, in line with the concepts outlined. Since the turns ratio is taken into account when referring the voltages and resistance and reactance values from one winding to the other, the transformer can be omitted R1

I1

X1

X2

2

I1

X1

2

N

1 R2 – N

X2 –1

N

N

1 V2 –

N2

(( 2

I1

V2

N2

N

R1

I2

N2

2

V1′=V2′ –1

V1

V2

(( (( (( (( N

R1

R2

V2′

V1′

V1

I2

1 R2 – N

X1

2

(( (( 2

N

N

1 X2 –

I2 –1

N2

N2

((

((

N

N

V1′=V2′ –1

V1

1 V2 –

N2

N2

(( (( (( (( (( 2

2

N

R1

I1

1 X2 –

N2

X1 Io

V1

N

1 R2 –

V1′

Iw Im Ro Xo

N2

N

2 I2 –

N1

N V2′ –1 N2

N

V2 –1 N2

▲ Figure 2.4

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30 • Advanced Electrotechnology from the third equivalent circuit illustration. The fourth and final illustration shows how the equivalent circuit is modified to allow for the load, resulting from the iron loss and magnetising magneto-motive force. A constant no-load current I0 will flow when the primary winding is energised, irrespective of the secondary load, like the current in a parallel circuit. Since I0 is made up from Im (a magnetising component) and Iw (a power component), respectively in quadrature and in phase with the applied voltage, the parallel circuit is treated as a resistor and a pure inductor. The equivalent circuit in the last illustration shows all aspects of transformer working, and is used for complex problems. However, for practical considerations, as the no-load current is omitted from the phasor diagram, the equivalent circuit parallel section can be omitted. The circuit is often used in its simplest form (figure 2.5), after the phasor diagram of figure 2.3. – Z1 – R1

– X1

(( N

V1′=V2 –1

V1

N2

V2′

V2

L o a d

▲ Figure 2.5

Figure 2.3 shows the transformer phasor diagram associated with the simplified equivalent circuit. Although treatment up to now has shown all resistance and reactance referred to the primary side, it can be referred to the secondary side. The simplified equivalent circuit and phasor diagram are shown in figure 2.6 with V1′ = V1, as all voltage drops are accounted for on the secondary side so that: 2

R2

2

⎛N ⎞ ⎛N ⎞ R2 + R1 ⎜ 2 ⎟ and X 2 = X 2 + X1 ⎜ 2 ⎟ and it follows that Z2 = R22 + X 22 ⎝N ⎠ ⎝N ⎠ 1

1

Voltage Regulation This expression refers to the ‘sitting down’ of terminal voltage on load, and is expressed as a percentage (%) or per-unit value. It is the ratio of change in secondary terminal

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The Transformer (Testing and Efficiency) • 31 – Z2 – R2

V1

– X2

( (

N2 V2′ = V1′ – N1

V1′

(( (( N

N

N1

N1

V2

L o a d

V2′ =V1′ –2 =V1 –2

– I2 X2

– I2 Z2

– I2R2

V2

φ2 I2

▲ Figure 2.6

voltage between no-load and full-load to the no-load terminal voltage, for a constant primary input voltage. Thus: No load voltage Full-load voltage or in terms of the No-load voltage

Voltage regulation (per unit) = secondary as a percentage: %=

Secondary voltage at no load − Secondary voltage on load ×100 Se econdary voltage at no load

It is noted that a numerical difference and not a phasor difference is involved. Since the definition for regulation is given in terms of the secondary then: Percentage regulation =

V ′ − V2 × 100 which is also: V′

=

⎛N ⎞ or V1 ⎜ 2 ⎟ = ⎝ N1 ⎠

9781408176030_Ch02_Final_txt_print.indd 31

⎛N ⎞ V ′ ⎜ 2 ⎟ − V2 ⎝ N1 ⎠ ⎛N ⎞ V ′⎜ 2 ⎟ ⎝ N1 ⎠

(

+

× 100 or =

) +( 2

⎛N ⎞ V1 ⎜ 2 ⎟ − V2 ⎝ N1 ⎠

× 100

⎛N ⎞ V1 ⎜ 2 ⎟ ⎝ N1 ⎠ +

)

2

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32 • Advanced Electrotechnology If R 2 and X 2 are known, then the regulation for any load current can be estimated if the transformer Open-Circuit (O.C.) terminal voltage is known. Example 2.2. A 660/220 V, single-phase transformer has a primary resistance of 0.29 Ω and secondary resistance of 0.025 Ω. The corresponding reactance values are 0.44 and 0.04 Ω. Estimate the percentage regulation for a 50 A secondary load current at 0.8 P.F. lagging (2 decimal places). ⎛N ⎞ 3 Turns ratio = and R2 = R2 + R1 ⎜ 2 ⎟ 1 ⎝ N1 ⎠

2

1⎞ ⎛ = 0.0 025 5 + 0.29 × = 0.0 025 5 + 0.0322 = 0.057 057 Ω ⎝ 9⎠ also 2

⎛N ⎞ 1 X 2 + X1 ⎜ 2 ⎟ = 0.04 0.44 × = 0.04 0.049 = 0.089 Ω N 9 ⎝ ⎠

X2

1

Substituting we have: 220 =

{(

× 0 8 ) + (50 × 0 057 05 )} + {( 2

2

2.85)

or 220 2 = (0.8V2

2

{(

= 0.64V22 = 0.64 6 V2

2

(2

× 0 6 ) + (50 × 0 089 )}

2

2

(0.6V2 + 4.45)2 0.8V2 × 2.85) 8.12) + (0.36 36V22 (2

0.6V2 × 4.45 45) 19.8

)}

4.56V2 + 8.12 0.36V2 + 5.34 3 V2 19.8 2

= V2 + 9.9V2 + 27.92 2

Thus V22

9.9V2 − 48 400 + 27.92 = 0

giving the quadratic equation V22 Solving V2 = =

9.9V2 − 48 372 = 0

−9.9 ± 9.92 + ( 4 × 48 372) 2 −9.9 ± 98.01+ (193 488 ) 2

−9.9 ± 193 586 2 −9.9 9 + 102 × 4 4 =

(

−9.9 ± 10 19.36 = 2 2 −9.9 ± 440 430.1 = = = 215.05 05 V 2 2 2

=

Voltage regulation =

9781408176030_Ch02_Final_txt_print.indd 32

)

220 − 215.15 4.95 × 100 = = 2.25% 220 2.2

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The Transformer (Testing and Efficiency) • 33 Although the estimation method for regulation is simple, it requires a tedious mathematical solution. For this reason an approximation formula for evaluating the internal voltage drop was developed. This formula gives a result very close to the method above, and is invariably used. The explanation of how the formula is deduced is now given and used to show its accuracy. Since the difference between the no-load voltage and the load voltage is due to the causes of internal voltage drops in both windings, it follows that, if this voltage drop can be estimated then: Voltage regulation =

Internal voltage g drop p × 100% No-load voltage

The approximation formula deduced for this internal voltage drop is given by I2 R 2 cos φ2 X 2 sin i φ2 and

(

)

Voltage regulation =

I2

(

± V2

) × 100%

Substituting the values obtained for the previous example: 50 (0.057 × 0.8 ± 0.089 × 0.6 ) × 100 220 50 (0.0456 ± 0.0534 ) = × 100 220 50 × 0.099 × 100 = 220 5 × 99 45 = = = 2.25% 220 20

Voltage regulation =

The formula uses a plus or minus sign and with the latter, a negative regulation results. With a capacitive load, secondary voltage rises with increased load, resulting in a −ve regulation value or regulation ‘UP’. Sitting down of terminal voltage means a +ve regulation value or regulation ‘DOWN’. Loading conditions, influenced by the P.F., are shown in figure 2.7. From condition 2.7(c), it is seen that the on-load terminal voltage phasor V2 is longer than the induced e.m.f. or O.C. voltage condition V2′. Thus a rise of terminal voltage with load is evident and a − ve or Regulation ‘UP’ figure can occur. Transformer internal voltage drop and regulation can be determined from the primary side, and with the aid of a phasor (figure 2.3), the treatment for obtaining a regulation expression can be repeated.

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34 • Advanced Electrotechnology (a)

V2′

(b)

(c)

I2

– I2X2 V2

V2′

– I2R2

V2′

– I2X2 I2

V2

Unity P.F.

– I2X2 V2

I2

– I2R2

Lagging P.F.

– I2R2

Leading P.F.

▲ Figure 2.7

⎛N ⎞ V1 − V ’2 ⎜ 1 ⎟ ⎝ N2 ⎠ Thus Voltage regulation = × 100 V1 or since V2 = V2′ all causes of voltage drop considered on the primary side, ⎛N ⎞ V1 − V2 ⎜ 1 ⎟ ⎝ N2 ⎠ = × 100% V1 ⎛N ⎞ To calculate V2 ⎜ 1 ⎟ proceed as before using the expression: ⎝ N2 ⎠ 2

V1

⎡ ⎛ N1 ⎞ ⎤ ⎡ ⎛ N1 ⎞ ⎤ i φ 2 + I2 X 1 ⎥ ⎢V2 ⎜ ⎟ cos φ 2 + I1R1 ⎥ + ⎢V2 ⎜ ⎟ sin ⎣ ⎝ N2 ⎠ ⎦ ⎣ ⎝ N2 ⎠ ⎦

2

It is convenient to use regulation expressions in this form, i.e. referred to the primary, as data are often given as R1 and X1. Data may also be obtained from O.C. and S.C. tests, described shortly. An expression for the internal voltage drop (approximate value) can be deduced for the primary side. Its application is illustrated by example 2.3, which uses the figures for the transformer from the secondary side. The formula for the approximate voltage drop is yet to be deduced, but in ‘primary form’ is written as: Voltage drop (referred to primary) = I1 (

±

)

Example 2.3. A 660/220 V, single-phase transformer has a primary resistance of 0.29 Ω and a secondary resistance of 0.025 Ω. The corresponding reactance values are 0.44 and 0.04 Ω. Estimate the percentage regulation for a 50 A secondary load current and 0.8 P.F. (lagging) (2 decimal places).

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The Transformer (Testing and Efficiency) • 35

Turns ratio =

3 1

2

R1

⎛N ⎞ R1 + R2 ⎜ 1 ⎟ = 0.29 29 ⎝N ⎠

(0.025 × 9 )

0.29 + 0.225 = 0.515 Ω

2

2

⎛N ⎞ X1 + X 2 ⎜ 1 ⎟ = 0.44 44 ⎝ N2 ⎠

also X1

Primary current will be

(0.04 × 9 )

0.44 + 0.36 0.8 Ω

50 = 16.67 A cos cos φ = 0 8 (lagging) 3

Voltage =

16.67 {(

×

)+(

×

660 16.67 67 (0.412 + 0 48 ) = × 100 660 16.67 × 0.892 14.87 = = = 2.25% 660 66

)}

× 100

This is the same value as that obtained by applying the formula to the secondary side.

Internal Voltage-Drop Formula (For Approximation) The reasoning for this expression is based on the phasor diagram of figure 2.3. It is assumed that the transformer internal voltage-drop triangle is small, and far from point ⎛N ⎞ O so that phasors V1 and V2 ⎜ 1 ⎟ are parallel. ⎝ N2 ⎠ Figures 2.8a and 2.8b show this triangle and the phasors referred to, drawn parallel and designated OA and OB (added constructions are shown by the dotted lines). Since the numerical difference between the no-load and the on-load terminal voltages is equal to the internal voltage drop, then on the phasor diagram, it is the difference in lengths of ⎛N ⎞ V1 and V2 ⎜ 1 ⎟ (the lines OA and OB) a difference given by length AD (figure 2.8b). ⎝ N2 ⎠ Now DA = DE + EA = BF + EA or BC cos φ1 AC ACssin i φ1 So the voltage drop = I1R1 cos φ1 + I1 X1 sin i φ1

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36 • Advanced Electrotechnology ⎛N ⎞ For a capacitive load( figures 2.8c and 2.8d), the voltage drop is = V1 − V2 ⎜ 1 ⎟ , or the ⎝ N2 ⎠ length of the lines OA and OB. As OA is shorter, this difference is given by the length – AD. Now – AD = – (AE – DE) = DE – AE = FC – AE = BC cos φ1 − AC sin i φ1 Voltage drop = I1R1 cos φ1 + I1 X1 sin i φ1 Example 2.4. A 20 kVA, 2000/220 V, single-phase transformer has a primary resistance of 2.1 Ω and a secondary resistance of 0.026 Ω. The corresponding leakage reactances are 2.5 and 0.03 Ω. Estimate the regulation at full load under P.F. conditions of (a) unity, (b) 0.5 lagging and (c) 0.5 leading (all 2 decimal places). Turns ratio =

(a)

2000 9 09 = 220 1 (b)

V1

((

A

N

1 V2 –

N2

Lagging P.F. E F φ1

D

φ1

I1

O

B

(c)

C

(d) O

A

I1

φ1

Leading P.F. D V1

E

F

((

φ1

N

1 V2 –

N2

B

φ1

φ1

C

▲ Figure 2.8

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The Transformer (Testing and Efficiency) • 37

Full-load secondary current I2 =

20 000 = 90.9 A 220

2

R2

⎛N ⎞ ⎛ 1 ⎞ R2 + R1 ⎜ 2 ⎟ or R2 = 0.026 + 2.1 = 0.026 + 0.0255 0.0515 Ω ⎝ 9 09 ⎠ ⎝N ⎠ 2

1

2

X2

⎛N ⎞ ⎛ 1 ⎞ X 2 + X1 ⎜ 2 ⎟ or X 2 = 0.03 2.5 = 0.03 0.0302 = 0.0602 Ω ⎝ 9 09 ⎠ ⎝ N1 ⎠ 2

(a) Voltage regulation at unity P.F. (cos φ2 = 1) = = =

I2 R2 cos φ2 + X 2 cos φ2 × 100 V2 90.9 9 {(

× )+(

×

)}

220 90.9 0.0515 × 100 = 2.12% (d 220

× 100 )

(b) Regulation at 0.5 P.F. (lagging) (cos φ2 = 0.5) =

90.9 9 {(

×

)+(

)}

×

220 90.9 (0.02575 + 0 0523) = × 100 220 90.9 0.078 = × 100 = 3.23% (d 220

× 100

)

(c) Regulation at 0.5 P.F. (leading) =

90.9 9 {(

×

)+(

×

220 90.9 (0.02575 − 0 0523 2 ) = × 100 220 90.9 0.02655 −241.34 = × 100 = 220 220 = −1.09% or 1.09 9% (up)

9781408176030_Ch02_Final_txt_print.indd 37

)}

× 100

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38 • Advanced Electrotechnology

Transformer Testing As for all electrical machines, transformer testing is central to checking overall performance. As direct loading involves wasted energy, methods enabling the design specification to be checked by assessment are used if possible. Such methods determine the size of losses, overall efficiency, regulation and, for a transformer, involve 2 complementary tests. These are (1) the Open-Circuit Test (O.C.) and (2) the Short-Circuit Test (S.C.). The technique measures iron and copper losses separately and the results are applied to appropriate formulae. Similar tests are used for other A.C. machines such as the alternator and induction motor. Variations of the basic O.C. and S.C. tests help to obtain further information where appropriate.

The open-circuit test Mention of this test was made in Chapter 1 when the conditions of working on noload were investigated. One winding is connected to a normal voltage supply and frequency while the other is left Open-Circuited (O.C.). Figure 2.9 shows the circuit with appropriate instruments. Input power P0 (in watts) is measured by a wattmeter used on low P.F. circuits. A low-range ammeter measures the no-load current I0 and suitable voltmeters measure the primary and secondary voltages. As the iron core’s temperature rises when the winding is energised, clearly some energy is drawn from the supply and converted into heat. Unwanted heat is a ‘loss’ due to the cyclic magnetisation of the iron core and to copper losses in the energised winding. As winding resistance is very small and the no-load current is usually less than a tenth of the full-load value, the copper loss, under this condition, less than a hundredth of the full-load value can be neglected. Thus input power is assumed to be due to the iron losses, PFe, only. P0 I0

V1

A1

V2

▲ Figure 2.9

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The Transformer (Testing and Efficiency) • 39 Furthermore, since applied voltage and frequency are assumed constant under all loading conditions then, the flux density remains fixed and iron losses, measured by the O.C. test, are considered constant for all loads. In the introduction to transformer efficiency, mention was made of the iron losses component that is the hysteresis loss (Рну ), and the eddy-current loss (PEC). Usually it is not necessary to know their relative magnitudes; however, mention is made of a technique used for determining these if needed. SEPARATION OF IRON LOSSES. From Volume 6, Chapter 6 (pp. 150–154), the iron loss, PFe = PH + PEC. If flux density Bm is held constant during testing, the expression can be written: PFe = HN + EN2 for a machine or PFe = Hf+Ef2 for a transformer. The quantity Hf is the hysteresis loss and Ef2 is the eddy-current loss at that frequency. If the test is made at a particular frequency and repeated at another frequency – preferably a multiple or sub-multiple of the frequency, enough information is obtained to determine the constituent losses at these frequencies. The next example shows the application of this technique. Example 2.5. When an O.C. test is made on the primary of a transformer at 440 V and 50 Hz, the iron loss is 2.5 kW. When the test is repeated at 220 V and 25 Hz, the corresponding loss is measured to be 850 W. Determine the hysteresis and eddy-current loss at normal voltage and frequency (1 and 2 significant figures respectively). At 50 Hz we can write 2500 = Hf + Ef2

(a)

f ⎛f⎞ At 25 Hz we can write 850 = H + E ⎝ 2⎠ 2

2

(b)

Solving by multiplying (b) by 4, we have: 2500 = Hf + Ef2 . . . . . . 2

and 3400 = 2Hf + Ef . . . . . .

(a) (c)

Subtracting (c) – (a): – 900 = – Hf Thus the hysteresis loss = 900 W at 50 Hz and so the eddy-current loss = 2500 – 900 =1600 W at 50 Hz.

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40 • Advanced Electrotechnology

The short-circuit test Figure 2.10 shows the test circuit. One winding is short-circuited (S.C.) through an ammeter and a reduced voltage is applied to the other winding. The applied voltage is just sufficient to circulate full-load current through the S.C. winding, and, since the only current-limiting factor is the transformer impedance (the resistance of the instruments being negligible), then the applied voltage will be very low. Under this test condition, the core flux, being proportional to applied voltage, is small and iron losses may be ignored. The wattmeter reading (PSC) indicates the copper losses. An ammeter and voltmeter are included in the energising circuit and readings are used to estimate efficiency and regulation. The procedure for using the O.C. and S.C. test results for calculation purposes is illustrated in the next example, with reference to phasor diagram (figure 2.11). Consider figure 2.11, since the secondary terminal voltage is zero (on S.C.), the applied voltage is used to drive current through the transformer’s equivalent impedance referred to the primary. Thus point В of figure 2.8 coincides with point О and: Psc A1 Reduced voltage

A2

V

▲ Figure 2.10 V1 – I1 Z1

– I1X1 – I1R1

– I1 X1 – I1 Z1 O

– I1 R1

(( N

V2 –1 N2

I1

▲ Figure 2.11

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The Transformer (Testing and Efficiency) • 41

Z1 =

Vsc P R1 = 2sc and X1 Isc I sc

Z12 − R12

The results of these expressions are in the appropriate efficiency and regulation formulae. Example 2.6. The following results were obtained in tests on a 50 kVA, single-phase, 3300/400 V transformer. O.C. Test Primary voltage 3300 V, second voltage 400 V, input power 430 W. S.C. Test Reduced voltage on primary to give full secondary current was 124 V, primary current 15.3 A, input power 525 W. Calculate: (a) The efficiency at full load and 1/2 full load both at 0.707 P.F. (lagging). (b) The regulation at full load for 0.707 P.F. (lagging and leading). (c) Full-load terminal voltage under the condition of 0.707 P.F. (lagging). (All 1 decimal place.) From O.C. Test. Iron loss = 430 W From S.C Test. Copper loss = 525 W Also Z1 =

124 = 8.12 12 Ω 15.3

Also R1

525 15.32

. 5Ω

d X1

8.122 − 2.25 52

7.78 Ω

Therefore: (a) Efficiency at full load, 0.707 P.F. (lagging) ηFL = =

kVA cos φ kVA cos φ + PFe + PCCu 50 × 0.707 = 97 7.34% (50 × 0.707) + 0.43 + 0.525

and η 1 FL = 2

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25 × 0.707 25 × 0 .707 707 ) + 0.43 0.131 (

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42 • Advanced Electrotechnology Note that on 1/2 full load, current is halved and copper loss α I2 is quartered. η 1 FL = Efficiency(half full load) = 96.9% 2

(b) Voltage regulation, 0.707 P.F. (lagging) =

I1R1 cos φ + I1 X1 sin i φ × 100 V1

=

15.3 (2.25 × 0.707) + (7.78 × 0.707) × 100 = 3.3% (d 3300

)

Voltage regulation, 0.707 P.F. (leading) =

15.3 (2.25 × 0.707) − (7.78 × 0.707) × 100 = 1.81% ( p) 3300

(c) Terminal voltage = O.C. voltage – voltage drop = 400 – (15.3 × 2.25 × 0.707) + (15.3 × 7.78 × 0.707) ×

400 3300

= 400 – 13.2 = 386.8 V or using the percentage regulation expression: 400 − 400 ×

33 = 400 − 13.2 = 386.8 V 100

Direct-Loading Test (Sumpner’s Test) The disadvantages of the O.C. and the S.C. tests lie in the fact that iron and copper losses do not occur at the same time and so the temperature rise of a transformer cannot be checked by tests alone. If more than one unit is available, to avoid waste of power by direct loading, a back-to-back test may be used. Sumpner’s test is performed with 2 transformers which should be identical (but may be similarly rated units provided the P turns ratios are the same). If the copper loss of one unit = Cu where PCu is the reading 2 of one wattmeter then the efficiency of a transformer can be calculated from the expression: η=

kVA cos φ kVA cos φ + Fe (kW ) +

Cu

(kW )

× 100

Since both transformers are loaded a temperature rise can be found with a minimum power loss.

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The Transformer (Testing and Efficiency) • 43

Percentage Resistance, Reactance and Impedance It is convenient to express these values as a percentage voltage drop in the O.C. voltage. Here the treatment is made in terms of the primary. Thus: Percentage resistance =

IR Resistance voltage g drop p × 100 1 = 1 1 × 100 O.C. voltage V1

Percentage reactance =

IX Resistance voltage g drop p × 100 0 = 1 1 × 100 O.C. voltage V1

Similarly,

and, Percentage impedance =

IZ Impedance p voltage g drop p × 100 0 = 1 1 × 100 O.C. voltage V1

A further useful relationship used by transformer manufacturers is derived from: Percentage voltage regulation = =

I R cos φ1 ± X1 sin i φ1 Voltage g drop p × 100 = 1 1 × 100 O.C. voltage V1 I1R1 cos φ1 I1 X1 sin i φ1 ± × 100 V1 V1

Percentage voltage regulation = percentage resistance × cos φ ± percentage resistance × sin φ Note also that since percentage resistance: =

I1 R1 IR Cu loss × 100 = 1 1 × 100 = × 100 V1 V1I1 Full-load VA

or Percentage resistance = Percentage copper loss of full load. Example 2.7. A 50 kVA, 3.3 kV/230 V, single-phase transformer has an impedance of 4.2% and a copper loss of 1.8% at full load. Calculate the percentage reactance and ohmic resistance, with reactance and impedance referred to the primary side (all 1 decimal place). Estimate the primary S.C. current assuming the supply voltage is maintained constant (3 significant figures).

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44 • Advanced Electrotechnology

Primary full-load current =

50 000 = 15.15 A 3300

Since Percentage impedance =

Then 4.2 = Or Z1 =

I1Z1 × 100 V1

15.15 × Z1 = 100 3300 4.2 × 33 = 9.15 15 Ω 15.15

Percentage resistance = Percentage full-load copper loss = 1.8% ∴Percentage resistance = 1 8 = or R1 =

I1R1 × 100 V1

1 8 × 33 = 3.92 Ω 15.15

Percentage reactance can be obtained in 2 ways: Z 2 % − R 2 % = 4.22 − 1.82

either X

= 17.64 − 3.24 = 4.4 = 3.8% or since X1 = Z12

R12 = 9.152

3.92

= 83.72 − 15.37 = 68.35 = 8.27 Ω Then percentage reactance =

8.27 × 15.15 × 100 = 3.8% 3300

Under S.C. conditions, the transformer’s impedance will limit the current. Therefore: Isc =

V1 3300 = = 361 A Z1 9.15

Efficiency The efficiency curve is shown, in figure 2.12a, with the effect of the power factor. The earlier statement that maximum efficiency occurs when copper loss = iron losses can now be proved (figure 2.12b).

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The Transformer (Testing and Efficiency) • 45 Unity P.F.

(a)

Efficiency η

0.8 P.F. (Lagging or leading)

Load (b) PCu

Losses

PFe Max η

Load

▲ Figure 2.12

Conditions for maximum efficiency

Maximum efficiency occurs when Now η =

1 is a minimum. η

kVA cos φ kVA cos φ + PFe + PCCu

Substituting current and voltage symbols and referring to the secondary side then: 1 V2 I2 cos φ + PFe + I22 R2 = η V2 I2 cos φ 2 =

9781408176030_Ch02_Final_txt_print.indd 45

PFe V2 I2 cos φ2 I22 R2 + + V2 I2 cos φ2 V2 I2 cos o φ 2 V2 I2 cos φ2

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46 • Advanced Electrotechnology PFe I22 R2 1 = 1+ + η V2 I2 φ 2 V2 I2 cos φ 2 = 1+

PFe + I2 −1 R2 I2 + V2 I2 cos φ 2 V2 I2 cosφ 2

Now differentiating with respect to I2 the load current, then: ⎛ 1⎞ d⎜ ⎟ ⎝ η⎠ P I −2 = 0 Fe 2 dI2 V2 cosφ φ2 The above expression is zero when

and the minimum value of

If therefore

PFe V2

2 2 I2

R2 V2 +

φ2

0

R2 =0 V2 cos φ2

1 obtained. η

PFe R2 = by simplification 2 V2 I2 cosφ2 I2 V2 I2 cos φ2 PFe = R2 or PFe = I22 R2 I2 2

1 or maximum value of η occurs when the copper loss = the iron η kW losses. Thus for a given kW output at constant terminal voltage, since I2 = then V2 cosφ2 1 I2 ∝ cosφ2 A minimum value of

Now copper loss = I22 R2 so copper loss ∝ I22 or

1 . cos2 φ2

Thus as the P.F. falls, copper loss rises and since η is adversely affected by losses, the efficiency falls as the load P.F. falls below unity (figure 2.12a). Figure 2.12b illustrates the condition of maximum efficiency from loss curves plotted against load.

Conditions for all-day efficiency It is usual for a transformer’s primary to be connected permanently to the supply voltage and for load switching to be carried out in the secondary circuit. Since the

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The Transformer (Testing and Efficiency) • 47 copper loss varies with load but iron losses are constant and continuous, it is clear that the efficiency depends on loading and losses, which varies throughout the day. For units continuously excited but supplying loads intermittently, a low iron loss is desired but a low copper loss is important where the load factor is high. For a transformer working on full load most of the day, maximum efficiency is arranged to occur around the full-load value but for a transformer unit whose full-load value may be supplied for as little as ¼ of the day and only lightly loaded the rest of the time, it is desired to arrange maximum efficiency to occur at about ½ of the full-load value. Consideration of such factors indicates that transformer efficiency is best estimated on an energy rather than a power ratio and leads to the term ‘all-day efficiency’ defined as: All-day efficiency =

Output in kWh for 24 hours Input in kWh for 24 hours

Example 2.8. A lighting transformer rated at 10 kVA has a full-load loss of 0.3 kW, made up equally from iron and copper losses. The duty cycle consists of full load for 3 hours, half full-load for 4 hours and no-load for the remainder of a 24-hour period. If the load operates at unity P.F., calculate the all-day efficiency (2 decimal places). Energy output at full load 10 kW for 3 hours = 30 kWh Energy output at half-load 5 kW for 4 hours = 20 kWh Total energy output = 30 + 20 = 50 kWh Energy input = Output + Losses = 50 kWh + Energy wasted in losses and, Energy wasted in losses = Iron losses + Copper losses Iron loss energy = 0.15 kW for 24 hours = 3.6 kWh Copper loss energy = 0.15 kW for 3 hours +

0 15 kilowatts for 4 hours 4

= 0.45 + 0.15 = 0.6 kWh Energy wasted in losses = 3.6 + 0.6 = 4.2 kWh All-day efficiency =

50 50 = = 0.9225 or 92.25% 50 + 4.2 54.2

Since copper loss is proportional to current2, then for ½ load, current will be halved and copper loss quartered; hence the figure

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0.15 kW for 4 hours. 4

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48 • Advanced Electrotechnology

Practice Examples 2.1 On open-circuit, a single-phase, marine, dry-type transformer gives 115 V at its secondary terminals when the primary winding is supplied with 460 V. The resistance and leakage reactance of the primary windings are 0.36 and 0.83 Ω respectively while those of the secondary windings are 0.02 and 0.06 Ω respectively. If the secondary terminals are accidently short-circuited while the transformer is connected to the supply, what would be the value of current flowing in the primary winding (3 significant figures)? Assume that the supply voltage is maintained at 460 V and the magnetising current value can be neglected. 2.2 For a 25 kVA, 450/121 V, single-phase transformer, the iron and full-load copper losses are respectively 165 and 280 W. Calculate (a) the efficiency at full load, unity P.F. and at half-full load 0.8 P.F. lagging (2 decimal places); (b) and the load at which efficiency is maximum in kVA (1 decimal place). 2.3 A marine, dry-type, 17.5 kVA, 450/121 V, 50/60 Hz, single-phase transformer gave the following data on test: O.C. Test 450 V, 1.5 A, 115 W at 50 Hz S.C. Test 15.75 V, 38.9 A, 312 W at 50 Hz Estimate (a) efficiency of the transformer and (b) the voltage of the secondary terminals (both 1 decimal place) and when supplying full-load current, at 0.8 P.F. (lagging), from the secondary side. Assume the input voltage is maintained at 450 V, 50 Hz. 2.4 A single unit of the 3 units making up a three-phase, marine, dry-type transformer, was subjected to a short-circuited test and the following results were obtained: Voltage applied to the H.V. side: 14.3 V at 60 Hz Current supplied: 55.6 A Power taken: 316 W The unit is rated at 25 kVA, 450/121 V, 60 Hz. Determine the approximate value of the secondary terminal voltage (3 decimal places) when the transformer is operating at full-load, 0.8 P.F. (lagging) and 0.8 P.F. (leading). 2.5 The resistances of the primary and secondary windings of a 27.5 kVA, 450/121 V, single-phase, marine, dry-type transformer are 0.055 and 0.00325 Ω respectively. The iron loss is 170 W. (a) At 0.8 P.F. (lagging), calculate the full-load efficiency (1 decimal place). (b) At 0.8 P.F. (lagging), calculate the kVA output (2 decimal places) at which efficiency is a maximum and find the maximum efficiency value (2 decimal places).

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The Transformer (Testing and Efficiency) • 49 2.6 A 50 kVA, 440/110 V, single-phase transformer has an iron loss = 250 W. With the secondary windings short-circuited, full-load currents flow in the windings when 25 V is applied to the primary, power input being 500 W. For this transformer determine (a) the percentage voltage regulation at full-load, 0.8 P.F. (lagging) (2 decimal places) and (b) the fraction of full load at which the efficiency is a maximum (3 decimal places). 2.7 A 10 kVA, 440/110 V, single-phase, marine, dry-type transformer was tested and gave the following results: O.C. Test – Primary applied voltage 440 V, power input 75 W. S.C. Test – Primary applied voltage 30 V for full-load current, power input 135 W. Draw the equivalent primary circuit indicating the values of the circuit constants, but neglecting no-load conditions, and then calculate (a) the secondary terminal voltage when the transformer is operating at full load, 0.8 P.F. (lagging) with the rated voltage applied to the primary, (b) transformer efficiency for (a) above (both 1 decimal place). 2.8 A 50 kVA, 440/230 V, marine, lighting-transformer has primary resistance and reactance of 0.09 and 0.19 Ω respectively and secondary resistance and reactance of 0.015 and 0.042 Ω respectively. Calculate the secondary terminal voltage (3 significant figures) when the transformer is supplying full-load current at 0.8 P.F. (lagging). If the secondary terminals were accidently short-circuited, what would be the current taken by the primary winding, assuming the primary supply voltage remained constant at 440 V (4 significant figures)? 2.9 A 17.5 kVA, 460/115 V, 60 Hz, single-phase transformer has primary and secondary resistances of 0.125 and 0.008 Ω respectively and primary and secondary leakage reactances 0.39 and 0.025 Ω respectively. The iron loss, when normal voltage is applied to the primary winding, is 300 W. Draw the equivalent circuit and calculate (a) for the full-load, 0.8 P.F. (lagging) condition, the voltage at the secondary terminals (3 significant figures) and (b) the efficiency of the transformer (1 decimal place). Neglect the no-load current. 2.10 A marine, dry-type, 50 kVA, 440/110 V, single-phase transformer gave the following data on test: O.C. Test – Primary applied voltage 440 V, secondary voltage 110 V, power input to primary 250 W. S.C. Test – Primary applied voltage 25 V, primary current 113.6 A, power input at primary 500 W. Calculate (a) the efficiency at half full-load, 0.7 P.F. lagging (1 decimal place) and (b) the voltage regulation at full-load, 0.8 P.F. lagging (2 decimal places).

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3

THE TRANSFORMER APPLICATIONS Speculations? I have none. I am resting on certainties. Michael Faraday

This chapter discusses various transformer types used for special applications, such as instrument transformers (current and voltage transformers), auto-transformers (fixed and variable ratios), saturable reactors, magnetic amplifiers, thyristor controlled devices, as well as ‘solid-state’ transformer devices. Transformer power levels range from low-power applications such as electronics power supplies to very high power applications such as power distribution systems. As treatment is mostly descriptive, several A.C. theory examples are introduced for revision and further instruction.

Instrument Transformers This section covers current and voltage transformers: the latter are sometimes called potential transformers. When current in a circuit is too high to directly apply measuring instruments a Current Transformer (C.T.) produces a reduced current, accurately proportional to the current in the circuit, and also isolates the measuring instrument from any high voltage in the circuit. Digital clamp meters use a C.T. for measuring A.C. current. Instrument transformers may also be used with indicating or recording instruments, protective equipment, alarms and control gear, and have several advantages. 1. Indicating and protective instrument movements are standardised, e.g. 5 A for current coils and 110 V for voltage coils, allowing for a flexible instrument system which is comparatively inexpensive using mass produced meters.

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The Transformer Applications • 51 2. Indicating and protection equipment may be isolated from the high or medium voltage sides of the system. The main insulation between the primary and secondary systems can be built into instrument transformers, and the meter system takes this approach. Thus instruments, relays, alarm units, etc., can be small, with minimal insulation and at a potential to earth considered safe. 3. Connections from the transformer secondaries can be ‘light’, allowing cheap and neater wiring and control boards, located conveniently remote from heavy power equipment such as generators, switchboards, transformers and motors. In relation to instrument transformers some new terms are introduced such as burden, class, terminal markings, phase-angle and accuracy. Since the secondaries feed into systems which are isolated from the main system, it is clear the total load of such systems must be known and that transformers must be of a size able to operate the systems efficiently, accurately and effectively. The transformer ratio is the prime requirement but its rating must suit the load or burden carried, a term unique to instrument transformers.

The current transformer (C.T.) The main C.T. requirements are ratio accuracy, small angle variation from the 180° phase displacement between the primary and secondary currents, a sufficient rating to operate the burden, high insulation strength, mechanical rigidity and an ability to withstand heavy transient currents. It is seen in figure 3.1 that the primary is part of the main series circuit. The C.T. is however dissimilar to all other transformers where the primary is connected across the system and subject to the supply voltage. Supply 6.6 kV 3000 A

Load

C.T. 3000/5 A

Special fuses V.T. 6600/110 V A

A2

A1

A1

w

V1

V

A2 P.F.I

V2

V1

V2

F

▲ Figure 3.1

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52 • Advanced Electrotechnology Because of the connection method, the primary has a conductor section which is determined by the main line current’s magnitude and not by the secondary line current. The ampere-turns, needed for the magnetic circuit, are decided by this current, and transformers of this type are shown in figures 3.2a and 3.2b. The former uses a few thick conductor turns (known as a wound primary), while the latter uses the main cable or busbar as the primary which itself constitutes a single turn, an arrangement known as a bar primary. Transformers are available in variations of these 2 basic construction types and are usually built by instrument makers or specialists rather than power transformer manufacturers. THEORY OF OPERATION. From basic transformer theory it is known that: I2N2 = I1N1. In practice I2N2 is approximately equal to I1N1. Reference to the phasor (figure 3.3) shows that this approximation is acceptable if I0, the no load current, is small. In C.T. design every attempt is made to minimise I0 (resolved into its components I m and Iw). P2

(a) P1 Primary of heavy gauge conductor (insulated) and wound on insulating sheet

Secondary of light gauge S2 insulated wire S1 Laminated iron core (b)

Laminated iron core

Busbar primary

Insulating tape Insulated tube or sheet

Secondary of insulated wire

S2 S1

▲ Figure 3.2

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The Transformer Applications • 53 I1 I1′

Iw

I0 Im

Φ

I2

▲ Figure 3.3

Iron used in current transformers is chosen for its magnetic characteristics. Mu-metals are used for their low flux-density values and high permeability. Mu-metals are a range of Ni-Fe alloys notable for high magnetic permeability, composed of about 77% nickel, 16% iron, 5% copper and 2% chromium or molybdenum. Their high permeabilities make them ideal for shielding against static or low-frequency magnetic fields. The Mu-metal high permeability also provides a low reluctance path for magnetic flux. Modern technology has also produced grain-oriented silicon irons (versions of Stalloy) with excellent characteristics, and if thin laminations of these are used with insulation between them, iron losses may be minimised so that Iw is kept small. For a magnetic circuit Magnetomotive Force = Flux × Reluctance or ImN (the energising Φl BAl Bl Bl ampere-turns) = = or Im N = and Im = μA μ A μ μN Im is decided by the factors shown and for a C.T. working flux density is typically 1–1.5 × 10−4 T compared with a usual power transformer figure of 14 × 10−4 T. The iron circuit length is made as short as possible using high-permeability iron. N may be large, but this is not possible if a single-turn or ‘bar-primary’ is used. Other construction features result from the following requirements. Since the iron circuit must have low reluctance, for the best C.T.s, laminations have the least number of joints, or are ideally complete rings. Windings are placed by hand and leakage reactance is reduced by superimposing or sandwiching windings. The primary circuit is largely unaffected by insertion of a C.T. The rated secondary current is commonly standardised to 1 or 5 A. For example, a 2000 5 C.T. provides an output current of 5 A when the primary passes 2000 A.

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54 • Advanced Electrotechnology PRECAUTIONS. Since the secondary low-voltage system is isolated from the ‘mains’, it is possible to work on a circuit provided appropriate precautions are taken, and after appropriate training. One rule, however, is always observed. Never open the secondary circuit of a C.T. while the primary is energised. The reason for the rule is obvious if reference is made to the phasor diagram. With correct functioning I2N2 = I1N1 the primary ampere-turns are nullified by the secondary ampere-turns, the main flux being produced by ImN1, whose design value is kept to an acceptable figure for the reasons mentioned. Thus Im may be about 500 mA for a C.T. ratio of 500/5. Assume the secondary develops a terminal voltage of 3 V under correct loading conditions. If the secondary circuit was opened and the demagnetising turns removed, the primary 500 magnetising current will become 500 A or , 1000 times greater than the 500 × 10 −3 normal value. The primary magnetising ampere-turns rise in proportion and, as the core magnetic material operates at a low flux density, it can rise appreciably. The core material is chosen for its magnetic characteristic, i.e. a steep graph and high saturation value. Transformer flux density rises in proportion with the new ampere-turn value and the secondary voltage could rise 3000 V in the example considered. It is clear that a transformer designed to operate with 3 V across its output terminals will likely breakdown due to insulation failure. If one terminal is earthed other parts of the secondary system, also subject to an unexpectedly high voltage, are likely to be damaged. The danger to an operator working on this circuit is now considerable – hence the rule regarding open-circuiting the secondary of a C.T. Unlike a normal transformer, no danger is introduced if the secondary is short-circuited since the maximum current flowing is 5 A or less, a value decided by the relationship: Secondary At = Primary At. Under emergency conditions it is possible to take an instrument out of a circuit by shorting the terminals to which a meter is connected, ensuring the secondary current still flows, so the instrument can be disconnected, leaving the shorted link in position. Similarly when inserting a current-coil of a meter into a circuit, it must be connected across the shorted link and then the latter opened. It should be clear from this that working on the secondary side of a C.T. is hazardous and must not be undertaken unless one is confident of the operation. Marine engineers may have watched specialist maintainers undertake this operation during ‘setting up’ or undertake testing procedures and may be tempted to make similar checks. Proceed with caution, and only if absolutely necessary. Often a special current-circuit switch, e.g. an ammeter switch, is provided on switchboards, so a current is noted in the 3 lines of a three-phase supply. The meter and 3 C.T.s, one in each line, and the specialist switch shorts out a line C.T. and then disconnects the instrument. It next connects the ammeter across the shorted link of the next line

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The Transformer Applications • 55 and then breaks the latter connection. It thus carries out the operation described in correct sequence. It is noted that if the secondary of a C.T. is disconnected from the load while current flows in the primary, the transformer secondary will try to continue driving current across effectively infinite impedance. This may produce a high voltage across the open secondary (up to kV levels) which can cause arcing, endangering both operators and equipment safety, potentially damaging the transformer. BURDEN AND CLASS. The secondary load of a C.T. transformer is called the burden to distinguish it from the load of a circuit whose current is being measured. The C.T. burden is the output VA value at the rated secondary current. Thus for a 15 VA burden at 5 A, terminal voltage is 3 V and the instrument impedance or relay circuit must not exceed 0.6 Ω, otherwise errors are introduced into the readings and operation. The appropriate British Standards (B.S.) specification: BS EN 60044-1:1999, IEC 60044-1:1996 for Instrument Current transformers covers C.T.s and sets out the various burden sizes together with further information. BS EN 60044-1:1999 also sets out the various C.T. classes. Those used for precision and industrial instruments are listed, together with the permitted errors for each class. A study of this specification is advised if more detailed information about instrument transformers is required. B.S. publications can be purchased at http://shops.bsigroup.com. C.T. burdens are typically given as 1.5, 3, 5, 10, 20, 30, 45 and 60 VA. Burden ratings are typically in the form B-0.1, B-0.2, B-0.3, B-0.4, etc. This means a C.T. with a burden rating of B-0.1 can tolerate up to 0.1 Ω impedance in the metering circuit before its output current is no longer on a fixed ratio to the primary current (essential for accurate proportionality). Consequently the knee-point voltage of a C.T. is the magnitude of the secondary voltage after which output current ceases to follow linearly the input current. The knee-point is defined as the voltage at which a 10% increase in applied voltage increases the magnetising current by 50%, above this point the magnetising current increases in a rapid non-linear manner even for small voltage increases across the secondary terminals. TERMINAL MARKINGS. For simple current meter conditions terminal markings are unimportant but for measurement of power, power factor and with three-phase instruments, the correct connection sequence is vital. To ensure connections are made in accordance with the instruments’ requirements, a transformer’s primary terminals are marked P1 and P2 while its secondary terminals are marked S1 and S2. Connections are then such that at the instant when current flows in the primary from P1 to P2 it also flows in the secondary from S1 through the external circuit to S2. A colour (red spot) may mark P1 and S1. For switchboard work the S2 side of the secondary is often connected to earth.

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56 • Advanced Electrotechnology PHASE ANGLE. A phasor diagram shows the secondary current phasor is almost in anti-phase, i.e. almost at 180° to the primary current phasor. If the angle was 180°, no phase-angle error would result. Components Im and Iw cause the angle to be a little less than 180° and methods to minimise these were considered. The angle between I1′ or I2 reversed and I1 is the phase angle. The angle is +ve if, when reversed, the secondary current leads the primary current. On a low P.F. circuit, the angle may be −ve. For current measurement I2 should be a specific fraction of I1 and phase-angle error is unimportant. For power measurements, current ratio must be accurate and the secondary current 180° out of phase with the primary current. As this condition does not exist, a phase-angle error may introduce an appreciable error into power measurements if precautions are not taken by the instrument and transformer makers. Example 3.1. For the bar-primary type of current transformer (figure 3.1), estimate the secondary turns and the burden if the impedances of the ammeter, wattmeter and P.F. indicator current-coils are 0.15, 0.3 and 0.3 Ω respectively (2 decimal places). The transformer ratio is

3000 600 or . If the primary Ats equals the secondary Ats then: 5 1

1 × 3000 = N2 × 5 or N2 = 600 turns (in inverse proportion to the current ratio). The 600 turns is insulated wire of cross-section able to carry 5 A. With 5 A flowing, the voltage drop across the ammeter is 5 × 0.15 = 0.75 V. The meter burden is 5 × 0.75 = 3.75 VA. The wattmeter and P.F.I. burdens are 5 × 0.3 × 5 = 7.5 VA. The total burden is then 3.75 + 7.5 + 7.5 = 18.75 VA and a C.T. rating equal or greater than this is needed. Thus a 30 VA unit is suitable. Instrument burdens add arithmetically, as they are due mainly to a resistance load, and any error introduced by this assumption is small. If a suitably rated unit is not available one should use 2 C.T.s. and 2 independent instrument circuits. Thus an ammeter may be energised from the first C.T. together with another instrument like an over current relay, and the second C.T. used for the wattmeter and P.F. indicator. However, most instrument and protective-gear circuits are separated rather than mixed. Shapes and sizes vary significantly depending on the end user, switchgear and manufacturer. C.T. accuracy is directly related to several other factors besides burden and class which are rating factor, load, external E.M. fields, temperature and the physical arrangement. The rating factor, the factor by which the nominal full load current of a C.T. can be multiplied to determine its maximum measurable primary current, depends largely on ambient temperature with most C.T.s rated for 35°C and 55°C working.

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The Transformer Applications • 57

The voltage transformer (V.T.) The theory of the voltage transformer (V.T.) follows that of the power transformer (P.T.), the main difference being that the secondary current, referred to the primary, may be of the same order of magnitude as the no-load current. From phasor theory V2′ is to V1′ as the turns ratio, and the desired approximation of V2 to V1 is achieved if the voltage drops, as mentioned, are reduced. The following precautions are usually taken in the use and building of transformers: 1. Copper on the windings is made of larger section than used for a commercially built P.T. Turns are the same but the area of the wire is much greater, meaning a reduced ohmic value for R1 and R2. 2. By careful winding distribution, leakage reactances X1 and X2 are much reduced. This precaution results in a further reduction of voltage drops. 3. A good quality iron core is used with careful construction and suitable insulation between laminations. The no-load current components, Im and Iw are thus reduced and I0 will be much smaller than that for an equivalent rated P.T. 4. Sensitive instruments are used as they require a smaller operating current and, if I2 is small, internal voltage drops are reduced. Unlike the C.T., the primaries are connected across the ‘mains’ and an S.C. on the secondary will be a short referred to the primary. S.C. protection is needed and highrupturing-capacity fuses are fitted in primary circuits. As this is a high-voltage supply, fuses are insulated and a specialist tool provided for replacement without switching the ‘mains’ off. A secondary circuit’s mid-point can be earthed so the maximum voltage to earth on the system is 55 V. As instruments are in parallel as for the C.T., disconnection or reconnection of a suspect meter is not difficult. There is no danger of a high induced voltage, as for the C.T., but short-circuiting of terminals must not be undertaken. BURDEN AND CLASS. Details are covered under B.S. EN 60044-2:1999, IEC 60044-2:1997 Instrument transformers. Inductive V.T.s specify a secondary voltage of 110 V when the rated primary voltage is applied. A range of primary voltages (110 V to 220 kV line to line) is specified with the classes and permitted errors. The standard rated burdens, from 15 to 200 VA, are listed with the duty type for each class, similar to those mentioned for C.T.s. TERMINAL MARKINGS. Single-phase. Primary terminals are marked: A and В or N. Secondary terminals are marked: a and b or n and the polarity such that, when A is positive with respect to В then a is +ve with respect to b.

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58 • Advanced Electrotechnology Three Phase. Standard P.T.s are star-connected on the primary and secondary sides. Neutrals may be added, especially for the secondary, and for special requirements. Primary terminals are marked А, В and С while secondaries are marked a, b and c. The primary neutral (if provided) is N and the secondary neutral is marked n. Transformer polarity is such that when A is +ve with respect to N, a is +ve with respect to n. PHASE ANGLE. Like the C.T., a V.T. can introduce metering errors in both voltage magnitude and phase. Ratio and phase-angle errors depend on the impedance voltage drops’ relative sizes, and it was explained how these can be minimised. Phase-angle error is only important when power measurement is required. Example 3.2. If the voltage transformer shown in figure 3.1 is rated at 15 VA, estimate if a frequency meter may be added. A manufacturer lists the meter burden at 110 V to be 2 VA and that of the voltmeter 4 VA. The wattmeter voltage windings and the P.F. indicator are rated 3.5 VA and 4.5 VA respectively at 110 V. Total burden of existing instrument voltage windings = 4 + 3.5 + 4.5 = 12 VA The transformer is suitable for a burden of 15 VA, so an extra 15 − 12 = 3 VA can be added. It is possible to add the frequency meter in the manner shown. Inductive V.T.s are used for power networks with insulated or grounded neutral points and variants are used in the terrestrial electricity generating industry for accurately measuring voltage on high-voltage power lines.

The Auto-Transformer Fixed ratio type The Auto-Transformer (A.T.) employs only one winding tapped to provide appropriate voltage. From the operational theory discussed here it is seen there is no electrical difference between the action of a normal double-wound transformer and an autotransformer, and so the related ratio of V and I to the number of turns on the winding is still true. Since there is only one winding, it is cheaper than a double-wound unit of the same kVA rating, and it contains less copper, so copper losses are less, while efficiency is higher. The A.T. is advantageous for certain applications, the chief being where output voltage is not dissimilar from input voltage; then it is operated as a voltage booster and connected into a transmission line where the load terminal voltage is below an acceptable value. In

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The Transformer Applications • 59 this case, its function is to raise voltage to the required value. For marine work, its main application relates to motor starters. Some starters for induction motors operate on the principle of lowering the voltage to a motor, to limit the starting-current surge taken from a ship’s mains. As a motor runs up to speed, voltage is increased by transformer tappings, until the full supply is reached when it is switched out of the circuit. As the transformer is used for only a brief period the unit must be as cheap as possible and auto-construction satisfies this requirement. A.T.s are often used to step up or step down voltages on land in the 110–117–120 V range and in the 220–230–240 V range, providing 110 V or 120 V (with taps) from a 230 V input allowing equipment designed for 110 V or 120 V to be used with a 230 V supply (US and Continental supply voltages). A.T.s have a disadvantage in that, secondaries are not isolated from the primaries and the benefits of a 2 V system are not realised. This applies to marine installations where the main power system is of the order of 440 V and the small-power system at 110 V. An earth fault on one line of either system is immediately noticed on the other. Of great importance is the fact that if a break occurs on the common winding section, the high voltage will become manifest on the low-voltage system, as seen, if figure 3.4b is studied and an O.C. imagined at point X. THEORY OF OPERATION. Consider figure 3.4a. As for a normal transformer, if one primary and secondary terminal are joined – with a pair always of the same polarity as each other – one can find a point on the primary winding with the same potential as the remaining secondary terminal. These points may be connected and a secondary winding removed. Advantage is taken of the fact that I2 and I1 are practically in phase opposition. If the secondary is formed at 2 tappings on the primary, current flowing away in this section is I2 − I1 and the winding conductor area can be reduced, making a further cost saving. The extent of the copper saving can now be investigated:

(a)

(b) I1

I1 N1–N2

N1 200 V

I2

100 V

V1 N2 100 V

0

L o a d

I2

N1

I2 –I1

N2 X

L o a d

0

▲ Figure 3.4

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60 • Advanced Electrotechnology SAVING OF COPPER. For both the double-wound and auto-transformer it is known that V1 N1 I2 = = =k V2 N2 I1 It is deduced that for the double-wound transformer: Weight of copper in Primary ∝ I1N1 Weight of copper in Secondary ∝ I2N2 Total weight of copper ∝ (I1N1 + I2N2) For the auto-transformer: The top section of (N1 − N2) turns carries I1 so weight of copper ∝ I1 (N1 − N2) Bottom section of N2 turns carries (I2 − I1) so weight of copper ∝ N2 (I2 − I1) or total weight of copper ∝ I1N1 − I1N2 + I2N2 − I1N2 ∝ I1 (N2 − 2N1) + I2N2 Thus

I (N − 2N2 ) + I2 N2 Wt of Cu in auto-transformer = 1 1 I1N1 + I2 N2 Wt of Cu in double-wound transformer

Dividing top and bottom by I1 N2

(N1 −

N2 )

N1 I −2+ 2 N2 N I1 = = 2 N1 I2 N1 I2 + + N2 I1 N2 I1 k − 2 + k 2k − 2 k − 1 1 = = = = 1− k+k 2k k k +

I2 I1

(1) If the primary to secondary voltage is 10:9 k =

10 1 9 so the ratio = 1 − = 9 10 10

The Wt of Cu in auto: Wt of Cu. in double-wound unit = 1:10 or Wt of Cu in auto =

1 of 10

that of a normal transformer. (2) If the primary to secondary voltage is 10:1 then k =

1 9 10 and the ratio = 1 − = 1 10 10

The Wt of Cu in auto: Wt of Cu in double-wound unit = 9:10 or Wt of Cu in auto is

9 that required for a normal transformer. 10

Examples (1) and (2) show that saving is effective when k is close to unity.

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The Transformer Applications • 61 Example 3.3. A single-phase, 400/440 V auto-transformer is used to step up voltage and supply a load of 8.8 kVA, operating at unity P.F. Neglecting losses and the magnetising current, find the output (2 significant figures) and input currents (2 significant figures), and the current in the transformer winding’s common section (1 significant figure). Load or output current =

8800 = 20A 440

Input current =

8800 = 22 A 440

Current in common section = 22 − 20 = 2 A. Example 3.4. Determine the core area in m2 (3 decimal places), number of turns and the position of the tapping point for a 500 kVA, 50 Hz, single-phase 6.6/5.0 kV autotransformer, assuming the following: e.m.f. per turn = 8 V, maximum flux density = 1.3 T. Using the transformer e.m.f. equation E = 4.44 ΦmfN V Then or Φ m =

E = 8 = 4.44 Φ m 50 N

8 W. Since Фm = BmA 222 8 8 × 10 6 m2 = mm2 222 × 1.3 222 × 1.3 8 000 000 = = 27 720 mm2 = 0.028 m2 288.6

∴ Area =

Primary turns =

6600 = 825 8

Secondary turns =

5000 = 625 8

The winding must be tapped 625 turns from the end, common to primary and secondary.

Variable ratio type These are used as automatic voltage regulators and in applications such as a lighting ‘dimmer’ without the E.M. interference associated with thyristor-based dimmers. A variable A.T. with a sliding-brush secondary connection is often called a Variable Transformer or Variac (the original US General Radio trade name), consisting of one winding wound on an iron core used for relatively low-voltage designs and as a variable A.C. transformer. The core is formed into a cylinder, of annular lamination. Stampings are insulated from each other in keeping with good transformer practice. The winding consists of enamelled wire in a toroidal form (figure 3.5). A movable carbon brush

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62 • Advanced Electrotechnology

Carbon brush Track Enamelled wire

Flexible lead from brush Supply

▲ Figure 3.5

is arranged on a radial arm, making contact with a track formed on the winding by cleaning off the enamel insulation. One construction weakness is that, full voltage is applied along the length of the track where the insulation between adjacent turns is a minimum. In addition, grime and carbon dust can lead to insulation breakdown between the turns, resulting in turns shorting and burn out or insulation failure along the whole track length. Modern construction techniques employ durable synthetic enamel and varnish impregnation or ‘potting’, together with a carbon brush of suitable type, grade and shaped tip. Such refinements removed the main causes of failure, resulting in the A.T. being a reliable electrical item used in laboratories and electrical factory test-rooms. The A.T. unit is built in sizes up to 2.5 kVA for 230 V working, and may be arranged in ‘ganged’ form for 440 V working. In the smaller sizes, it is easily operated by a small sensitive motor and can be built into voltage regulators. Example 3.5. A 250/200 V A.T. is rated at 2 kVA. What is the economy in copper compared with the equivalent double-wound transformer? Estimate the copper section of the winding (1 decimal place). From the deduction made already V Wt of Cu in auto-transformer 1 = 1 − where k = 1 Wt of Cu in double-wound transform s er k V2

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The Transformer Applications • 63

Here k =

250 5 = 300 4

So ratio = 1 −

Therefore A.T. Wt of Cu =

1 4 1 = 1− = 5 5 5 4

1 4 that of the double-wound unit, a saving = or 80% in 5 5

copper required. Input current =

2000 =8A 250

Output current =

2000 = 10 A 200

Current in common section = 10 − 8 = 2 A Working on a current density of 1.55 A/mm2 the common conductor section should be at least 1.29 mm2 (say 1.3 mm2), the top section should be 4 times this, i.e. 5.2 mm2. In essence a Variable transformer is a versatile electricity control device, providing stepless and undistorted A.C. voltage variation, available in the United Kingdom under the registered trade marks of ‘Regavolt’ and ‘Varatran’.

Three-phase transformation For high-power applications, three-phase transformers are often used. There are many different combinations of flux paths linking windings possible so it is usual to treat three-phase transformers as 3 independent single-phase transformers. Production of a three-phase output supply from a three-phase input supply of the same frequency but different voltage value is effected in two ways: (1) by use of a composite three-phase unit and (2) by interconnecting 3 similar, but separate single-phase units. (1) THE THREE-PHASE TRANSFORMER. The magnetic circuit for this unit provides 3 cores on which are wound the primary and secondary of each phase. Figure 3.6 shows the arrangement. It is noted that the 3 windings are not wound on a single limb, as the resultant flux of the 3 primaries would be zero because, being 120° displaced from each another, fluxes would cancel. The construction technique is similar to that for the single-phase unit except that windings for each phase are confined to one core. The usual arrangement favours connection of either the primary or secondary in delta to allow passage of harmonic currents. As yet no mention has been made of transformer harmonic current generation, so the condition is summarised here. Assume a sinusoidal applied voltage. Since Ф α V, the flux waveform will also be sinusoidal but not so the no-load current, which is distorted for two reasons. (1) Since the B/H curve for the core is not a straight line a ‘peaked’ current wave results, especially

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64 • Advanced Electrotechnology Primary (in delta) B

A

S

S

S

F

F

a

F

S

S F

C

F

S F

b c Secondary (in star)

o

▲ Figure 3.6

if the iron is worked near saturation. (2) Due to hysteresis, the B/H curve is dissimilar on its ascending and descending values so the current waveform is distorted, being asymmetrical about the maximum value (figure 3.7). Fourier’s theorem shows that, on analysis, the no-load current waveform may be treated as made up of a fundamental waveform and its 3rd, 5th, 7th and 9th harmonics (respective odd multiples of the fundamental frequency). These are present in relatively decreasing magnitude values and their effects minimised with appropriate precautions. A delta connection for a three-phase winding enables the 3rd harmonic e.m.f.s – the

Current

Flux

Time

▲ Figure 3.7

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The Transformer Applications • 65 most important, produced in each phase, to circulate 3rd harmonic currents. These currents add to the fundamental to produce the required sinusoidal flux wave and resultant sinusoidal induced e.m.f.s. (2) THREE SEPARATE SINGLE-PHASE UNITS. Three similar transformers are interconnected into the three-phase primary and secondary lines, an arrangement favoured for marine practice, since only one spare unit must be carried. By a system of selector switches, the spare unit can be connected into the circuit if one of the other units (unit 1 here) fails. The rating of each of the single-phase unit only needs to be one-third of the total output (figure 3.8.)

Main busbars

R Y B

3-ph 440 V power loads 1

2

3

4

Stand-by transformer

Switched as no. 1 mechanically interlocked for use in one position only

Small power 3-ph loads r y b

1-ph loads

▲ Figure 3.8

Methods of connection 1. Star/Star. Figure 3.9a shows the arrangement and connection diagram. Phase V windings must withstand so that less insulation material is needed, a space saving 3 and a simpler design. Windings carry the full line current and if one-phase winding fails, the whole three-phase arrangement will be out of action (failing catastrophically). For this connection, 3rd harmonic currents do not flow in the line and distortion of secondary e.m.f. may be expected. If a neutral line is run from the star point, 3rd harmonic current will flow along the wire – which is unwanted.

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66 • Advanced Electrotechnology (a)

C EOA

A B1

A1

C2

B2

A2

c2

b2

a2

c1

b1

a1

Eoa

P

P O o

S

EOB

EOC

B C1

Eoc

S

Eob c

b

C

(b)

a

C1

B B1

A A1

C2

B2

A2

c2

b2

a2

P ECA

P

EAB

Eca

EBC

S

Eab

S

Ebc c1

(c)

b1

a1

c

b

a

C C1

B B1

A A1

C2

B2

A2

c2

b2

a2

c1

b1

a1

Eoa ECA

P EBC

EAB

P

S

Eoc

o S

Eob c

b

a

▲ Figure 3.9

2. Delta/Delta. Figure 3.9b shows that with this connection, each phase winding must 1 be insulated to the full line voltage value but windings are only designed to carry 3 times line current. In the event of a winding failure the unit need not be shut down since, by disconnecting the faulty winding, supply can be maintained at a reduced output (graceful degradation). This working method, known as open-delta, is favoured by Continental and American marine practice. As no neutral point is available, a 3-wire supply is used. Any 3rd harmonic currents, set up by 3rd harmonic e.m.f.s in the phases, circulates around the closed mesh and won’t flow out into the lines. The connection avoids distortion of secondary e.m.f. waveform by internally generated harmonics.

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The Transformer Applications • 67 3. Delta/Star or Star/Delta. The Delta-Star (Europe), Delta-Wye (North America), is a type of three-phase electric power transformer design employing delta-connected windings on its primary and star-connected windings on its secondary. These are common in commercial, industrial and high-density residential areas, to supply threephase distribution systems, e.g. for a distribution transformer with delta primary, running on three 11 kV phases with no neutral or earth needed, and a star secondary providing a three-phase supply at 400 kV, with the domestic voltage of 230 V available between each phase and an earthed neutral. Figure 3.9c stresses the important point that, the transformation ratio is not merely the transformer turns ratio but involves the connection type. Thus for a Delta/Star arrangement, transformation ratio V Vph1 N1 N = 1 = where 1 is the phase-turns ratio. V2 3 Vph2 N2 3N2 Irrespective of whether the delta/star or star/delta connection is used, no trouble is experienced with 3rd harmonic currents as they circulate around the mesh. A starpoint is available if a 4-wire arrangement is needed for earthing. For marine work, star/ delta working provides a good arrangement as a star-point on alternator is usual and a neutral introduced if necessary. For common polyphase transformer connections (figure 3.9), three- to two-phase and three- to six-phase arrangements are possible for special applications. For the examples in figure 3.9, variations are possible where windings per phase are split into 2 sections and combinations: star/inter-star, star/double-star, delta/double-delta combinations are used for special purposes. The only marine application for these is connected with static rectifiers. Example 3.6. The bow docking-propeller of a large liner is driven by a 600 kW, 3.3 kV, three-phase, 60 Hz delta-connected induction motor supplied from the ship’s 440 V auxiliary power mains through a step-up transformer unit, connected in delta, on the L.T. side. Assuming a full-load efficiency 85% and 0.8 P.F. (lagging) respectively for the motor and a 98% efficient transformer, find the motor phase and line currents, and the phase and line currents taken by the transformer’s primary winding. If the secondary is in star, find the turns ratio and transformation ratio. Output power to motor

600 kW 600 Input power to motor = × 100 = 705.8 85 705.8 Apparent power input to motor = = 882 kVA 08 882 000 8820 Motor current = = = 154 A 3 × 3300 57.2

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68 • Advanced Electrotechnology Power output from transformer 705.8 kW 705.8 Power input = × 100 1 = 720 kW 98 720 Apparent power input to transformer = = 900 kV k A 08 900 000 90 000 Mains current = = 1181 A 76.2 3 × 440 Motor line current = 154 A Primary line current = 1181 A The current ratio is given by

Phase current 88 9 A Phase current = 682 A

682 . Note these are the transformer phase currents. 154

The turns ratio is in inverse proportion to the currents, i.e. The transformation ratio is

154 1 = i.e. 1 : 4.43 682 4.43

440 4 1 = = i.e. 1 : 7.5 3300 30 7 5

Note: delta/star transformers introduce a 30°, 150°, 270°, or 330° phase shift and cannot be paralleled with delta/delta transformers.

The Saturable Reactor If the B/H curve of the iron material, used in the reactor core, is examined (figure 3.10a), it is seen there is a relatively flat portion where B α H and permeability μ = B/H is large and substantially constant, the part of the curve used by the design engineer. At higher values of H as the ‘knee’ of the curve is reached, the characteristic ceases to be linear and the B/H or μ value varies from point to point as the graph flattens out. The curve shows the iron core has passed into saturation, tending towards the horizontal. Thereafter the μ value decreases rapidly. Variation of μ is referred to as the incremental permeability and, if plotted against H, N2 I gives figure 3.10b. Now from Volume 6, Chapter 6 – reluctance S = and L = S μA μA AN 2 A so that: L = or rather L α μ. Thus for an iron-cored reactor, inductance varies l with μ vs H, the curve is similar to that shown for μ. The flux density a reactor core can carry into a region is saturated if the energising coil is subject to an over-voltage. This may occur due to a badly designed transformer or if an increased no-load current

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The Transformer Applications • 69 (a)

(b) B

L

L

µ µ

H

H

▲ Figure 3.10

results, accompanied by output voltage waveform distortion. In both cases, reactance is reduced if inductance is reduced by saturation, but this may be of benefit if the incremental permeability μ or inductance can be controlled. Usually saturation is achieved by providing the iron reactor with an initial biasing flux-density value from an extra winding fed from a D.C. source. The extra ampereturns or initial H value results in biasing В, seen as a movement of the H axis upwards. Irrespective of the D.C. current polarity or direction, biased flux-density variation allows the inductance to vary from zero to its maximum value. When the biasing or control coil current is maximum, inductance will be minimum and when the former is zero, inductance will have its normal value. Control is illustrated by the ‘butterfly’ curve (figure 3.11). The two curves are due to the B/H curve hysteresis, while symmetry about the μ axis is due to current direction reversal in the D.C. control coil. If a saturable reactor is connected as shown in figure 3.12, inductance variation by the control winding affects the reactance and the total impedance ( Z ) offered to the μ

H–

H+

▲ Figure 3.11

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70 • Advanced Electrotechnology Load D.C. voltage source

Control winding

Load winding

A.C. voltage source

▲ Figure 3.12

A.C. supply source. By changing impedance the control winding produces a load current change and power variation supplied to the load. The arrangement was first used to fade theatre lights but unfortunately an A.C. voltage was induced into the control circuit by the transformer. Early improvements resulted in a more effective unit known as the transductor.

The transductor This device was similar in construction to a power transformer. It was a type of magnetic amplifier used in power systems for compensating reactive power, consisting of an iron-cored inductor with 2 windings – a main winding through which an A.C. current flowed from the power system, and a secondary control winding carrying a small D.C. current. By varying the small secondary D.C. current, a transductor’s iron core can be controllably saturated at different levels thus varying the reactive power absorbed. Transductors were widespread before the advent of modern solid-state electronics and have largely been replaced.

The magnetic amplifier The magnetic amplifier (Mag amp) is an electromagnetic device for amplifying electrical signals which was invented in the early 20th century, and used as an alternative to vacuum tube amplifiers when high-current capacity was needed. World War II Germany perfected this amplifier in the V-2 rocket, with the Mag amp at its zenith between 1945 and 1960 for power control and low-frequency signal applications. After this the transistor started to replace it and is now largely replaced by transistor-based amplifiers. For the figure 3.13 arrangement, load windings are connected in parallel while control windings are in series. Load windings are connected in opposition so that on alternate half-cycles of the supply voltage, the control ampere-turns drive an opposite core into saturation. Each load winding is designed to support the full supply voltage when the control current is zero.

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The Transformer Applications • 71 Load

A.C. supply

D.C. control input

▲ Figure 3.13

Consider core 1. For a given control current condition, for a +ve supply voltage halfcycle, flux increases until a point is reached when the core saturates and the winding no longer supports the voltage. Core 2, at the same time, is driven away from saturation and would support the supply voltage except for the fact that the core 1 winding is across it. Since core 1 has saturated, its winding effectively short-circuits the core to load winding, so the supply voltage is impressed across the load, resulting in a partial half-cycle of current. For the −ve supply voltage half-cycle, the operation is repeated. Core 1 comes out of saturation but its winding is short-circuited by core 2 which has gone into saturation. Control is obtained over a full cycle by the D.C. control winding with its current maintained continuously in one direction. Control current reversal only changes the order in which saturation of the cores occurs, and the controlling effect is as it was before. Figure 3.14 shows the waveforms associated with a simple mag amp. Output is alternating and pulsed, and may be converted for D.C. operation with a suitable rectifier. CURRENT GAIN. This is a measure of the amplifying property of the arrangement. Under controlled conditions, i.e. when one core is saturated and the effect of the other offset by short-circuiting its winding, there is no flux change. Any load ampere-turns increase is balanced by an equal and opposite increase of control At, so the resultant core m.m.f. remains the same. As high-quality iron is used and worked at high permeability, the m.m.f. needed to cause saturation is small and is neglected. Thus, under saturated conditions, we write: IcNc = IN or

I Nc = where Nc = the number of turns of the control winding and Ic the Ic N

control current. N = the number of turns of the main winding and I the load current. The relationship shows the current gain is proportional to the turns ratio of load and control windings.

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72 • Advanced Electrotechnology A.C. supply voltage

Load current

Control current

▲ Figure 3.14

I c Nc , load current depends on the turns ratio and the control current, N independent of the supply voltages or load resistance, and the amplifier is regarded as a constant current source. Before leaving the current waveform (figure 3.14), it is noted that, since the control current is fed from a D.C. source its value might be assumed constant (from the control circuit resistance). However, if flux is constant, IcNc = IN and if I varies, Ic will vary. Thus a reflection of the load-current peaks is seen in the control current.

Since I =

When cores are unsaturated, load winding impedance is large and only a small magnetising current flows, lagging supply voltage by nearly 90°. When control is Supply voltage ineffective, I is determined by , and as we have the relationship I = kIc, Load resistance the magnetising current is neglected. The characteristic in figure 3.15 can be deduced, indicating that a mag amp may operate where a large load current is varied linearly by a small control current. FEEDBACK. Another way of showing mag amp construction is shown in figure 3.16. The A.C. windings’ inductance is controlled by a D.C. winding consisting of 2 sections

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Load current

The Transformer Applications • 73

Load current =V – R

Magnetising current Control current

D.C. input

A.C. supply

Load

▲ Figure 3.15

in series opposition to eliminate induced A.C. voltages. By proportioning the ratio of D.C. to A.C. turns, the amplifier has an output practically independent of load circuit resistance. When D.C. amplification is needed, a rectifier is placed in series with the A.C. windings. However, the amplifier cannot discriminate between D.C. input signals’ polarities. To increase device gain and to make it polarity sensitive, positive feedback is used. A second winding is placed on the centre limb and connected to the D.C. terminals of a bridge rectifier in series with the load rectifier. With the input signal in one direction a current through the feedback winding aids the control winding and increases gain. With a reversed signal, current through the feedback winding remains in the same direction and opposes the control current,

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74 • Advanced Electrotechnology D.C. input

A.C. supply

Load

▲ Figure 3.16

reducing gain. The new characteristic is shown in figure 3.17. Current gains of up to 300 and power gains of 104 are possible with this arrangement. Thus a bias winding can be added to move the no-signal operating point and if two amplifiers are used a reversible output with no standing current is achieved. Amplifiers can also be used in cascade for increased gain. Magnetic amplifiers have been used in various control schemes for marine applications, providing early automatic voltage regulating systems used to monitor large D.C. currents and to protect A.C. systems. Semiconductor technology device developments such as the Thyristor or Silicon Controlled Rectifier (S.C.R.) were developed in the 1960s by General Electrics to have characteristics similar to the mag amp. The semiconductor alternative is now preferred due to its small size and weight as their production costs are much reduced. Example 3.7. A coil of unknown inductance and resistance is connected in series with a 25 Ω, non-inductive resistor across 250 V, 50 Hz mains. The p.d. across the resistor

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The Transformer Applications • 75

Load current

Max Range of control

Min

–

Control current

+

▲ Figure 3.17

is 150 V, and across the coil 180 V. Calculate the value of coil resistance (3 decimal places), reactance (1 decimal place) and inductance (3 decimal places). Find also the P.F. of coil (3 decimal places). From figure 3.18 the following mathematical solution is possible. (a)

180 V

150 V 25 Ω 250 V

(b) Vy

Vc = 180 V

V = 250 V

φc VX

I V = 150 V

▲ Figure 3.18

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76 • Advanced Electrotechnology Let x = the resistive voltage drop across the coil у = the reactive voltage drop across the coil Then x2 + у2 = 1802 (a) and (x + 150)2 + у2 = 2502 (b) or, on further simplification and rearranging of (b) and (a), we have x2 + 300х + 1502 + у2 = 2502 and x2 + у2 = 1802 Then by subtraction 300х + 1502 = 2502 − 1802 or 300х = (250 + 180) (250 − 180) − 1502 giving 300х = 430 × 70 − 1502 So 300х = 30 100 − 22 500 = 7600 or x =

76 = 25.33 V 3

Then у2 = 1802 − 25.332 = 32 400 − 641.6 = 31 758.4 or y = 102 3.176 = 178.2 V The circuit current =

150 = 6A 25

Resistance of coil =

25.33 = 4.221 Ω 6

Reactance of coil =

178.2 = 29.7 Ω 6

Inductance of coil =

29.7 = 0.095 095 H 2 × 3.14 × 50

P.F. of coil cosφL = 0.141 (lagging). Example 3.8. A three-phase, 440 V, 50 Hz, 90 kW ‘circulating water pump’ motor has an 82% efficiency and operates with 0.8 P.F. (lagging). Calculate (a) the kVA input to the motor (2 decimal places) and (b) the load current (2 decimal places). If the motor is connected to the main switchboard by a 3-core, 95 mm2 cable, 100 m long, calculate the cable line-voltage drop (2 decimal places). Copper resistivity = 17 μΩ mm. Motor power input = 90 ×

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The Transformer Applications • 77

Apparent power input = Load current =

kVA 3kV

=

kW 109.76 = = 137.2 kVA cosφ 0.8 137.2 3 × 0.44

= 180 A

(a)

(b)

pl 17 × 10 −6 × 100 × 103 = A 95 17 × 10 −1 = = 0.018 Ω 95

Resistance of cable core =

The voltage drop in one cable core = 180 × 0.018 = 3.24 V So cable voltage drop in the three-phase line = 3 × 3.24 = 5.6 V Note the introduction of the 3 when considering the three-phase conditions. Example 3.9. While in port a tank-ship obtained its ‘shore-main’ supply from a threephase, 3300/440 V, delta-star transformer. For lighting purposes on board ship the voltage is stepped down by three, 440/110 V, single-phase transformers connected in delta/delta. If the total lighting load, comprised of tungsten filament and fluorescent lamps, is balanced to 15 kW at 0.85 P.F. (lagging), calculate the currents in the connecting cables and the phase currents of the transformer windings. What will be the kVA supplied from the high-voltage supply? It is assumed the transformer loss is negligible and only lighting is supplied (2 decimal places). Lighting transformer. Secondary line current =

15 000 3 ×110 × 0 ⋅ 85

Secondary phase current =

93

Primary phase current = 54 ×

3

= 93 A

= 54 A

110 = 13.5 A 440

Primary line current = 13.5 × 3 = 23.3 A Supply transformer. Secondary line current = 23.3 A Secondary phase current = 23.3 A (since the winding are connected in star) Primary phase current = 23.3 ×

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78 • Advanced Electrotechnology

Primary line current = 1.8 × 3 = 3.1 A Assuming no losses kVA input from supply = kVA output from lighting transformer =

15 kW = = 17.65 kVA. cos φ 0.85

Solid-State Transformers Although hopefully obvious to the reader, the historic context of the transformer has been in terms of high-power handling transmission with the power grid delivering A.C. power (for reasons discussed), with every appliance, from electrical equipment such as vacuum cleaners and hair dryers, to washing machines, P.C.s, notebooks, electric motors, batteries, etc. running on D.C. power. It is only recently that transformer interest has shifted in weighting towards the lower power ‘personal computing power requirements’ end of the spectrum, generating developments in computer based Solid-State Transformer (S.S.T.) applications, an area rather unsuited to the high-power requirements of the majority of marine and terrestrial electrical power distribution systems. The term S.S.T. is a little inaccurate because it is not a transformer in the traditional sense, but an array of high-power semiconductor silicon solid-state components, control circuitry and digital software employed to flexibly convert the voltage and provide control to the power distribution networks, an arrangement referred to as a smart transformer. However, things are rapidly changing. Power electronic converter S.S.T.s have become a focus of recent research. Combined with high-frequency modulation, the volume and weight of S.S.T.s can be significantly smaller than a conventional 60 Hz (US) transformer. Critically S.S.T.s provide active power flow control output for efficient distribution generation from sources, an essential requirement for future renewable energy power grids (whether from solar, wind or otherwise). However, S.S.T.s must have power efficiencies similar to conventional passive transformers (typically 97.5%). In a distribution system overall efficiency is comparable to a conventional passive transformer but until now is still not a power efficient device method, as individual components are only just over 90% efficient, meaning 10% of the A.C. to D.C. power conversions result in wasted heat because silicon is too unreliable for large-scale use at high voltages.

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The Transformer Applications • 79 The US Defense Advanced Research Projects Agency (DARPA) currently supports Project Cree, a project to develop a silicon carbide (SIC) transistor for use in S.S.T.s capable of meeting the needs of the emerging smart grid. SICs are different from earlier Si computer transistors as they can handle high-frequency switching of grid-scale voltages and are suited to the intermittent nature (e.g. energy derived from sources such as waves, tide and wind), typical of renewable energy generation. Although most work has focused on low-voltage converters (400 V), exchanging Si for SiC components increases efficiency well above 90%, with some solar inverters reaching 99%, paving the way for even more efficient, and much smaller, grid transformers. Other materials such as gallium nitride (GaN) may allow operation at even higher voltages. Project Cree, run by DARPA and the US Office of Naval Research (ONR), has built a 10 kV, 100 A power module with advanced SiC MOSFET. These power modules were integrated into equipment from General Electrics to operate a 1 MW transformer at 20 kHz. By comparison the Cree transformer (16 inches high, weighing 75 lb) operates at 250 kVA, while a conventional 330 kVA transformer (55 inches high, weighing 2700 lb) operates at 60 Hz only. Project Cree is developing 15 and 20 kV Insulated Gate Bipolar Transistors (IGBTs), more advanced than the MOSFET, out of SiC to provide 50 and 100 kW S.S.T.s. SiC 12 kV IGBTs achieve 4 times the switching speed and 4 times lower losses than IGBTs fabricated out of Si. The 20 kV IGBT will provide a ten-fold loss reduction compared with silicon. Most S.S.T. advantages come through their reduction in size. Faster scaled-device switching speeds will make it easier to handle multiple power sources feeding into the grid because you have many more transformers controlling and fine-tuning power levels. The high-speed digital software architecture already exists for these applications, having been demonstrated to great effect in the control of microwave active phased array radar systems organising the power output from many thousands of low-power sources. In this way the ‘Victorian’ power grids of the world can be reorganised into something more like the Internet for the 21st Century – a network channelling energy not from just a few centralised power stations but from any number of sources to a selected destination by the most efficient routes (as in communications packet switching) balancing both supply and demand, minimising power plant production waste in terms of waste power and pollution. Conventional transformers handle A.C. power relatively well but require bulky electromechanical switches to redirect energy. Compact S.S.T.s can process D.C. as well as A.C. signals and can be electronically controlled to respond almost immediately to fluctuations in supply and demand. In parallel with the S.S.T. (A.C. to D.C. power converters and vice versa), A.C. to A.C., and D.C. to D.C. circuits are being engineered. If applied in urban centres and megacities use of smart transformers and the smart grid may save many TeraWatt-hours of energy each year (e.g. New York City consumes

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80 • Advanced Electrotechnology about 50 TeraWatt-hours of energy a year) with the real benefits being realised from innumerable small generators provided from renewable and sustainable energy sources (such as the Atlantic Ocean Wave Hub).

Practice Examples 3.1 A direct-reading wattmeter with a 5 A current-coil and a 110 V voltage-coil is used to measure the power in a single-phase, 6.6 kV circuit, carrying a maximum current = 100 A. State the appropriate ratios for the instrument transformers (2 decimal places), and calculate the constant by which the wattmeter reading must be multiplied to measure the power consumed (2 significant figures). 3.2 A 75 kW, 415 V, three-phase induction motor has 80% full-load efficiency. The input line current is measured by a C.T. operated ammeter. Suggest an appropriate ratio for the C.T. if ammeter full-scale deflection is 5 A (2 decimal places). State the expected ammeter reading (1 significant figure). The motor P.F. on full load = 0.87 (lagging). 3.3 A single-phase wattmeter with 5 A and 250 V ranges is used in conjunction with a 25/5 C.T. to measure the power of one phase of a balanced, three-phase, starconnected load. If the load absorbs 12 kW from a three-phase, 415 V, 50 Hz, supply and 0.8 P.F. (lagging), calculate (a) the wattmeter reading (2 decimal places), (b) the load’s impedance per phase (2 decimal places) and (c) if a phase impedance consists of a resistor and reactance in series, calculate their equivalent phase values (3 decimal places). 3.4 The output power of a 415 V, three-phase alternator supplying a balanced load is measured by one wattmeter. If the current transformer has a 25/5 ratio and the wattmeter reading is 7 kW, what is the total power (3 significant figures)? 3.5 For recording the input to a three-phase, 7.5 kW induction motor, which is rated to take a line current of 14 A at 415 V, the following instruments are to be used: Ammeter 0 to 5 A, of resistance 0.08 Ω. Voltmeter 0 to 120 V, of resistance 3636 Ω. Three-phase wattmeter (two element type): current coils 0 to 5 A, of resistance 0.1 Ω; voltage coils 0 to 120 V, resistance 4 kΩ. Estimate the ratio and burden rafting of suitable current and voltage transformers (1 decimal place). 3.6 The primary and secondary voltages of an auto-transformer are 500 V and 400 V respectively. Find the windings’ current distribution if the secondary current is 100 A (2 decimal places), and calculate the possible copper saving percentage (2 significant figures).

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The Transformer Applications • 81 3.7 A double-wound, 6 kVA, 250/150 V, single-phase transformer, is used as an auto-transformer on 400 V mains, to supply a load of 12 kVA at 250 V. Check the windings’ current load and with a practical connection method determine the arrangement’s suitability. 3.8 Find the values of the currents flowing in the branches of a three-phase, starconnected A.T. loaded with 400 kW and 0.8 P.F. (lagging) with a ratio of 440/550 V (3 significant figures). Neglect voltage drops, magnetising current and all transformer losses. 3.9 A 440 V, three-phase induction motor started with a delta-connected A.T. provided with a 70% tapping on each phase winding. Find the voltage applied across the motor terminals at the starting stage when the transformer tappings are in the circuit (4 significant figures). 3.10 A three-phase, 440 V, 40 kW induction motor has 82% efficiency and operates with 0.85 P.F. (lagging). When direct-on started, the motor takes a current of 6 × full-load current and produces a torque of 1.5 × full-load torque. Calculate the current taken from the supply and the ratio of starting to full-load torque if a motor is started through a star-connected A.T. with a 75% tapping (2 decimal places).

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4

D.C. MACHINES For an engineer it is not how full or empty the glass is, but what design size you require. Chris Lavers

In Volume 6, the D.C. generator and motor were the subjects of separate chapters. It was stressed that further work was required to develop particular aspects of theory. Our electrical study, biased towards the marine engineer’s requirements, has in recent years been limited by the scope of the electrical syllabus, to the exclusion of D.C. systems. Certainly a book devoted to marine electrotechnology must relate to modern developments and trends. Current practice has seen considerable change and new electrical equipment installed aboard ships, but D.C. systems are still present on marine and terrestrial platforms. It should be remembered that the dynamo was the first electrical generator to produce D.C. with the advent of the commutator, and were the first electrical generators able to deliver power for industry, the foundation upon which other electric-power conversion devices were based, including the electric motor, alternator and rotary converter. A commutator-based dynamo has, however, many disadvantages, while converting A.C. to D.C. with semiconductor power rectification devices is proven effective and economic. Never before has such a high standard of electrical knowledge of both A.C. and D.C. systems been required by the working marine engineer. Automated engine-rooms (and bridges) are areas for which the engineer is trained where knowledge of relevant control systems must be based on correctly learned fundamental concepts.

Testing of D.C. Machines Before considering test methods, let’s examine briefly the need for, and test types made by, electrical machine manufacturers. The waveforms from small marine generators

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D.C. Machines • 83 are more erratic compared with their land counterparts, and often change under load. This chapter looks from an electrical view point but considerations must be given to fuel consumption, frequency stability, emissions (NOx, CO) and noise levels to provide full generator (or motor) characterisation. A wide range of marine generators exists on the market: Kohler, Mastervolt, Onan, etc., and interoperability is an issue, e.g. take frequency stability, this is critical for ship induction cookers – even a small variation from a nominal 50 Hz generator output is critical. Small machine development typically starts with prototype construction, incorporating any desired features. The machine is subject to a series of direct loading tests, in accordance with the design team. Results help check design effectiveness, with machine output measured to obtain overall efficiency. The final temperature rise, behaviour on load, commutation, vibration, noise and response to speed and excitation control are checked. Test results help confirm or amend the empirical design formulae, and are examined in detail to extract appropriate information. When the prototype is proven to meet the intended rating and purpose, planned production may commence. At the earliest production stage, minor modifications are introduced to speed manufacture and assist mass production. The first machine for sale is comprehensively tested to provide guaranteed performance figures and results for tests made on specimen sample machines as they leave the production line. The development procedure described applies to mass produced machines not required for special duties. A machine is considered special due to size, requirements, dimensions or enclosure which warrants different to usual construction methods. In such cases designers rely heavily on experience and skill to building machines to a customer’s specification. Only limited testing may be possible at the manufacturer’s works because of the machine size or operating requirements, and assessment tests are devised. Such tests are made and results used to check the design before a machine is installed. The final testing, certification and acceptance tests are then made on site under the correct operating conditions.

Output Machine output is measured by direct loading. For motors this is achieved with a brake, calibrated generator or electrical dynamometer. For generators, loads can be water resistance tanks or calibrated variable resistors. Waste energy and its dissipation are key factors if large machines are tested.

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84 • Advanced Electrotechnology

Efficiency Efficiency of an electrical machine may be expressed as: Efficiency =

Output Input Losses Output or or Input Input Output Losses

Efficiency (η) is usually expressed as a percentage and for a generator can be written as: ηG =

Output (W or kW) Output (W or kW) + Losses (W or kW)

Similarly for a motor: ηM =

Input (W or kW) Losses (W or kW) Input (W or kW)

Losses The losses which occur in electrical machines are considered under 2 main headings: (1) Electrical Losses and (2) Mechanical Losses. (1) ELECTRICAL LOSSES (PE). These comprise the Copper Losses (PCu) and the Iron Losses (PFe). Copper losses occur in the armature conductors and connections due to the passage of current through the material’s ohmic resistance, and are determined from the expression Ia2Ra. A copper loss also occurs in the field windings and is 30–50% of the full-load loss. Copper loss is often minimised using the maximum amount of copper on the machine consistent with commercial design, and is usually calculated at the working temperature. Iron losses occur in the armature, pole-tips and other parts of a magnetic circuit, subject to cyclic magnetisation or flux change, consisting of a Hysteresis Loss (РHY ), and an Eddy-Current Loss (PEC) (dealt with in detail in Volume 6, pp. 152–154, 352). Hysteresis loss is proportional to flux density to the power 1.6 and speed or frequency (РHy α Bm1.6N), and, eddy-current loss is proportional to flux density squared and speed squared, or frequency squared, i.e. PEC α Bm2N2. (2) MECHANICAL LOSSES (PM). These comprise Windage and Friction Losses (bearing and brush) with ventilation needed to reduce electrical machine temperature rises. Windage loss is a big issue, some machines have a comparatively smooth armature with little fanning effect and most of a machine’s heat dissipates through the yoke, the case for

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D.C. Machines • 85 a low-speed machine reliant on radiative heat loss and basically of ‘open’ construction. High-speed machines have smaller dimensions and rely on forced ventilation, i.e. an associated fan. Windage loss is proportional to the cube of speed but is comparatively small even when a fan is fitted. Bearing friction is minimised using the correct bearing type for the duty undertaken by a machine. Small and medium size high-speed machines are often provided with ball or roller bearings. Low-speed machines of such sizes use well-lubricated sleeve bearings. In marine work this bearing is chosen because of its superior vibration performance. For large machines, special bearing arrangements involve forced high-pressure lubrication. Rotor weight and speed, drive type, position and machine mounting are factors which decide the bearing type. Bearing friction loss is roughly proportional to speed. Brush friction depends on the friction coefficient, the total pressure and peripheral commutator speed. It must be appreciated that a specific machine brush grade should be chosen in accordance with the design requirements and must not be altered. In summary: Mechanical losses are proportional to speed and attention will be drawn to this relationship as required. Example 4.1. A 220 V shunt motor takes 10.25 A on full load. Armature resistance is 0.8 Ω and the field resistance is 880 Ω. Losses due to friction, windage and iron = 150 W. Find the output power and motor efficiency on full load (3 significant figures). Motor input on full load = 220 × 10.25 = 2255 W Field current =

220 = 0.25 A 880

Armature current = 10.25 − 0.25 = 10 A

Armature copper loss on full load = 102 × 0.8 = 80 W Field copper loss = 220 × 0.25 = 55 W

Friction, windage and iron loss = 150 W

Total losses = 80 + 55 + 150 = 285 W Full-load output = input – losses = 2255 − 285 = 1970 W = 1.97 kW Full-load efficiency =

1970 = 0.874 or 87.4 % 2255

ROTATIONAL AND CONSTANT LOSSES (PR and Pc). The mechanical losses (friction and windage) are proportional to speed and the iron losses are proportional to speed (for constant flux density). The term Rotational Loss (PR) includes both PM and PFe. Another description of machine losses are the Constant Losses (Pc) and the variable losses. In this manner the field copper losses are constant for most working conditions and included

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86 • Advanced Electrotechnology with rotational losses (for a constant speed) to give Pc. Variable losses are the armature copper losses and an efficiency expression is developed thus: Output or, for example, as a generator a Output + Losses Output ηG = Output + Arm. Cu loss + Constant loss

Efficiency =

CONDITION FOR MAXIMUM EFFICIENCY. If field copper loss is included in the constant loss, the field current is considered negligible compared with the armature current on load, with the efficiency expression written as above. Thus for a generator: ηG =

VIa VIa + Ia2 Ra + Pc

Dividing by VIa ηG =

1 Ia Ra PC 1+ + V VIa

or

IR P 1 = 1+ a a + c ηG V VIa

If calculus is used and the proof taken in a simple manner, then: I R P I −1 1 = 1+ a a + c a , and differentiating with respect to Ia: ηG V V ⎛ 1⎞ d⎜ ⎟ ⎝ η⎠ R P I −2 Then =0+ a − C a dIa V V This expression will be zero when

Thus if

Ra PC Ia −2 1 = and the minimum value of obtained. V V ηG

Ra P = c then by simplifying: PC V VIa2

Ia2 Ra and the minimum value of

1 or maximum ηG

value of ηG occurs when: Armature copper loss = Constant loss.

As Ia2 Ra = PC

Ia =

PC , the current at maximum efficiency. Ra

Note: the field copper loss is included in Pc. The same condition for maximum efficiency can be devised for a motor.

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D.C. Machines • 87 Example 4.2. A 220 V shunt generator is rated to have a full-load current of 200 A. Its armature resistance is 0.06 Ω and its field resistance is 55 Ω. Rotational losses, at correct speed and excitation, are measured to be 3 kW. Find the output power rating of the prime-mover, and the load current for maximum efficiency (both 4 significant figures).

Field current =

220 =4A 55

Armature current = 200 + 4 = 204 A

Armature copper loss = 2042 × 0.06 = 204 × 12.24 = 2497 W Field copper loss = 220 × 4 = 880 W

Rotational loss = 3000 W

Constant loss = Rotational loss + Field Cu loss ∴PC = 3000 + 880 = 3880 W Required generator input = Output + Arm. Cu loss + Constant loss = 220 × 200 + 2497 + 3880 = 50 377 W Thus prime-mover rating = 50.38 kW Current for maximum efficiency =

3880 38 8 = 102 = 102 6.466 466 = 254 A 0 06 6

This is Ia, so load current will be 254 − 4 = 250.0 A. STRAY LOSSES. When machines operate on load, small extra losses occur, arising from iron loss increases (due to field distortion in the armature reaction). Usually these are ignored.

Testing methods Electrical machine efficiency and output are determined by 3 common methods. DirectLoading Methods which use various types of brake for loading I.C. Engines and Turbines. Assessment Tests estimate output and efficiency. On-load performance cannot be checked because results are obtained from the no-load running conditions. RegenerativeTest Methods, so-called ‘back-to-back’ methods, are favoured because large machine fullload performance can be checked, and testing made with minimal power waste. DIRECT LOADING. A brake measures the machine’s output power and is applied to a motor operated at the correct voltage, speed and load. If brake measurements are made in newton metres with speed in rev/min, output power is determined from the expression:

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88 • Advanced Electrotechnology

P=

2πN NT N with N in revolutions per minute and T in newton metres. 60

So Efficiency is obtained from

2π NT 60 VI

(1) The Friction Brake. This can be a rope, band or drum type (taking the form shown in figures 4.1a and 4.1b). (a)

(b)

W w

R

W

R

▲ Figure 4.1

For type (a) the force recorded by spring balances and acting on the drum equals (W − w) newtons. If the radius of the drum is R metres, torque T is given by (W − w) R newton metres. For type (b) the force, recorded by a spring balance, is W newtons. If the brake arm radius is R metres, the torque T = WR newton metres. A disadvantage of friction brakes is the difficulty in dissipating heat generated at the drum. Excessive load placed on the bearing nearest the brake may reduce efficiency. The method is suited to small motors. Example 4.3. In a motor brake-test, the effective load on the brake-drum = 222.7 N, the effective drum diameter = 0.5 m and the motor speed = 960 rev/min. The motor input = 30 A at 230 V. Calculate the motor output and efficiency (2 significant figures). Braking torque on motor 222.7 × 0.25 Nm = 55.68 2 × 3.14 × 960 × 55.68 Motor power outpu p t= 60 = 3.14 × 32 × 55.68 = 5594.7 W or 5.6 kW 5 5595 η= 30 × 230 or Efficiency = 0 81 81%

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D.C. Machines • 89 (2) The Calibrated Generator. This is a generator pre-calibrated by testing held in the test-room. Its output is fed into a resistive load, which is dissipated as heat, or fed into a common busbar system. Figure 4.2 shows the motor tested, directly coupled to the calibrated generator and its electrical arrangement. Let P be the power transmitted through the coupling. Then η M

P IMVM

d ηG =

IGVG IV IV Whence η M η G = G G or η M = G G P IMVM IMVM η G

A

A P

Test supply V

M

G

V

Load

▲ Figure 4.2

Generator efficiency η

Generator efficiency is known from the calibration tests, and determined for any output on the calibration graph (figure 4.3) with these values substituted into the efficiency expression.

Output VC IC

▲ Figure 4.3

This method’s advantage is that energy output can be put to useful purpose, i.e. (Internal Combustion) I.C. engine-makers frequently use the generator initially as a motor for running-in engines, taking the supply from other loaded engines and fed into common busbars, or energy supplied from load tests can be fed back into the factory and used.

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90 • Advanced Electrotechnology (3) The Electrical Dynamometer. This machine is constructed as a generator or motor (figure 4.4). Since the machine stator tends to rotate when an armature is loaded (due to magnetic coupling between field and armature), a restraining torque must be provided to give the reaction normally at the feet of a loaded machine. A torque-arm is fitted and the reaction measured by a spring balance, so T = WR newton metres. Motor output, or the input to the generator under test, is measured directly and no calibration curve is needed.

w

R

▲ Figure 4.4

A dynamometer is constructed as a motor held in a test room for generator testing or as a generator to test motors. Its output as a motor (or input as a generator) is checked and used with the machine measurements to which it is coupled. Electrical energy generated during test is used for the calibrated generator. If effective use or feeding energy back into the ‘mains’ is not worthwhile, a dynamometer can be modified to waste heat. The armature is constructed like an induction motor cage and water cooled. Cage bars are heated by eddy currents and cooled by water flowing continuously, which runs to waste. The main advantage of direct-loading methods is that machine performance on load can be checked. Output and efficiency are determined, commutation and temperature rise is observed and relevant design specifications are noted. ASSESSMENT TESTS. Two testing methods are considered. The first is well-established and named after Sir James Swinburne (1858–1958), a baronet and pioneer in electrical machinery development, and is the best technique for assessing efficiency without loading a machine and should be appreciated. A second lesser method is used for determining the size of various machine component losses. Tests are made with the aid of a small auxiliary motor and little waste power involved. Limitations of the assessment testing method are noted when the procedure is taken and calculations made. (1) The Swinburne Method. This is suitable for large machines where loading either as a generator or as a motor might present difficulty. For a D.C. shunt motor change of speed from no load to full load is small. Mechanical loss is assumed to stay constant from no load to full load. If field current is held constant during loading, core loss is assumed to remain constant. To make a test, a machine is run ‘light’ as a motor irrespective of its

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D.C. Machines • 91 final function. Armature and field resistances are measured before the no-load test and ambient temperature are noted. During the test, the supply voltage and input current are recorded after the speed is adjusted to the value at which the machine is required to operate on load. Efficiency is then assessed (figure 4.5). A A

Mains

V

Ra

Rf

▲ Figure 4.5

When a machine is run light no output is taken from the shaft (net mechanical output power is zero), only sufficient input power is needed to supply the machine no-load losses (the copper and rotational losses). Thus Input Power = No-load copper losses + Iron losses + Mechanical losses = No-load armature Cu loss + Field Cu loss + Rotational loss or VI0 = Ia20 Ra + If20 Rf + PR Whence: PR

VII0 V

Ia20 Ra

If20 Rf

Once rotational loss is determined for a particular speed and excitation condition, one can assess the efficiency at any load (speed and excitation is assumed constant). Thus for a generator we use the expression: Efficiency = or η G =

Output Output = Output + Losses Output + Arm. Cu loss + Field Cu loss Rotational loss

VI where Ia and If are the load values. VI − I R + If2 Rf + Rf 2 a a

For a shunt-connected machine If = I0 so calculation is easier for determining the constant loss. The predictive efficiency procedure is illustrated by example 4.4. Example 4.4. A 50 kW, 500 V shunt generator is tested by the Swinburne method. When run light as a motor it takes a 10.1 A no-load current at the correct voltage and speed. Ra = 0.25 Ω and Rf = 500 Ω. Assess efficiency at full load as a generator (1 decimal place). No load.

Supply current 10.1 A

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If0 =

500 = 1A 500

So Ia0 = 10.1 1 = 9.1 A

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92 • Advanced Electrotechnology Input = Cu losses + Rotational loss or 500 ×10.1 = 9.12 × 0.25 + 500 × 1 + PR Thus PR = 5050 − 20.7 − 500 = 5050 − 520.7 = 4529 W Full load. Line current =

50 000 = 100 A 500 If

∴η = =

500 500

Ia = 100 + 1 = 101 A

50 000 50 000 + (Cu loss + PR ) 50 000 +

(

50 000 × + (1× 500 ) + 4529

)

50 (worked here in kW ) 50 + 2.55 0.5 4.529 50 = = 0.868 868 or Efficiency = 86.8% 57.58 =

This problem can also be solved by finding the constant loss thus: No load. Input = Armature Cu loss + Constant loss or Constant loss = 5050 − (9.12 × 0.25) = 5050 − 20.7 Pc = 5029.3 W or 50.3 kW Full load. η=

Output Output + Arm. Cu loss + PC

50 50 + 2.55 + 5.03 50 = or Efficiency f = 86.8% 57.58 =

With the Swinburne test, a closer approximation to actual efficiency is obtained if Ra and Rf values are raised to a ‘hot’ value, e.g. 50°C rise. Regarding assumptions made in such problems, no speed and flux variation between the no-load and full-load tests is permitted, so constant iron and mechanical losses are assumed. (2) Summation of Losses Method (by Auxiliary Motor). Figure 4.6 shows the arrangement with 2 machines coupled together by a belt. In practice to avoid belt-slip or losses due

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D.C. Machines • 93 Field excitation supply

Auxiliary motor

Main machine

Supply

▲ Figure 4.6

to belt friction, machines are fixed in line and directly coupled. The smaller machine is the auxiliary motor and tests are made in stages. Let P0 be the input power to the auxiliary motor when running light and at the speed required to drive the large machine at its operational speed of N rev/min. Let P1 be the input power to the auxiliary motor when coupled to the large main machine and driving its armature in an un-excited field at an operational speed = N rev/min. So: P1 − P0 = PM the mechanical losses (windage and friction) of the large machine. Let P2 be the input power to the auxiliary motor, coupled to the main machine and driving its armature in an excited field at a speed of N rev/min. Then P2 − P1 = Iron losses (PFe) of the main machine under the correct test conditions of speed and flux. Since PM and PFe = PR rotational loss (PR) is determined from P2 − P0. If Ra and Rf are measured the main machine efficiency can be assessed for any load (like the Swinburne test). The test method, using an auxiliary motor, further interests the designer as any effects of small alterations can be determined. Input power readings are taken first with the brushes lifted off and again with them at the correct pressure. Losses due to brush friction are then obtained. Tests can be repeated with variations of bearings, ventilation openings, fans, etc., and data obtained for improving performance. The test set-up is useful for determining the constituent iron losses parts. The procedure is as follows. Separation of Iron Losses. From Volume 6, iron losses were made up of hysteresis loss and eddy-current loss or PFe = PHy + PEC. If flux density Bm is kept constant during testing, iron losses are written: PFe = Hf + Ef 2 or for a machine: PFe = HN + EN2 with HN representing hysteresis loss and EN2 the corresponding eddy-current loss. Examination of the equation PFe = HN + EN2 shows that a solution is possible if PFe values are obtained at 2 speeds. Tests are made by keeping flux density constant and varying the driven armature speed. If the auxiliary motor method is used iron loss is measured at normal

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94 • Advanced Electrotechnology speed and excitation. With excitation constant, testing is repeated at half or double speed and a solution of the equation is possible, illustrated by example 4.5. Example 4.5. A test is made on a 230 V shunt motor to determine the constituent iron losses. The shunt field is separately excited at 230 V and its excitation kept constant. Measurements were made of the iron losses at full speed and half speed. At 1500 rev/min iron losses = 1100 W, at 750 rev/min losses = 450 W. Find the hysteresis and eddy-current losses at normal speed, i.e. 1500 rev/min (1 significant figure). Substituting in the equation PFe = HN + EN2 we have 1100 = HN + EN2 450 =

HN ⎛ N⎞ +E ⎝ 2⎠ 2

(a) 2

(b), or

1100 = HN + EN2

(a), and

1800 = 2HN + EN2

(c)

Subtracting (a) from (c) we have: 700 = HN so the hysteresis loss = 700 W The eddy-current loss is thus: 1100 − 700 = 400 W At half speed losses are: Hysteresis

700 400 = 350 W , and the Eddy-current = 100 W . 2 2

REGENERATIVE TESTS. Direct machine loading is the best method of checking performance because efficiency, commutation and temperature rise can be noted. Loading difficulties are overcome by operating the test machine with a similar machine which provides electrical power or absorbs it. These two machines operate ‘back-toback’ and just enough energy is taken from the mains to supply both machine losses. The method was adapted to shunt machines by John Hopkinson, once Professor of Electrical Engineering at London University. Circuit modification allows the technique to be adapted for compound machines. Testing of series machines presents complications but can be achieved, provided appropriate circuit modifications and protective arrangements are made. With mass production obtaining a similar machine is not difficult to achieve. (1) The Hopkinson Method. The test is performed and used for machines which are similarly rated. Machines are mechanically and electrically connected (figure 4.7). The machine operating as a motor is started by cutting out the starting resistance and speed adjusted by variation of the field regulator, controlling IfM . The generator selfexcites and its voltage is controlled by the regulator varying the current I f G . A change-over

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D.C. Machines • 95 I

+

A A If

IG

IM

A

A

V

M

Mains

A

C

G

M 1

–

If

2

Paralleling switch

▲ Figure 4.7

switch is operated from position 1 to position 2 and the generated voltage EG adjusted and compared with the mains voltage V. When EG = V the parallel switch is closed since no current can circulate between G and M for this condition. If EG is increased by raising If G , a generator can be loaded, i.e. it supplies current. The motor consequently loads and requires an increased current from the generator. By manipulating the field regulators of both machines a condition is achieved when either the motor or generator is fully loaded. Power taken from the supply will be a minimum and then supplies both machine losses. Under this condition: VI = PRM

PRG + Ia2M RaM + IfM V + Ia2G RaG + IfG V where IaG = IG + IfG and IaM

IM − IfM

From this expression PRM PRG is found and if they are assumed to be similar the rotational loss of each machine can be found. Resistance Ra and Rf are measured ‘hot’ after the test and efficiency calculated for any load under normal working conditions either as a generator or motor (example 4.6). Note: approximate efficiency is found on the test-bed for checking from the following. Let P be the power transmitted through the coupling between motor and generator. Then η M

P VIM

d ηG =

VIG P

∴η G η M =

VIG I P × = G P VIM IM

If the efficiency of each machine is assumed equal then: η 2

IG IM

η=

IG IM

Example 4.6. Two shunt motors on a Hopkinson test take a current of 15 A from 200 V mains. The current to the motoring machine is 100 A and the field currents are 3 A and 2.5 A. If the resistance of each armature is measured to be 0.05 Ω estimate the efficiency of each machine under the particular test loading conditions (1 decimal place).

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96 • Advanced Electrotechnology Total input = 15 × 200 = 3000 W. The generated e.m.f. of the generator must be higher than the motor back e.m.f. and both armatures are rotating at the same speed, then the field current of 3 A must refer to the generator and the 2.5 A to the motor. Since IM = 100 A Then IaM = 100 – 2.5 = 97.5 A and IaG = (100−15) + 3 = 88 A ∴3000 = 2 PR + Motor armature Cu loss + Motor field Cu loss + Generator armature Cu loss + Generator field Cu loss or 3000 = 2PR + (97.52 × 0.05) + (2.5 × 200) + (882 × 0.05) + (3 × 200) = 2PR + 475.3 + 500 + 387 + 600 Then η M = =

So 2PR = 3000 – 1962.3 = 1037.7 W or PR = 518.8 W

Input Losses Input

(100 × 200 ) − (518.8 + Arm. Cu loss + Field Cu loss)

20 000 (100 × 200 ) − (518.8 + 475.3 + 500 ) = 20 000 20 000 0 − 1494.1 18 505.9 = = × 100 or Efficiency = 92.4% 20 000 20 000 and η G = = =

Output Output + Losses 85 × 200

s s (85 × 200 ) + (518.8 + Arm. Cu losse

Field Cu losses )

85 × 200 (85 × 200 ) − (518.8 + 387 + 600 )

17 000 0 17 000 + 1505.8 17 000 = × 100 or Efficiency = 91.7% 18 505.8 =

Check. Approximate machine efficiency =

85 = 0.85 100

η = 0.922 or 92.2%

The Hopkinson test can be made for compound machines if care is given to the correct series field connection. If 2 compound generators are tested, since one machine ‘motors’, the series field must be reverse connected as armature current direction will reverse and differential compounding for a motor otherwise results. The motor, if differentially connected, could not be loaded effectively and correct testing would be

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D.C. Machines • 97 impossible. Assuming a satisfactory load test is achieved, rotational loss and efficiency are determined as described for shunt machines, not forgetting the added series field copper losses. (2) Field’s Method. Due to the tendency for series motors to race on no-load and difficulties of operating a series generator in parallel with a constant voltage supply, regenerative test method is not possible without taking precautions and giving the test-rig continual attention. In its essential form Field’s test is not strictly a regenerative method. Two similar series motors are mechanically coupled, one machine acting as a motor and the other as a separately excited generator. Both series fields are connected in series with the motor armature and the generator output absorbed in a resistance load with no switches in the circuit to eliminate any possibility of a detached load. The method of calculating efficiency is given in example 4.7. Example 4.7. The following results were obtained from a Field’s test on 2 series motors: Motor current 20 A; Generator output current 17.5 A; Generator voltage 160 V; Supply voltage 220 V; Resistance of each armature 0.6 Ω; Resistance of each series field 0.4 Ω. Estimate full-load efficiency of a machine as a motor (2 decimal places). Input to test rig = 220 × 20 = 4400 W Output of generator = 160 × 17.5 = 2800 W Total losses P = 4400 − 2800 = 1600 W Motor armature copper loss = 202 × 0.6 = 240 W Motor and generator field copper losses = 202 (2 × 0.4) = 400 × 0.8 = 320 W Generator armature copper loss = 17.52 × 0.6 = 183.75 W Rotational loss of 2 machines = 1600 − Total copper losses or 2PR = 1600 − (240 + 320 + 183.75) = 1600 – 743.75 = 856.25 W and PR = 428.125 W. Then η M =

Input Losses Input

Now voltage drop across the generator series field = 20 × 0.4 = 8 V ∴Voltage applied to motor = 220 − 8 = 212 V

Input = 212 × 20 = 4240 W

Losses = PR + Motor armature Cu loss + Field Cu loss = 428.125 + 202(0.6 + 0.4) = 428.125 + 400 = 828.125 W or η M =

42 040 − 828.125 3411.875 = = 0.8047 or Efficiency = 80.47% 4240 4240

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98 • Advanced Electrotechnology

D.C. Generators in Parallel The benefits of operating generators in parallel are known to marine engineers and we will consider the theory for satisfactory running and load sharing. The technique of paralleling 2 D.C. generators involves basic conditions and safeguards built into the switchboard to ensure a correct operational sequence. Thus ‘to put a machine on the bars’, the generator must be run up to its correct speed and its voltage adjusted until it equals that of the busbars. The circuit-breaker of the ‘incoming’ machine is then closed, when it floats on the bars. If the excitation is then increased with the field regulator, an output current is then noted indicating the generator is taking load. The busbar voltage increases and adjustment of the field regulators of all the machines in parallel is needed before the correct busbar voltage and load sharing between generators is achieved. Essential conditions for correct paralleling of D.C. generators are assumed and the polarity of the incoming machine correct, i.e. it corresponds to that of the busbars which is ensured with moving-coil voltmeters. If the polarity reverses, because of a fault, the ‘machine’ voltmeter will indicate this. Compound generators are often used so an equalising connection is required. Such a connection must be made before an incoming machine circuit-breaker is closed. Modern equipment ensures mechanical interlock between switch and circuit-breaker, typically a 3-pole circuit-breaker. The equalising connection is usually the centre pole and given a ‘lead’, i.e. its contacts ‘make’ first, before the +ve and −ve poles ensuring the equalising connection is automatically assured first.

Parallel operation The possibility of D.C. generators operating in parallel is investigated by considering their load characteristics. It is reminded that for a shunt-connected generator, terminal voltage falls with increasing load current, so the characteristic is a falling one. For a series machine terminal voltage rises with load current until saturation of the magnetic system occurs, a characteristic described as rising. For compound generators, the shape of the characteristics is decided by the series field strength, which requires special attention in parallel operation. (1) Shunt Generators. Such machines are easily paralleled and operate with stability for the following reasons. If the prime-mover of No. 1 (see figure 4.8) slowed down momentarily, the generated e.m.f. would fall and the output current, I1, given by the E V would fall. expression 1 Ra1

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D.C. Machines • 99 I

I1

I2

2

+

E2

E1

Ra

Ra

2

1

V

1

–

▲ Figure 4.8

Stability is assured by the fact that a reduced input means a reduced output. E1 might fall below V when current I1 reverses so the machine ‘motors’. Sharing of load with a similar machine can be considered. Take 2 similar machines in parallel and operating correctly. Next there is a load increase, i.e. a demand for extra current. The voltage of each machine falls with the result that each tries to ‘reject the load’. Since both machines tend to ‘sit down’, eventually a new load is taken jointly by both machines, the final proportion on each decided by the load characteristic slope (influenced by the prime-mover characteristics), the sizes of armature resistance and excitation. Stable working conditions are a feature of parallel shunt generators or a shunt generator in parallel with a battery. (2) Series Generator. A series generator will not self-excite until the load circuit is completed and its resistance made less than the critical value. Such a machine cannot be paralleled onto ‘live busbars’. To investigate рагаllеl operation, assume 2 similar series generators in parallel, started together from rest with the load circuit completed. If the latter is adjusted until self-excitation results, operational stability can be examined (figure 4.9). I

Ic

I I2= – + Ic 2

+

I I1= – – Ic 2 V

Ra

2

2

1

Ra

1

–

▲ Figure 4.9

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100 • Advanced Electrotechnology Let E1 and E2 be the e.m.f. of each generator and R1 and R2 the resistance (armature and field). Generators share load as long as E1 = E2. If there is a momentary slow-down, E1 becomes less than E2, and a current circulates whose value will be: Ic =

E2 E1 R1 R2

As the load current is I, generator No. 2 now supplies a current = a current =

I + Ic and No. 1 supplies 2

I − Ic . 2

The effect of the new current conditions strengthens field No. 2 but weakens field No. 1. The generated e.m.f. follows in response, the machine currents and their effect is cumulative until No. 2 takes all the load and current No. 1 reverses, once E1 falls below busbar voltage V. No. 2 now tends to ‘motor’ No. 1 which when field polarity is reversed, tends to brake in order to reverse. Generator No. 2 under this condition supplies a large current which could burn out both machines. Failing this occurring and if the primemover of No. 1 had picked up speed and sufficiently powerful to drive the generator in the original direction, the machine polarity reverses, resulting in 2 generators in series on a short-circuit − a disastrous condition. Series machines operate in parallel in a highly unstable state and to avoid this, the current in field No. 1 must be made independent of the e.m.f. The method provides a connection between the armatures, an equalising connection (dotted in figure 4.9) and should have negligible resistance. If this fault occurs, the circulating current Ic will not pass through the field winding, avoiding polarity reversal. The fields are now parallel and their currents substantially equal, even though the armature currents are different. Stability is introduced with fixed polarity, if the No. 1 armature current reverses, the machine will motor and the prime-mover speeds up. (3) Compound Generators. Such generators have a series field and rising characteristics. When operating in parallel, if due to momentary slowing down of a generator its e.m.f. falls, a current imbalance occurs with consequent variation of field strengths caused by the series fields. Once the e.m.f. of No. 1 falls below V, the current in the series field reverses and differential compounding results in more severe faulty conditions which are cumulative and may result in polarity reversal. The machine is liable to an instability common to series generators. Figure 4.10 shows how an equalising connection is introduced, shown dotted. Stability is assured because the current directions in the series fields are independent of the reversed armature currents in No. 1 and polarity reversal is avoided. The equalising connection is made between the armature and series field. If commutating or interpoles are fitted they are considered part of the armature circuit. Marine regulations require series fields to be fitted on the −ve side for uniformity, and ammeters to be in the opposite pole.

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D.C. Machines • 101 +

I A

2

A I2

E2

Ic

I1 E1

1 V –I 2

–I 2

–

▲ Figure 4.10

Load sharing For D.C. generators as for batteries, no difficulty in sharing load is encountered provided stable working in parallel is achieved. Consider a battery of e.m.f. 4 V connected in parallel with another of e.m.f. 4 V. No current flow is observed, which is explained by the fact that no e.m.f. acts around the complete circuit formed by the 2 batteries. The 4 V of No. 1 is opposed by the 4 V of No. 2, resulting in no effective e.m.f. for a current. If a resistor is connected between the common +ve and −ve lines a current will flow, part of which is due to battery No. 1 and the rest from battery No. 2. The proportion of shared current is decided by the batteries’ internal resistances and by application of basic circuit laws. For D.C. generators similar reasoning is applied but since machine characteristics are not always straight lines load sharing problems are best solved by graphical rather than mathematical methods. Both methods are illustrated next. Example 4.8. Two shunt generators A and B, each with straight line load characteristics operating in parallel. The characteristics are: Machine

Open-circuit voltage

Terminal voltage at 50 A

A

460 V

420 V

B

440 V

410 V

Determine how both machines share a 100 A load and find the common busbar voltage (1 decimal place).

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102 • Advanced Electrotechnology Mathematical Solution. For a generator since E = V + IaRa then: For machine A: 460 = 420 +

(

×

460 6 − 420 40 = = 0.8 Ω. Thus 50 50 For machine B: 440 = 410 +

(

×

aB

) so R

aB

aA

aA

=

)

so RaA =

460 − 420 and, 50

=08Ω 440 − 410 30 and thus u RaB = =06 Ω 50 50

When in parallel let IA = current of machine A and Iв = current of machine B. Let IA + IB = I (the total load current). Let V = the common busbar voltage. V + IA RaA

Substituting in equations E A or 460 = V + 0.8 IA and, EB

(a)

V + IB RaB

440 = V + 0.6 IB

(b)

Using equation (b) then 440 = V + 0.6(I − IA) = V + 0.6 (100 − IA)

or

440−60 = V − 0.67IA

giving 380 = V − 0.6IA

(c)

Using (a) and (c). Then 460 = V + 0.8IA 380 = V – 0.6IА Subtracting 80 = l.4IA

or

A

=

80 = 57.1 A 1.4

Also IB = 100 − 57.14 = 42.89 A And the common busbar voltage V = 460 − (0.8 × 57.14) = 460 – 45.712 = 414.2 V. Graphical Solution. Two methods are possible and both are applied to this problem. Method 1. The two characteristics are plotted (figure 4.11a) with voltage ordinates spaced 100 A apart. The common terminal or busbar voltage is given by the intersection of the lines. The required answers can be read off the graph. The method is suited for problems where load sharing for one current value is needed. If a problem requires solution for more than one load current method 2 is more suitable.

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D.C. Machines • 103 460

460 M

ac

hin

440

e

440 A

420

420 414 V

eB

hin

400

c Ma

380

400

57 A

360

0 100

20 80

380

43 A

40 60

60 40

Terminal voltage – machine B (V)

Terminal voltage – machine A (V)

(a)

360 100 0

80 20

Load current (A) (b)

460

Terminal voltage (V)

440

Com

bine

420 33 A

d ch

arac

tic

50 A

Ma

400

teris

Ma

ch

ine

ch

B

ine

A

380 360

0

20

40 60 80 Load current (A)

100

120

▲ Figure 4.11

Method 2. The procedure is to draw both characteristics to a common voltage ordinate (shown in figure 4.11b). The ‘combined characteristic’ assumes the generators are in parallel and supply currents satisfying the common busbar voltage condition. Assume the busbar voltage = 420 V. Machine A supplies 50 A and machine В 33 A. The total current supplied = 83 A. Plot this point and repeat the procedure. For a busbar voltage of 410 V, machine A supplies 63 A and machine В supplies 50 A. Total current is 113 A. One plots this and obtains the combined characteristic. This latter values determine the busbar voltage and currents supplied by each machine for any load current value. Thus a load current of 120 A consists of 66 A given by machine A, and 54 A given by machine B. The busbar voltage is then 408 V.

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104 • Advanced Electrotechnology

Commutation and Armature Reaction Before considering the full treatment of Commutation and Armature Reaction for D.C. machines, the reader should revise the basics of Volume 6, Chapter 7. The modern D.C. armature benefits from refinements such as the equalising ring, duplex winding, etc. In its simplest form, the pattern follows by connecting the coil elements to each other and to the commutator segments and is called a winding. Coil elements can be preformed from several turns in series or of multiturn coils connected in parallel. Coil elements are connected to form a closed circuit and the method used results in the winding which are classed as either ‘lap’ or ‘wave’. The description of lap and wave windings is not repeated here, as this was covered previously (Volume 6, Chapter 12).

Commutation For both lap and wave windings as a brush passes from one segment to the next, a coil element of a single coil or a series of coils is short-circuited briefly during the commutation process. Brushes are placed at points on the commutator where currents converge from and diverge to parallel paths of the armature. At these points brushes contact adjacent segments connected to coil elements in the interpolar gaps, optimally placed for short-circuit. In summary: the currents in the armature coil elements on either side of the brushes are in opposite directions and as the commutator turns and a complete S.C. element passes under a brush from one side to the other, current in the element must reverse. In the arrangement of figure 4.12 adjacent segments pass under a brush, and current in connected coil element X must be stopped and restarted in the opposite direction. A width of brush equal to that of a commutator segment is assumed. 50 A

50 A

50 A

50 A

X

50 A

50 A

X

50 A

50 A

50 A

t=1

t=0

100 A

50 A

100 A

▲ Figure 4.12

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D.C. Machines • 105 t

t

Current

+

Time

–

▲ Figure 4.13

Figure 4.13 shows the current/time curve for coil element X being commutated. The time marked ‘t’ is the period when current direction reverses, i.e. the time taken for a brush to pass completely from one segment to the next, is perhaps 1/1000th second or less. The ±ve signs of the current ordinate are arbitrary and indicate current reversal. The graph illustrates ‘straight-line commutation’ where each brush contact area section takes its correct proportion of current, illustrated in figure 4.14. 50 A

25 A

50 A

X

25 A 75 A

50 A 0 A 50 A

50 A 25 A 50 A

X

X

50 A 50 A

– t =1 4

– t =1 2

75 A 25 A – t =3 4

100 A

▲ Figure 4.14

The commutation condition here is for the ‘ideal case’ and is not achieved in practice. For the coil element being commutated, because of self and mutual inductance effects, due to the rapid collapse of flux associated with a changing current, an induced e.m.f. is set up called the reactance voltage. According to Lenz’s law, this e.m.f. opposes and tends to slow the rate of current change. Thus current in the S.C. coil element changes more slowly and the condition is shown by a new current/time graph (figure 4.15). It is seen from the actual working conditions graph that although the coil current has reversed, it has not reached its full –ve value when commutation finishes. The effect

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106 • Advanced Electrotechnology 50 A Act

ual

Id

ea

+ Current

l

3 – 4

1 – 2

1 – 4

25 A

– 50 A t

▲ Figure 4.15

3 1 of ‘under-commutation’, where for the instant considered, i.e. when t = , the last 4 4 section of a brush carries 50 A instead of the 25 A for ideal straight-line commutation (figure 4.16). The effect of concentrating current into the brush trailing edge can result in sparking, burning and overheating. Furthermore, at t = 1, when the segment is due to break contact with the brush, the coil element current being commutated is 25 A, so the remaining 25 A must pass between the segment and brush as an arc, until coil current X rises to its correct value. Arcing and heating results in commutator damage. The arcing condition is shown in figure 4.17. 50 A

0A

50 A

X

50 A

50A – t =3 4

100 A

▲ Figure 4.16

Since for the coil element X inductance and resulting reactance voltage are the main causes of poor commutation, it is clear these must be minimised. If an e.m.f. can be generated in coil element X to produce a current in the same direction as occurs when the element enters the armature-current path after it has been commutated, the effect of

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D.C. Machines • 107 50 A

25 A

50 A

X

50 A

25 A

25 A

Arc

t=1

100 A

▲ Figure 4.17

under-commutation can be cancelled. This is the main means of improving commutation but commutation can benefit by increasing brush resistance. ‘Resistance commutation’ is used in small machines and is one of the methods used for improving commutation. Example 4.9. Consider a machine where the following commutation conditions occur. If a 100 A current is collected by a brush and the commutator speed = 1200 rev/min, calculate the reactance voltage (1 significant figure). There are 50 commutator segments. The coil element inductance connected to adjacent segments = 10 μН. Speed of commutator

1200 60

20 rev/s. Thus 1 revolution =

The time taken for a segment to pass under a brush = Current in a coil element being commutated = Rate of change of current in the coil = So reactance voltage = L

1 s 20

1 = 1× 10 −3 s 20 × 50

100 or 50 A 2

50 − ( −50 ) = 100 × 103 = 1× 105 A/s 1× 10 −3

di = 10 × 10 −6 ×1 × 1 × 10 5 = 1 V . di

METHODS OF IMPROVING COMMUTATION. These are considered under 3 distinct headings but a combination of 2 methods is often used. (1) Resistance Commutation. Carbon (with a little copper added) is the material most commonly used for D.C. machines, whose resistivity is much higher than copper. However, even small changes in composition (a few percentage) can significantly alter brush properties. The effect of a high-resistance brush is considered in figure 4.18.

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108 • Advanced Electrotechnology 50 A

50 A Y

X

A a b

▲ Figure 4.18

As segment A contacts and moves past a brush, current from coil element Y can pass along 2 paths. It can flow directly down brush section a or it can continue through coil element X into the other brush section b. As a result of the commutator movement under the brush, section b gets smaller and its resistance rises while section a gets larger and its resistance falls. The overall effect of brush resistance on commutation is now seen. At the first instant segment A contacts with a high-resistance brush, little current passes and the unchanged value of current continues to flow through coil element X. As more of brush a is contacted, brush resistance path a falls inversely in proportion to the movement, and current diverts to take the falling resistance path. Coil element X current falls in accordance with the commutator segments’ movement, following the condition of straight-line commutation. When current in X has reversed and increases, correct commutation is aided by the rising resistance of section b and the falling resistance of section a. The rising reversed current is forced through the coil in accordance with the ideal requirements. (2) E.M.F. Commutation. This method improves commutation by generating a reversing e.m.f. in the S.C. coil element, such a reverse e.m.f. opposes the reactance voltage and is achieved by (a) giving brushes a ‘lead’ or (b) using Commutating Poles – often called Interpoles. (a) Brush Displacement. For a drum winding to operate correctly, brushes are placed in a neutral position to contact segments connected to coil elements in the interpolar gaps. For this condition, no e.m.f. is induced in coils being short-circuited by brushes but if, for a generator, brushes move forward in the rotation direction to a position slightly in advance of the neutral position, commutation takes place in a reversing field. Since the coil element, being short-circuited, is now no longer in the interpolar gap but in the field of the main pole ahead, an e.m.f. is induced because of armature rotation and in a direction opposite to that resulting from the coil element passing under the

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D.C. Machines • 109 poles prior to commutation. The new e.m.f. opposes the reactance voltage and tends to stop the original current, accelerating its build-up in the reverse direction to satisfy the ideal commutation condition. The method needs a different ‘angle of lead’ of the brushrocker for each load current, but for small machines a compromise is achieved by fixing brushes at the ‘lead’ required, about 3/4 load. This position, together with resistance commutation, gives a reasonably sparkless performance over working load range. For a motor, brush displacement is in the opposite direction, i.e. against the rotation direction with a ‘lag’ angle. The armature motoring current will be in a direction opposite to that for generating and the reversing e.m.f. must be opposite, i.e. obtained from the pole behind. (b) Use of Commutating Poles. Instead of moving forward a coil element being commutated to be under the field of the pole ahead, the same effect is achieved by moving the magnetic field back by adding small electromagnetically energised poles midway between the main poles, whose polarity and flux density are decided by current passing through the exciting coils. Commutating poles (compoles) are run at low flux densities to avoid saturation, and energised by coils of a few turns of wire or strip conductor, connected in series with the armature. Energising ampere-turns vary commutating flux density in accordance with the armature current. The commutating e.m.f. generated in the S.C. coil will vary with the reactance voltage which in turn varies with load, i.e. the coil current being commutated. Although armature reaction has yet to be studied figure 4.19 shows the revolving armature acting as a stationary electromagnet. When conductors carry current with small poles placed as shown, flux from the armature ampere-turns passes through the armature and round the yoke. Polarity is given to the compoles, i.e. flux passes from

Armature flux

Commutating flux S

– N

S + N

▲ Figure 4.19

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110 • Advanced Electrotechnology a N pole through the armature to the S pole. Since flux due to the armature reaction must be cancelled by the commutating poles’ magnetic fields, if the latter are to achieve this, the compoles’ ampere-turns must produce the induced commutation e.m.f. This arrangement means the brushes need not move from the neutral position and the ampere-turns per pole of the commutating poles is made to exceed the armature ampere-turns by some 25%. For a generator, interpoles have the same polarity as the main poles ahead of them and for a motor, the same polarity as the main poles behind them. The direction of rotation is taken as the reference.

Armature reaction We will now look further at the armature reaction effect. The armature conductors produce flux when a machine is supplying or taking current. This armature reaction effect provides the characteristics of various D.C. generator and motor types. Armature reaction may be defined as: ‘the effect on the main-field flux caused by armature flux, which due to the armature current, distorts and weakens the main-field flux’. BRUSHES ON NEUTRAL AXIS. Consider the simple 2-pole machine of figure 4.19 but without compoles (figure 4.20). According to basic theory, brushes must be on the neutral axis midway between the main poles, i.e. on the Magnetic Neutral Axis (M.N.A.) which coincides with the Geometric Neutral Axis (G.N.A.). If an armature is loaded as a generator and carries current, if no interpoles are provided, associated armature flux takes the path of least reluctance and cross the air-gap orthogonally (figure 4.20).

G.N.A

– N

S +

G.N.A

▲ Figure 4.20

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D.C. Machines • 111 Armature flux is at right angles to the main flux and termed a cross-magnetising flux crossing each air-gap twice, i.e. if the N pole-piece is considered, armature flux is opposite to the main flux at the leading pole tip and in the same direction at the trailing pole tip. The terms leading and trailing are used in this regard with the armature rotation taken as the reference for this distinction. The net effect of a cross-magnetising flux is seen in figures 4.21a and 4.21b where the total flux twists and the M.N.A. moves through an angle of lead. G.N.A. and M.N.A.

(a)

S

N

Main flux Armature flux (b)

Strengthening Trailing pole tip

N

Leading pole tip Weakening

Weakening G.N.A. M.N.A. Leading pole tip

S

Trailing pole tip Strengthening

▲ Figure 4.21

Flux effects are illustrated by an m.m.f. vector diagram (figure 4.22) showing the effect of armature reaction developed so far. Fm is the main field m.m.f. and Fa is the armature or cross-magnetising m.m.f. As the M.N.A. moves forward, brushes must be on this axis to function properly and move onto the new M.N.A. Since the lead angle changes as the armature current changes the brush position must change. This is a poor arrangement and attempts to cancel the cross-magnetising flux are beneficial.

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112 • Advanced Electrotechnology Fm

Fa

F

▲ Figure 4.22

G.N.A. M.N.A.

θ N

S

θ

▲ Figure 4.23 Fm

Fb

1

Fm

θ

Fa

Fc

F

▲ Figure 4.24

BRUSHES GIVEN AN ANGLE OF LEAD. The arrangement of figure 4.23 has the brushes in optimum position for a generator, i.e. contacting the commutator at appropriate tapping points, while allowing correct commutation of the S.C. coil element. With the passage of current, the armature electromagnet axis turns through θ° and the m.m.f. vector Fa is as shown (figure 4.24). Armature flux combines with the main flux to give the resultant flux, represented by a new m.m.f. vector diagram. The resultant F is found by resolving Fa into quadrature components Fc and Fb. Fb is a demagnetising component which weakens Fm to give a new reduced value Fm1 . Fc is the cross-magnetising component which acts with Fm1 to give the new resultant F. To ensure satisfactory commutation, the new armature reaction condition results in the 2 effects mentioned previously, namely: (1) the main

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D.C. Machines • 113 field weakens and (2) the main-field flux density in the air-gap distorts. The forward movement of brushes or angle of lead is only required for generators. For a machine operating as a motor, the armature current direction is reversed for the same rotation direction and main field polarity. As a result armature flux reverses and brushes are given an angle of lag for correct operation. The main flux is distorted and weakened as before, and if brushes move when a motor runs, speed can vary. This is not a recommended method of speed variation as optimum commutation conditions are desired, but the cause of speed variation is due to flux-density variation. CROSS-MAGNETISING AND DEMAGNETISING AMPERE-TURNS. In Figure 4.23, the armature ampere-turns is considered to consist of 2 parts. One a band of conductors between lines XX and YY constitutes an energising electromagnet whose effect is represented by the m.m.f. Fb of figure 4.24. These current carrying conductors form ampere-turns which oppose Fm, the main-field ampere-turns (and are demagnetising ampere-turns) giving a final main field Fm1. The demagnetising ampere-turns are proportional to the angle of lead or lag, and corresponds to the vector diagram, i.e. Fb = Fa sin θ. Conductors outside the lines XX and YY constitute an ampere-turns band or energising solenoid when carrying current, resulting in a cross-magnetising m.m.f., Fc from the total armature ampere-turns, if brushes are not moved off the G.N.A. Since armature reaction reduces the main field, armature demagnetising ampere-turns set up by the brushes’ movement should not be too big when related to the ampere-turns needed for the main field under full-load. The ratio of the latter to the former is always made greater than 1, ensuring armature reaction effects are not too big and machine characteristics not unduly altered. Example 4.10. A 150 kW, 250 V, 6-pole D.C. generator has a lap-wound armature with 432 conductors. If the brushes are given a 5° angle of lead from the G.N.A., estimate the added ampere-turns per pole needed to neutralise an armature’s demagnetising effect when a machine is on full load. Neglect the shunt field current. A solution is considered with 2 pole-pitches. Full-load current =

150 000 = 600 A 250

Effective conductors in two pole-pitches =

Current per conductor 432 = 144 3

600 100 A 6

Effective armature turns = 72

Thus armature ampere-turns = 7200 360 or 60 mechanical degrees are equivalent to 180 6 electrical degrees or 1° (mechanical) = 3° (electrical).

Since this is a 6-pole machine then

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114 • Advanced Electrotechnology If θ is the brush shift in electrical degrees then θ = 15° (electrical). The proportion of armature demagnetising ampere-turns to the total armature ampere-turns is given by 2θ . 180 2 15 Thus demagnetising ampere-turns per pole pair × 7200 = 1200 180 or demagnetising ampere-turns/pole = 600. The cross-magnetising ampere-turns can also be determined thus: Total armature ampere-turns = 7200 Demagnetising ampere-turns = 1200 ∴Cross-magnetising ampere-turns = 7200 − 1200 = 6000 Cross-magnetising ampere-turns/pole = 3000. FLUX-DENSITY DISTRIBUTION AROUND THE ARMATURE. The armature and field poles have been drawn in developed form (figure 4.25).

N

S Main field flux-density distribution (no armature current)

Armature field flux-density distribution (armature carrying current – no main field)

Flux-density distribution due to the combined effects of main field and armature current

▲ Figure 4.25

The flux-density distribution under the main poles and in the interpolar gap, caused by the main-field ampere-turns and the armature ampere-turns respectively, is shown separately and combined. The armature-current m.m.f. acts with the main-field m.m.f. to produce a resultant m.m.f. and flux-density distribution pattern which is uneven

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D.C. Machines • 115 over the pole arc. The e.m.f.s generated in the armature-coil elements are non-uniform as they pass across the pole face and voltages between commutator segments may, in some positions, be much higher than the mean value. If field distortion is pronounced there is a danger of a ‘flash-over’ across the commutator. The effect of the armature m.m.f. can be mitigated against partly by fitting a compensating winding, which is an expensive solution only used in large machines subject to heavy and sudden current peaks, and is not discussed here as use is limited. Examples of such applications include motors for steel-rolling mills and special generators such as those used in electric locomotives.

Special D.C. Machines Some D.C. machines must perform special duties which may have unique construction features or operating characteristics. Many smaller machines have application in the marine field and are of interest to the marine engineer.

The rotary transformer This machine is a D.C. generator and D.C. motor combined consisting of an armature with separate windings on the same core and operating in a common field system. In essence it is a variable frequency transformer that aids the process of allowing electrical signals associated with 2 rotating components near one another to ‘connect’ without generating undesirable activity, helping reduce components’ wear, and are commonly found in recording equipment, e.g. VCR recorders and washing machines. Each winding is connected to its own commutator and its physical dimensions chosen when considering D.C. machine construction. Old ship’s depth sounders used to operate off a 24 V D.C. supply from 220 V D.C., so a static transformer could not be used. A small rotary transformer was used instead. If a sounder needs 36 W one winding is wound 36 for = 1 5 A serving as the generator. Assuming 60% overall efficiency the input will 24 100 60 3 be × 36 = 60 W. The other winding is rated: = = 0.273 A or 273 mA and 60 220 11 functions as a motor winding. Field flux is provided by a permanent magnet or shuntconnected field coils energised from the motor supply. Commutation requirements are simple as the armature reaction effect is reduced by the fluxes of motoring and generating currents tending to cancel each other out. There is no voltage control of the field on the generating side, as this alters the common flux and motor speed. The

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116 • Advanced Electrotechnology machine can be installed for a specific duty and operates at predetermined speed and started like a D.C. motor.

The rotary converter Although at one time widely used, the A.C. to D.C. rotary converter is now largely replaced by the semiconductor rectifier, but is still found in service. Even as late as 1999, large terrestrial substations using synchronous rotary converters operated in the New York city subways. Semiconductor devices do not need daily maintenance, manual synchronisation for parallel operation, skilled personnel, requiring only periodic visits for inspection and maintenance. Rotary converters have an advantage that they can be run ‘inverted’, i.e. supplying A.C. from a D.C. source, but because there is a fixed relation between A.C. and D.C. voltage, and that varying the field alters speed and frequency, the more common motor-alternator arrangement is preferable for ship work.

The rotating amplifier This device was a cross-field generator (C.F.G.) given the brand name Amplidyne, a special-purpose D.C. generator which still supplies large precisely controlled D.C. currents to large D.C. motors used to drive heavy physical loads, such as gun turrets or missile launchers. It is really a motor and generator which for most maritime purposes has been replaced in low-power applications by the MOSFET and Insulated-Gate Bipolar Transistor (IGBT).

Practice Examples 4.1 A D.C. shunt motor has an armature resistance of 0.9 Ω and takes an armature current of 18 A from 230 V mains. Calculate the motor’s power output and overall efficiency (3 significant figures). Rotational losses are measured to be 112 W and shunt-field resistance is 300 Ω. 4.2 Calculate the light-load current taken by a 100 kW, 460 V shunt motor assuming the armature and field resistances to remain constant and to equal 0.03 Ω and 46 Ω respectively. Efficiency at full load is 88% (2 significant figures). 4.3 A 200 V, 15 kW motor when tested by the Swinburne method gave the following results: Running light, the armature current was 6.5 A and the field current 2.2 A.

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D.C. Machines • 117 With the armature locked the current was 70 A, when a potential difference of 3 V was applied to the brushes. Calculate the efficiency on full load (1 decimal place). 4.4 A Hopkinson ‘back-to-back’ test made on two shunt generators rated at 150 kW, 220 V, required 140 A to be taken from the mains. The field currents of the two machines were 5 A and 7 A and the armature resistance measured to be 0.02 Ω. Estimate the efficiency of each machine when loaded to 150 kW (1 decimal place). 4.5 The results of a test made to determine the open-circuit characteristic of a ship’s shunt-connected generator are set out below. The test was made with the machine being separately excited and driven at 700 rev/min. Terminal voltage (V)

10

20

40

80

120 160 200 240 260

Field current (A)

0

0.1

0.24

0.5

0.77

1.2

1.92 3.43

5.2

(a) If the generator is coupled to a 850 rev/min engine and to be made selfexciting, find the range of the field rheostat required to vary the O.C. voltage between the limits of 200 V and 250 V. The resistance of the shunt field was measured to be 52 Ω (1 decimal place). (b) If the armature resistance was 0.04 Ω, estimate the O.C. voltage and rheostat setting if the armature is to deliver 200 A at a terminal voltage of 220 V. Neglect the effects of armature reaction and brush-contact voltage drop (1 decimal place). 4.6 A ventilating fan in a ship’s galley is driven by a 220 V, 10 kW series motor and runs at 800 rev/min at full load. The total armature circuit resistance is 0.6 Ω. Calculate the speed (4 significant figures) and percentage change in torque (2 significant figures) if the current taken by the motor reduces by 50% of the full load value. The efficiency of the motor is 82%. Assume flux is proportional to the field current. 4.7 An engine-room ventilator fan motor is a series machine with a total resistance of 0.5 Ω and runs from a 110 V supply at 1000 rev/min when the current is 28 A. What resistance in series with the motor will reduce the speed to 750 rev/min? The load torque is proportional to the square of the speed (e.g. for a fan) and the field strength can be assumed to be proportional to the current (2 decimal places). 4.8 A 230 V, 10 kW shunt motor with a stated full-load efficiency of 85% runs at a speed of 1000 rev/min. At what speed should the motor be driven, if it is used as a generator to supply an emergency lighting load at 220 V? Armature resistance is 0.2 Ω and the field resistance is 115 Ω. Find the kW ‘rating’ of the machine under this condition (2 decimal places).

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118 • Advanced Electrotechnology 4.9

A series D.C. motor is run on a 220 V circuit with a regulating resistance of R ohms for speed adjustment. The armature and field coils have a total resistance of 0.3 Ω. On a certain load, with R zero, the current is 20 A and the speed is 1200 rev/min. With another load and R set at 3 Ω, the current is 15 A. Find the new speed and also the ratio of the two values of the power output of the motor. Assume the field strength at 15 A to be 80% of that at 20 A (2 decimal places).

4.10 A D.C. shunt machine generates 220 V at 800 rev/min on open circuit. The armature resistance including brushes is 0.4 Ω and the field resistance is 160 Ω. The machine takes 5 A running as a motor on no-load at 220 V. Calculate the speed and efficiency of the machine taking 45 A at 220 V (1 decimal place). Assume armature reaction weakens the field by 3%.

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5

D.C. RECTIFICATION Edison and Tesla – the real transformers of electricity. Chris Lavers

Transformers, by their design, only operate with the application of a continuously varying input voltage, varying from +ve to –ve and then back again to +ve. The requirement of some electrical applications on a D.C. (+ve only) supply requires a method of turning the A.C. output of a transformer into a D.C. waveform. Thus ‘rectification’ describes the conversion of an A.C. voltage or current waveform into a constant or D.C. value. The basic principles of rectification were discussed when the action of a commutator, as fitted to a D.C. generator, was first introduced, but the more widely accepted meaning is in connection with operation of static devices such as the semiconductor diode. The diode is a device which only permits a current to flow in one direction, a device with zero resistance when a +ve voltage is applied but with infinite resistance when a –ve voltage is applied across it. Mention will be made of the one-way conducting property of the diode semiconductor (Volume 7, Chapters 11 and 12) and although other types are named in this chapter, it is sufficient to point out that rectification can be achieved by any of these devices irrespective of how they operate. The general relationships between input (A.C.) and output (D.C.) values hold good for any specific devices used to achieve rectification. To understand D.C. rectification some revision of electromagnetic induction is needed. Electromagnetic induction was first described in Volume 6, Chapter 7, and the reader will find it useful to review this subject. For the more able student however, sufficient reference is made here to basic principles to enable study to proceed. The subject of static rectification is new but knowledge of terms, such as r.m.s. and average values, together with their relationships for a sinusoidal waveform, is assumed from Volume 6. The integration method will also be introduced here.

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120 • Advanced Electrotechnology

Electromagnetic Induction In Volume 6, both Faraday’s and Lenz’s laws were introduced. These laws stated the magnitude and direction of the e.m.f. which is self-induced in an inductor when its associated flux-linkages are changing. A basic formula was given for the average induced e.m.f. =

Flux-linkages NΦ or Eav = Time of change t

Eav gives the value of the induced e.m.f., when N is the number of turns of the coil, Φ is the flux value (in weber) associated with the coil, and t is the time taken for the flux to change. This formula is useful, if a steady rate of change occurs but if the change occurs in a short interval or varies from instant to instant the condition is best expressed as NdΦ e= . dt Note: it is the flux that is changing and not the number of turns. Now the calculus method was introduced and is suitable here for finding the rate of change of flux as instantaneous values are involved. Here e gives the instantaneous magnitude of induced e.m.f. resulting from a rate of change of flux, which differs from instant to instant.

Inductance A changing current in an inductor results in a changing flux which is responsible for an dΦ induced e.m.f., which is proportional to the rate of change of flux-linkages or e N . dt Now any value of flux is given by: (the flux set up by 1 ampere) × (the value of current in amperes). Thus Φ = (Flux set up by 1 ampere) × I and, The rate of change of flux = (Flux set up by 1 ampere) × (rate of change of current). Or

dΦ di = (Flux set up by 1 ampere) dt dt

Thus e

N

dΦ di = (N × Flux set up by 1 ampere) dt dt

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D.C. Rectification • 121 If the value within the bracket is called L, the coefficient of self-induction, or more simply di the circuit inductance, then the expression becomes: e L which was introduced in dt this form in Volume 6, Chapter 9. NΦ or I L = Flux-linkages per ampere. The unit of inductance is the Henry defined as: ‘A circuit has an inductance of 1 Henry, if an e.m.f. of 1 volt is induced in it when the current changes at the rate of 1 ampere per second.’

The assumption L = N × Flux set up by i amperes can be developed to L =

A further development from L = the flux linkages per ampere is obtained thus: Since Flux =

m.m.f F NI NI μA A then Φ = = = Reluctance S l μA l

Note: l is used for the length of the magnetic circuit to avoid confusion with L, the inductance. Since Φ =

μ ANI μ AN then the flux set up by 1 ampere = l l

Remember that μ μ r μ o and since = ∴L N×

NΦ = I

× Flux set up by 1 ampere

μ AN μ AN 2 or L = l l

The formula determines a coil’s inductance, if the area and length of the magnetic circuit and its permeability are known. For a given inductor working at a constant permeability, the inductance is proportional to the coil turns squared.

The Direct Current LR Circuit Consider a coil of resistance RL and inductance L is carried on an iron core and connected in parallel with a non-inductive resistor R as shown in figure 5.1. The arrangement is energised from a D.C. power supply. When the switch is closed, current I2 through the resistor R rises to its maximum value instantaneously as shown in figure 5.2. Current I1 through the inductor L takes time to grow to its maximum value. The maximum final V V value of each current is given by Ohm’s Law, i.e. I2 = and I1 = as a coil has an R RL electrical inertia due to its inductance.

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122 • Advanced Electrotechnology

+

A

I1

A V

RL

I2

L

R

–

▲ Figure 5.1

Current (I )

(a)

(b) I2

I1

Time (t )

▲ Figure 5.2

Growth of current An explanation for the shape of the I1 curve (graph b) is as follows. An increase in current is accompanied by an increase of flux which, because it is changing, induces an e.m.f. in the coil. By Lenz’s law the direction of the e.m.f. opposes the change taking place, i.e. it tends to counter the applied voltage and halt the growth of current. The e.m.f. can be di described as a back e.m.f. and its value at any instant can be stated as: e L . dt From earlier work on induction and inductance, it was seen that, the effects of induction are always present, even under a circuit’s steady running conditions. For our study we are concerned with transient conditions, i.e. effects which occur when an inductive circuit is switched ‘on’ or ‘off ’. Our main interest here is in D.C. circuit conditions but mention will be made of corresponding A.C. conditions. SHAPE OF THE CURRENT CURVE. This is considered for a condition of inductance with and without resistance. The first condition is theoretical but helps understanding of the second.

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D.C. Rectification • 123 (1) CIRCUIT WITH INDUCTANCE (L) AND NEGLIGIBLE RESISTANCE. In accordance with theory so far, we know that the induced back e.m.f. created when a circuit is switched on will oppose the applied voltage V. There is no voltage drop due to resistance and e di Ldi di V or = must equal V. Thus V e. but e L . ∴V = dt dt L dt V amperes per second. This L current will not reach a steady value since there is no resistance and continues to grow (figure 5.3a).

With V volts applied, current increases at a steady rate of

Current (i )

(a)

Time (t ) L t0 = – R

Current (% of I )

(b)

75 63.2

L t = 3 × 0.693 – R L t = 2 × 0.693 – R V I= – R

50 25 0

0.693 t0

Time (t )

▲ Figure 5.3

(2) CIRCUIT WITH INDUCTANCE (L) AND RESISTANCE (R). At any instant after the circuit switch is closed, the current will be i amperes. The voltage condition is given by: Applied voltage = Resistance voltage drop + Induced back e.m.f. ∴V = iR + e or iR + L

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di dt

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124 • Advanced Electrotechnology

Rewritten, the change of current at the instant considered is Thus

V

iR L

di V iR = and this, being a differential equation will solve to i dt L

Rt − ⎞ ⎛ I ⎜1− e L ⎟ . ⎝ ⎠

Here e is not the symbol for induced e.m.f. but is the base of Naperian Logarithms, where e = 2.718 and log10e = 0.4343. Differential equations of this type are derived at appropriate points and the solution method will be of interest to students. This method is set out as follows for the above equation. Since V = iR + L or

di then V dt

iR L

di =0 dt

V L di L di −i − = 0 and I i = R R dt R dt

Rearranging

di I i

=

R dt and by integration L −loge (I − i ) =

R +K L

К is a constant of integration and its value is obtained thus: When t = 0

i = 0 and К = –logeI Rt L

Using this value for K loge (I i ) or

Thus

loge (I i ) loge I = − Rt − I i =e L I

loge I

Rt I i Rt and loge =− L I L

i=I(

e

−

Rt L

).

This is often known as the Helmholtz equation and shows that the growth curve is of exponential form. Further investigation of the mathematics leads to the term timeconstant and its meaning. Example 5.1. A 20 H inductor with a resistance of 20 Ω is connected to a 200 V D.C. supply. Calculate the value to which current will rise 0.1 seconds after switching-on and how long it takes to grow to a value of 5 A (both 3 decimal places). Here

20 × 0.1 − 200 10 A and = 10 (1 − e 20 ) 20

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D.C. Rectification • 125

or i

(

e −0 1 ) = 10 10 − 10 10e −0.1

let x 10 e −0.1 then log x = log 10 – 0.1 log e or log x = 1 – 0.1 × 0.4343 = 1 – 0.04343 = 0.95657 giving x = 9.048 Thus i = 10 − 9.048 = 0.952 A Rt − ⎞ ⎛ I ⎜1− e L ⎟ ⎝ ⎠

Similarly substituting in i

5

Thus

1 1 = 1 − e − t or = e 2 2

t

t

) Here

R 20 = =1 L 20

whence 2 = e t giving t log e log 2

or 0.4343 t = 0.301 Thus t =

0.301 = 0.693 seconds. 0.4343

THE TIME CONSTANT. If the initial rate of growth of current was sustained, the time for Current value the current to attain its full value will be given by: Initial rate of change of current V L Let this time be t0 then t 0 = R = V R L This time t0 is the circuit time constant. In this time, current rises to a value given by RL − ⎛ ⎞ ⎛ 1⎞ i I ⎜1− e L R I −e I 1− ⎝ e⎠ ⎝ ⎠

(

Thus i

I

)

⎛ 2.718 − 1⎞ or i ⎝ 2.718 ⎠

1.718 i = 0.632 I 2.718

The value of i will be 63.2% of I, shown in figure 5.3b. The time constant is defined as the L ratio and is equal to the time taken for the current to attain 63.2% of the full value. R The following deduction will be of interest.

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126 • Advanced Electrotechnology

When the current increases to half its value in a time then Thus e

−

Rt L

Rt 1 or 2 e L 2

Rt log e L

So log2 = and

Rt Rt − I I = I I e L or = I e L 2 2

Rt log2 0.301 L = = = 0.693 and t = 0.693 L l g e 0.4343 R

L seconds or 0.693 t0 seconds. R 3 L Similarly the current reaches of its final value in a further 0.693 seconds. 4 R The current reaches half value in 0.693

Example 5.2. A lighting circuit is operated by a relay of which the coil has 5 Ω resistance and an inductance of 0.5 H. The relay coil is supplied from a 6 V D.C. source through a pushbutton switch. If the relay operates when the current in the relay coil reaches a value of 50 mA, find the time interval between pressing the button and the closing of the lighting circuit (3 significant figures). Determine the circuit’s time constant (1 decimal place). Here I

6 5

dI =

.

50 = 0.05 A 1000

5t − ⎛ ⎞ 05 Then 0.05 = 1.2 ⎜ 1 − e 0.5 0 5=12(1 ⎟⎠ or 0.5=12(1 ⎝

10 t 0.5 = 12 −12 12e −10

and 10t log e log 1 1.043 043 or t =

10t

) giving:

12ee −10 t = 11.5 whence e10 t = 12

24 = 1.043 23

0.0183 0.0183 = 10 × 0.4343 4.343

giving t = 0.00421 or 4.21 ms Also t 0 =

L 05 = = 0.1 1s R 5

Decay of current Assume for simplicity that the inductor, after connection to the supply, is short-circuited at the same instant as the supply is switched off. As the field collapses the energy put

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D.C. Rectification • 127 into the magnetic field must be released through a current with heat generated in the circuit’s resistance with the arrangement shown in figure 5.4. The current decays in a circuit with the same constants as before. +

R V

i

K

L

–

▲ Figure 5.4

For the condition considered V = 0 and the induced e.m.f. at any instant opposes the decay of current. The voltage condition at any instant is given by: 0 = iR + L

di dt

This is a differential equation which can be solved to i The solution will be:

Ie

−

Rt L

di iR di R = − or = − dt dt L i L

Integrating both sides loge i = −

Rt +K L

К is the constant of integration and its value is obtained thus: When t = 0 i = I ∴K = logeI giving loge or loge

Rt L

loge I = −

loge I Rt L

Rt − i which can be written as = e L I

i Ie

−

Rt L

SHAPE OF THE CURRENT CURVE. The decay of the current is of exponential form and a reflection of the growth curve. It is shown in figure 5.5 where t0 is the time constant,

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128 • Advanced Electrotechnology

e=V e

Current (%)

I= V – R 75

i L t = 0.693 – R x

50

L – x t = 2 × 0.693R

25 0

– t0 = L R

Time (t )

▲ Figure 5.5 Rt LR

I I = or i 0.368 I . As before e 2.718 L the current decays to half the original value in a time of 0.693 seconds. R the current falls to a value equal to i

Ie

−

Ie −1 =

The e.m.f. or voltage-drop curve is as shown, since e at any instant is responsible for the discharge current and equals the circuit voltage drop v where v = iR.

The field switch and discharge resistor If the field circuit of a large machine such as an alternator or D.C. generator is opened di quickly, a high and potentially dangerous e.m.f. may be induced. Since e L dt theoretically if current falls to zero instantaneously, it follows that e will be infinitely high. In practice this does not happen. As current begins to fall an e.m.f. is induced which tends to maintain the current at its original value. The direction of the e.m.f. assists the mains voltage and the current flow is prolonged, sometimes generating an arc across the switch contacts. Added resistance is introduced into the circuit by the resistance of the arc and a condition of balance exists during the circuit-breaking period, when the value of applied voltage plus that of the induced e.m.f. equals the voltage drop across the field resistance plus that across the resistance of the arc. The arrangement is undesirable due to the fact that voltage developed across the field depends on the speed of opening a field circuit. Thus opposite condition may arise, that for safety purposes, it is desired to ‘kill’ the field quickly with a fast opening switch. This creates a large voltage induced across the field which results in insulation breakdown

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D.C. Rectification • 129 di di tends to a large value so e L also tends to a dt dt large value). The alternative is to make the switch open relatively slowly and tolerate a di long arc which would persist and burn the contacts (since will tend to a small value dt di so e L tends to a small value). The field dies slowly but the induced voltage will be dt kept at a safe value. and damage to the machine (since

A compromise arrangement provides an extra contact on the field switch which connects in a non-inductive ‘discharge resistance’ RD in series with the main field coils, at the instant they are disconnected from the supply. The circuit is as shown in figure 5.4 except that instead of the direct link at X, resistor RD is included. The arrangement allows current to decay in a closed circuit and the magnetic energy stored in the field discharges as heat and the e.m.f. and arcing at the main contacts is minimised. If the field was short-circuited there will be no voltage rise, induced e.m.f. being only sufficient to momentarily maintain the same current, i.e. e = iR. Since no sudden alteration of current value is desirable to maintain the e.m.f. condition, i, at the start of discharge, should be equal to I, the full field current. To avoid a short-circuit on the supply when the switch is operating and the main arc conducts, discharge resistance RD is needed and has a high value. The maximum value of induced e.m.f. will be given by i(RD + R), i.e. I(RD+R). If the field has to be ‘killed’ rapidly, the time constant must be small. L cannot be reduced, so RD is increased. A working value is RD = R. The time constant is thus half of that for the field circuit alone and the induced voltage is, at most, twice the working voltage, an acceptable figure. Example 5.3. The 6 field coils of a D.C. generator, each of 1500 turns, are connected in series to a 500 V D.C. supply. The field current is 5 A and the flux per pole 0.05 Wb. Calculate (a) the field circuit inductance, (b) the value of induced e.m.f. if the circuit is broken in 0.1 second, (c) the value of non-inductive discharge resistor connected across the field terminals, if the induced e.m.f. on opening the switch must not exceed 750 V and (d) the time taken for the field current to decay to 10% of its steady value under these conditions (all 2 significant figures). (a) L = N (Flux set up by 1 A) = 1500 ×

0 05 = 15 H per coil = 6 × 1 15 5=9 90 0 H for the complete field 5

(b) e = L × Rate of change of current = 90 ×

5 = 4500 V or 4.5 kV 01

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130 • Advanced Electrotechnology

(c) Let RT be the total resistance of the circuit. Then 5RT = 750 But RT = R + RD. As R =

500 = 100 Ω and since RT = 150 Ω 5

So RD = 150 − 100 = 50 Ω (d) Here = ∴0.1

−

5t 3

150 t − 10 × 5 = 0.5 A. So 0 5 = 5e 90 100 5t 3

g

5 g log10 = t × log l ge 3

or 1 = 1.666 t × 0.4343 0 4343 and dt=

1 = 1.4 s 0.7235

The Direct Current CR Circuit Because the LR circuit behaviour was considered in some detail, it is appropriate to give attention to that of a circuit of resistor R and a capacitor of capacitance C connected to a D.C. supply. In Chapter 8 of Volume 6 the capacitor was introduced and its ability to store electricity discussed. Charging action was described, in that, a current will pass on closing a switch to the supply, and the current falls to zero as the potential difference across the capacitor plates rises to eventually equal the applied e.m.f. supply. Discharging conditions were also described and attention given to a thorough investigation of charging and discharging, noting that the quantity of electricity which can be stored is proportional to charging voltage. CAPACITANCE. Before deducing the mathematical expressions for current and voltage growth, we remember that C, the capacitance of a capacitor, depends on the number and dimensions of the plates, upon their spacing and upon the dielectric. The unit is the Farad and given in terms of unit quantity and unit voltage. Thus a capacitor has a capacitance of 1 Farad if 1 Coulomb of electricity is stored when 1 Volt is applied across the plates. Note: the expression is deduced for series and parallel capacitor arrangements. 1 1 1 1 Thus if C is the equivalent capacitance, for a Series connection: = + + + ... C C1 C2 C3 and for a Parallel connection С = C + C + C + . . . . 1

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2

3

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D.C. Rectification • 131

Growth of current Charging is considered and as stated above, Q ∝ V, so we rewrite Q = CV where C is V a constant, termed the capacitance. Since Q = It it follows that It = CV or I C . The t V current, at any instant, depends on or the rate of change of voltage. As for an t inductor, since we are more interested in instantaneous values, we can write the relation dv as i C . This expression shows that, if the rate of change of voltage is uniform for a dt period of time, a constant current flows. This condition was considered in Volume 6, Chapter 8, in the absence of resistance. However, resistance is present and the effect of a high value is considered. If the current and p.d. across the plates are noted at various intervals after closing the switch and plotted to a time base, the graphs shown will be of exponential form (figure 5.6). The reason for this exponential curve shape is accounted in the mathematics in figure 5.6. V I= – R

t0 = CR sec v

0.632 V V

i

Time (t)

▲ Figure 5.6

SHAPE OF THE CURRENT CURVE. Consider a circuit of С farads and R ohms, connected as shown in figure 5.7. On switching on, there is no p.d. across the plates and maximum V current flows, given by I = . R Let v = the capacitor p.d. after t seconds. Then I = Also i is the rate of charge build up. ∴ I =

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V

v R

Cdv dt

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132 • Advanced Electrotechnology R +

C

V

–

▲ Figure 5.7

∴

Cdv V − v dv dt = or = dt R V − v CR

This is a differential equation which solves to v

V( − e

−

t

)

The solution will be by integration. − log(V − v ) + K =

t where К = a constant of integration. CR

When t = 0, v = 0 and K = logeV So loge

or

V

⎛ V ⎞ ⎝V v⎠

v V

and v

=e

−

t CR

V − Ve

−

t t V whence = e CR CR V v

giving V − v t CR

or v

V Ve

V( − e

−

−

t CR

t CR

)

V , is to be maintained R and the time taken to charge the capacitor to V volts is t0 seconds, the quantity of V QR CVR V electricity stored is Q It 0 = t 0 . ∴t 0 = = = CR seconds, which is the circuit time R V V constant or t0 = CR. In this time, voltage increases to 63.2% of its maximum value. As before, voltage reaches its half value in 0.693 CR seconds, etc. During charging, current decays exponentially and its value at any instant can be found thus: THE TIME CONSTANT. If the initial rate of current build up, i.e. I =

Let i = the current value at any instant

∴i =

V − CRt e R

g

g i = Ie

9781408176030_Ch05_Final_txt_print.indd 132

−

V

v R

V V( − e or i = R

−

t CR

)

t CR

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D.C. Rectification • 133 Example 5.4. A 1 MΩ resistor is connected in series with a 20 μF capacitor. The arrangement is connected across a 2.5 kV D.C. supply. Find the value of charging current 10 seconds after closing the circuit switch. Find also the circuit time constant (both 2 significant figures). Time constant t0 = CR = 20 × 10–6 × 1 × 106 = 20 s The maximum value of charging current = then i or log

. e

−

t CR

Here −

V 2 5 × 103 = = 2.5 × 10 −3 = 2.5 mA R 1 × 10 6

t 10 = = −0.5 so i CR 20

e −0 5

log2.5 − 0.5log2.718 = log2.5 − 0.5 × 0.4343

= 3.3979 − 0.2173 = 0.1807 giving I = 1.52 mA.

Discharge conditions The circuit switch of figure 5.7 is operated so the capacitor is disconnected from the supply and discharges through a short-circuit. As no resistance is present, the discharge current will be very large and may damage the capacitor. It is usual to include resistance in a circuit for this reason which may be achieved by moving R to the hinged side of the switch. The circuit time constant will be the same as for the charging condition, but may be altered by adding a further discharge resistance RD into the short-circuiting link. DECAY OF CURRENT. The current and voltage curves are shown (figure 5.8). The current graph is drawn below the horizontal to remind the reader that it is a reversed current. Let capacitor C be charged to V volts and discharged through a resistor of R ohms. V The maximum value of current is I = . The voltage across the resistance after a time t R

+ V

v

0

Time I=V – R

i

–

▲ Figure 5.8

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134 • Advanced Electrotechnology

seconds is v volts and the current at that instant i =

v . The current is caused by the R

voltage change or i=−

Cdv . The negative sign shows current reversal due to a decrease of voltage dv dt

Thus −i = ∴−

Cdv dt

v Cdv = R dt

so

dv 1 = − dt v CR

This is a differential equation which solves to ν = Ve

−

t CR

. As before the time constant t − ν V −t is CR and the current follows the same exponential law since = e CR i = Ie CR . R R Solution for the differential equation is similar to those already detailed and is not repeated here. Example 5.5. A voltmeter having a resistance of 30 MΩ is connected across a capacitor charged by a D.C. supply to a p.d. of 500 V. If after 3 minutes the reading is 100 V, calculate the capacitor value (3 significant figures). Since ν = Ve

−

t CR

then

t ν V −t t = e CR or e CR giving l gV − log ν = log e V ν CR

and log500 − log100 = Thus 0.699 =

180 6 × 10 −6 log e or 2.699 2 699 − 2.0 2 0 = × 0.4343 C × 30 × 10 6 C

2.6058 6058 2 6058 × 10 −6 and dC = × 10 −6 = 3 73 × 10 −6 F or C = 3 73 μF . C 0.699

Example 5.6. A 2 μF capacitor is connected across the terminals of an electrostatic voltmeter of negligible capacitance and a resistor of 50 MΩ is placed in series with the parallel arrangement. Determine the voltmeter reading 1 minute after connection to a 500 V D.C. supply. The voltmeter of negligible capacitance is assumed to have no effect on the circuit (1 decimal place). The charging current 1 minute after switching on is given by: i= and i

V − CRt e R

60

500 −6 6 e 2×10 ×50×10 or i 50 × 10 6

10 −106 e 10 6

e −0 6 μA. So log i = log 10 − 0.6 log e

= 1 − 0.6 × 0.4343 = 1 − 0.26058 = 0.7394

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D.C. Rectification • 135 giving i = 5.488 μA. For this charging current, the voltage drop across the resistor will be 5.488 × 10–6 × 50 × 106 = 274.4 V. ∴the reading on the voltmeter will be: 500 − 274.4 = 225.6 V.

The Alternating Currents LR and CR Circuit It is not intended to give a detailed treatment of conditions occurring at the instant of switching on or off an inductive or capacitive A.C. circuit, but rather to draw attention to their importance in relation to the switching duty of circuit-breakers and fuses. Circuit conditions and terms used for the non-resistive A.C. circuit during transient periods have not yet been considered as only steady-state operating theory has received attention. This introduction is to the basic fundamentals which are applicable to both LR and CR circuits. For convenience only the former are considered since corresponding relationships can be deduced for capacitive-resistive circuits.

Current asymmetry Consider an A.C. circuit in which the ohmic values of inductive reactance and resistance are comparable, i.e. the operating P.F. is in the region of 0.7 (lagging). When connected to a sinusoidal supply voltage, with current starting from zero no matter at what point in the voltage cycle the switch is closed. Figure 5.9 illustrates the resulting current condition. After an interval of time the current settles down to a steady or permanent V Vm condition given by I = this is the r.m.s. value, or instantaneous value i1 = sin(ω t −φ ) Z Z where φ is the circuit phase angle and the applied voltage expressed as v = Vm sin ωt. Instant of closing switch i (Actual current) i2 (Transient current)

v

i1 (Permanent current)

▲ Figure 5.9

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136 • Advanced Electrotechnology Let the actual current after closing the switch at any instant be i = i1 + i2. The simplest condition to understand is shown, which assumes the supply voltage is going through its normal value when the switch closes. Since the current for the normal closed circuit condition has a definite value and phase angle (but is actually zero when the switch is closed), it follows that a component of D.C. current is induced which decays exponentially to zero. This is shown in the diagram and i2 is assumed to be the instantaneous value of the transient current which may or may not be present, according to the circuit conditions at the instant of switching. The expression for the transient current is obtained from the following deduction where i1 is the permanent current. Instantaneous applied voltage = Instantaneous voltage drop + Induced back e.m.f. or v = Ri + L

di since i dt

or v = Ri1 L

di1 di + Ri2 + L 2 dt d

i1 + i2 . Then v = R ( i1 + i2 ) L

d ( i1 + i2 ) dt (a)

Under steady conditions i2 = 0 and v = Ri1 L

di1 d

(b)

Subtracting (b) from (a) 0 = Ri2 + L

di2 dt

so

di2 R = − dt i2 L

Rt

By integration i2 I2 e L where I2 is the initial maximum value of the D.C. transient current at the instant the switch closes. The current flowing in a circuit at any instant after closing the switch is the result of both the A.C. and D.C. components. With an asymmetrical current: (1) the alternate ‘loops’ of current are dissimilar in magnitude consisting of major and minor loops and (2) the time intervals between the loops are unequal, unlike those for the permanent current. The condition of current asymmetry is of interest when considering a circuit under fault conditions. Thus if a short-circuit developed on an alternator, at the instant of fault, the short-circuit current rises to a high value, limited only by the alternator’s resistance and reactance which are comparatively low. If the reactance was substantially greater than the resistance then the fault current, due to the low circuit’s P.F., will lag voltage by nearly 90°. If the fault occurred at the condition of zero voltage complete asymmetry of current results, and the initial maximum value of fault current in the first loop may be 20 times the final steady value, which could take several seconds to fall to the final

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D.C. Rectification • 137 permanent value. If a fuse or circuit-breaker was made sufficiently fast acting, it would operate immediately the fault occurred to break a current substantially greater than that which is switched normally. It is hoped this brief introduction to A.C. transients provides sufficient knowledge for the student to appreciate the importance of equipment which is designed to function under fault conditions. Testing and certification of circuit-breakers and fuses for particular duties is a requirement of both British and foreign specifications, compiled to ensure safe operational characteristics.

Rectification Several terms should be first understood before the half-wave and full-wave rectification methods are considered in detail.

Terms (1) Peak Inverse (or Reverse) Voltage (P.I.V.). This is the maximum value of reversed voltage which a rectifier unit must withstand during non-conducting periods. (2) Peak Current Rating. When a rectifier is used with a ‘reservoir’ capacitor smoothing system, it can pass pulses of current which are greater than the load current. The greatest current value, which can be safely passed, is the device Peak Current Rating. (3) Ripple. This is the output voltage variation which produces a current in the load consisting of a steady value and superimposed sinusoidal fundamental and harmonic values. The ripple is expressed as the ratio of r.m.s. ripple current to the mean current, or r.m.s. ripple voltage to the mean voltage.

Rectifier arrangements The property of a diode to conduct when anode voltage is made positive with respect to a cathode and isolated when voltage is negative is used in rectifier circuits. The commonest arrangements are considered. HALF-WAVE (ONE PHASE). The simplest use is the half-wave connection (figure 5.10a). For all rectifier arrangements, a straight line characteristic is assumed, i.e. ‘forward

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138 • Advanced Electrotechnology (a)

+

(b)

–

Input A.C. voltage

+ 1-ph supply

π

2π

3π

Load – Rectified voltage Rectified current

▲ Figure 5.10

resistance’ – if the conducting direction is constant and current will, at all times, be directly proportional to applied voltage. The applied and rectified voltages have the waveforms shown (figure 5.10a). The latter is half sinusoidal as is the current if load is resistive. If the load is inductive or capacitive the current waveform shape is altered. By suitable choice of L and С values, smoothing is introduced. To estimate the effect of current obtained from half-wave rectification one must consider the process for which current is used. For electrolysis or batterycharging, current is unidirectional and used in half-wave pulses, as shown. The effect of current is equivalent to that of a smaller D.C. flowing at a constant value over the time 2 of a complete cycle. The average value of a sine wave is or 0.6365 times the maximum π value but since one half of the wave is suppressed, the average output current ID must 1 be taken over a complete cycle and is thus: or 0.318 times Im (the peak value). π If the current is used for its heating or electro-magnetic effect, the r.m.s. value over a full cycle should be considered. The r.m.s. value for a sinusoidal waveform was Im2 I deduced mathematically from = m = 0.707Im . For a full cycle, half of which is 2 2 non-conducting, the r.m.s. value is

Im2 2 2

=

Im = 0 5Im. The relation of r.m.s. and average 2

values to maximum or peak value for a sinusoidal half wave has not been deduced from first principles, but forms a useful exercise and should be attempted by a student for either a mathematical or graphical solution. Example 5.7. A half-wave rectifier is connected in series with a 18.8 Ω resistor and copper-sulphate voltameter of negligible resistance. The rectifier is assumed to have a constant resistance of 1.2 Ω in the forward direction. Determine the equivalent values

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D.C. Rectification • 139 of the D.C. currents which would deposit the same amount of copper and produce the same heating effect as the rectified current. The supply voltage has an r.m.s. value of 40 V. Neglect the effects of polarisation of the voltmeter (3 decimal places). Peak value of applied voltage =

40 = 56.56 V 0.707

Average value of rectified current =

1 56 56 1 × = × 2.828 = 0.9 A π (18 8 + 1 2) π

The same amount of copper will be deposited by a D.C. current of 0.9 A flowing for the same time as a rectified current. Similarly r.m.s. value of rectified current = 1 × 2.828 = 1.414 A. Similarly the same amount of heat is produced in a resistor by a 2 D.C. current of 1.414 A flowing for the same time as a rectified current. During the nonconducting period the P.I.V. applied to the rectifier is equal to the maximum supply voltage Vm. To compare different rectifier arrangements, it is usual to evaluate the P.I.V. in terms of average output voltage. Thus for a half-wave arrangement the ratio is V Peak inverse voltage = m = 3 14 Average output voltage Vm π FULL-WAVE (ONE PHASE BI-PHASE CONNECTION). Here 2 diodes conduct on alternate half cycles of supply voltage, and the output voltage and current waveforms are illustrated in figure 5.11 – a resistive load is assumed. The average value (ID and VD) 2 for the current and voltage waves is the same as for a full sine wave, (0.6365) times π maximum value. Similarly the r.m.s. value (I and V) is 0.707 times the maximum value. Thus: VD = 0.6365 Vm and ID = 0.6365 Im, also V = 0.707 Vm and I = 0.707 Im. From these can be deduced: VD = 0.9 V and ID = 0.9 I

where V and I are the r.m.s. values.

It is noted that as Vm is the peak value of the forward voltage applied to a diode, it is the voltage of each half of a transformer secondary winding and so the maximum P.I.V. which a diode must withstand is 2 Vm. This P.I.V. is applied to 2 rectifiers in series, but since one is 2V Peak inverse voltage = m. conducting the other must withstand the full voltage or Average output voltage 2Vm π The half-wave arrangement employs a centre-tap transformer, and is used for older valve diodes where the 2 anodes and a single cathode are accommodated in one glass

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140 • Advanced Electrotechnology (a)

D1 + 1-ph supply

–

Load

– +

+

–

D2

(b)

A.C. voltage on diode D1 π

2π

3π

Rectified voltage D1

Rectified current

A.C. voltage on diode D2

Rectified voltage Rectified current

D2

Load voltage Load current Load

▲ Figure 5.11

envelope. A transformer is essential, although its use is uneconomic, since each half of the secondary is used only half the time, while insulation for twice the voltage of a half-winding is needed. The D.C. side can however be completely isolated from the A.C. mains with such a transformer. Example 5.8. A half-wave rectifier is connected in series with a moving-iron ammeter and a permanent magnet moving-coil ammeter. The supply is sinusoidal. The reading on the moving-iron instrument is 10 A. Find the reading on the other ammeter. Estimate the readings on both meters if full-wave rectification is used. The moving-iron meter

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D.C. Rectification • 141 indicates the r.m.s. value of current while the moving-coil meter indicates the average value of the rectified current (2 decimal places). For a full wave, r.m.s. value =

Max. value 2

= 0.707 Max. value

Average value Max. value × For a half wave, r.m.s. value =

2 = 0.6365 Max. value π

Max. value = 0 5 Max. value 2

Average value Max. value ×

1 = 0.318 Max. value π

On half wave, maximum value = 2 × 10 = 20 A Average value = 20 × 0.318 = 6.36 A Thus the moving-coil ammeter registers 6.36 A On full wave, since maximum value = 20 A Then r.m.s. value = 20 × 0.707 = 14.14 A The moving-iron meter will indicate 14.14 A Also, average value = 20 × 0.6365 = 12.73 A The moving-coil meter will indicate 12.73 A. FULL-WAVE RECTIFICATION (ONE-PHASE BRIDGE CONNECTION). The arrangement is illustrated in figure 5.12 showing a full-wave system whose average and r.m.s. values of voltage and current are the same as for full-wave rectification. Thus, e.g. VD = 0.6365 Vm or 0.9 V where VD is the D.C. output voltage. Vm the peak and К the r.m.s. value of the applied A.C. voltage. The P.I.V. is the full transformer voltage since in relation to the applied potential the two rectifiers are in series but only one is conducting. Then

V Peak inverse voltage 3 14 = m = = 1.5 57 Average output voltage 2 V 2 m π

A transformer generally provides the desired output voltage and isolation from the mains. Example 5.9. A permanent magnet moving-coil voltmeter which requires 10 mA for full-scale deflection is used on a 110 V A.C. system. A suitable bridge-connected

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142 • Advanced Electrotechnology (a) D1

D4 –

1-ph supply

D3

D2

– Load +

(b)

A.C. voltage

π

2π

3π

Rectified voltage D1

&

Rectified current

D3

Rectified voltage D2

Rectified current

& D4

Load voltage

Load current

Load

▲ Figure 5.12

instrument rectifier with a forward resistance of 2.5 Ω per unit is available. Find the necessary series resistor value (4 significant figures). For a full-wave sinusoidal supply, since the ratio of r.m.s. to average values is the formr.m.s. value factor, this value can be used. Thus = 1.11. Average value 110 = 99.1 1 V . Let R be the value of the 1.11 required series resistor. There are two rectifier units in series when conduction occurs The average value of rectified direct voltage =

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D.C. Rectification • 143 during a half cycle. Thus the average value of D.C. current required by an instrument for full-scale deflection, i.e. to read 110 V A.C., is 10 × 10–3 A. So 10 × 10 −3 =

99.1 R + 2.5 + 2.5

or R+5=

99.1 = 9910 10 × 10 −3

Required series-resistor value = 9910 – 5 = 9905 Ω HALF-WAVE (THREE PHASE). Although single-phase systems often have an advantage of cost in low-power arrangements, three-phase systems can achieve large power smoothing without expensive components. Because for given load a large capacitor reduces ripple but costs more and creates higher peak currents in the transformer secondary and supply feeding it. With a three-phase supply and 3 diodes (figure 5.13) the pulsing nature of the unidirectional current is decreased – nearing a D.C. current. Current conduction occurs at the diode whose anode is at a +ve potential with respect 1 to the cathode. Figure 5.14 shows that any one diode conducts for of a cycle. Assume 3 diode D1 conducts, as the potential across it falls and that across D2 rises, conduction must transfer when the load potential equals that across D1. As the red phase voltage R

Y

B

B ph Yph

R

ph

Yph

Transformer primary

Transformer secondary

B ph R

ph

– Load +

D2

D3

D1

▲ Figure 5.13

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144 • Advanced Electrotechnology Rph

Yph

Bph

Rph

Yph

Bph

R

A.C. voltage π

2π

3π

D.C. voltage D.C. current D1

D.C. voltage D.C. current

D2

Load voltage D1

D2

D3

D1

Load current D2

D3

▲ Figure 5.14

decreases, the potential across D1 falls further and the load potential now acts as a back voltage, so D1 stops conducting. The load is now supplied by the yellow phase via D2 and current passes from one diode to the next and its value never falls to zero. For a resistive load the output current and voltage waveforms are similar and represented by rectangular blocks. The circuit was used widely for plating equipment as the arrangement has a higher efficiency and a lower voltage regulation than a three-phase bridge rectifier. It can be shown mathematically that there is a definite relation between the D.C. output and A.C. applied voltages. This relation is written as N π VD Vm sin where N is the number of diodes or rectifying circuits. VD is the average π N value of the output D.C. voltage and Vm is the maximum or peak value of the applied forward voltage – here a phase voltage. Thus for the three-phase arrangement in figure 5.14 there are 3 rectifying circuits and: VD

Vm

3 180 3 sin = Vm i 3 14 3 3 14

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3 3 14

3 Vm. Thus VD = 0.827 Vm 2

8/22/2014 4:28:08 PM

D.C. Rectification • 145

or VD

0.827 V = 1.17V where V is the r.m.s. value of the applied voltage. 0.707

The P.I.V. is the full peak line voltage applied across a rectifier unit when it is not conducting. Here the ratio is: 3Vm Peak inverse voltage 1.732 = = = 2.1 Average output voltage 0.827Vm 0.827 An alternative method of achieving full-wave rectification involves a 2-diode system with a centre tap on the transformer, known as bi-phase rectification. Such systems provide D.C. output for a load but are not steady in D.C. voltage. These waveforms may be adequate to supply a load such as a D.C. motor, but some loads require a steady D.C. voltage achieved by introduction of smoothing components (C and L). In this regard D.C. supply circuits may be referred to as ‘unsmoothed’, the level of smoothing being achieved by the size and cost of the capacitors used to ‘filter’ the output waveform. FULL-WAVE (THREE-PHASE BRIDGE CONNECTION OR 6-PULSE RECTIFICATION). A threephase rectifier operating on a transformer output will, as a result of the overlapping phases, produce a smoothed D.C. output. For this arrangement 6 diodes are used, connected into the three-phase lines, as shown by figure 5.15. For an explanation of circuit action consider the voltage between the R–Y lines (figure 5.15) with Red +ve with respect to Yellow. Disregarding the effect of the other line voltages, it is seen that diodes D1 and D5 conduct until the voltage falls to zero and reverses with the Yellow line becoming +ve with respect to Red. Diodes D2 and D6 will conduct and full-wave rectification will occur. The D.C. load voltage and current conditions for one cycle of supply voltage are shown by the second illustration, where diodes 1, 2, 5 and 6 are involved. The advantage of three-phase rectification over single phase is its ease of achieving a reasonable level of smoothing for large power loads without expensive smoothing components. If the conditions for the Y–B and B–R lines are considered the appropriate diodes will be conducting and the illustrations of figure 5.15 can be studied. If all 3 voltage conditions between the lines are considered, it is seen that a level of output voltage is reached so that, as the voltage across any conducting diode circuit falls below this value, the conduction ceases but D.C. continuity is maintained through the pair of diodes in a parallel circuit across which voltage rises. The concept of a commutating switch is useful here with each diode circuit conducting for a sixth of a cycle, with the switch having 6 positions. In the general formula, the number of rectifying diode circuits is 6 – 2 for each voltage cycle, one for forward and one for reverse. Thus N = 6 and if Vm is the maximum value of applied line voltage then:

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146 • Advanced Electrotechnology (a)

R

Y

B

D2

D1

D3

+ Load

D4

D5

D6

(b)

R–Y

–

Y–B

D1–D5

B–R

R–Y

Y–B

D1–D5

D2–D6

B–R

D2–D6

Line R–Y

D3–D5

D2–D4

D3–D5

D2–D4

Line Y–B

D1–D4

D3–D6

D1–D4

Line B–R

D3–D5 D1–D5 D1–D4 D2–D4 D2–D6D3–D6 D3–D5 D1–D5 D1–D4 D2–D4 D2–D6

▲ Figure 5.15

VD

or VD

6 180 6 sin = Vm i 3 14 6 3 14 3 = Vm Vm 3 14 Vm

Vm

6 3 14 × 2

0.954 V = 1 35 V where V is the line r.m.s. value. 0.707

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D.C. Rectification • 147 Note that, as for the single-phase bridge arrangement, there are 2 diodes in series but Vm Peak inverse voltage as one is conducting, the ratio of: = = 1 05 Average output voltage 0.954Vm The expression for VD was developed in terms of a line voltage, i.e. with no transformer present. In practice a transformer is used as it provides a way of obtaining the required D.C. voltage value from a fixed A.C. supply. Such a transformer may be delta or star connected and VD developed in terms of the transformer phase voltage – as was done for the three-phase, half-wave condition. If Vm

Vphhm then h VD

or VD

Vph ×

Vpphm h = 1.654Vphm hm

0.954 = 2 34Vph 0.707

Vphm is the maximum value of a phase voltage and Vph is the phase r.m.s. value. For a very high power rectification some methods of three-phase rectification use Star and Delta outputs to produce ‘12 pulse rectification’ where each output is rectified with its own three-phase bridge, the rectifier outputs being connected in parallel or series to reduce ‘ripple voltage’ further (typically 2.5%) and the effects of current harmonics on the supply. Such methods are generally not used on lower power (usually single-phase) systems. Example 5.10. For a three-phase, half-wave rectifier express (a) the output voltage variation as a percentage of the mean value (1 decimal place). (b) If the latter is 2000 V calculate the P.I.V. (2 significant figures). (c) If the supply is at 400 V and the transformer primary is in delta with the secondary in star, find the step-up ratios for each phase. (a) For three-phase, half-wave rectification, any one diode conducts for a third of a cycle. The conducting period is for 120°, spread 60° either side of the maximum value. The change-over occurs when the voltage is at the 30° value or 1 ν = Vm ° = Vm × = 0.5 Vm . 2 Thus output voltage varies from Vm to 0.5 Vm = 0.5 Vm and mean value of D.C. output = 0.827 Vm Voltage variation =

0 5 Vm × 100 = 60.5% 0.827 Vm

(b) If VD = 2000 Then P.I.V. = 2000 × 2.1 = 4200 V Note: the ratio of

Peak inverse voltage = 2.1 as already developed. Average output voltage

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148 • Advanced Electrotechnology

(c) Also VD = 1.17 V where V is the r.m.s. value of a phase applied voltage. So V =

2000 . 1.17

This is a secondary phase voltage but the primary voltage per phase = 400 V. So step-up ratio = 400 :

2000 5 = 1: or 1:4:28 1.17 1.17

Battery charging by rectifier When a rectifier operates in the forward direction, the supply voltage is countered by the battery e.m.f. which reduces the voltage producing the charging current. The action is illustrated by a half-wave arrangement (figure 5.16).

(a) A

Supply

Rb

Rx (b)

Rectified D.C. voltage Battery e.m.f.

Charging current

▲ Figure 5.16

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D.C. Rectification • 149 When the rectifier output voltage exceeds the battery, output current flows into the battery to charge it. During the complementary half-cycles, a rectifier offers no conduction and a reverse current does not flow. Thus charging is possible and an actual effective current value can be determined by making a graphical solution in accordance with figure 5.16. The rectified voltage wave is plotted, the battery e.m.f. set off and the resultant voltage waveform drawn. Instantaneous values of current are deduced v from i = where Rb is the battery internal resistance and Rx the variable control Rb Rx resistance value. The average for the complete cycle is determined in accordance with i +i +i i procedures previously established. Thus i = 1 2 3 n where n is the number of midn ordinates for a complete cycle. The deduced value of current is that indicated by an ammeter. Rectifiers have a characteristic which drops part of the input voltage (which for silicon devices is typically 0.7 V) plus a non-linear equivalent resistance. At high frequencies this voltage drop distorts the waveform and dissipates power. One aspect of most rectifications from the peak input voltage to the peak output voltage is caused by this built in voltage drop (which can be as low as 0.3 V for the Schottky diode). Halfwave rectification and full-wave rectification with a centre-tapped secondary will have peak voltage loss of 1-diode drop (0.3 to 0.7 V). Bridge rectification will have a 2-diode loss drop (0.6 to 1.4 V). These will reduce output voltage and limits the available output voltage if very low A.C. voltages must be rectified. For high-voltage A.C. systems this is less of a problem.

Practice Examples 5.1 A D.C. supply of 100 V is applied to a coil of resistance 10 Ω and inductance 10 H. Find the current 0.1 s after switching on, and the time for the current to reach 5 A (both 3 decimal places). 5.2 A resistor is connected across the terminals of a 20 μF capacitor previously charged from a D.C. supply to 500 V. If the p.d. falls to 300 V in 0.5 min, determine the resistor value (3 significant figures). 5.3 The field winding of a separately excited D.C. generator has an inductance of 10 H and a resistance of 50 Ω. There is a discharge resistance of 50 Ω in parallel with the coil. The coil is energised by a D.C. supply at 200 V which is suddenly switched off. Find the field current 0.04 s after the instant the supply is switched off (3 decimal places).

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150 • Advanced Electrotechnology 5.4

A resistor of 100 kΩ is connected in series with a 50 μF capacitor to a D.C. supply of 200 V. Calculate the voltage across the capacitor and the current at a time 0.2 s after switching on (both 2 decimal places).

5.5

The time constant of a coil is 2.0 and the inductance is 15 H. Determine the current 0.2 s after connecting the coil to a 300 V, D.C. mains. Find the time taken for the current to reach half its maximum steady value (3 decimal places).

5.6

A 1 μF capacitor is charged from a D.C. supply to 50 V, then discharged through a 5 MΩ resistor. After 5 s another 5 МΩ resistor is connected in parallel with the first. Determine the voltage after a further 5 s have elapsed (2 decimal places), and find the capacitor current (1 significant figure).

5.7

A relay coil has a 1 kΩ non-inductive resistor connected across it and the parallel arrangement is connected to a 50 V D.C. supply. The relay data is: Resistance 2 kΩ, Inductance 100 H, Operating current 10 mA, Release current 1 m A. Find the time taken by the relay to (a) operate when the supply is applied (3 significant figures) and (b) release when the supply is removed. Assume instantaneous operation when the current reaches both the operate and release values (4 significant figures).

5.8

A circuit has a 2 μF capacitor in series with a 100 kΩ resistor used to control another device so that if the p.d. across the capacitor reaches 63.2 V its instant discharge is effected and the voltage, applied to the circuit, recharges the capacitor as before. If the applied voltage is 100 V, find: (a) the initial current (1 significant figure) and (b) the charging current at the instant before the capacitor is discharged (3 significant figures). Sketch a current/time curve for the current in the resistor throughout the charging and discharging sequence and (c) determine the frequency of the sequence.

5.9

A metal rectifier, a moving-coil ammeter and a thermo-junction ammeter are connected in series across a 2 V supply. The rectifier has a forward resistance of 20 Ω, an infinite reverse resistance and the resistances of the ammeters are neglected. When a D.C. supply is applied, the ammeters give identical readings. With an A.C. supply at 50 Hz and 2 V (r.m.s.) the readings are different. Calculate the actual meter readings (1 significant figure).

5.10 Given that a three-phase, 50 Hz supply is at 450 V between lines and feeds a deltastar transformer with primary to secondary turns ratio 1:5, determine the D.C. output voltage on no-load from a suitably connected full-wave rectifier, and the ripple voltage size (1 decimal place) and frequency (1 significant figure).

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6

THE A.C. GENERATOR It was found that the dynamos in a power house six miles away were repeatedly burned out, due to the powerful high frequency currents set up in them, and which caused heavy sparks to jump through the windings and destroy the insulation! Credited to Nikola Tesla regarding the Colorado tests of 1898–1900

A.C. generators or alternators (as they are also called) operate on the same fundamental principle of electromagnetic induction as do D.C. generators. In Chapter 7, Volume 6, we introduced the basic laws of electromagnetic induction, and considered the principles of e.m.f. generation by dynamic induction. Generation of an alternating or A.C. voltage is accomplished by a relatively simple arrangement but, if the A.C. generator is to function as a machine, distinct from the D.C. generator, it is clear that further work is needed. The advantages of A.C. over D.C., from a power generation and applications view point, have been discussed but if both generators are compared, the A.C. generator wins out because it does not need a commutator (as there is no need to invert the opposite negative half-wave cycle) and remembering our discussion of the transformer (Chapters 1–3) it can easily step up or step down voltage so that a low-voltage A.C. current can be made into a high-voltage A.C. current and vice versa. If it is just a relative movement of conductors with respect to the magnetic field it is irrelevant whether the field moves and the conductors are stationary or vice versa. Thus we have 2 operational modes available to the machine designer but the arrangement of moving field with fixed armature is usually preferred due to reasons to be discussed. A rotating magnetic field which continuously changes direction is also a key principle to A.C. motor operation. The basics of machine construction are here considered before the main A.C. generator theory is given.

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152 • Advanced Electrotechnology

The A.C. Generator The A.C. generator or alternator is a machine that converts mechanical energy to electrical energy in the form of alternating current. The term ‘synchronous A.C. generator’ or ‘A.C. generator’ is used as a direct comparison with the ‘D.C. generator’. The term alternator is also used by engineers, although this usually refers to small rotating machines driven by yachts, automotive and other combustion engines. Marine alternators, typically 12 or 24 V, are used in boats and yachts and are similar to vehicle alternators but with adaptation to the harsh salt-water environment. They are designed to be explosion proof so brush sparking will not ignite explosive gas mixtures in an engine room. The main types of A.C. generators in common use today are now considered, generating electricity using the same principle as D.C. generators. When the magnetic field around a rotating conductor changes a current is induced in the conductor. Typically a rotating magnet turns within a stationary set of conductors wound in coils on an iron core. The field cuts across the conductors generating an induced e.m.f. as the mechanical input causes a rotating magnet to turn. A rotating magnetic field will induce an A.C. voltage in the windings. Often there are 3 sets of windings, physically offset so the rotating magnetic field produces a three-phase 1 current, displaced by period with respect to each other. 3

Rotating-armature type A.C. is always produced in a basic generator. The provision of a reversing switch – or commutator, results in D.C. but, if A.C. is needed slip-rings are used. A machine of this type is constructed like a D.C. generator with a fixed-field system and a moving armature or ‘rotating-armature’ alternator is common for small machines up to 40 kVA and voltages of 450 V. It is a cheaper machine than an A.C. generator constructed on the rotating-field principle, because a D.C. generator or suitably sized motor uses sliprings, replacing the commutator. A D.C. armature winding is retained and tappings are brought out to the slip-rings, as required for single- or three-phase supply. The number of poles is chosen to suit the speed of the prime-mover and the required frequency. Yoke and field system construction follows D.C. practice and separate excitation. The generator voltage waveform is not exactly sinusoidal but is acceptable for applications where this alternator type is used.

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The A.C. Generator • 153

Rotating-field type The stationary portion is called the stator and carries the armature conductors while the field system moves and is called the rotor. The magnetic field is produced by energising the pole windings with D.C. through slip-rings from a D.C. generator (the exciter), usually driven off the main shaft with D.C. voltage below 110 V, although 220 V or 440 V may be used. A stator arrangement has an advantage in that it can be uniformly slotted and wound for high voltage working. As no centrifugal forces are involved, a greater insulation thickness can be used and there is more room for windings and endconnections than if these were on a moving rotor. Bracing end-connections against electro-magnetic forces, evident under S.C. conditions, is easier and more effective. A rotor can be made compact, with a construction allowing a low-voltage, high-current winding to withstand the large centrifugal forces encountered. The stator core is built from laminations insulated from each other, like the armature of a D.C. machine, but for small A.C. generators each lamination may be a complete ring. The rotating-field machine is built in 2 main forms under the general category of the Salient-Pole Rotor or the Cylindrical Rotor. SALIENT-POLE ROTOR. The salient-pole type of field construction is used for slow and medium speeds and may be driven by steam-engines, water-wheels or turbines and slow-speed diesel engines. Construction is distinguished by a machine having a large diameter compared to its axial length. The basic arrangements are shown in figure 6.1. Pole-pieces are similar to those used in D.C. machines and bolted to a flanged magnet wheel of solid or spoked construction. Smaller machines may have poles fitted to a solid hub. The number of poles varies from 4 to 40, usually made from steel laminations or cast2 iron with laminated pole-shoes. The pole-shoe width is of a pole-pitch and its shape 3 chosen to provide a sinusoidal flux-density distribution across the air-gap. Field windings follow D.C. machine practice, except that effective bracing is needed; to achieve this a copper strip wound on edge, suitably insulated, is used. In rotor construction for a three-phase A.C. generator, slots are uniformly distributed around the inside periphery of the stator and spaced to allow 3 or multiples of 3 to be accommodated in a polepitch. Assuming a 16-pole machine and 3 slots per pole per phase, 9 slots are needed for a pole-pitch, giving 144 slots in total. The number of slots/pole/phase is decided by voltage and waveform design requirements, as is the shape of the stator slots which can be open-rectangular or semi-enclosed.

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154 • Advanced Electrotechnology (a)

Yoke (cast iron or welded steel)

Laminations

2 slip-rings (b)

Space

Laminations

N

S

(c)

Armature conductors

S

S

N

Steel hub integral with shaft

▲ Figure 6.1

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The A.C. Generator • 155 CYLINDRICAL ROTOR. This consists of a steel forging, out of which rotor winding slots are milled. The shaft may be integral with the field system or, for a large rotor, made from 2 identical parts bolted to the ends of the magnet unit. The number and spacing of slots is fixed by the number of poles. Figure 6.2 shows 2- and 4-pole rotor systems 2 1 where of a pole pitch is slotted, leaving for the pole centre. Field windings are made 3 3 up from coils of copper strips, wound as for the salient-pole rotor. (a) 4-pole rotor

N

2 slip-rings S S

(c) (b) Mica

Air duct

N

Wedge (steel or bronze)

Copper strip

S 2-pole rotor

▲ Figure 6.2

Steel or manganese-bronze wedges are driven into the slots to hold the winding against the centrifugal forces and special methods secure the end-connections, since the mechanical stress which the copper experiences can be high. Non-magnetic steel retaining rings or end-caps, which screw on to the rotor, are usually provided to cover the end-connections, once clamped and braced between bakelised-paper and hardwood blocks. Because of the rotor’s smooth surface, cooling is achieved by forced ventilation, as there is virtually no fanning effect, unlike the salient-pole machine, so the rotor has suitable air-ducts. Construction permits it to be driven at high speeds and the generator is associated with steam or gas turbines. Because of speed considerations, the diameter is limited to 1 to 1.3 m and, to achieve high electrical output, axial length is large. This machine is characterised by its shape: length is 4 to 5 times its diameter. Stator construction follows the salient-pole machine except that because of its length, cooling issues are more important and influence design. Laminations are punched and

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156 • Advanced Electrotechnology assembled into a fabricated steel frame so that axial air-vents and radial ducts allow the air streams provided by rotor or external fans to circulate. The A.C. generator is enclosed in a steel case, and air is fed into the air-gap and cooling passage-ways at the ends of the stator, discharging through radial ducts into the space between the stator core and casing. Air is filtered and a closed-circuit system is common with air passed through a cooler. For marine work, a composite A.C. generator and sea-water cooler unit is often used. Complete machine enclosure reduces noise, and for large power-station units, further development uses hydrogen instead of air as the cooling medium. Hydrogen cooling reduces windage loss and increases heat removal, permitting a 1% efficiency increase and a 20% rating increase. Hydrogen cooling also results in a windage noise reduction, less insulation oxidation and reduced fire risk. Disadvantages include the requirement of the cooling system, gas-tight shaft-seals and the need for an explosion-proof construction.

Excitation Arrangements Modern methods of energising the A.C. generator field are placed under 2 headings: (1) Rotary excitation and (2) Static excitation systems. At first the field current was obtained from a separate D.C. generator but with development of the semiconductor rectifier, other excitation methods were added to give desired characteristics. One such characteristic for a marine alternator is a fast response speed to the Automatic Voltage Regulator (A.V.R.). It is usual practice to ‘direct-on’ start large induction motors which may have start currents comparable with the full-load current of the supply alternator. A.V.R. controls the field current to keep output voltage constant. If output voltage from stationary armature coils drops due to an increase in demand, more current is fed into the rotating field coils through the voltage regulator. This increases the magnetic field around the field coils which induces a larger voltage in the armature coils, so the output voltage is brought back up to its original voltage. Both analogue and digital A.V.R. systems exist, where in the former the terminal voltage of an A.C. generator is fed back by a feedback circuit and the difference between the terminal voltage and a reference voltage circuit’s output. In the latter, the output voltage of an A.C. generator is produced through a full-wave rectifier and converted to a digital signal by an Analogue to Digital (A/D) converter by thyristor with field current control. The thyristor remains non-conductive until a firing pulse is applied through the gate to cathode junction (so controlling the firing pulses). If the anode is +ve with respect to cathode when the pulse is applied a silicon-controlled rectifier will conduct and stay conducting until

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The A.C. Generator • 157 the anode voltage goes –ve. Thus current goes to zero. A.V.R. regulates the voltage applied to the generator as a generator can draw a large amount of power from the source, especially for diesel generators and avoids the changing load causing voltage fluctuations that could damage a generator. Transient voltage performance is important and as the voltage dip is a function of alternator reactance, this must be accounted for, even at the expense of machine size and cost. In relation to excitation, due to the high inductance of the field systems of the exciter and main alternator, large time constants may be involved. Many excitation designs use field-forcing techniques, with the 2 main excitation methods now considered.

Rotary excitation systems The D.C. generator or exciter is usually mounted on an extension of the main alternator shaft. For large A.C. synchronous generators, as used in power stations, 2 or more exciters may be used in series (cascade). The first exciter energises the field of the next, the output of which supplies a second larger exciter field or alternator field. Smaller marine machines do not require this cascade exciter arrangement but a small pilot exciter may be used, of a permanent-magnet field system rotating within a stator containing the armature coils. Generated A.C. voltage is fed into a rectifier system, the D.C. output feeding the field of the main exciter. Excitation reliability is ensured as a D.C. generator by itself may fail to excite due to residual magnetism loss. With the advent of silicon semiconductor diodes, able to handle large powers, D.C. generator exciters are now largely replaced by an arrangement dispensing with commutator, slip-rings and associated brush gear. Thyristor or silicon-controlled rectifier static excitation systems are common as they require little maintenance. THE BRUSHLESS A.C. GENERATOR. In this machine the exciter consists of an A.C. armature mounted on the main alternator shaft rotating within a normal D.C. field system. The armature A.C. output is rectified by diodes mounted on and revolving with the shaft (figures 6.3a and 6.3b). The diode assembly D.C. output is fed to the rotating field of the main alternator by leads taken through or secured to the drive shaft. Exciter and alternator can be mounted in a common frame and air space, forming a brushless generator. A 2-bearing arrangement is used and maintenance requirements are reduced as there is no commutator, slip-rings or brush gear. The brushless A.C. generator illustration (figure 6.3a) was provided originally by the Brush Electrical Company of Loughborough, England. The machine can be made self-exciting as long as sufficient residual magnetism is present. As no commutator is required, interpoles are replaced by permanent magnets

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158 • Advanced Electrotechnology (a) Up to 250 kVA, 0.8 P.F., 3-ph, 4-pole, 1800 rev/min, 60 Hz

Diode boss

Main generator rotor assembly Main generator stator assembly

Diodes Exciter rotor assembly Exciter stator assembly

(b)

Bus bars L1

L2

D.C. supply L3

– Alternator stator windings (fixed)

+

Exciter field (fixed)

+ Shaft

– Alternator field

Rotating parts

Exciter armature Rectifiers

▲ Figure 6.3

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The A.C. Generator • 159 fixed within the interpolar spaces. Alternatively the exciter main poles are built up with permanent-magnet laminations. Adequate residual voltage at the main alternator terminals is ensured to overcome the impedance of any components of the exciter field circuits. Unlike a conventional alternator-exciter arrangement, since power supply for the A.V.R. cannot be obtained from the exciter armature, it is obtained from a pilot exciter or reliable stationary D.C. supply, or from an A.C. generator output using rectifiers. The disadvantage of this last method is that if a three-phase ‘short-circuit’ occurs the main machine voltage collapses. Alternator current has an initial high value but decays rapidly so the machine circuit-breaker does not ‘trip’. Under-voltage or instantaneouscurrent protection must be fitted, but if these are not desired auxiliary S.C. transformers must be provided. Transformers ensure an added power supply to the exciter field or A.V.R. and enable the main alternator to maintain sufficient S.C. current to trip associated circuit-breakers. ‘Field suppression’ introduces a further complication for brushless machines which can be solved with thyristors on the main alternator rotor but exciter field suppression is used and although this results in a longer period to achieve complete and correct machine ‘shutdown’ it is generally satisfactory. Rectifier diode rating depends on the full-load excitation current and the cooling arrangements, i.e. amount of air and heat-sink areas. As transient voltages are induced in the rotor during a stator three-phase short circuit, the P.I.V. rating must be sufficient to withstand prolonged induced voltages and currents. Diodes are arranged in a bridge connection with a combination of units in series or parallel to avoid a machine ‘shutting down’ since diode failure by internal S.C. breakdown clears to an open circuit due to over current. Diodes must be checked periodically but a rotor insulation test should be made with the diodes disconnected and isolated or with a temporary short circuit applied across each such unit.

Static excitation systems This heading covers self-regulating and thyristor excitation. For these methods, an A.C. generator requires slip-rings and brush gear but no rotating exciter. There are various systems developed by different manufacturers to provide a desired voltage regulation performance for ‘direct-on starting’ large motors. Self-regulating systems are divided into those relying on excitation power taken from the main alternator output and conveyed to the field by a combination of transformers, inductors and rectifiers, using A.V.R. As close voltage regulation cannot be obtained by transformers and inductors alone A.V.R. provides closer regulation – typically ±1%. Without A.V.R., ±5% regulation is achieved. Figure 6.4 shows a typical arrangement.

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160 • Advanced Electrotechnology Compounding C/T’S

Stator windings (fixed) Rotor field

G

S

Sliprings and brushes

Voltage build-up circuits and A.V.R. circuit

Static exciter

▲ Figure 6.4

Because of added compound transformers or inductors, a field-forcing effect is achieved when a large alternator load current passes. With the A.V.R., in series with the alternator field, the voltage dropped across it can be lowered by the control current and an increased voltage applied across the field, resulting in an added field-forcing effect. Like the brushless machine, the need for sufficient generated terminal voltage under S.C. conditions is clear. Compound transformers achieve this and currents up to 5 times normal can operate machine circuit-breakers. Initial voltage build-up when the set is run up to speed is provided by a pilot exciter, permanent-magnet exciter field inserts or a special starting circuit with a relay which shorts out exciter-circuit impedance until voltage builds up. Thyristor excitation systems were developed to exploit the benefits of electronic control. High-efficiency silicon rectifiers and thyristors result in small size and weight without large iron cores and rotating exciters. The thyristor acts as rectifier for which conduction can be controlled. In addition to a normal diode anode and cathode, it has a 3rd electrode, ‘the gate’. When a thyristor is forward biased by a half-cycle of applied A.C. voltage, it conducts when the gate is pulsed with the correct polarity but stops conducting at the end of the half-cycle. If the next half-cycle is to conduct a second thyristor is needed with its gate pulsed accordingly. Thyristors are arranged like diodes used in rectification, although for a bridge assembly only 2 thyristors need be used with

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The A.C. Generator • 161 diodes in series in opposite arms. The instant a thyristor conducts or ‘fires’ is controlled during the half-cycle and thus switch-on is delayed, so the power level averaged over several cycles is less than when no delay is added. Variation of delay length varies average power, with the system known as ‘phase shift’ control. An alternative system employs ‘burst triggering’ control with both systems controlling conducted power. When excitation thyristors are used, irrespective of the control method employed, the A.C. field current is varied by the static exciter. Figure 6.5 shows a typical thyristor-controlled excitation arrangement. A.V.R. is built into the system but as the electronics involved require specialist knowledge, the equipment and its operation is treated in block form. A three-phase voltage-sensitive measuring unit is shown with a transistor error amplifier, triggering circuit and halfwave thyristor output stage. As for excitation systems, provision is made for some of the field power to be obtained from excitation current transformers (not shown) in the A.C. generator output. This arrangement ensures that, in the event of a short-circuit, excitation is maintained.

Voltage build-up relay

Voltage trimmer

X Reference and measure unit

Error amplifier

Trigger and pulse unit

Field S.C.R. XX

R

Y

B

R

Y

Voltage from generator terminals

▲ Figure 6.5

System operation can be followed with reference to figure 6.5. Three-phase A.C. generator terminal voltage is applied to the measuring unit where it is rectified and applied across a resistor/zener-diode reference bridge. The bridge signal output, a measure of generator voltage error, passes to the amplifier and is fed into a trigger circuit. The bigger the amplified signal the greater the delay time before the trigger fires the thyristor to allow half-wave rectified current-flow through the field winding. Thus if the terminal voltage rises due to load reduction, the thyristor firing delay time is extended.

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162 • Advanced Electrotechnology The greater the delay, the smaller the average field current, with consequent lowering of machine-generated voltage. Conversely, terminal voltage reduction reduces delay time, so firing commences earlier in the rectified half-wave and average current rises with consequent generated voltage increase. A voltage build-up relay unit is shown which operates as other static-excitation systems, building up voltage by residual magnetism during the initial starting up. This system controls voltage regulation to an accuracy of ±1% from no-load to full-load.

The Speed-Frequency Equation Before considering the conditions for generation of e.m.f. by an alternator, we consider the basic relationship between: alternating voltage frequency, the speed at which the machine is driven and the number of poles, since output frequency depends on the number of poles and the rotational RPM speed. The synchronous speed corresponds to a particular frequency, e.g. for a 2-pole system at 50 Hz it is 3000 rev/min, but at 60 Hz may be 3600 rev/min. The term synchronous speed used in connection with motors is introduced here since speed directly relates to frequency, and the equation developed relates to both alternators and motors. From basic theory an A.C. waveform voltage is generated as a coil passes through 2 pole-pitches, so the waveform frequency is the number of cycles generated per second. If ƒ = number of cycles per second or number of hertz, then the time for one cycle 1 = second f A coil moving through 2 pole-pitches generates one cycle of e.m.f. and if driven at a 60 speed of N rev/min then the time for one revolution = seconds. Hence the time taken N 60 to pass through 1 pole-pitch = seconds if P is the number of poles. Consequently the N 2 × 60 time taken to pass through 2 pole-pitches = seconds , so: PN The time taken to move through 2 pole-pitches = time for one cycle giving 120 1 PN = or f = Hz with N the synchronous speed for a frequency of ƒ HERTZ. PN f 120 Example 6.1. Find the speed at which an 8-pole, salient-pole A.C. generator must be driven in to generate voltage at a frequency of 60 Hz (3 significant figures).

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The A.C. Generator • 163

Since N =

120f 120 × 60 then N = = 900 rev/min. P 8

E.M.F. equation This is deduced in a similar way to that used for D.C. generators. Consider figure 6.6 where the pole system and stator conductors are shown. Every revolution one conductor is cut by РΦ weber, where P is the number of poles and Φ the flux per pole (in weber). Conductors per phase Stator Rotor N

S Φ Weber Pole pitch τ

τ

▲ Figure 6.6

The time for one revolution = =

60 seconds, so the average rate of cutting lines of force N

Total lines cut P Φ P ΦN = = Time taken to cut the lines 60 60 N

From Faraday’s law, the average e.m.f. generated in volts is given by the average rate of P N cutting lines of force so the average e.m.f. generated in a conductor = volts = E av 60 For A.C. work, r.m.s. values are required and if sine-wave working is assumed, the relation: Form factor =

r.m.s value may be used. average value Form factor =

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1 2 Maximum value π = = 1.11 2 π Maximum value 2 2

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164 • Advanced Electrotechnology

So the r.m.s. value of e.m.f./conductor = 1.11× value of e.m.f./conductor = 1.11×

p N 120f volts and as N = the r.m.s. 60 P

p 120f × = 2.22Φf volts. 60 P

However, if the generated waveform is not sinusoidal the 1.11 form-factor cannot be used. K is used for a general case so the e.m.f./conductor = 2KΦƒ volts. CONCENTRATED WINDING. Figure 6.6 shows all the conductors of one-phase Zph, concentrated in one slot per pole, and is a theoretical arrangement only. However, since conductors are spaced 1 pole-pitch apart, then for every one, flux rises and falls simultaneously, so the e.m.f.s rise and fall together, i.e. they are in phase. If all the conductors under a pole-pair are connected in series, the total e.m.f. is the phasor sum – here the arithmetical sum of the e.m.f.s per conductor. Thus for a concentrated winding the e.m.f. of winding per phase = 2.22 × Zрh × Φ × ƒ volts. DISTRIBUTED WINDING. A winding of one slot per pole per phase needs a broad deep slot which alters the air-gap flux distribution and the generated waveform. The winding would be difficult to accommodate and a more practical arrangement is the distributed winding which uses more than one slot per pole per phase – usually 2 or 3. Too many slots would mean the smaller teeth would be mechanically weak and, along with other possible faults, would vibrate, adding to mechanical noise. Figure 6.7 shows a working condition, where the conductors in one slot have e.m.f.s generated in them which are not in phase with those generated in adjacent conductor slots. If all the conductors per phase are connected in series the resultant e.m.f. will be the phasor sum, the resultant being slightly smaller in magnitude than if a concentrated winding were used, adding another multiplication constant in the e.m.f. equation called the Distribution or Breadth Factor. This term is written as KD and defined as the ratio of the e.m.f. in a distributed winding to the e.m.f. in a concentrated winding. Conductors per phase

Stator Rotor N

S

τ

τ

▲ Figure 6.7

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The A.C. Generator • 165

Thus K D =

e.m.f. in distributed winding O = e.m.f. in concentrated winding d O

OA and OB are shown in figure 6.8. with KD always less than 1. If the same number of conductors for each type of winding is considered, for a Distributed Winding, the e.m.f. per phase = KD × e.m.f. for a concentrated winding, or Eph = 2.22 × KD × Zph × Φ × ƒ volts. In summary the distribution of the phase winding in several slots for each pole-pitch improves the waveform but reduces the resultant e.m.f. value slightly. B

α

A

e.m.f. of distributed winding

α

e.m.f. of concentrated winding

O

▲ Figure 6.8

Example 6.2. Find the no-load terminal voltage of a three-phase, 4-pole A.C. generator. Flux per pole (sinusoidally distributed) = 0.14 weber (3 significant figures). Slots per pole per phase = 2. Conductors per slot = 2. Distribution Factor = 0.966. The machine is driven at 1500 rev/min and is star connected. Here f =

PN 4 × 1500 = = 50 Hz i.e. Eph 120 120

.

K D Z ph f volts

= 2.22 × 0.966 × ( 4 × 2 × 2) × 14 × 10 −2 × 50 = 2.22 × 0.966 × 16 × 14 × 10 −2 × 50 = 2.22 0.966 × 8 14 = 240.19 volts/phase

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166 • Advanced Electrotechnology 19 = 416 V . Terminal voltage = 3 × 240.19 = 1.732 × 240.19 DETERMINATION OF BREADTH FACTOR. To determine KD for a specific stator-slot arrangement, the following expression is used. Figure 6.9 shows voltages of conductors in successive slots: AB, ВС, CD, etc. are all equal with a phase displacement of α (degrees). Points А, В, C, D, etc. lie in a circle of centre O. Drawing OP perpendicular to AB, then angle AOP = angle РОВ = α/2 since angle AOB = α. E α D

α Q

C α

O

α

α — 2

B P

A

▲ Figure 6.9

Now AP = AO sin α/2 and AB = 2AO sin α /2 The resultant e.m.f. = AE and angle AOE = nα, if there are n slots per pole per phase. Now AQ = AOsin

nα nα and d AE = 2AO 2AOsin i 2 2

nα nα nα 2AO AOsin i 2AOsin 2 2 2 and kD = = = α n(voltage/slot) nAB n2AOsin 2 nα sin 2 or kD = nα nsin 2 2AOsin

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The A.C. Generator • 167 Example 6.3. From example 6.2, find the no-load terminal voltage of a three-phase, 4-pole alternator having 2 slots per pole per phase with 2 conductors per slot. The flux per pole is 0.14 Wb, the machine speed 1500 rev/min and the connection of the phases is in star (3 significant figures). As before f= nα 180 2 Here α = = Also K D = nα 3×2 n sin 2 sin

or K D =

4 × 1500 = 50 Hz 120 2 × 30 2 ° and K D = 30 0 2 sin 2

sin 30 0.5 0.5 = = = 0. 2 sin si 15 2 × 0.2588 0.5176

sin

(asgiven)

so Eph = 2.22 × 0.966 × 16 × 14 × 10 −2 × 50 = 240.19 volts/phase Terminal voltage = 3 × 240.19 = 416 V COIL SPAN, PITCH OR CHORDING FACTOR. To improve output waveform, the span of a coil may be less or greater than a pole-pitch. In figure 6.10 the phase difference between the e.m.f.s in the coil sides and the resultant e.m.f. is seen to be further reduced. The two sides of the coil are not 180° apart, the span being less than a pole-pitch. From θ phasor diagram (figure 6.10c), the resultant e.m.f. is AC or = 2AB cos and the Pitch 2 Factor, Ks defined as the ratio of the actual e.m.f. to that obtained from a fully pitched AC coil, or K s = 2AB θ 2 = cos θ 2AB 2

2ABcos Thus K s =

The effect of the Pitch Factor is observed if the span is 140° or less, i.e. θ = 40° or more. For θ = 40° Ks = 0.94. The term ‘chorded’ is used in connection with coils pitched less or greater than the pole-pitch. Occasionally the term Winding Factor is used to include the Distribution and Coil-span Factors. Thus Winding Factor = Distribution Factor × Coil-span Factor.

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168 • Advanced Electrotechnology (b)

(a)

N

N

S

S

θ Coil span fully pitched winding

Chorded or under pitched winding

(c)

C θ — 2 θ

A

B

▲ Figure 6.10

Example 6.4. The e.m.f. generated in a generator’s coil is 20 V. Calculate the e.m.f. between the ends of two such series connected coils, separated on the generator core by 30° electrical (2 decimal places). Since the voltages of the coils are out of phase by 30° (figure 6.10) the resultant is given by: θ where θ is the external angle of an isosceles triangle = 2 × internal angle. 2 ∴ Resultant voltage = 2 × 20 × cos 15° = 2 × 20 × 0.9659 = 4 × 9.659 = 38.64 V. 2 V cos

Example 6.5 Find the no-load terminal voltage of a 4-pole, star-connected, marine, turbo-alternator from the following data: Flux per pole (sinusoidally distributed) 0.12 Wb, 4 slots per pole per phase, 4 conductors per slot, coil-span 150°, connection – star, speed 1500 rev/min (4 significant figures).

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The A.C. Generator • 169 nα 4 15 sin 2 = 2 Distribution Factor K D = α 15 nsin 4 sin 2 2 sin

Note α =

180 180 sin 30 05 05 = = 15° so K D = = = = 0.958 3×n 3×4 4 sin7.5 4 × 0.1305 0 522

Pitch Factor K s = cos or K s = cos

θ where θ = 180 − 150 = 30° 2

30 = cos15 = 0.966 2

Generated voltage per phase Eph Here f =

.

KD × K s

Z ph × Φ × f

4 × 1500 = 50 Hz 120

∴Eph = 2.22 × 0.958 × 0.966 × ( 4 × 4 × 4 ) × 0.12 × 50 = 2.22 × 0.958 × 0.966 × 64 × 6 = 789 V Terminal voltage = 3 × 789 = 1367 V or 1.367 kV .

Waveform of generated e.m.f. For A.C. current theory and practice, ideal sinusoidal working is assumed. As the A.C. generator is the source of e.m.f. to achieve this condition, it is vital the generated voltage waveform approximates closely a sine wave. For modern power-stations, synchronous A.C. generators of 300–500 MW are used, where it is assumed that ‘mains’ voltage is sinusoidal. For marine work, although A.C. generator sizes are smaller, every effort is made to achieve sine-wave voltage conditions as generating capacity is both appreciable and independent. Non-sinusoidal conditions create errors in metering, circuit protection and plant rating and adverse effects such as excess voltages due to unwanted harmonic resonances. In Chapter 3, mention was made of harmonics connected with a transformer’s no-load current. Fourier showed that an A.C. waveform of any shape can be analysed into a fundamental sine wave and several sine waves of harmonic frequencies. Figure 6.11 shows the resultant of a sine wave combined separately with 2nd and 3rd harmonic sine waves. With an even harmonic, the second half-cycle of the wave is not an exact inverted reflection of the first, while with an odd harmonic this condition is fulfilled.

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170 • Advanced Electrotechnology (b)

(a) Resultant

Fundamental

Fundamental

Resultant 2nd harmonic 3rd harmonic

▲ Figure 6.11

For an A.C. generator, a North magnetic pole followed by a South pole sweeps past a conductor so the e.m.f. generated varies instant to instant with flux density. Since air gap flux-density distribution is similar for both poles, even though the South-pole flux is opposite to the North-pole flux, the generated waveform has a –ve half cycle exactly the same as the +ve half cycle. Thus only odd harmonics can be generated. Earlier study showed that for a circuit with inductance and capacitance, there is a frequency for which the circuit resonates. For a series circuit this creates high currents and voltages across the inductive and capacitive circuit parts which may be much greater than the supply voltage. For a parallel circuit the ‘mains’ current is reduced to a minimum but large currents arise in the inductive and capacitive branches, again undesirable. Now a circuit not resonant at 50 Hz may be resonant at 150 Hz. If a supply voltage with a 3rd harmonic, of amplitude less than the fundamental (but still appreciable), is applied to a circuit then, due to the strong 3rd harmonic e.m.f., disagreeable effects result which wouldn’t be seen if only a fundamental sine wave voltage of 50 Hz were used. Other generated harmonic effects disturb and interfere with associated communication or control circuits through induction. Designer methods to achieve sine wave generation with minimum harmonic distortion are considered, explaining some A.C. generator construction features. THE AMPERE-TURN OR M.M.F. WAVE. To understand a common treatment method let’s look at a procedure for drawing out the m.m.f. or ampere-turn wave. Take a simple 3-turn, flat coil with conductors arranged as in figure 6.12. If conductors a and d form one turn, the m.m.f. available with a current of 1 ampere to set up flux is 1 ampere-turn

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The A.C. Generator • 171 a

b

c

d

e

f

▲ Figure 6.12

and is constant between the current-carrying conductors. Each time a current-carrying conductor is passed over, the m.m.f. reverses. If the resultant m.m.f. wave for the set of 3 turns ad, be and cf is drawn from the series of rectangular waveforms, a stepped m.m.f. distribution wave is obtained. The m.m.f. increases gradually across each slot so the ampere-turn wave is more like that shown dotted, i.e. a trapezium. ALTERNATOR WAVEFORM IMPROVEMENT. The ideal method of obtaining a sine wave of induced e.m.f. requires armature conductors to cut or be cut by a flux whose density is sinusoidally distributed over a pole-pitch. In practice this is not possible but working methods, differing for the cylindrical and salient-pole types, closely approximate this. Cylindrical Rotor. Since the rotor of a turbo-alternator is smooth, the machine airgap is uniform. The field winding is distributed over the cylinder surface, resulting in a flux wave in space which is more or less sinusoidal than if the field winding were concentrated in one slot for each coil side. Consider the rotor of figure 6.13a spread out (figure 6.13b) in this 2-pole field system. The graphs show the size, location and direction of the effective resultant m.m.f. resulting from rotor current in the coil sides. If a smooth curve is drawn, ignoring the steps, a trapezium is obtained which represents the m.m.f. acting at any point along the surface of the air-gap. The trapezium is made of several sine waves – 3 are shown, the fundamental, with the 3rd and 5th harmonics. Flux is directly proportional to m.m.f., and flux-density distribution over a pole-pitch is sinusoidal so a sine waveform voltage is induced if harmonic e.m.f.s are eliminated. This is achieved by added corrective measures such as chording of the stator windings, etc.

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172 • Advanced Electrotechnology (a)

(b) Rotor m.m.f. wave (theoretical)

m.m.f. wave (actual)

Fundamental

5th harmonic

3rd harmonic

▲ Figure 6.13

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The A.C. Generator • 173 Salient-Pole Rotor. In this case, as independent poles exist – each with its own field winding, a sinusoidal flux-density waveform in the air-gap across the pole face cannot be achieved by ‘distribution of the field winding’. However, as the poles are small and independent they are shaped and modified to allow one or more of the following methods to be effective. Method (1) Shaping of Pole-Shoes. Magnetic circuit reluctance is varied by shaping the pole profile so the air-gap is graded, being least at the centre of the pole-shoe, increasing towards the edges. In practice the profile is determined by theory and the final shape achieved by trial and error. An approximation is obtained by making the pole-shoe radius of curvature 0.7 times the stator core inside radius. Figure 6.14 illustrates the method and again corrective methods are usual.

Po lepit ch

Sinusoidal flux-density distribution

τ

Insulation Winding usually copper strip on edge

Graded air gap Field winding (may be tapered)

▲ Figure 6.14

Method (2) Skewing of Poles or Armature Slots. If a pole face is sinusoidally shaped as in figure 6.15 the armature conductors’ active lengths vary as they are cut by the field, the air-gap being uniform. Pole edges are sine shaped and the arrangement is impractical. One approximation is the third arrangement and an e.m.f. wave of trapezium form results (figure 6.16). This can be corrected to give the required fundamental sine wave.

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174 • Advanced Electrotechnology Skewing shown is exaggerated

N

N

S

N

S

Induced e.m.f. (e)

▲ Figure 6.15

Time (t )

▲ Figure 6.16

The same result is obtained by fixing poles axially and skewing the armature conductors. Method (2) is less favoured than (1) because it does not permit easy modification once a machine is built. Example 6.6. A 4-pole alternator, on open circuit, generates 200 V at 50 Hz, with a 4 A field current. Determine the generated e.m.f. at a speed of 1200 rev/min and field current of 3 A, neglecting iron saturation (3 significant figures). The original alternator speed is given by f =

PN f 120 5 × 120 or N1 = = = 1500 rev/min 120 P 4

The generated e.m.f. E α ΦN and assuming Φ is proportional to the exciting current then: E α IfN so we can write

E2 KIf 2 N2 EI N 200 × 3 × 1200 200 3 4 = and E2 = 1 f 2 2 = = E1 KIf 1N1 If 1N1 4 × 1500 4 5

Thus generated e.m.f. = 40 × 3 = 120 V.

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The A.C. Generator • 175

Stator Windings Mention was made of the types of slots into which stator or armature conductors are placed. From a construction view, ‘open’ slots are preferred (figure 6.17), one can preform and insulate the coil side, made of one or more conductors, inserted into the slots. The main ‘open’ slot disadvantage results in teeth which encourage flux to ‘stray’ as it crosses the machine air-gap. Local magnetic leakage from the slot is high and leakage reactance is low. Flux ‘jumps’ tooth to tooth so a ‘ripple’ frequency is generated (figure 6.17). The ‘closed’ slot overcomes this effect but makes winding hard and gives the machine a high leakage reactance.

Open type slot

Closed type slot

Sta

Ripple effect

tor

Po le

sur

fac e

Inside surface of stator

Semienclosed slot

Generated e.m.f.

Tufting of flux

Skewing of stator slots

▲ Figure 6.17

A ‘semi-enclosed’ slot is a compromise. Skewing of conductor slots reduces the e.m.f. wave ripple. A ‘semi-enclosed’ slot increases leakage reactance but reduces air-gap reluctance and is preferred for induction-motor construction. The ‘semi-enclosed’ slot is best suited to a multi-turn coil where, as for a D.C. winding, hand-wound or preformed coil conductors can be inserted singly into the available space after slot insulation is positioned. An ‘open’ slot arrangement is suited to bar-type conductors or insulated coils, a method favoured for high-voltage machines.

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176 • Advanced Electrotechnology

Types of windings Winding techniques are not considered in detail. Only basic examples are described but generally for A.C. machines ‘open-type’ windings are used in contrast with D.C. machines where ‘closed’ windings are needed. It is usual for smaller and rarely used rotating-armature A.C. generators to follow D.C. practice with a closed winding provided with tappings, brought out to the appropriate slip-rings. For a common rotating-field alternator, the armature or stator winding is of ‘open’ type with 2 ‘free ends’ per phase. (a) One coil side

Two coil sides

Wedge Single layer slot

(b)

Wedge Double layer slot Stator core Bent end connection

Straight end connection

▲ Figure 6.18

A.C. windings may be of (a) the ‘single-layer’ or (b) the ‘double-layer’ type (figure 6.18). SINGLE-LAYER WINDINGS. For a single-layer winding each slot is fully occupied by a conductor or coil side, an arrangement suited to high-voltage machines or coils with a large number of conductors per slot (figure 6.19). The 3rd diagram of figure 6.19 shows one phase of a three-phase, single-layer winding made from concentric, half-coiled coils. Mean coil span equals a pole-pitch and the number of coil groups equals the number of pole-pairs. This figure shows how the winding explanation can be followed from the simple skeleton bar winding. In each case one phase is shown and diagrams set out logically to follow on. The bar winding is extended into the 2nd diagram which shows a full-pitched concentrated winding. This in turn is developed to show a distributed arrangement. In connection with the distributed winding, as coils are in series, the order in which conductors are connected is

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The A.C. Generator • 177

N

S

N

S

N

S

N

S

N

S

N

S

N

S

N

S

N

S

N

S

x

▲ Figure 6.19

not important so other arrangements are possible, such as concentric whole coils, shown by the 4th diagram where the number of coil groups equals the number of poles. Both half-coiled and whole-coiled concentric windings use coils of different spans. It can be difficult to make the ‘overhang’ or end connections, as these pass each other. An accepted method is to bend appropriate coils through 90° or 45° if needed, as shown by the last diagram in figure 9.18. Another method ‘cranks’ the overhand of the end-connections but a better alternative winding sequence as shown (5th diagram in figure 6.19). This creates a regular arrangement with coils of the same pitch and shape, coil-span being equal to a pole-pitch. A ‘distributed’ effect is achieved and a lap or wave effect obtained as a winding progresses round the stator. The need for the necessary end-connection X between coil groups is noted. The lap-wave arrangement may be called a ‘lattice’ winding, from its symmetrical woven appearance.

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178 • Advanced Electrotechnology DOUBLE-LAYER WINDINGS. For D.C. machines, the 2 winding types, lap and wave, are the result of the way the coils per pole-pair are connected. To assist the ‘lay’ of the coils at the ends and for symmetrical assembly with all coils pre-formed the winding is arranged in 2 layers with one coil side in the upper half slot and the other coil side in the lower half slot, one pole-pitch apart. For an A.C. winding, coils similar to those used for D.C. armatures are arranged in lap or wave windings (figure 6.20). Coil span is often equal to a pole-pitch but can be made slightly different if ‘chording’ is needed. Such windings are called ‘lattice’ windings, due to their symmetry, and are common for medium-sized, salient-pole machines for marine work. For example, only one phase of a three-phase winding is shown, as the other phase winding repeats the first. Pitch

Pitch

Pitch

Lap Rs

Ys

Bs

Rf

a

a Rs

Ys

Bs

Wave

Rf

▲ Figure 6.20

Example 6.7. An 11 kV alternator provides a three-phase, 50 Hz supply if run at 3000 rev/min. It has 48 slots with 20 conductors per slot. Each coil spans 18 teeth. Find the flux per pole, assuming it to be sinusoidally distributed (3 decimal places). From f =

PN 50 × 120 6000 P= = =2 120 3000 3000

48 = 24 teeth, constituting 180° 2 18 540 = = 135° so 18 teeth = 180 × 2 4 A fully pitched coil spans

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The A.C. Generator • 179 nα 2 2 . Also α = 180 where n slots/pole/phase = 24 KD = =8 α 3 n 3 n sin 2 8 × 7.5 sin 180 2 α= = 7 5° and K D = 75 3×8 8 sin 2 sin

Thus K D = Also K s =

sin 30 0.5 0.5 = = = 0.956 8 sin i 3.75 8 × 0.0654 0.5232

cos(180 − 135) 45 = cos = cos 22.5 = 0.924 2 2

Now Eph =

1100 3

and since Eph

= 6360 V (assuming star connection) .

K D K s Z ph f

then 6360 = 2.22 × 0.956 × 0.924 × Here Z ph = or Φ =

960 × Φ × 50 3

48 × 20 960 6360 = = 320 and Φ = 3 3 111× 0.956 × 0.924 × 320

636 = 0.203 Wb 3552 × 0.883

The Alternator on Load As for any e.m.f. source, A.C. generator terminal voltage falls when current is supplied. The difference of terminal voltage between open-circuit and load conditions depends on the current and the causes of voltage drop inside a machine. These causes are due to: (1) machine impedance and (2) the armature reaction effect. The terminal voltage variation is affected by the P.F. at which a machine operates (figure 6.21) which shows the ‘on-load’ characteristic. Before the load condition is covered in detail, the term Regulation is introduced, which defines a key machine performance aspect, estimated without direct loading, by developing the phasor diagram of a machine with figures obtained from assessing tests.

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180 • Advanced Electrotechnology

D 0.8 P.F. leading

Terminal voltage (V )

C

B

A 1.0 P.F. 0.8 P.F. lagging

0

Load current (I L)

▲ Figure 6.21

VOLTAGE REGULATION. The variation of terminal voltage between full load and no load, expressed as a percentage or per unit value of full load voltage, is termed the regulation. Rise in terminal voltage when full load is removed Thus: Voltage regulation = ×100 Full-load terminal voltage (refer to figure 6.21). AC × 100% at unity P.F. OA AC = × 100% at 0.8 P.F. (lagging) OA .

Voltageregulation =

Since regulation is influenced by P.F. and load current, these are specified when stating percentage regulation. Due to magnetic circuit iron hysteresis, a rise of p.d. when load is thrown off will be below the fall in p.d. when a load is applied. One method of standardised regulation is defined by the rise in p.d. when load is thrown off, sometimes termed regulation ‘up’ or ‘down’. Example 6.8. Terminal voltage of a three-phase alternator is 440 V, by adjusting the field excitation when the speed is correct and the full-load current supplied at 0.8 P.F. (lagging). When the machine circuit-breaker is opened and the load thrown-off, terminal voltage rises to 506 V. Estimate the voltage regulation (2 significant figures). Voltage regulation =

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506 − 440 66 6 × 100 = = = 15% (up) 440 4.4 0.4

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The A.C. Generator • 181 PHASOR DIAGRAM. The causes of internal voltage drop have been mentioned and when considered against the generated e.m.f. and terminal voltage, the best approach is through a phasor diagram as phase relationships are examined and appropriate phasor procedures followed. Unless otherwise stated, phase conditions are assumed. When an A.C. generator supplies current, internal voltage drops occur due to the following reasons: (1) Armature Resistance. The voltage drop IR, caused by this factor, is always small and usually neglected. It is proportional to the current causing it and is in phase with the current. For the basic phasor diagram, current is taken as the reference since the build up of the diagram follows the procedure for a simple series circuit. (2) Armature Reactance. The voltage drop due to this factor is caused by individual inductive effects associated with the stator winding. Any A.C. winding has selfinductance when carrying current, but the consequent reactance is often attributed to the joint reactance of individual parts, so mention is made of slot-reactance, end-connection reactance, etc. Overall reactance is proportional to current and is in quadrature with it. This reactance is one constituent part of the Xs shown on the phasor (figure 6.22). E

IXS IZS V IR

O

φ X

Y

I

▲ Figure 6.22

(3) Armature Reaction. The magnetic field due to the armature current modifies the main field, altering the generated e.m.f. For a lagging P.F., armature reaction weakens the main field, resulting in a reduced generated voltage. For a leading P.F., the main flux is strengthened and generated voltage rises. The overall effect of armature reaction is represented by a reactance voltage drop again proportional to current and in quadrature with it. This reactance is the other constituent part of Xs shown on the phasor diagram.

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182 • Advanced Electrotechnology Armature reactance and armature reaction produce similar effects and are combined in so-called synchronous reactance Xs. This is not a true reactance, but treated as such for practical purposes and shown on the phasor diagram. If combined with the armature resistance, synchronous impedance Zs is obtained. The effect of Zs is represented by a voltage drop on the phasor diagram illustrating the on-load voltage conditions. Thus for the diagram, I is the load current, OV the terminal voltage and OE the generated voltage. The armature resistance voltage drop (IR), the synchronous reactance voltage drop (IXB) and the synchronous impedance voltage drop (IZB) are shown. The phasor diagram helps determine the A.C. generator regulation if a simple construction is made (dotted lines in figure 6.22). In terms of the large triangle having hypotenuse OE: or V cos φ + IR

Horizontal side = OX + IR

Vertical side = VX + IXs or V sin φ + IXs ∴OE2 = (ОХ + IR)2 + (VX + IXs)2

(V cosφ + IR )2 + (V

or Generated voltage E and Voltage regulation =

IIX

)2

E −V × 100% V

The following example illustrates use of figure 6.22 and its expressions, but the regulation value obtained from this method is higher than the actual practical value. Here we become familiar with the terms Synchronous Impedance and Synchronous Reactance and use of the Voltage Equation developed from the phasor diagram. Example 6.9. An alternator has the following values: effective resistance 0.2 Ω, synchronous reactance 1.1 Ω. The load operates with 0.8 P.F. (lagging). Find the generated e.m.f. for a 100 A load current and a phase terminal voltage of 250 V (3 significant figures). + IR ) + (V

(V

Eph

IIX s )

2

2

=

{(

=

(200 + 20 )2 + (150 + 110 )2

=

(220

×

2

)+(

×

)} + {(250 × 0.6 ) + (100 × 1 1)} 2

2

)

+ 2602 = 102 2.22 + 2.62 = 102 4.84 + 6.76

= 102 11.60 = 100 × 3.4 = 340 V

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The A.C. Generator • 183

Practice Examples 6.1

A three-phase, 1000 rev/min A.C. generator supplies a 50 Hz current to a threephase, 120 rev/min synchronous motor. Calculate the number of poles on both the alternator and motor.

6.2

A 440 V, three-phase, 8-pole marine alternator generates a 50 Hz supply when driven at normal speed. If the system is converted to 60 Hz working, find the speed at which the prime-mover must run. Assuming the same excitation value find the new system voltage (3 significant figures).

6.3

The e.m.f. generated in a coil of an A.C. generator is 20 V. Calculate the e.m.f. between the ends of 2 such coils connected in series, if they are displaced on the A.C. generator core by 30º electrical ( 2 decimal places).

6.4

Calculate the breadth factor for a three-phase, 6-pole alternator having 72 slots (3 decimal places).

6.5

A three-phase, 8-pole alternator has 96 slots. One side of a coil of a 2-layer winding lies in slot 1, while the other side lies in slot 9. Calculate the pitch factor for the winding (3 decimal places).

6.6

A 3.3 kV, 50 Hz, three-phase alternator is driven at 750 rev/min with a single-layer winding contained in 120 slots. The flux per pole is 0.0448 Wb. If the machine is to have a full-pitched winding and is star connected, calculate the number of conductors per slot. 3 For the machine of example 6.6, if the stator winding is short-chorded to pitch 4 and mesh connected, what would be the required number of conductors per slot to the nearest whole number.

6.7

6.8

The three-phase, 2-pole, star-connected alternator of a turbo-electric vessel supplies a 10 000 kW, mesh-connected, 50-pole synchronous motor operating at 0.95 P.F. (lagging) and 90% efficiency at a full-load propeller speed of 110 rev/min. Calculate the phase currents of the alternator (1 decimal place) and the motor (4 significant figures) when the line voltage = 3.3 kV. Find also the kVA alternator output when supplying the motor at the above output power (5 significant figures). Find the supply frequency of the system on full load (1 decimal place).

6.9

When a reduced excitation current is applied to a single-phase A.C. generator the O.C. voltage is 45 V. With the same excitation current and the alternator terminals

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184 • Advanced Electrotechnology short-circuited, the current is now 30 A. The stator resistance is 0.4. Ω. Estimate the synchronous reactance of the machine (2 decimal places). 6.10 A 750 kVA, 3.3 kV, three-phase, star-connected alternator has a resistance of 0.5 Ω per phase and a synchronous reactance of 3 Ω per phase. Find the percentage voltage regulation when full load at 0.8 P.F. (lagging) is switched off (1 decimal place).

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7

THE MARINE ALTERNATOR A cord of three strands is not quickly broken. Ecclesiastes 4.12

Having looked at the A.C. generator in Chapter 6 we now consider the more complex rotating magnetic field produced when polyphase currents flow in polyphase windings. For the induction motor and synchronous motor, the rotating magnetic field is due to currents arising from application of externally applied polyphase e.m.f.s. For the alternator, this field is due to the currents through the machine itself, currents resulting from the load circuit being completed and supplied by the alternator’s generated voltage. Passage of such currents through the machine windings produces an armature reaction effect similar to that found in D.C. machines. Here, the polyphase machine is described, with a review of the current conditions and their overall effect. We will focus on the viewpoint of a marine alternator, as to most mariners, installation of a suitably powered alternator will likely provide the greatest benefits, safety and comfort than any other single electronics component. In many cases alternator performance is the foundation of modern high-power systems. A typical marine diesel may have a standard alternator with a rating of 55 A, and typical sustained output of perhaps 30 A, so addition of a 100 A high-output alternator can more than triple the useable power on board. Note that modern alternator RPM (e.g. BMW, Volvo, Motorola, Hyster, Petter, etc.) may operate at 2–3 times the engine RPM, e.g. an alternator fully output rated at 5000 RPM can spin up to 10 000 or more, without any added increase in output. However, when considering alternator size its output should be about 0.25–0.4 times the size of the battery banks in amp hours, so a 60 A alternator will work well with a 240 Amp hour battery bank.

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186 • Advanced Electrotechnology

The Rotating Magnetic Field This is best described by the simple diagram of figure 7.1, showing 3 coils displaced from each other by 120° in space and supplied by three-phase currents themselves displaced from each other by 120° in time. If a compass needle is mounted at the centre of the coil system, it will rotate, and if 2 connections to the supply are interchanged the needle rotation direction will reverse. R

3-ph supply

B

Y

▲ Figure 7.1

Needle rotation shows that a rotating magnetic field results from the combined effect of the 3 separate single-phase alternating coils’ magnetic fields. In practice the 3 single-phase windings are laid out along the stator circumference as described for the alternator. By choosing a suitable coil pitch, a machine can be wound for any number of poles, provided the number is even. The magnetic field rotation speed depends on the number of poles and the supply frequency. Explanations of the rotating magnetic field effect are here given by 3 different methods. One alone is likely to be sufficient, but comprehension of all 3 will allow a better understanding of armature reaction effects when these are introduced. Method 1. By Consideration of the Resultant Magnetic Field Distribution. A basic approach to understanding the combined magnetic field produced by a threephase winding is made by considering a simple 2-pole stator with one conductor per pole per phase. In figure 7.2, conductors A and В constitute the‘go’and‘return’conductors of a single turn and are supplied from one phase of a three-phase transformer. Similar conductors С and D are supplied from the yellow phase and EF from the blue phase. Currents flowing in the stator windings are illustrated by the figure 7.3 waveform and by considering conditions at different instants of time. For a suitable starting time

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The Marine Alternator • 187 A R Rph

B1 F

D Y1

Bph E B

3-ph supply

Yph

Y C R1 B

▲ Figure 7.2

1 Rph

2

3 Yph

Bph

Current

+

Time

–

▲ Figure 7.3

consider Instant 1, when the Rph current is maximum. Then the conditions shown (figure 7.4a) exist and the resultant magnetic field is as shown. Note: the currents in the Yellow and Blue phases are –ve with respect to the Red phase current and represented by dots in the respective ‘go’ conductors to illustrate their direction and are half their maximum value. The stator functions as a solenoid with its axis along the horizontal and the resultant magnetic field is as depicted. For Instant 2, the time is 30° later. The 3 current in the Red phase is still +ve but falls to of its maximum value. The current 2 3 in the Yellow phase falls to zero, while the Blue phase rises to maximum value and 2 is negative. The resultant magnetic field is shown in figure 7.4b and its axis will have rotated through 30°. For Instant 3, the Blue phase current will be at negative maximum. The Red and Yellow phases current will be +ve but half of maximum value. The resultant magnetic field is as shown (figure 7.4c) and its axis will rotate through a further 30°.

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188 • Advanced Electrotechnology (a)

(b)

Im

3 – Im 2

Im/2

Im/2 Y′

R

N

B′

B

3 – Im 2

30°

S

N Y

R′

Im/2

Im/2

3 – Im 2

Im

(c)

3 – 2 Im

Im/2

N

Im/2

S

Im

60°

Im/2

Im

Im/2

S

▲ Figure 7.4

On examining steps (a), (b) and (c) we observe that the three-phase currents flowing through coils, suitably displaced from each other by 120° (electrical), produce a resultant magnetic field which rotates in time and in phase with the currents, i.e. rotating at a speed decided by the frequency, giving rise to the synchronous speed, as seen if alternative methods are considered. Method 2. By Mathematical Derivation of the Rotating Field Conditions. For the mathematically minded student, the following method of treating the phenomenon of a rotating magnetic field will be appreciated. Figure 7.5 is similar to figure 7.4a, except that three-turn phase windings are shown. For simplicity a 2-pole arrangement is treated, the 3 individual windings produce fluxes ΦR, ΦY and ΦB each varying in magnitude according to a sine law, but not rotating. Each is a purely alternating flux. We can write: ΦR

Φ m sin ω t

ΦY

Φ m sin( i (

−

o

)

ΦB

Φ m sin(ω t − 240°)

The resultant is found by resolving appropriate flux vectors into horizontal and vertical components.

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The Marine Alternator • 189

R

Y1

ΦB

B1 ΦR

B

Y

ΦY R1

▲ Figure 7.5

Thus, ΦH

ΦR − Φ Y ΦB cos 60 = Φm i Φ m sin( i (ω (ω t − 120) cos 60 − Φ m sin(ω t − 240) cos 60 = Φ m [sin ω t coss {sin( { iin((ω (ω t − 120) cos 60} + sin(ω t − 240)] = Φ m sin ω t cos 60 sin( iin((ω (ω t − 180) ∗

*Note the trigonometric formula set out as follows: S T S T cos 2 2 continuing ΦH = Φ m {sin ω t + cos 60 sin sin S + i T

2 i

}

sin ω t ⎫ ⎧ = Φ m ⎨sin si ω t + ⎬ 2 ⎭ ⎩ 3 = Φ m sin ω t 2 Also, ΦV

ΦB cos 30 − Φ Y cos 30 = Φm i ( ) = Φm =

9781408176030_Ch07_Final_txt_print.indd 189

3 { 2

− 240))

Φ m sin( iin((ω (ω t − 120) cos 30 3 0)}

3 Φ m {2 co coss (ω t − 180 ) − sin 60} * 2

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190 • Advanced Electrotechnology *Note the trigonometric formula set out as follows: S T S T sin i 2 2 ⎧ ⎛ 3 ⎞⎫ 3 3 or Φ v = Φ m ⎨2( cos ω t ) ⎜ − ⎟ ⎬ = Φ m cos ω t 2 ⎝ 2 ⎠⎭ 2 ⎩ sin S

2 cos

Resultant flux Φ

3 Φm 2

Φ H2 + Φ v 2

3 i 2 ω t + cos2 ω t = Φ m 2

3 × the maximum value produced by any one phase 2 winding. If θ is the angle the resultant field makes with the horizontal, then: The resultant flux has magnitude

3 Im sin i ωt 2 tanθ = = tan ω t . Thus θ ω t 3 Im cos ω t 2 so the resultant field is rotating at a constant angular velocity, i.e. at synchronous speed, since θ = ωt = 2π ft Method 3. By Deduction of Magnetomotive Force or Ampere-Turn Field-Producing Patterns. Consider figure 7.6a to show one pole of a three-turn phase winding. The procedure for drawing out the m.m.f. distribution patterns associated with a winding was considered earlier. T T a

b

T

c

d

e

f

(a)

(b)

▲ Figure 7.6

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The Marine Alternator • 191 Let’s consider the ‘go’ and ‘return’ conductors of a single-turn phase winding. The m.m.f. available to set up flux is constant in space between the current-carrying conductors a and d, b and e, с and f. Each time a conductor is passed over, the m.m.f. reverses and, if the distribution for the set of phase conductors is drawn as in figure 7.6b we get a series of rectangular waveforms, which when added gives the resultant stepped distribution waveform. The m.m.f. increases gradually across each slot, so the distribution pattern is shown dotted. This procedure is used to find the m.m.f. distribution over one pole (a) Instant when the current in Rph is a maximum. R

T B1

Y

Imax

1 – Imax 2

1 – Imax 2

(a)

R1

T B

Y

Imax

1 – Imax 2

1 – Imax 2

R

1 – Imax 2

B

Y

1 – Imax 2

m.m.f. wave – sharp pointed

(b) Instant when the current in Yph is zero. (b)

R

B1

3 – 2 Imax

3 – 2 Imax

Y

R1

B

3 – 2 Imax

3 – 2 Imax

Y1

R 3 – Imax 2

Y 3 – Imax 2

B

m.m.f. wave – flat topped

(c) Instant when the currents in Rрh and Yph are +ve and half the maximum value. (c)

R

R

B1

Y

R1

B

Y1

1 – Imax 2

Imax

1 – Imax 2

1 – Imax 2

Imax

1 – Imax 2

Y

1 – Imax 2

1 – Imax 2

B

m.m.f. wave – sharp pointed

▲ Figure 7.7

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192 • Advanced Electrotechnology pitch due to three-phase currents in their respective windings. Three instants in time are considered and the associated phasor time diagram drawn to illustrate current magnitudes and their relationships occurring for each consideration. The relevant diagrams are figures 7.7a, 7.7b and 7.7c. Examination of the 3 m.m.f. distribution waveforms shows the following: (1) Pole pitch is constant. (2) Movement in time, e.g. 30° of the phasors, corresponds to a 30° space wave movement. Thus this m.m.f. wave moves at synchronous speed and in the same direction as the phase rotation. (3) The m.m.f. distribution wave is in time phase with the current for a particular phase winding. If the current is a definite value at any instant in the winding, the position and form of the m.m.f. wave at that instant can be found. The two waveforms are extreme shapes (figure 7.8). Both are approximated by a fundamental constant height sine wave with 3rd and 5th harmonics. The latter are not very important and, if care is taken in design to minimise all the harmonics the m.m.f. wave can be considered sinusoidal, with a maximum value with a mean between the maximum values of the 2 limiting cases. Note that as the magnetic path reluctance is fairly uniform, especially for a cylindrical rotor alternator and induction motor, the flux and m.m.f. distribution waveforms are similar.

Fundamental sine wave

▲ Figure 7.8

Also note the rotating m.m.f. or flux wave is always in a position where it can combine with maximum linkages (illustrated by figure 7.9). Thus for figure 7.9a if, e.g. current in the red phase is maximum, the m.m.f. and flux wave is symmetrically distributed across that phase winding and maximum flux-linkages occur. Similarly for figure 7.9b when current is zero the m.m.f. or flux distribution is such as to have its zero value at the centre of the winding – zero flux-linkages occur, i.e. peak value is displaced by 90°. Reference is made to this fact when armature reaction effects are considered.

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The Marine Alternator • 193 (b)

(a)

Im

I=0

Im

I=0

▲ Figure 7.9

Armature reaction The work done, in connection with a rotating magnetic field, shows that armature currents produce a field of constant strength rotating at synchronous speed. In figures 7.10, 7.11 and 7.12, the e.m.f.s are generated in stator conductors by the movement of the main field poles. Conductors of only one-phase winding are shown, those above the N pole constituting a three-turn coil with those above the S pole. The main flux path is shown by full lines and that of the armature current-produced flux by dotted lines. When such current flows, an extra magnetic effect is produced which modifies the main field, this being the armature reaction described previously. The

e.m.f. current

+ +

+ +

N

+ +

+ +

+ +

Conductor motion

+ +

S Field motion

Main field m.m.f. & flux

Armature m.m.f. & flux Resultant flux

▲ Figure 7.10

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194 • Advanced Electrotechnology exact effect of such armature reaction depends on the current phase relative to the e.m.f.s generated by the main field. CASE 1. CURRENT IN PHASE WITH GENERATED E.M.F. When the phase winding coil is in the position shown by figure 7.10, the flux linking with it changes at the greatest rate and maximum e.m.f. is generated. The armature currents produce flux which passes across the pole shoes without directly affecting total pole flux. The effect is purely cross-magnetising, the leading pole-tip being weakened and the trailing pole-tip strengthened. The effective pole centre thus moves backwards. Since the armature m.m.f. is cross-magnetising, field flux is reduced only slightly due to saturation effects of the field system. The generated e.m.f. is reduced by a small amount but the waveform may be distorted. In the diagram, use is made of both e.m.f. and current conductor ‘bands’. Since E and I are in phase, the bands are in line and similar. CASE 2. CURRENT LAGGING GENERATED E.M.F. BY 90°–ZERO P.F. CONDITION. Here 1 current bands lag cycle behind the e.m.f. bands to produce fluxes which have a 4 direct demagnetising effect on the main poles (figure 10.11). The armature m.m.f. is entirely demagnetising and the flux produced is large since the coil centre is opposite the poles. The resultant flux in the air-gap reduces considerably and generated e.m.f. is accordingly smaller.

e.m.f. current

+

+

+

+

+

N

S

Resultant flux

Armature m.m.f. & flux

Main field m.m.f. & flux

▲ Figure 7.11

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The Marine Alternator • 195 CASE 3. CURRENT LEADS GENERATED E.M.F. BY 90°–ZERO P.F. CONDITION. The diagram is similar to the previous one except the current in the armature conductors is in the opposite direction and armature flux is reversed (figure 7.12). The current bands lead 1 the e.m.f. by cycle to produce fluxes with a direct magnetising effect on the main 4 poles. The armature m.m.f. is entirely magnetising but the flux increase is not as big as the decrease in Case 2, due to iron circuit saturation effects. The resulting flux increase increases the generated voltage. Resultant flux Main field m.m.f. & flux

Armature m.m.f. & flux

▲ Figure 7.12

The phasor diagram (continued) An introduction has been made to the simple alternator phasor diagram (figure 6.22) and the causes of internal voltage drop was mentioned. These are considered in greater detail and the full diagram developed. ARMATURE RESISTANCE (R). ‘Effective resistance’ includes the actual ohmic value and an extra component, arising from the uneven current distribution in the conductors caused by irregular and alternating fluxes across the slots. This extra resistance is usually taken as 10% of the ohmic value and added to the true resistance to give the effective resistance. Even after making this allowance, the resistance voltage drop is tiny and often neglected in calculations. ARMATURE REACTANCE (XS). The reactance, as shown on the simple phasor diagram (figure 6.22) is repeated in figure 7.14, and termed synchronous reactance which, from investigation of voltage regulation, is greater than that due to the machine’s inherent inductance. This reactance is considered to have 2 components (1) true leakage reactance and (2) a reactance effect from the armature reaction.

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196 • Advanced Electrotechnology (1) Leakage Reactance (XL). The armature currents give rise to a magnetising field which varies with frequency and cuts the conductors to generate a small back e.m.f. This is the normal understanding of inductance and the machine is known to have this leakage inductance which results in a reactance value. This leakage reactance is considered by designers under 2 main headings: (a) Slot and tooth-tip reactance and (b) Endconnection reactance (figure 7.13) and are roughly equal. The former varies with coil position relative to the poles and a mean value is taken.

Slot Leakage Flux

▲ Figure 7.13

(2) Armature-Reaction Reactance (Xa). This is not true reactance but considered as such for convenience. In this context it is stressed that armature reaction does not cause reactance, it is responsible, as stated earlier, for a change in generated voltage, a change which is likened to a voltage drop. Since the voltage drop is proportional to current, it can be represented by a reactance value multiplied by a current value. Further explanation is made in considering the complete phasor diagram. THE SIMPLE PHASOR DIAGRAM. This, as already developed, is redrawn to assist in the solution of the first example (figure 7.14). Example 7.1. A 50 kVA, 500 V, single-phase alternator has an effective resistance of 0.2 Ω and a synchronous reactance of 1.688 Ω. Find the synchronous impedance and the voltage regulation when a load of 100 A operates at 0.8 P.F. (lagging) is removed (1 decimal place). Zs

R 2 + X s2 = 0.22 + 1.6882 = 0.04 + 2.85 = 2.89 = 1.7 Ω

Synchronous impedance = 1.7 Ω From figure 7.14 the generated voltage E is represented by phasor OE.

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The Marine Alternator • 197 IXs

V IR

φ O

▲ Figure 7.14

Thus E

cos φ IR )2

( =

{

×

= (

+

(OV O sinφ IX I s )2 ×

)2 + (

}2 + {(500 × 0..6)

((100 100 × 1.

)}

2

.8 )2 = 4202 + 4692

= 100 4.22 + 4.692 = 100 17.64 + 21.996 = 100 39.64 = 100 × 6.3 = 630 V Voltage regulation =

630 − 500 130 × 100 = = 26% 500 5

This is described as 26% ‘up’, since a voltage rise is involved. Note the synchronous reactance, resistance and impedance are often given in percentages rather than definite figures. This practice was given in relation to transformers and the next example illustrates this. Note: the percentage value refers to a voltage drop and not to an ohmic figure. Example 7.2. A 600 kVA, 3.3 kV, 50 Hz, three-phase marine alternator is star connected and operated at full-load 0.8 P.F. (lagging). If the resistance and synchronous reactance values are 1% and 6% respectively, find the percentage voltage regulation value for the full-load condition described (2 decimal places). Load terminal voltage per phase =

33

Resistive voltage drop per phase = 1.91×

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3

1.91 kV

1 = 0.019 kV 100

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198 • Advanced Electrotechnology If E is the generated voltage per phase and V is phase terminal voltage then: (V cos φ + IR )2 + (V sin inφ IX I s )2

E

{

=

×

}2 + {(1.91× 0.6) + 0.1146}2

+

= (1.528 + 0.019 )2 + (1.146 6 + 0.1146 )2 = 1.547 .5472 + 1.2612 = 2.39 3 + 1 59 = 3.98 1.995 kV k THE COMPLETE PHASOR DIAGRAM. In line with the treatment of A.C. circuits and machines, a phasor diagram can be derived to illustrate alternator operating conditions. An introduction was made to such a diagram, with reference made to the causes of internal voltage drops such as: resistance R, and the combined leakage reactance and armature reaction effect (or synchronous reactance), Xs being used. Since ‘on-load’ the main field is altered in magnitude and phase by the armature flux, this effect is shown on the diagram and illustrates the related effects on generated e.m.f. The diagram can be drawn to show voltage and current phasors and the flux (the m.m.f. vectors). In this form, figure 7.15 depicts more completely machine operation as described. Since the m.m.f. vectors are added to those shown on the diagram as deduced up to now, these warrant attention. Ea Φ0

IXa

Φ F0

E0 IX s IXL

F V IR

φ Fa Φ a

I

Eb

▲ Figure 7.15

Flux in the Air-Gap. When ‘on-load’, in the alternator air-gap, there are 2 sinusoidally distributed fluxes caused by the m.m.f. distribution waves. These fluxes rotate at synchronous speed in the same direction. They remain in the same position relative to each other and produce a resultant flux which is assumed sinusoidal. The fluxes and their producing m.m.f.s are shown on the left-hand side. Thus Φ0 is the no-load flux set up by F0 the main field m.m.f. or ampere-turns. Φа is the armature-reaction flux and

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The Marine Alternator • 199 the corresponding m.m.f. or ampere-turns is Fa. Φ is the resultant flux caused by F, the resultant m.m.f. or corresponding ampere-turns. Since the ‘on-load’ or resultant flux is the original flux altered by the armaturereaction flux, the e.m.f. E0 as originally generated is considered to be weakened and retarded by an internally generated e.m.f. Eb as shown. The final generated e.m.f. is E1. Since Eb is proportional to armature flux which is proportional to current I, the effect is similar to a reactance voltage drop, in quadrature with current. The machine is credited with Xs – a larger reactance than the leakage reactance XL, the extra amount Хa taking care of the armature reaction e.m.f. The term ‘synchronous reactance’, Xs is now treated in depth and must be appreciated by the reader, and is shown on the phasor diagram. The phasor diagram is drawn for one phase only, although the armature-reaction effect is due to the combined action of all the phases in a polyphase machine. From figure 7.7 it was shown that the armature m.m.f., due to all 3 phases, produces its maximum value, relative to any one phase, when the current in that phase is a maximum. It is represented by a phasor in phase with the current in that phase. Generated e.m.f. is produced by a change in flux linked with the associated coils, and the e.m.f. phasors are shown lagging fluxes by 90°. Summarising the diagram, we see it is drawn for a load current I lagging the terminal voltage V by an angle φ. The main field ampere-turns are shown by F0, and the e.m.f. generated by the associated no-load flux Φ0 is E0. The armature ampere-turns Fa cause the reaction effect or an associated e.m.f. Eb equivalent to a voltage drop IХa. The on-load flux Φ, caused by F ampere-turns – the resultant of F0 and Fa – is responsible for the onload generated e.m.f. E1. The resistance voltage drop is IR in phase with the current and the leakage reactance voltage drop is IXL in quadrature with the current. IХa and IXL are added to give IXS the synchronous reactance voltage drop from which the synchronous impedance voltage drop IZS is derived and stated as earlier: Z s R 2 + X s2

Prediction of voltage regulation Because of difficulties in loading and driving large alternators, performance on load is usually determined from assessed tests. Modern power-station generators operate up to 500 MW and marine machines of 300–500 kW rating are common. Although the latter are small when compared with corresponding land units, direct loading is seldom undertaken until machines are installed on board ship, when performance is checked against predicted values. Two performance predicting methods are considered but both require an open-circuit and short-circuit test to be made on machines.

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200 • Advanced Electrotechnology Open-Circuit (O.C.) Test. This is made like that for obtaining the magnetisation or opencircuit characteristic (O.C.C.) for a D.C. generator. The machine is driven at correct speed and terminal voltage noted for various field current values. Test results when plotted generate a curve of the type associated with a generator O.C.C. and a check made to ensure that 50% overvoltage is obtained without saturating the field system. This precaution covers the requirement for a machine to generate full voltage under overload and poor P.F. conditions. Short-Circuit (S.C.) Test. This is made with the machine terminals short-circuited, with a suitable current measuring instrument included in the circuit. The machine is run at normal speed and a reduced excitation applied. Readings of line current are noted for appropriate field current values, the former being raised to as large a value as possible, while keeping in mind the danger of damage overheating to windings. Plotted test results give a straight line, so only 2 or 3 readings are needed. The O.C. and S.C. test results are often plotted to the same base of field current or field ampere-turns as they are used jointly for making assessment tests as described shortly. THE SYNCHRONOUS-IMPEDANCE METHOD. This is also referred to as the Belm Eschenberg or Pessimistic method – for the reason set out below. The effect of armature reaction is ignored and the alternator credited with a greater reactance value than it actually has. In relation to the phasor diagram, the effect of Fa is accounted by the Хa value from the windings. As the terminal-voltage value V is zero it is assumed that all the voltage generated is used in overcoming voltage drops IR and IXS when current circulates through a machine. Synchronous impedance is thus obtained and used in conjunction with the phasor diagram and regulation formula. The reasoning appears acceptable but for the fact that ZS is measured from a S.C. test and, if its values, as determined by the procedure considered, are plotted to a base of excitation current it is seen to decrease as excitation is strengthened due to field system saturation. For calculations a value of ZS is chosen as near as possible to the excitation under load conditions. For the S.C. test, field current is low and the field is weak, i.e. the magnetic circuit is unsaturated. On full-load, field current is high and the iron nearly saturated. Under the latter condition greater opposition is offered to the flux causing armature reaction than under the former or test condition. For the S.C. test, more flux-linkages is associated with the armature than occurs on load and thus Xs, as obtained from the S.C. test, is larger than its value under normal load conditions. A larger assessment test voltage drop occurs than in practice, giving a pessimistic figure for voltage regulation. Figure 7.16 shows the curves obtained from the O.C. and S.C. tests, illustrating how test results are used.

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The Marine Alternator • 201

Terminal voltage (V )

Armature current (I )

IL

If

Full load current

L

▲ Figure 7.16

The voltage generated, when making the S.C. test, drives current through the armature impedance. For the full-load current value IL the field current is IfL as shown. From the O.C.C., for IfL a voltage of VL generated. A current of IfL in the field gives a voltage of VL generated and an armature current of IL results. Therefore Z s =

VL for that field current value. R is measured for a hot armature condition IL

and XS is found from X s

Z s2 − R 2 . From the basic phasor diagram E is obtained, and

voltage regulation deduced. The procedure is illustrated by the next example. Example 7.3. A single-phase marine alternator required an excitation current of 1.9 A to produce a short-circuit full-load current of 100 A. The same excitation current results in an O.C. voltage of 430 V. Armature resistance is measured to be 0.77 Ω. Find the full-load voltage regulation percentage at unity P.F. when the terminal voltage is 500 V (2 significant figures). Zs =

430 = 4 3 Ω Xs 100

.32

0.

2

8. 9 0.59

.9

. 3Ω

(V cos φ + IR )2 + (V sin inφ IRs )2

E =

{

× +

×

2 1 × 4 23 } }2 + {(500 × 0) + (100

= (500 + 77)2 + (0 + 423)2 = 5772 + 4232 = 100 5.77 7 2 + 4.23 = 100 0 33.29 + 17.89 = 100 51.18 = 100 × 7.15 = 715 V Voltage regulation=

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715 − 500 215 × 100 = = 43% 500 5

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202 • Advanced Electrotechnology Example 7.4. A three-phase, 5 MVA, 6.6 kV alternator is star connected and has O.C. and S.C. test results as set out in the following table: Field current (A)

Terminal voltage (V)

Line current (A)

Phase voltage (V)

100

4800

690

2770

150

6500

1020

3750

200

7400

–

4270

250

7900

–

4560

Estimate percentage voltage regulation at full load, 0.8 P.F. (lagging) by the synchronous impedance method. Assume stator resistance is negligible (1 decimal place). The graphs of figure 7.17 show the appropriate values plotted to a common base of field current. The figures in the last column are not test results but deduced from column 2. 4800 Thus = 2770 V . This is a star-connected machine and the problem is worked in 3 phase values. The largest measured S.C. current is used to obtain ZS, as this gives the maximum field strength leading to saturation, with a reduced ZS value error.

4800 .C O.C

1500

3000

Line current (I ) amperes

Phase voltage (V ) volts

4000

.C S.C

2000

1000

1000

500

0

50

100 150 200 Field current (It) amperes

250

▲ Figure 7.17

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The Marine Alternator • 203

Full-load current =

5000 × 1000 6.6 × 1000 × 3

Normal phase voltage V =

=

5000 = 437.4 A 6.6 × 1.732

6600 3

Generated voltage per phase = E R = 0 Ω, also X s

= 3820 V 6600

3 68 Ω and Zs

3 × 1020

Xs

Then (V cos φ )2 + (V sin inφ

E

= (

I s )2 IX

× .8 )2 + {(3820 × 0.. ) + (

× .68 )}

2

= 30562 + (2292 + 1610 )2 = 30562 + 396022 = 103 3.0562 + 3.9022 = 103 9.3 15.2 = 103 24.5 103 Voltage regulation =

4.95 = 4950 V

4950 − 3820 × 100 = 29.6% 3820

Synchronising torque It is useful to consider the condition when a machine is connected in parallel with ‘live’ busbars. An ‘incoming’ alternator may be paralleled onto infinite busbars, i.e. a system so large such as the ‘grid’ system, whose stability is unaffected by the speed and size of the incoming machine. Alternatively, the alternator may be connected to a machine of equal rating, which is already on load. As the procedure of synchronising involves bringing the machine up to the correct voltage value and frequency, it is assumed this can be done accurately by noting the voltmeter, tachometer and frequency-meter readings. However, a further condition to be satisfied for correct synchronising is the phase sequence and phase angle of the incoming alternator at the instant of closing the machine circuit-breaker. In the event of a machine ‘synchronised’ partially out of phase with the busbar voltage, the incoming alternator experiences a ‘pull-in’ torque which will either instantaneously speed it up or slow it down relative to the system. If paralleled onto an infinite busbar system, the incoming machine will be affected, but if it is connected to a small independent system, as occurs on board ship, the system frequency and stability may change. If the incoming machine lags, the system will sustain a shock, as the machine

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204 • Advanced Electrotechnology is pulled up into step and the loaded machine momentarily slowed down. Once synchronised, the incoming alternator will run light or ‘float’ on the busbars. Similarly if the incoming machine is leading it will sustain a momentary shock while pulled back and the system frequency may rise until the arrangement attains steady-state conditions with the machine floating on the bars. During the transient conditions described, a circulating current flows between the incoming machine and the busbars. The current size is determined by the ‘out of phase’ magnitude when the parallel circuit-breaker is closed. This current is extra to that supplied by the system and may well ‘trip out’ closed circuit-breakers, causing their over-current or reverse-power protective devices to operate. Such a current may be dangerous, as the synchronising torque which it produces can result in damage to the incoming machine or system. Factors affecting this current and synchronising torque are considered and illustrated by example. SYNCHRONISING POWER AND TORQUE. The position of the incoming and ‘running’ machines is regarded from the common circuit point of view, i.e. the busbars and the phasor diagram (figure 7.18).

α Er

Is

▲ Figure 7.18

Let E1 be the generated voltage of machine (No. 1) or machine already on the ‘bars’ and E2 the e.m.f. of the incoming machine (No. 2). E2 is in advance by an angle of α or its equivalent mechanical displacement. Due to this displacement, a resultant e.m.f. Er is created which causes synchronising current flow IS. IS is limited mainly by reactance of the closed circuit, so the machines’ resistances being negligible, then IS lags Er by 90°. Now Er 2 = sin α E2 2

9781408176030_Ch07_Final_txt_print.indd 204

Er

E2 siin

α 2

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The Marine Alternator • 205 α 2E2 sin Er 2 where X is the synchronous reactance per phase of each and IS = = s/ph 2 X s ph 2 X s ph machine, which is assumed to be identical. α Power flow from machine No. 2 = E2 IS cos 2 α α α α E2 E2 sin cos E22 sin cos 2 2 2 2 Synchronising power/ph = = X s ph X s ph Example 7.5. A single-phase, 10 kV, 50 Hz alternator runs at 1500 rev/min with a synchronous reactance of 4 Ω. Assuming the machine is paralleled on to an infinite busbar arrangement, calculate the synchronising power in kW for 1 mechanical degree of phase displacement (3 significant figures). 120f 120 × 50 = = 4 poles. Total number of electrical degrees contained on N 1500 4 rotor circumference = × 360 = 720°. Number of equivalent mechanical degrees = 360° 2

Here P =

For this machine 1 mechanical degree 720 / 360 = 2 electrical degrees. Synchronising current IS =

Er Er for two small machines but for infinite busbars IS = 2 X s ph 2 X s ph

The X s ph of the busbar system being negligible. Here Er

E2 sin

α 2

000 × sin siin

Synchronising power = E2 IS cos

α 350 = 350 V IS = = 87.5 A 2 4

α 10 000 × 87.5 × 0.9998 = = 874 kW 2 1000

If this was a three-phase machine, the total power would be 3 × single-phase power. Synchronising torque. This is derived from the relation: Synchronising power =

2π NT 6 Synchronising power × 60 or T = = newton meters. 60 2π N

Hunting or Phase-swinging. For small phase differences, the synchronising current IS and thus the synchronising torque are proportional to the phase-difference angle. The parallel working arrangement can have a free period of oscillation about the maintained speed which is known as ‘hunting’ or phase-swinging. After synchronising, on light load or when load is thrown off, the oscillation may increase the circulatory

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206 • Advanced Electrotechnology current amplitude until a machine falls out of step and is ‘tripped off ’ the busbars. All alternators designed for parallel operation have a damping winding set into the pole faces. When hunting occurs, relative motion takes place between the damping-winding bars and the field flux, resulting in large currents in the bars producing a damping or stabilising effect. Damping windings are referred to as damping grids and as hunting occurs, particularly with engine-driven salient-pole machines, where cyclic oscillations are prone to be present, an arrangement as in figure 7.19 is common. For extreme cases, special precautions are needed and short-circuiting damping rings may extend around the rotor periphery. Damping bars

Short-circuiting end strap

▲ Figure 7.19

For a cylindrical rotor, retaining wedges may be made of brass and bonded at the ends to serve as damping grids. Considerable damping already exists since the rotor is solid and augmented by the effect of metal wedges in the rotor slots. For special cases, sufficient damping is provided by an extra bar in the slots, connected to end rings to act as a damping winding. The next 2 examples are included for interest and revision purposes. Example 7.6. In a 1 MVA, 11 kV, three-phase, star-connected alternator the resistance of the winding per phase is 2 Ω. The iron core loss is 15 kW, the friction and windage loss is 10 kW and the input required to the field excitation circuit amounts to 5 kW. Estimate the percentage machine efficiency when operating at full load and 0.8 P.F. (lagging) (1 decimal place).

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The Marine Alternator • 207 Output power at 0.8 P.F. = 1000 × 0.8 = 800 kW Full-load current =

1000 × 1000 3 × 11 000

=

1000 1000 = = 52.48 A 1.732 × 11 19.05

Copper loss per phase 52 482 × 2

52.48 × 104.96 1000

5 51 kW

Three-phase copper loss = 16.53 kW Total loss = 16.53 + 15 + 10 + 5 = 16.53 + 30 = 46.53 kW Input power = 800 + 46.53 = 846.53 kW Efficiency =

52.48 × 104.96 800 800 = × 100 = = 94.5 % 1000 846.53 8.465

Example 7.7. An industrial load of 4 MW has a supply voltage of 11 kV with 0.8 P.F. (lagging). A synchronous motor must meet an added 1130 kW and at the same time raise to 0.95 P.F. (lagging). Find the motor’s kVA rating (4 significant figures), the P.F. (3 decimal places) at which it must be operated and the motor’s line current (1 decimal place) if it is star-connected. Assume the motor efficiency is 93%. Original condition. Active power of basic load = 4000 kW Apparent power of basic load = Also cosφ = 0 8 so sinφ

4000 = 5000 kVA 08

0.6

Reactive power of basic load = 5000 × 0.6 = 3000 kVAr Additional load. Motor active power output = 1130 kW Motor input power =

1130 = 1215 0 93

Final condition. Total active power of load = 4000 + 1215 = 5215 kW Apparent power of combined load =

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5215 = 5500 kVA 0 95

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208 • Advanced Electrotechnology Also cosφ

0.95 ∴ φ = 18.11 so sinφ

0.312

Reactive power of combined load = 5500 × 0.312 = 1715 kVAr By deduction: Reactive power rating of motor = 3000 − 1715 = 1285 kVAr The motor is to operate at a leading P.F. so as to raise the P.F. of the combined load. The original load was working under a lagging P.F. condition. Apparent power rating of motor = 12852 + 12152 = 103 12852 + 12152 = 103 1.65 + 1.48 = 10 103 3.13 = 1770 kVA P.F. of motor cos φ = Line current of motor =

1215 = 0.687 (leading) 1770

1770 × 1000 3 × 11 000

=

1770 = 92.5 A 1.732 × 11

Practice Examples 7.1 A 4-pole alternator, on open circuit, generates 200 V at 50 Hz when its field current is 4 A. Determine the generated e.m.f. at a speed of 1200 rev/min and a field current of 3 A, neglecting saturation of the iron (3 significant figures). 7.2 A single-phase alternator has an effective resistance of 0.2 Ω and a synchronous reactance of 2.2 Ω. Find the generated e.m.f. when the load current is 50 A at a terminal voltage of 500 V. The load operates at 0.8 P.F. (lagging) (3 significant figures). 7.3 Find the synchronous impedance and reactance of an alternator in which a given field current produces an armature current of 250 A on S.C. and a generated e.m.f. = 1500 V on O.C. Armature resistance is 2 Ω. Calculate terminal voltage in kV when a load of 250 A at 6.6 kV and 0.8 P.F. (lagging) is switched off (2 decimal places). 7.4 A 250 kVA, 3000 V, three-phase, star-connected alternator has a synchronous reactance of 20 Ω per phase. The armature resistance measured between two

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The Marine Alternator • 209 terminals is 2.8 Ω. Determine the value to which the terminal voltage will rise when full load at (a) 0.8 P.F. (lagging) and (b) 0.9 P.F. (leading) is switched off (both 4 significant figures). 7.5 A marine type 9 kVA, 230 V, single-phase alternator has an armature resistance of 0.25 Ω. The results of the O.C. and S.C. tests are as follows: Field current (A)

1

2

3

3.5

4

5

6

7

Terminal voltage (V)

78

144

198

220

237

265

284

296

S.C. current (A)

11

22

34

40

46

57

69

80

Estimate, by means of the synchronous-impedance method, the percentage voltage regulation of the machine on a full-load, 0.8 P.F. (lagging) condition (1 decimal place). 7.6 Using the data of example 7.7, estimate the machine’s percentage voltage regulation by the ampere-turn method (1 decimal place). 7.7 A 500 kVA, 440 V, three-phase, star-connected alternator has a resistance of 0.01 Ω per phase. The results of O.C. and S.C. tests are as follows: Field current (A)

0

20

40

60

80

100

120

160

200

Terminal voltage (V)

0

130

253

336

393

440

478

524

545

260

500

–

1000

Line current (A)

Calculate by means of both the ampere-turn and the synchronous-impedance methods the value to which the terminal voltage will rise when full load at 0.8 P.F. (lagging) is switched off (both 3 significant figures). What is a more likely voltage value for this machine? 7.8 A marine-type, 30 kVA, 440 V, three-phase, star-connected alternator has a resistance of 0.15 Ω per phase. Results of O.C. and S.C. tests are as follows: Field current (A)

2

4

6

7

8

10

12

14

Terminal voltage (V)

156

287

396

440

474

530

568

592

S.C. current (A)

11

22

34

40

46

57

69

80

Find the percentage voltage regulation of the alternator on a full-load and 0.8 P.F. (lagging), by the synchronous-impedance (2 decimal places) and ampere-turn methods (1 decimal place).

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210 • Advanced Electrotechnology 7.9

A three-phase, 3.3 kV, 50 Hz alternator having an equivalent armature reactance of 5 Ω per phase and negligible resistance is connected to busbars to which an identical alternator is already connected. A circuit-breaker was closed at an instant when the rotors of the 2 machines ran at 1500 rev/min but was out of phase by 1° mechanical. Find the value of the synchronising current (2 decimal places) and the synchronising torque in Nm at this instant (1 decimal place).

7.10 A 2 MVA, three-phase, 8-pole alternator runs in parallel with other machines on 6 kV busbars. If full load at 0.8 P.F. (lagging) is supplied, calculate the total three-phase synchronising power in kW per mechanical degree of displacement (3 significant figures). The machine’s synchronous reactance is 6 Ω per phase.

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8

THE INDUCTION MOTOR The validity of all the Inductive Methods depends on the assumption that . . . the beginning of every phenomenon, must have some cause; some antecedent, upon the existence of which it is invariably and unconditionally consequent. John Stuart Mill (1843)

An induction or asynchronous motor is an A.C. electrical motor where the current in the moving rotor needed to produce mechanical torque is induced by electromagnetic induction from the changing magnetic field of a fixed stator. An induction motor does not require mechanical commutation or separate-excitation of energy from stator to rotor as is the case in D.C. and some synchronous motors. An induction motor’s rotor can be either wire-wound or of ‘squirrel-cage’ type and is often referred to as an asynchronous machine since it operates at a speed below that of the rotating magnetic field set up within it. It is this asynchronous feature which determines the motor’s simple action and robust construction. It is the most common A.C. motor since A.C. working is almost universal for shore work and much marine practice, and as such is an important machine. The main disadvantage in its use however, lies in the fact that its speed cannot be varied if it is used in its basic cage form and that, if variable speed is required a special type of A.C. or induction motor with built-in speed control characteristics must be used. The latter are more expensive and an alternative is to use the basic single-speed cage motor and provide power output adjustment by mechanical means, e.g. for pumps and fans adjustment is achieved with valves and dampers. Other methods use gears and/or adjustable pulleys with belt drives, methods well known to the marine engineer. Here attention is only given to induction motors allowing limited speed control. Three-phase squirrel-cage induction motors are often used in industry because they are reliable and

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212 • Advanced Electrotechnology robust, while single-phase induction motors are used widely for smaller loads, such as many common household appliances.

Principle of Operation In its basic form the machine is known as a squirrel-cage or simply a cage motor. The theory of the rotating magnetic field was introduced in Chapter 7 and it is now appreciated how, if a polyphase motor’s stator is suitably wound with an appropriate number of conductors and energised from a polyphase supply, a rotating magnetic field is produced. It is the effect of this field on the motor’s rotor which is considered next to understand the operational theory. It can be shown the rotating magnetic field will drag around a rotatable solid iron cylinder when placed concentrically in the field. For efficiency, copper conductors are embedded in the face of the cylinder to yield 2 forms of rotor known as (1) the cage and (2) the wound-rotor types. The cage type is more simply constructed, consisting of copper bars inserted without insulation into slots spaced evenly around the circumference of a rotor built from iron laminations. Bars are connected to copper rings at each end of the rotor and winding and looks like a cage – hence the name. The winding consists of single turns connected in parallel. The woundrotor construction employs insulated conductors placed in coil form, in a similar way to that of the stator. There are thus 3 insulated phase windings, correctly positioned and spaced relative to each other. Phase windings are usually star-connected and the 3 free ends are brought out to slip-rings insulated from the shaft. Contact is made to the rings with brushes mounted on a brush-arm and the external circuit completed through 3 star-connected variable resistors. Control gear, built into the starter, allows resistors to be cut out and the slip-rings ‘shorted’. Thus the winding, when the motor runs, is shortcircuited and resembles the cage winding except there are more turns per phase coil. The theory set out as follows assumes conditions for a wound-rotor machine. THEORY OF ACTION. Figure 8.1, drawn for an instant in time, shows conductors on a rotor situated in a rotating magnetic field produced by a suitably wound stator fed with polyphase currents. Some stator conductors are shown but current direction is not given since rotating flux is the result of the effect of currents of different magnitudes in 3 phases. Rotor conductors are considered to constitute a winding which is shortcircuited. If the field is considered to move, the relative cutting between flux and rotor conductors induces e.m.f.s in the rotor, producing currents in the direction shown as when the ‘right-hand generator rule’ is applied, conductors move in the direction opposite to that as shown for the flux and the latter considered to be stationary. Currents circulate

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The Induction Motor • 213 Direction of flux rotation

Rotating stator flux Induced current Force (in direction shown)

▲ Figure 8.1

in paths provided by the end-rings and the conductors under the opposite stator poles. Applying first principles for the action between the flux and rotor currents, a force results in the direction shown, i.e. the rotor conductors follow the rotating field because of the torque produced. The rotor starts to revolve and accelerates up to speed N2 approaching the synchronous speed – the speed of the rotating field N1. There is always a difference in speed between N1 and N2; otherwise there is no relative cutting, no rotor e.m.f.s induced and hence no rotor currents or torque. Further consideration of the process shows that transformer action is also involved. Taking again the ‘standstill’ condition, i.e. a stationary rotor, the rotating magnetic field is of sinusoidal form and as it rotates, the resulting induced e.m.f. in the rotor and stator conductors will also be sinusoidal. Rotor currents are also of sine-wave form and, with transformer working in mind, the rotor acts as a short-circuited secondary to the stator (the primary). With the transformer concept we see that if the secondary (rotor) is opencircuited, rotating flux will cut the stator conductors to produce a back e.m.f. almost equal to the supply voltage and so negligible current is taken – like a transformer on no-load. If the rotor winding is completed to form a short-circuit, rotor currents flow and transformer on-load conditions occur. A secondary (rotor) flux is produced which interacts with the stator flux to produce a resultant smaller than the original. The stator induced back e.m.f. is reduced and the supply voltage forces a larger current through the stator – so stator current depends on the rotor current. Up to now standstill (stationary) conditions were considered but if the rotor rotates, the speed with which the field cuts the rotor conductors is reduced, and the rotor e.m.f. is reduced in size. Consequently rotor current is reduced and the stator or supply current reduces, giving a condition with a large starting current which falls as the machine ‘runs up to speed’. Earlier we saw the need for a difference in speed between rotor and magnetic field, and the first term associated with the induction motor, namely slip or slip speed, is now introduced.

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214 • Advanced Electrotechnology Slip speed. This is the speed at which the rotating field cuts the rotor conductors, expressed as a fraction or percentage of synchronous speed. Thus if NS is the slip speed: NS = N1 − N2. The ratio

Ns N N2 or 1 is called the slip s N1 N1

As a fraction we have Fractional slip s =

Ns N1 N2 = N1 N1

Or percentage slip s =

N1 N2 × 100% N1

Example 8.1. A 4-pole, 400 V, 50 Hz induction motor operates with 4% slip. Find the motor speed. Synchronous speed N1 is given from the relation f = ∴ N1 =

PN1 120

1500 − N2 50 × 120 = 1500 rev/min ∴ 4 = × 100 giving 4 × 15 = 1500 − N2 4 1500

or N2 = 1500 − 60 = 1440 rev/min Alternatively, for every 100 rev/min the slip is 4 revolutions or rotor speed = 96 rev/min. Thus for 1500 rev/min the slip is 4 × 15 revolutions so rotor speed = 96 × 15 = 1440 rev/min. When a motor ‘runs light’, i.e. unloaded, the torque is only that required to overcome friction and windage so the speed is nearly synchronous and the stator draws little current from the supply. As load torque is increased, i.e. the motor is loaded, speed falls slightly causing an increase in the relative cutting of the rotor conductors by the field, so rotor e.m.f. and current are increased. This effect gives the increased torque required and speed is determined by this condition when power developed equals the power required. The fall in speed from no-load to full-load is about 4–5% for small motors and 1.5–2% for large motors (with greater inertia). The speed-load characteristic is similar to the D.C. shunt motor.

Rotor to stator relationships These can be deduced and provides the basic theory of induction motors.

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The Induction Motor • 215 (1) FREQUENCY. For both standstill and running conditions, the stator frequency is the supply frequency or f1 = ƒ. At standstill the rotating field also cuts the rotor at synchronous speed, so f2 = ƒ. When running at speed N2, the speed at which the rotor conductors are cut by the field is N1 − N2. Thus f2′ = But fractional slip s = and f2′ =

N2 ) P

(N1

120

N1 N2 Therefore rearranging: sN1 = N1 − N2 N1

sN1P = sf or f2′ 120

ffs

So Rotor frequency = Supply frequency × Fractional slip. The introduction of the dash, e.g. f2′ , should be noted and is used to distinguish ‘running’ conditions against stand-still conditions when the dash is omitted. Example 8.2. If the supply voltage to an 8-pole induction motor is at 50 Hz and the induced e.m.f. in the rotor is 1.5 Hz, find the speed at which the motor runs and the slip. Since f2′

sf then s =

15 3 or i.e. Fractional slip is 0.03 or Percentage slip is 3%. 50 100

Synchronous speed N1 and N2 = 750 −

120f1 P

120 × 50 30 25 = 750 rev/min 8

750 × 3 = 750 − 22 22.5 5 = 727.5 rev / min. 100

(2) E.M.F. The rotating magnetic field cuts the stator and generates an e.m.f. given by the formula E1 = 2.22 KD1 KS1 Zph1 Φf1 volts. Neglecting voltage drops due to resistance and reactance E1 can be considered as a back e.m.f. approximately equal to the supply voltage V. For a stationary rotor the rotating field cuts the rotor at the same speed as the stator and an e.m.f. E2 is induced where E2’ = 2.22 KD2 KS2 Zph2 Φf2 volts. Here f2 = ƒ the supply frequency. With the rotor revolving the induced e.m.f. varies depending on the rotor frequency since f2′ sf . ∴ E2′ = K D2 K S2 Z ph2 Φsf = E2 s Rotor induced e.m.f. at any speed = Standstill e.m.f. × slip. (3) REACTANCE. A rotor has reactance and resistance. The reactance is due to the circuit’s inductance formed by the cage conductors embedded in iron. The characteristics of the machine can be altered by the degree the conductors are embedded in the iron, i.e.

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216 • Advanced Electrotechnology the slot depth. Since reactance is directly proportional to frequency the rotor reactance will vary as the slip changes. Accordingly we can write X 2′

f2′ L2

s2π f2 L2 = X 2 s

Rotor reactance at any speed = Standstill reactance × slip. From these relationships the following is deduced: For any speed. Rotor Impedance = Z 2 = R22 + ( sX 2 )

2

Rotor current I2′ =

E2′ Z 2′

sE2

= R2

2

2 s 2) ( sX

and,

Rotor phase angle cos φ 2′ =

R2 Z 2′

=

R2 R22 + ( sX 2 )

2

Example 8.3. A 6-pole, three-phase, 50 Hz induction motor runs at full load with a 4% slip. The rotor is star-connected and its resistance and standstill reactance are 0.25 and 1.5 Ω per phase respectively. The e.m.f. between slip-rings at standstill is 100 V. Find the full-load conditions for (a) the e.m.f. induced in each rotor phase (1 decimal place) (b) the rotor impedance per phase (3 decimal places) and (c) the rotor current, and P.F. assuming the slip-rings are short-circuited (2 decimal places). The rotor phase e.m.f. at standstill Eph2 =

100 3

= 57.7 V

The rotor phase e.m.f. at full-load = Eph ′ 2 = sEph2 = 0.04 × 57.7 = 2 31 V Rotor reactance per phase on full load X ph ′ 2 = sX ph2 = 0.04 × 1.5 = 0.06 Ω Rotor impedance per phase on full load Z ph 52 + 0.062 = 0.257 Ω ′ 2 = 0.25 Rotor current =

2 31 =9A 0.257

Rotor P.F. cos φ 2′

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0 25 0.257

0.97 (lagging).

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The Induction Motor • 217

Relation between rotor loss, rotor input power and rotor output The power taken from the supply creates a rotating magnetic field or flux (as in the transformer analogy). Copper and iron losses occur in the stator winding, so the power put into the air-gap flux or the stator output = stator input − stator losses. As no losses occur in the air-gap we assume the stator output power is the rotor input power thus: Stator output = Rotor input 2π NT watts, with torque T in newton metres. 60 Returning to the induction motor relationships, we know copper and iron losses occur in the rotor. On load, because of the very low frequency, the latter are usually negligible so we can write: The machine power output is given by:

Rotor input = rotor output + rotor copper loss or rotor copper loss = rotor input − rotor output = stator output − rotor output =

2π NT 2π N2T 2π T (N1 − N2 ) − = 60 60 60

T is in newton metres and N is in revolutions per minute. Thus

Rotor copper loss 2π T (N1 − N2 ) N1 − N2 = = =s Rotor input 2π N1T N1

(a)

∴Rotor copper loss = s × Rotor input Since the above relation can also be written as: Slip = ∴s =

Rotor copper loss Rotor input

3I22 R2 from which it follows that if rotor resistance is large, the slip will be Rotor input

large and vice versa. A further deduction is: Rotor copper loss 2π T (N1 − N2 ) N1 − N2 = = Rotor input 2π N2T N1

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218 • Advanced Electrotechnology

Since Thus

N1 N2 =s N1

∴N1 − N2 and N2 = N1 (1 − s)

sN1 Rotor copper loss N1 − N2 s = = = Rotor input N2 N1 ( − s ) 1 − s

or Rotor copper loss =

(b)

s × Rotor output 1− s

From (a) and (b) we have: Rotor input: Rotor copper loss = 1 : s and, Rotor copper loss: Rotor output = s : 1 − s Rotor input : Rotor copper loss : Rotor output = 1 : s : 1 − s. Example 8.4. The input power to a three-phase induction motor is measured to be 50 kW. Stator losses are 800 W. Find the rotor copper loss per phase and the mechanical power developed, if the slip is 3%. Stator input 50 kW

Stator output = 50 − 0.8 = 49.2 kW

Rotor input = 49 200 W and Rotor Cu loss = s × Rotor input or Rotor Cu loss = 0.03 × 49 200 = 492 × 3 watts Rotor Cu loss per phase =

492 × 3 492 W 3

Mechanical power developed = Rotor input − Rotor Cu loss = 49 200 − (492 × 3) = 49 200 − 1476 = 47 724 = 47.72 kW.

Torque conditions Earlier theory considered how torque is produced by the rotor but it is clear further examination of the conditions resulting in torque is needed. A machine ‘brake test’ produces a mechanical speed/torque characteristic of the type shown in figure 8.2, where the normal operating region is shown for each motor type, as a shaded area. The comparison with a D.C. shunt motor is given and from these it is seen that the induction motor torque is limited to a maximum value. Since torque and slip are directly related it is usual to give the characteristic of the machine as a torque/slip or torque/speed curve (figure 8.3). The maximum torque value is decided by the motor’s physical dimensions

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The Induction Motor • 219

Induction motor N

N

D.C. shunt motor

T

T

▲ Figure 8.2

Max torque

T

Unstable region Stable region

0 100

Slip (%) Speed (%)

100 0

▲ Figure 8.3

and cannot be exceeded even by overloading, but can be adjusted to occur at a desired slip value. Such adjustment is achieved by altering the rotor-cage winding resistance and is exploited in wound-rotor machines. STANDSTILL TORQUE. For a stationary rotor condition, Ts is the standstill torque in newton metres, produced on the rotor by the magnetic field rotating at a speed of N1 rev/min. When the motor is not running, there is no mechanical output so the rotor input power must be converted into rotor electrical power, i.e. a copper loss. Here rotor iron loss being small is neglected for this assumption. Thus Rotor input = Rotor copper loss

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220 • Advanced Electrotechnology

or

2πN1Ts = 3I 2 R 2 or 3E 2 I 2 cos φ2 60 3E22 R2 R = 3E22 2 2 2 2 Z2 R2 X 2

Thus Ts =

3E22 × 60 R × 2 2 2 E2 and N1 are constants whose sizes depend on motor 2πN N1 R2 X 2

design. We thus have Ts

K

R2 R22

X 22

or Ts ∝

R2 R22

X 22

Starting. Since the rotor is usually coupled to machinery, maximum starting torque is a vital feature, and we will investigate how this is obtained. The expression above shows torque is affected by rotor resistance, and the following mathematical deduction indicates the direction in which control can be made. The result and not the method is important to the engineer but is given here for the benefit of mathematically minded students. The torque expression has a maximum value R2 found by differentiating the expression with respect to R2 and equating to zero. Since TS =

KR2 then: R2 X 22 2

(

)

R22 X 22 − R2 ( R2 ) dTTS =K =0 2 dR2 R22 X 22 or K

(

X 22

R22

R22

X 22

)

2

from which X 22

(

)

R22 or X2 = R2

This is the condition for maximum starting torque and explains why a woundrotor is used to ensure a good starting torque. At starting, rotor frequency is ƒ, so X2 is large compared with R2. Normally the rotor-circuit P.F. is also low and TS will be small. Improvement can be made, if cos φ2 is raised by (1) decreasing X2 – not easily achievable, (2) increasing resistance – by putting in external resistance as for a woundrotor machine, or using a high resistivity material for the rotor cage. The latter does not allow for good running conditions and high efficiency, so as a result special cage forms were developed to give improved overall performance. As mentioned for the woundrotor motor, as the rotor speeds up, external resistance is cut out. Before considering actual running conditions, it is vital to point out the effect of supply voltage on starting torque.

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The Induction Motor • 221

The expression Ts α

R2 R2

2

X 22

was deduced on the assumption that E2 and N were

constant. It is noted that E2 depends on the machine flux Φ which is in turn proportional to the supply voltage V, so the torque expression may be written as: Ts =

CV 2 R2 , where C is a constant. R22 X 22

We now see a change of supply voltage will considerably alter starting torque and this must be kept in mind when choosing from various methods used for motor starting. Example 8.5. An induction motor with star-connected stator takes 45 A from a 110 V supply with the rotor stationary, and 0.3 P.F. (lagging). If the ratio of stator to rotor turns per phase is 2, determine the resistance and reactance per phase of the rotor winding. What extra resistance per phase is required to make the starting torque maximum (3 decimal places)? Stator voltage per phase =

100 3

= 63.5 V

Stator current per phase = 45 A Impedance Z 2 of rotor per phase

Rotor voltage per phase = 63.5 ×

1 = 31.75 V 2

Rotor current per phase = 45 × 2 = 90 A 31.75 90

0.353 Ω

cos φ2 = 0.3

∴ R2 = Z 2 cos φ2 = 0.353 × 0.3 = 0. 06 Ω X 2 = Z 22 − R22 = 0.3532 − 0.1062 = 0 1246 4 − 0.01124 = 0.11336 = 0.337Ω For maximum starting torque R2 = X2 ∴R2 must be 0.337 Ω per phase or 0.337 − 0.106 = 0.231 Ω must be added in each phase. RUNNING TORQUE. With the rotor up to speed, as seen earlier, the rotor e.m.f. varies with rotor frequency. Thus E2′ sE2 . Since mechanical power is produced at the shaft, rotor input = rotor output + rotor copper loss. ∴

2πN1T 2πN2T = + 3EE2′ I2′ cos φ2 60 60

or

2πN1T

2πN2T E ′2 R = 3E2′ I2′ cos φ2 or 3 2 × 2 60 Z ′2 Z ′2

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222 • Advanced Electrotechnology 2πT (N1 N2 ) R = 3s 2 E2 2 × 2 2 60 R2 ( ssX 2 )2 2πT ( sN1 ) R = 3s 2 E2 2 × 2 2 2 2 60 R2 s X 2 or T =

3sEE22 × 60 R sR × 2 2 2 2 = K 2 22 2 2πN N1 R2 s X 2 R2 s X 2

Whence T α

R22

sR2 ( sX s 2 )2

A better understanding of the torque/slip characteristic is now possible. From the sR work above, it is seen that, on examining the expression T α 2 2 2 , when the R2 ( sX s 2) rotor just starts to move, the slip is large and the reactance X 2′ is large. By comparison s 1 resistance R2 is small and the expression can be modified to T α 2 2 = . Thus at s X2 sX 22 starting, torque varies inversely with slip at low speeds. Thus when slip is high, torque is small and the relation explains the unstable right-hand side of the torque/speed curve shown in figure 8.1. At high speeds, i.e. small slip, rotor frequency is low and the reactance is small compared with the resistance. Again with reference to the expression sR sR s T α 2 2 2 , for approximation under the condition of small slip T α 22 Tα . R R R2 ( sX s 2) 2 2 Thus T α s or torque is directly proportional to slip and this explains the stable left-hand side of the characteristic (figure 8.3). Summarising, the torque/slip curve first rises steeply showing an increase of load produces increased torque. This linearity continues until maximum torque is reached when the curve flattens and turns back. Beyond this the curve fails to follow an inverse law. Here an increase in load results in a decrease in torque and the slip increases. Maximum Running Torque. The technique for determining this condition is the same as for the standstill condition. Thus by differentiating with respect to slip and equating to zero, we have: KsR2 Since T = 2 2 R2 s sX then

{

R dT = dS

(

)

s ) } KR K 2 {K KsR sR R ( sX

{

R

9781408176030_Ch08_Final_txt_print.indd 222

( ssX ) }

2

× sX

}

=0

8/22/2014 4:07:03 PM

The Induction Motor • 223 For the maximum condition: ∴KR22 + Ks2X22R2 = 2KR2s2X22 or R22 + s2X22 = 2s2X22 and R22 = s2X22 or R2 = sX2. R2 . R and X2 are X2 2 normally constants but by increasing R2, as in the wound-rotor machine, maximum torque can be made to occur at any slip value. Thus maximum starting torque is arranged for starting with heavy loads and this value is maintained through the accelerating period by cutting out external series resistance. Maximum torque is unaffected by resistance, but it is cut out when the motor gets up to speed to reduce loss and ensure a smaller slip. Figure 8.4 shows the effect of rotor resistance on the torque/speed or torque/slip curve. Curves are drawn with speed as the base and they appear reversed with respect to those shown on figure 8.3. Either method of representation is acceptable although slip as a base is more usual. The above shows that maximum torque occurs when the slip s =

R23

R22

R21

Max torque

T

R24

For R21 motor cannot start

a

For R2 1 motor can start

b

0

20

100

80

60 40 Speed (%) 40 60 Slip (%)

80 20

100 0 Operating slip 4% (approx.)

▲ Figure 8.4

For figure 8.4, the following points are considered for load torque. For line (b) and curve R21 , starting torque produced by the motor is greater than the load torque, so the motor starts and accelerates the load. The machine runs up to speed, operating at about 4%

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224 • Advanced Electrotechnology slip. For line (a) and curve R21 , the load requires a greater starting torque than that produced by the motor which cannot thus start and accelerate unless load torque is reduced. However, if resistance is introduced into the rotor circuit to operate on curve, R21 , starting is possible and the motor will run up to speed and operate at about 6%. Example 8.6. The rotor current per phase of an induction motor is 50 A, with the rotor stationary, and 5 kW power per phase. The lag angle in the rotor is 75°. Find the resistance of a rotor phase, then calculate the current per phase and the angle of lag when the slip is 10%. At what percentage slip (1 decimal place) and speed (4 significant figures) will torque be maximum? Assume a 4-pole machine and 50 Hz supply. Since P = I22R2 i.e. R2 = Also P = E2I2 cos φ2 So E2 = Then Z2

5000 =2Ω 502

i.e. 5000 = E250 cos 75°

5000 = 386 V/ph (at standstill) 50 × 0.259 386 50

7.72 Ω X 2 = 7.722

22 = 59.7 4 = 55.7

7.47 Ω

At 10% slip, e.m.f. E2′ = 386 ×

10 = 38.6 V 100

Impedance Z2

22 + 0.7472 = 4+0 4+0.557 557 = 4.557 = 2.13 Ω

Note: X’2 = sX2 = 7.47 × 10/100 = 0.747 Ω Current I2′ = cos φ2′= ′

38.6 = 18.1 A 2.13

2 = 0.939 (lagging) (l i ) i.e. i φ2′ = 20° 2.13 2 13

Maximum torque occurs when sX 2 = R2

s=

2 7.47

= 0.268 or 26.8%.

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The Induction Motor • 225 Synchronous speed N1 =

50 × 120 = 1500 rev/min. 4

∴Motor speed = 1500 − 0.268 × 1500 = 1500 − 402 = 1098 rev/min.

The phasor diagram Since the motor functions like a transformer, with the stator as the primary and the rotor as the secondary, the phasor diagram will be similar to that of the transformer and the associated reasoning can be applied, provided the following points are noted. (1) Transformer action is set up by a rotating magnetic field. (2) Because of the air-gap between the stator (primary) and rotor (secondary), the magnetic circuit has a high reluctance and the magnetising component of current Im is larger than for a transformer. Due to the large Im, the no-load current I0 is at a low P.F., so the motor should not be run light, but instead should be switched off. (3) Since the rotor circuit is closed there is no secondary terminal voltage V2. The rotor e.m.f. E2′ is responsible for the rotor current I2′ and overcoming rotor resistance and reactance voltage drops. The secondary impedance voltage triangle will thus be larger than for a transformer and constituted by all the secondary voltage drops. (4) Standstill conditions relate directly with the transformer but for running conditions slip must be considered since it influences the induced secondary e.m.f. and secondary reactance. Figure 8.5a shows the motor circuit with the rotor open-circuited, the condition resembles a transformer on ‘no-load’. Figure 8.5b is drawn for rotor standstill conditions, i.e. rotor locked and circuit complete, a condition similar to a transformer ‘on-load’ and phasor relations, as discussed earlier for the transformer, apply. Thus for the ‘locked rotor’ condition, the main flux generates stator and rotor e.m.f.s of E1 and E2 respectively. I0 is the no-load current consisting of components Im and Iw. Im is the current component which produces the ampere-turns resulting in the main flux Φ, and Iw is the energy component supplying the iron loss. The stator resistance voltage drop is I1R1 and the reactance voltage drop is I1X1 and these when added by phasors with the back e.m.f. E1 equal the supply voltage V. The rotor e.m.f. E2 produces a rotor current I2 which is limited by the resistance R2 and reactance X2 of this circuit. The rotor

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226 • Advanced Electrotechnology (a)

V

(b)

V

I0X1 E1

I1X1

I0R1 I1R1 E1 I 1′

I0

Iw

Iw

Φ

Im

I2R2

I2 I2X2

E2

(c)

V

I1

I0 Im

φ2

Φ

E2

I1X1 I1R1 E1 I1

I ′1

φ1 I0

Iw

Im

Φ

I2R2 φ2 E ′2 = sE2 I2 sX2

I2

▲ Figure 8.5

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The Induction Motor • 227 secondary resistance and reactance voltage drops, I2R2 and I2X2 respectively, by phasor addition equal the induced e.m.f. E2 or E2 = I2Z2. When the rotor rotates, we get the conditions shown by the phasor (figure 8.5c). Here the secondary e.m.f. is now reduced to E2′ or sE2 and the secondary reactance falls to X 2′ or sX2. Since rotor resistance is unaffected by slip, R2 is unchanged and the P.F. rises as the sX2 value falls. As mentioned under point (2), the P.F. is low if the motor runs light since component I is large in relation to X1′ . When a motor is loaded I and hence I1′ 0

2

increases I0 being constant, the result is that I1 rises and the P.F. condition improves. It is acceptable to relate rotor or secondary conditions at slip frequency to those of the stator or primary which are at the ‘mains’ frequency as when running. Although the rotor frequency is low, the field produced by the rotor current rotates at a speed relative to the rotor itself in the same direction as the stator field. Thus the rotor field relative to the stator is at synchronous speed, and the rotor quantities can be combined and related to stator quantities, provided appropriate adjustment is made for the turns ratio and slip. Example 8.7. Full-load torque is obtained from a 3 kV, 24-pole, 50 Hz star-connected induction motor at a speed of 240 rev/min. The slip-ring motor has a resistance of 0.02 Ω per phase and a standstill reactance of 0.27 Ω per phase. Calculate (a) the ratio of maximum to full-load torque and (b) the speed at maximum torque (1 decimal both). Since f =

∴Slip s

PN1 120 × 50 = 250 rev/min i.e. N1 = 120 24 250 − 240 250

0.04 or o 4%

Let T = the full load torque Then T =

KsR2 K × 0.04 × 0.02 = 2 R2 ( sX s 2) 0.02 + (0.04 × 0.27)2 2

=

K 8 04 8K 10 −4 8K = = 2 −4 0.0004 + (0.0108 ) 10 ( 4 + 1.1664 ) 5.1664

Also let Tm = maximum torque. Now Tm occurs when smX2 = R2 or when sm 0.27 = 0.02 ∴For maximum torque, sm =

9781408176030_Ch08_Final_txt_print.indd 227

0 02 = 0.074 0 27

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228 • Advanced Electrotechnology

Then at this slip sm, Tm =

∴ Tm =

K × 0.074 × 0.02

(0.02)2 + (0.074 × 0 27)2 K×

.8 × 10 −4

0.0004 + (0.01998 )

2

=

.8K 10 −4 14.8K = −4 10 ( 4 + 3.992) 7.992

14.8K Tm 14.8 × 5.1664 76.46 ∴ = 7.992 = = 8K T 8 × 7.992 63.94 5.1664 or

Tm T

12 (approx.) 1

For maximum torque, slip = 0.074 or 7.4%. ∴ 0.0 074 =

250 − N2 i.e. N2 = 250 − 250 × 0.074 250

= 250(1 − 0.074) = 250 × 0.926 = 250 × 0.2315 × 4 = 231.5 rev/min. Note that the information relating to motor-voltage rating and connection (3 kV, starconnected) is superfluous and unnecessary for the solution.

The Circle Diagram Circle or Locus diagrams are still used in electrical engineering to determine circuit working conditions when variables are altered in a known way. It is not intended here to give a detailed discussion of the circle diagram, for further discussion consider reading further textbooks on electrical engineering, e.g. Electrical Machines, Second Edition by Smarajit Ghosh, Pearson Education India, 81-3176-090-1. The basics upon which the induction motor circle diagram is based are simply introduced here. The diagram helps predict motor performance under varied operating conditions and can be developed to show current, efficiency, P.F., slip, speed and torque. The diagram can be constructed from results of simple tests which do not involve loading a machine and this performance assessing method is similar to that used for transformers and alternators. Since the circle diagram is developed from the phasor, mathematical operations appropriate to the latter can be applied to the former, and the results of

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The Induction Motor • 229 measurements made depend entirely on the accuracy of graphical construction, so the largest possible scale should be used. E

A

φ

φ

B

C

▲ Figure 8.6

Consider figure 8.6 which shows a circle of diameter ВС. Since A is a point on the circumference, the angle ВАС is a right angle and AB = BC sin φ or AB = K sin φ where К is the diameter of the circle. Next consider the rotor expression where the rotor sE2 If we multiply the bottom by X2 and rearrange, current is given by: I2 2 2 R2 ( sX s 2) we have: sE2

I2 = R2 =

2

s 2) ( sX

2

×

X2 sX 2 E = × 2 2 2 X2 X 2 R2 ( sX s 2)

E2 sinφ2 (from the rotor-circuit relationships) X2

= K sin φ2, since E2 and X2 alter in proportion. Comparing this with the expression AB = K sin φ2 it is seen the locus of I2 is on a circle with a diameter of constant value at right angles to E2 as shown by figures 8.7a and 8.7b. I1′ is the reflection of I2 in the primary or stator circuit. Thus I1′ lies on a circle of appropriate diameter. Again, since the primary current is the phasor sum of I0 and I1′, the relevant parts of the primary diagram can be adapted to give the Circle Diagram. It is unnecessary to know the diameter since the circle can be built up from current values obtained from simple motor tests. The diagram, once deduced, may be used for performance assessment and is illustrated by example.

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230 • Advanced Electrotechnology I 1′

(a)

φ2 φ2

φ2

E2 (b)

V

I1 I 1′

I 1′

I0

▲ Figure 8.7

Explanation of the circle diagram Reference is made to figure 8.8 which is the circle diagram extracted from the phasor. Both stator and rotor conditions are given, the latter in terms of appropriate stator values. OB is the no-load current. OA is the stator-load current value, made up of the no-load current OB and the rotor-load current BA reflected into the stator circuit (I1′). Thus current phasors, originating from В lying on a semicircle, are rotor values referred to the stator, while phasors originating from point О are stator values. Note: all current phasor vertical components are in phase with the voltage, and are energy components – a measure of input power, i.e. output power plus losses, which will be explained as the diagram is detailed. The semicircle is constructed if 2 phasors originating from О are known in length and in their angular relation to an axis. Such current phasors are obtained from simple machine tests performed under conditions requiring the minimum of loading.

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The Induction Motor • 231 V

A

φ0

B Φ

O

▲ Figure 8.8

Testing procedure Since the machine is a three-phase motor, three-phase measurements are made. Data can be used in this form but it is more conveniently worked in phase values as is evident from the problem worked. NO-LOAD TEST. The machine is run light at normal voltage and frequency. Measurements are made of line voltage, line current and total power. Referring to figure 8.9 length OB and phase angle φ0 are obtained. LOCKED-ROTOR OR STANDSTILL TEST. A reduced voltage is applied to the stator which is raised until full-load current is passing. The rotor is clamped to prevent rotation. Line voltage, line current and total power are measured. Referring to figure 8.9, length OAs and phase angle φs are found. Note: data from this test must be adjusted to full voltage V A AS

Rotor copper loss

e

lin put

t

Ou 1.0 0.5

C

ne rque li

To

φ0

O φ S

J

Stator copper loss

K G

B

L

M

F Copper loss

E H

D

Φ

Constant loss

▲ Figure 8.9

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232 • Advanced Electrotechnology values, so if Vs is the reduced voltage and V is the normal voltage the Standstill current V (at normal voltage) = measured current × Vs ⎛V⎞ Power (at normal voltage) = measured value × ⎜ ⎟ ⎝V ⎠

2

s

Construction of the circle diagram A P.F. quadrant is useful for ‘laying off ’ and measuring P.F.s and the procedure given here. An engineering drawing can be a highly stylised representation of an engineering idea, where ‘laying off ’ refers to measuring off a set distance. Along the vertical axis (figure 8.9), starting from point 0, a scale of 10 units is drawn, which is of convenient length to give accuracy and a quadrant drawn. Scale the vertical 0 to 1. For example, if a line drawn horizontally through the 0.2 mark to cut the quadrant and this point of intersection is joined to the origin, the line obtained must subtend with OV an angle whose cosine is 0.2. No-load Test. This test gives I0 and φ0. Thus no-load P.F. cos φ0 =

P0

where P0 = the 3VI0 no-load three-phase power. To start the circle diagram, using a suitable scale, lay off OB equal to I0 at an angle φ0. Load Test. The P.F. for this test is given by

Ps 3Vs Is

= cos φ s .

Next adjust the locked-rotor or S.C. test current to the correct current value at full voltage. This deduced value of current gives OAs, drawn to the chosen current scale at angle φ0. Construction now proceeds by joining BAs and bisecting it perpendicularly. The bisector is drawn to cut the horizontal through В at point M, which now becomes the centre of the circle. This diagram (figure 8.9) aids P.F. estimation for any load current and the optimum P.F. condition. It can be extended to estimate power output, losses, slip, etc. Consider any load-current value OA. AL will be the power component of OA and the input power to the motor is 3V × AL. This input power supplies the output power developed at the motor shaft, and the machine losses: the iron (hysteresis and eddy current) losses, the rotational (friction and windage) losses and the copper (stator and rotor) losses. Consider length GL. This is the power component of OB and is the no-load power input which makes up the rotational and stator iron losses and a very small stator copper loss.

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The Induction Motor • 233 The last is neglected but GL = EH can be taken as a measure of the constant machine losses. For the locked-rotor test all input power is converted into losses, i.e. copper and iron losses. Here there is no rotational loss but since the rotor frequency is now at full ‘mains’ value, it is assumed that constant loss component EH, under this condition, is made up of the iron losses, i.e. rotational loss is nil but the rotor iron loss increases to replace it. Component AsE is a measure of the rotor and stator copper losses under the lockedrotor condition. Next consider any stator current value such as OA. Since the stator current is proportional to the rotor current then Total copper loss α (rotor current)2 Now cos ABD = ∴

i.e. α (AB)2

AB BG also cos ABG = BD AB

AB BG = or AB2 = BD × BG BD AB

But BD is a constant i.e. AB2 = К × BG or AB2 α BG or the copper loss α BG We can thus deduce

Copper loss for any current BG KG = = Copper loss at standstill BE A SE

(since similar triangles are involved) Thus a line joining В to As cuts off the copper losses (to scale) for the current considered KG, to scale, is a measure of total copper losses. Output Line. Consider the power component AL of any current such as OA. Length GL represents the constant losses. Length KG represents the copper losses, and AK represents motor power output. Line BAs is called the Output Line and any length of a power component above this line is a measure of output (in watts). Note: current must be multiplied by voltage and 3 . Maximum output is obtained by drawing a tangent to the circle parallel to BAs or the bisector of ВAs to cut the circle to give this tangent point. Length of the vertical between this tangent point and the output line gives the maximum output possible. Example 8.8. The following test data refer to a 500 V, three-phase, 50 Hz induction motor. No-load Test: With 500 V applied, the current is 18 A and the power 1200 W. Locked-Rotor Test: With 250 V applied, the current is 100 A and the power 11 kW. Estimate maximum power output, the current and P.F. under this condition (2 decimal places). The problem is worked for a three-phase condition, so motor connection is immaterial.

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234 • Advanced Electrotechnology

1200

3VI cos φ 0 1200 or cos φ 0 =

Thus on no-load

11 000

Standstill test cos φs =

3 × 250 × 100

3 × 500 × 18

= 0.077

= 0.254

Note. The solution is drawn with a 10 mm = 10 A scale. This is not apparent in printed form due to photographic reduction. Students should draw this solution for themselves. Using the origin O, draw the P.F. quadrant (5 mm is used for 1 division). Draw OB = 18 A, i.e. 18 mm at 0.077 P.F. The horizontal line cutting the quadrant is shown. Draw OAs = 200 A, i.e. 200 mm at 0.254 P.F. The horizontal line cutting the quadrant is shown. The locked500 rotor current value is deduced for full-voltage value, i.e. OAs = 100 × = 200 A. 250 Join BAs. Bisect this line and produce bisector to give point M on BD. Draw the circle. Produce MS to cut the circle at T. Drop TU − this is the output component, with TU = 73 mm = 73 A and output power =

3 × 73 × 500 = 63.2 kW. 1000

Join ОТ. The estimated line current given by ОТ is 128 mm, i.e. 128 A at 0.73 P.F. (lagging). Torque Line. Further work on the circle diagram (figure 8.10) allows the determination of other values for any machine. To draw the Torque line, stator and rotor copper losses are separated. For the locked-rotor test, since voltage is reduced, iron losses are negligible and the test is assumed to give the copper losses – deduced for the full voltage condition. Furthermore, if Rs is the stator resistance per phase then the stator copper loss during the test is 3 IS2 Rs and, Rotor copper loss = Ps − IS2 Rs. T

Power factor

1.0 0.73

7.3 cm

12 .

8

cm

V

0.5

AS

S U M

O

B

D Scale: 1 cm = 10 A

▲ Figure 8.10

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The Induction Motor • 235 Thus the length ASE can be divided in the ratio A S C Rotor copper loss = and if ВС is joined, we have the torque line, which is explained CE Stator copper loss as follows: AL is a measure of the power input to the machine. GL is a measure of the constant loss. GJ is a measure of the stator copper loss for the line current OA being considered. It is deduced from the locked-rotor condition, using the proportion given by similar triangles. AJ is a measure of the power passed across the air-gap from stator to rotor, i.e. is the rotor input power and the torque can be deduced. Any power intercept up to the torque line such as AJ, when multiplied by the appropriate factors, gives the torque in ‘synchronous watts’. Since we are concerned with the power put into the rotor, i.e. produced by a rotating magnetic field, the synchronous speed N1 must be substituted 2πNT for N when using the expression . Other data can be obtained from the circle 60 diagram such as estimated slip, overall efficiency, optimum P.F., etc. The following example shows how these can be deduced and the student should again undertake the practical solution themselves. Example 8.9. A 4-pole, 56 kW, 440 V, 50 Hz, three-phase induction motor gave the following figures on test. No-load Test. Applied voltage 440 V. Line current 18.25 A. Input power 3.19 kW. Locked-rotor Test. Applied voltage 110 V. Line current 55.2 A. Input power 3.17 kW. Construct the circle diagram and find (a) the line current, (b) P.F., (c) full-load percentage slip and (d) the efficiency. At standstill rotor and stator losses are equal. For this solution (figure 8.11) a scale of 10 mm = 10 A is used. Construction commences by drawing the P.F. quadrant, with 5 mm for 1 division. From the No-load Test. P.F. cos φ 0 =

3190 3 × 440 × 18.25

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= 0 23

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236 • Advanced Electrotechnology

Y

A V

AS X Power factor

1.0

0.82 C

K 0.5 J O

M

G

B

E

D

L

▲ Figure 8.11

Using the P.F. quadrant, draw in the line for an angle whose cosine is 0.23. Next on this line measure off OB = 18.25 mm. From the Locked-rotor Test. P.F. cos φ S =

3170 3 × 110 × 55.2

=03

Using the P.F. quadrant draw in the OAs line by adjusting the locked-rotor current to full voltage value, i.e. OA S = 55.2 ×

440 = 220 220.8 8 A = 221 mm. 110

Join As to В giving the output line, bisect this so the bisector cuts BD at M. M is the centre of the circle, draw this in. For a 56 kW output the power component 56 000 = = 73.5 A = 73.5 mm. 3 × 440 From the output line erect two verticals X and Y = 73.5 mm and draw the parallel to cut the circle at point A. Then OA = 114 mm or line current = 114 A. The operating P.F. will be 0.82 (lagging). The next construction involves drawing the torque line. Since Rotor loss = Stator loss. Bisect AsE = 62 mm ASC = 31 mm = CE. Join ВС. Draw in the vertical from point A, marking points: K, J, G and L. Next Rotor copper loss = s × Rotor input or in terms of the proportions of the power component of load current:

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The Induction Motor • 237

Slip =

Length KJ K 0.8 = = 0.099 = 9.9% Length AJ 8.1

Efficiency =

Length AK 7.3 = = 0.785 or 78.5% Length AL 9.3

The answers show the machine is overrated, i.e. slip is excessive and efficiency low. Line current is also excessive and the P.F. should be higher, a 37 kW rating being better.

Practice Examples 8.1 The input to a three-phase induction motor is 50 kW. The stator losses amount to 0.8 kW. Calculate the rotor copper loss per phase (3 significant figures) and the total mechanical power developed in kW, if the slip is 3% (1 decimal place). 8.2 A 500 V, 6-pole, 50 Hz, three-phase induction motor develops 20 kW inclusive of mechanical losses, when moving at 990 rev/min, and 0.85 P.F. (lagging). Calculate (a) the percentage slip, (b) the frequency of the rotor e.m.f. (1 decimal place), (c) the rotor copper loss (3 significant figures), (d) the input to the motor in kW if the stator loss is 2.5 kW (1 decimal place) and (e) the line currents (2 decimal places). 8.3 A 4-pole, three-phase, 60 Hz induction motor is rated at 15 kW. Its full-load slip is 4%. The stator loss is measured as 950 W and the mechanical rotor losses = 830 W. Find the rotor copper loss in kW (2 decimal places) and machine efficiency (2 significant figures). 8.4 A 6-pole, 50 Hz, three-phase, wound-rotor induction motor running on full load develops a useful torque of 162.7 Nm. A moving-coil ammeter in the rotor circuit fluctuates at 90 complete ‘beats’ per minute. The mechanical torque lost in friction is 13.56 Nm and the stator losses are 750 W. Calculate (a) speed of motor (3 significant figures), (b) brake power input (3 decimal places), (c) rotor copper loss (watts) (1 decimal place), (d) motor input in kW (2 decimal places) and (e) motor efficiency (2 significant figures). 8.5 A 75 kW, 440 V, three-phase, wound-rotor induction motor has a rotor resistance and standstill reactance of 0.02 and 0.27 Ω/ph respectively. The stator to rotor phase turns ratio is 3:1 and the stator windings are connected in delta. If the motor is started by means of a resistance starter with a resistance of 0.25 Ω/ph, calculate the current taken by the motor from the supply (a) at start (3 significant figures) and (b) under full-load running conditions (1 decimal place) if full-load slip is 4%. What would be the current taken from the supply if the motor was accidently

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238 • Advanced Electrotechnology started with the starting resistance in the ‘run’ position (3 significant figures)? Neglect the no-load current, and stator winding resistance and reactance. Assume the rotor to be star-connected. 8.6 Full-load torque is obtained with a 440 V, 12-pole, 60 Hz, three-phase, deltaconnected induction motor, when driving a main circulating water pump at a speed of 576 rev/min. The slip-ring rotor has a resistance of 0.02 Ω/ph and standstill reactance of 0.27 Ω/ph. Calculate (a) the ratio of maximum to full-load torque and (b) the speed at maximum torque (revs/min) (1 decimal place). 8.7 A 440 V, three-phase, 10-pole, 50 Hz, boiler induced-draught fan motor has a delta-connected stator winding and a wound rotor. The resistance per phase of the rotor winding is 0.018 Ω and the ratio of the stator conductors per phase to rotor conductor per phase is 4:1. If the motor develops maximum torque when running at a speed of 540 rev/min, calculate the rotor current (3 significant figures) and P.F. (3 decimal places) for the machine when running at full load, with a slip of 4% and slip-rings short-circuited. Calculate the developed full-load output power in kW (3 decimal places), assuming mechanical losses are negligible. What starter resistance per phase must be connected into the rotor circuit to give maximum torque at start (3 decimal places)? Neglect stator winding resistance and reactance. 8.8 A 37 kW, 12-pole, three-phase, 50 Hz squirrel-cage induction motor as used aboard a tankship for driving a main circulating water pump, gave the following test results: No-load Test. Applied voltage 440 V. Line current 19 A. Input power 2.17 kW. Locked-rotor Test. Applied voltage 100 V. Line current 70 A. Input power 3.88 kW. The ratio stator/rotor copper loss is 4:5. Construct a circle diagram and find for full-load conditions: (a) Input line-current (3 significant figures) and P.F. (2 decimal places), (b) percentage slip (2 decimal places) and (c) percentage efficiency (1 decimal place). 8.9 During tests on a three-phase, 440 V at 60 Hz, 8-pole, 22 kW induction motor, the following figures were obtained: No-load Test. Applied voltage 440 V at 60 Hz. Input line current 10.5 A. Input power 1.82 kW. Locked-rotor Test. Applied voltage 113 V at 60 Hz. Input line current 50 A. Input power 3.92 kW.

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The Induction Motor • 239 Estimate for full-load conditions: (a) Input line current (3 significant figures) and P.F. (1 decimal place), (b) percentage slip (2 decimal places), (c) percentage efficiency (1 decimal place) and (d) developed torque of the motor in Nm (2 decimal places). For this motor the rotor copper loss under standstill conditions was found to be the same as the stator copper loss. 8.10 Draw the circle diagram for a three-phase, 415 V, 37 kW induction motor from the following test results: No-load Test. Applied voltage 415 V. Input current 40 A. Input power 3.6 kW. Locked-rotor Test. Applied voltage 83 V. Input current 80 A. Input power 3.45 kW. Under standstill conditions rotor copper loss and stator copper loss are equal. From the circle diagram deduce (a) the full-load current (1 significant figure) and P.F. (1 decimal place), (b) the full-load efficiency (2 significant figures), (c) maximum output in kW (1 decimal place) and (d) the ratio of full-load torque to maximum torque.

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9

A.C. MACHINES – OPERATION ‘O tell me, when along the line From my full heart the message flows, What currents are induced in thine? One click from thee will end my woes’. Through many an Ohm the Weber flew, And clicked the answer back to me, ‘I am thy Farad, staunch and true, Charged to a Volt with love for thee’. Credited to James Clerk Maxwell regarding telegraphy

This chapter discusses several common control and operation methods for A.C. power equipment. In considering the various A.C. machines of previous chapters, no discussion was made of operational procedures, or the practical problems of parallel operation of alternators and load sharing. In shipboard work, it is common to run more than one machine for ‘electrical power supplies’, and so relevant theory must be considered, especially as there are differences between the operating procedures of A.C. and D.C. generators. In considering the overall aspects of operating A.C. machinery, starting and speed-control methods for A.C. motors were not mentioned. However, shipboard practice requires such attention which is discussed in this chapter.

A.C. Generators A.C. generators can be connected in parallel because they control each other electrically by a circulating current which keeps them in phase. Series connection of A.C. generators

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A.C. Machines – Operation • 241 is not practicable unless the machines are coupled together mechanically. Without such coupling, since no provision for a circulating current exists, it is not possible for machines to control each other. As series running is not a feasible arrangement it will not be considered here.

A.C. generators in parallel The reader is reminded that an A.C. machine, like its D.C. counterpart, can function as either generator or motor if connected to a supply source. Such a supply would be ‘live busbars’ which are themselves supplied by an electrical power source, e.g. other alternators. D.C. conditions for parallel running were discussed in Chapter 4 (D.C. Machines) where if the machine generated voltage under consideration was higher than the busbar voltage it would feed electrical power out, and be driven by the power source, contributing power into the ‘common pool’ of the busbar system, to operate in parallel with others already ‘on the bars’. If the machine generated voltage was lowered below that of the busbars it would take electrical power to function as a motor. In this condition it provides mechanical power for driving associated machinery. Similar conditions arise for A.C. generators, which may operate as an alternator or synchronous motor. Control of a machine’s generated voltage alters the operating mode but if parallel operation is needed, unlike D.C. conditions, further adjustment of the prime-mover throttle valve must be carried out on each machine of the parallel arrangement, before load sharing is achieved. SYNCHRONISING. As discussed previously an A.C. generator or synchronous motor can only be connected to a live busbar system by ensuring correct synchronisation prior to closing the paralleling switch or circuit-breaker. If one machine is slightly out of phase it should ‘pull’ the other into step, but if the phase difference is too large there can be huge damaging voltage and current variations. The requirements for correct synchronising have been mentioned and are now considered. To ensure correct synchronisation the conditions are as follows: (1) The voltage of the incoming machine must equal that of the busbars. (2) The frequency of the incoming machine must be the same as that of the busbars. (3) The e.m.f. of the incoming machine must be in phase with the busbar voltage. (4) The phase sequence of an incoming machine must be similar to the busbars, and the switching arrangements made so that like phases of the machine and busbars are connected when the paralleling procedure is completed.

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242 • Advanced Electrotechnology

VR

VB

VR

1

VY Busbars

VY

VB

1

1

Incoming alternator

▲ Figure 9.1

If conditions for the busbars and incoming machine are represented by the figure 9.1 phasors, synchronising requirements are seen, i.e. (1) the respective voltage phasors must be equal, (2) phasors must rotate at the same speed and frequency, (3) they must be in phase with each other, i.e. corresponding line to line voltages must reach their positive maximum values at the same instant and (4) phasors must rotate in the same direction–same phase sequence. On land switchboard installations are provided with a synchronising panel of voltmeters, frequency meters and a synchroscope. In marine work it is common to add ‘synchronising lamps’ as well.

The synchroscope A synchroscope is an instrument used to establish phase angle and frequency synchronisation between A.C. current power supplies. This is a vital safety measure when A.C. power networks or generator outputs are connected together. Synchroscopes are commonly available in two arrangements: electro-mechanical or electronic. ELECTRO-MECHANICAL TYPE. Electro-mechanical synchroscopes use a digital microprocessor to indicate the degree of synchronisation with digital or Light Emitting Diode (LED) displays. When combining 2 or more A.C. power supplies, it is of vital importance to ensure both are at the same voltage, frequency and in phase with each other as discussed. Connecting A.C. supplies when they are not balanced may lead to severe damage to the network or connected A.C. machines. The electro-mechanical type incorporates a pointer and LED to indicate the synchronisation conditions. If the generator frequency is higher or lower than that of the mains the pointer rotates clockwise or anti-clockwise respectively. Speed and rotation direction depends on difference in frequencies which guides the operator to speed up or slow down the generator. The synchronisation condition is determined by a dark lamp method, where

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A.C. Machines – Operation • 243 the pointer coincides with a synchronisation mark and the LED is unlit, but the LED glows in accordance with phase difference, voltage frequencies and amplitudes. The main benefits of this system are its rugged construction, easy servicing and wide operating voltage range. ELECTRONIC TYPE. This type incorporates a circle of LEDs which glow sequentially in a clockwise or anti-clockwise direction at variable speed in accordance with the frequency difference in the 2 power systems being linked. Synchronisation is indicated by the glowing LED in the ‘SYNCHRO ON’ position. Anti-phase is indicated by the glowing LED in the ‘SYNCHRO OFF’ at the bottom. The advantages of this system are continuous operation, no moving parts, low power consumption and the fact that it is visible in the dark. SYNCHRONISING LAMPS. The error in the frequency of the incoming machine as compared with the busbar frequency is shown by the rate at which LEDs ‘darken’ or ‘brighten’. To understand synchronisation operation, it is best to first consider the synchronising of a single-phase A.C. generator where equality of voltage and phase results in zero voltage across an arrangement of lamps which are then totally dark. A difference of frequency causes lamps to brighten and darken, the difference in frequency being shown by the rate of pulsation. This ‘dark lamp’ synchronisation method makes it hard to judge the mid-point of the dark period – the instant of closing the paralleling switch. In practice, an engineer may find it difficult to adjust the incoming machine speed, so modern use microprocessors does this virtually instantaneously and automatically with synchronisation relays, with an advantage over manual attempts. Thus the moment for synchronising is determined exactly.

Parallel operation There is a difference between parallel operation of A.C. and D.C. generators and this speed difference is of great importance to marine engineers as they may have to operate either A.C. or D.C. electrical generating and distribution systems. The procedure for putting A.C. machines on the ‘bars’ was discussed here and in Chapter 7 (The Marine Alternator). Torque production to ensure a machine runs in synchronism was also considered. Attention is now given to the action required to make a paralleled alternator take its load share and to study the effects of the 2 possible adjustments, namely: (1) field regulator operation, i.e. excitation control and (2) throttle valve operation, i.e. speed control. With correctly designed D.C. generators, parallel operation and load sharing are achieved by excitation adjustment. As field excitation of a machine is increased, the

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244 • Advanced Electrotechnology machine generated e.m.f. rises so it takes a larger portion of load. The increased load slows the D.C. generator down and allows the power source to draw extra fuel, since a governor is actuated, and the extra load condition is carried. With 2 alternators in parallel, an excitation increase of one machine raises the generated e.m.f. and tends to make it take more load. However, the machine cannot slow down since it is tied synchronously to the system, and the governor of the power source is unaffected, thus no action results to cause the machine to take more load. Operation of the excitation control causes a wattless current to circulate in the paralleled machines and busbar system, with a current which lags generated e.m.f. by angle φ so the load = EI cos φ. The kW load remains constant, maintaining an unvaried governor setting. To change the load distribution between A.C. generators in parallel, the throttle valves must be altered, so that for 2 alternators in parallel, as the speeds (frequencies) must be identical, the kW loading on each machine must be related to the power source input, i.e. the amount of operation of the throttle valve, and cannot be controlled by the excitation. The excitation effect and throttle control are considered next, but 2 problems here revise the basics of A.C. load summation and P.F. control. Example 9.1. A factory takes 200 kW at 660 V from a three-phase supply at a lagging 0.707 P.F. A synchronous motor is installed to take an extra 100 kW. What must be the motor kVA rating to raise the system P.F. to 0.866 lagging (1 decimal place)? Total active power = 200 + 100 = 300 kW Total apparent power at 0.866 P.F. (lagging) =

300 = 346.4 kVA 0.866

Here cos φ = 0.866 and sin φ = 0.5 Total reactive power = −346.4 × 0.5 = −173.2 kVAr Original apparent power of load =

200 = 282.9 kVA 0.707

Here cos φ = 0.707 and sin φ = 0.707 Original reactive power = −

200 × 0.707 = −200 kVAr 0.707

With the motor load, reactive power is reduced by: −200 − ( −173.2) = −26.8 kVAr Thus the motor reactive power rating = 26.8 kVAr (leading), i.e. +26.8 kVAr ∴Apparent rating of motor = 1002 + 26.8 = 102 + 26.82 = 10 × 10.35 = 103.5 kVA. =10 100+7.18 = 10 107.18

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A.C. Machines – Operation • 245 Example 9.2. Two three-phase, 660 V, star-connected alternators in parallel supply the following loads: 40 kW at 1 P.F., 40 kW at 0.85 P.F. (lagging), 30 kW at 0.8 P.F. (lagging) and 80 kW at 0.7 P.F. (lagging). The armature current of one machine is 100 A at 0.9 P.F. (lagging). Determine the armature current (3 significant figures), power output (2 significant figures) and P.F. of the other machine (3 decimal places). Load

kVA

kW = kVA cos f

kVAr = kVA sin f

1

40

40

0

2

47.1

40

3

37.5

4

114.3

cos f

sin f

1

0

−24.8

0.85

0.527

30

−22.5

0.8

0.6

80

−81.5

0.7

0.714

190

−128.8

Machine No. 1 Apparent power = 100 × 660 × 3 = 114 kVA Φ1 = 25°51 sinφ1 = 0.436

Active power = 114 × 0.9 = 103 kW

Reactive power = 114 × 0.436 = −49.8 kVAr

Machine No. 2 Active power = 190 − 103 = 87 kW

Reactive power = −128.8 − (−49.8) = −79 kVAr

Active power = 872 + 79 = 10 8 8.72 + 7.92 = 10 75.5 + 62.5 = 10 138 = 117.5 kVA Current of machine No. 2 =

117.5 × 103 3 × 660

=

1175 = 103 A 1.732 × 6.6

Active power output of machine No. 2 = 87 kW P.F. of machine No. 2 =

87 = 0.742 (lagging) 117.5

Parallel operation of A.C. generators may be studied under 2 distinct headings. The first is parallel working with ‘infinite busbars’, as found in shore-based power stations joined together through a national transmission grid system. An ideal case of infinite busbars is one where the system is so large, in comparison with a single alternator, that its voltage and frequency are totally unaffected by an alternator’s behaviour. The second consideration is of key importance for marine engineers, as it relates to shipboard working. Here busbar voltage and frequency can be altered by local conditions and

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246 • Advanced Electrotechnology the more common case of 2 or more alternators run in parallel is given attention in the study which follows. (1) EXCITATION CONTROL. Assume 2 alternators paralleled correctly. Voltage, frequency and phase of each are the same and this condition is represented by the figure 9.2 phasors. V is the busbar voltage, i.e. that produced by machine No. 1 generating an e.m.f. of E1 volts and supplying the busbar load. E2 is the e.m.f. of machine No. 2. The e.m.f.s of the 2 machines in parallel are in phase relative to the external circuit, but in opposition when considered with respect to each other as shown for the local circuit. (a)

(b) V

(c) V

V

ER I

I ER E2′ E2

E2

E2

E2′

▲ Figure 9.2

In figure 9.2a, voltages are equal and opposed. As a result no current flows in the local circuit between the alternators since the resultant voltage is zero. In figure 9.2b, machine No. 2 excitation is increased. The generated e.m.f. E2 is increased to E2′ , creating a resultant voltage ER around the local circuit which is mainly reactive and the resultant current I lags ER by nearly 90°. This current represents no power flow either ‘from’ or ‘to’ the busbars and the alternator No. 2 power source is unaffected leaving the governor unchanged. The machine supplies a lagging current to or takes a leading current from the busbars which tends to neutralise the effect of a lagging load current drawn from the busbars, so over-excitation of an A.C. generator improves the paralleled system P.F. Figure 9.2c shows the effect of decreasing machine No. 2 excitation, so altering excitation does not really alter the busbar voltage either. This is explained by the reactive current I, for condition (b) having a lowering effect on the e.m.f. E2′ with lagging current having a demagnetising effect on the alternator field strength. Hence E1 the generated e.m.f. of machine No. 1 is increased and the circulating current makes E2 and E1 equal to V – the busbar voltage.

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A.C. Machines – Operation • 247 Example 9.3. Two three-phase alternators in parallel supply 125 kVA at 0.8 P.F. (lagging). Each machine supplies the same power at the same current. If excitation of each machine is changed so the P.F. of one falls to 0.5 (lagging), find the P.F. of the other. Note: as this is reactive excitation control, power output is unaltered. Original apparent power = 125 kVA

cos φ = 0.8 and sin φ = 0.6

∴ Total active power = 125 × 0.8 = 100 kW The active power supplied by each machine = 50 kW Under the new condition. For machine No. 1 cos φ1=0.5 and sin φ1 = 0.866 Apparent power =

50 = 100 kVA 0.5

Reactive power = −100 × 0.866 = −86.6 kVA Original reactive power = −125 sinφ = −125 × 0.6 = −75 kVAr For machine No. 2 Reactive power = −75 − (−86.6) = 11.6 kVAr Active power = 50 kW Apparent power = 502 + 11.6 6 = 10 0 52 + 1 162 = 10 26.35 = 10 × 5.15 = 51.5 kVA and cos φ2 =

50 = 0.97 (leading).) 51.5

With the lagging kVAr of one machine made larger than the total lagging kVAr, the kVAr of the other machine must be in anti-phase to produce the final result. Thus as No. 1 alternator operates at a lagging P.F., No. 2 alternator must operate with a leading P.F. (2) THROTTLE CONTROL. Assume the governor control is altered so machine No. 2 fuel valve is opened. Alternator No. 2 tends to speed up and phasor E2 tries to overtake V (figure 9.3). In connection with the local circuit, a resultant voltage ER immediately produces a current I lagging almost 90° and nearly in phase with E2 so alternator No. 2 develops power =E2I cos φ2. When power output equals the input power increase (by throttle valve actuation through a governor) power source No. 2 tends to speed up, and the alternator delivers power to the load. Alternator No. 1, relieved of load, speeds up

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248 • Advanced Electrotechnology V

ER

α

φ2 I

E2

▲ Figure 9.3

slightly until its power source governor reduces input power and brings about stable speed conditions. The final load distribution on each alternator is achieved by alternate operation of both machine throttle controls until the required loading is shown by each alternator wattmeter, and the voltage and frequency of the system achieve the desired condition. If the driving power of alternator No. 2 is removed, because of a mechanical fault (e.g. a fuel stoppage), the conditions are as shown (figure 9.4). Voltage E2′ drops behind the true synchronisation position by angle α ’. There is now a resultant voltage ER′ acting round the local circuit, producing a current I’, almost 90° behind ER′ − a circuit comprising the machine armature, being mainly reactive. Busbars now supply power VI′ cos φ′ to the machine to keep it running as a synchronous motor. The drop back of E2′ from the synchronous position is momentary and the machine then accelerates back into synchronism. If an increased mechanical load is added to alternator No. 2 while motoring, the machine e.m.f. E2′ would drop back more ER′ and I’ would increase so the total power supplied from the busbars increases. This is the basis of synchronous motor operation. Little work will be done on an A.C. machine operating in this way, however it can be

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A.C. Machines – Operation • 249 V

I′ φ′ E ′R

α′

E′2

▲ Figure 9.4

used as a propeller motor for marine A.C. electric propulsion systems, and warrants some mention. Reference was made to the synchronising current and this is now taken further. Consider parallel operation (figure 9.5). Alternators generate Е1 and E2 volts to maintain a busbar voltage of V volts. Although these voltages are in phase with respect to the load, they are in direct opposition to each other. Suppose the excitations and powers developed by the power sources cause currents I1 and I2 at P.F.s of cos φ1 and cos φ2 respectively. The total load is the phasor sum of I1 and I2, which are omitted from the diagram for clarity. Assume the power input to No. 2 is increased which tries to accelerate, advancing a small angle α. New load conditions are set up, E2′ and E1, to produce ER acting round a local circuit. This causes a circulating current, which under no-load conditions is the synchronising current IS, lagging ER by 90°. IS can be added by phasors to the original currents and combined with I2 to give a new machine current I2′ . Is is received by machine No. 1 and reduces current output to give I1′ the resultant of I1′ and Is. An increased No. 2 input makes it take more load so its speed settles to that decided by the governor-actuated throttle-valve opening. Meanwhile machine No. 1, now

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250 • Advanced Electrotechnology E1 I1 I ′1 φ1 φ′1

ER

φ2

α Is φ′2

I2

I ′2

E2

E2′

▲ Figure 9.5

relieved of load, accelerates to a new speed (frequency) determined in the last stage by the overall system load. IS is a short-duration circulating current, created by transient conditions arising from controls adjustment. Once the overall paralleled system settles down, operating conditions are similar to those found originally, except that I1, I2, cos φ1 and cos φ2 have new values.

Load sharing In summary increased machine excitation produces a wattless circulating current so a generated voltage change relative to the busbars changes the reactive kVA which a machine supplies. Varying power input tends to speed up a machine with power output E2 I2 cos φ2. Load sharing is now considered from 2 view points: (1) Sharing of kW and (2) Sharing of reactive kVA. kW LOAD SHARING. The speed regulation, i.e. governor characteristics of the complete power source and alternator set, determines the load proportions taken by paralleled machines. The frequency/load characteristics of 2 sets are shown in figure 9.6. The characteristics of sets Nos 1 and 2 are assumed to be unequal but since the 2 frequencies are tied together, the total load delivered at any frequency is found by adding the individual loads at this frequency. Here OR = OP + OQ. By repeating this addition for other frequencies, a common frequency curve is obtained which indicates the frequency drop with increase of total load and division of any total load,

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Speed or frequency

A.C. Machines – Operation • 251

Common frequency 2 Load of machine 1 Load of machine 2

1

Total characte ristic

Total load O

P

Q

R

kW load

▲ Figure 9.6

Speed or frequency

such as OR, into OP (machine No. 1) and OQ (machine No. 2). Thus the greater load proportion is borne by the machine having the flatter governor characteristic. As for load-sharing problems with D.C. generators, the characteristics can be plotted backto-back (figure 9.7). The intersection point X, shows the load sharing and the common frequency. Altering the governor settings of machine No. 1 raises curve L to L1 if curve L is raised to L2, the intersection point lies to the right of the right-hand vertical and the condition is represented when machine No. 2 is motored by No. 1, supplying both the total load and the motoring power as well.

L2 2

X

L1 1

Total load (kW)

▲ Figure 9.7

Assuming straight line characteristics, a mathematical solution can be found as follows. Assume 2 alternators Nos 1 and 2, of similar ratings, with prime-mover power sources (having straight line characteristics), so one drops a% and the other b% between no

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252 • Advanced Electrotechnology load and full load (in kW). Assume their speeds equal N at some total loading P (kW). Then, if load is increased by Q (kW), the load increase will be shared so both alternators fall by the same speed. Conditions are shown in figure 9.8.

100

Q –X

N

b X

a

Speed (%)

N1

Q

Load P

Full-load rating of each machine (kW)

▲ Figure 9.8

Let the increased load share taken by machine No. 1 be X kW, so No. 2 takes (Q−X) kW. Hence

aN bN ×X = (Q 100 100

X ) or aX = b(Q − X )

∴ Load gained or lost by No. 1 is X =

bQ and, a b

Load gained or lost by No. 2 correspondingly is Q

X=

aQ a b

Example 9.4 Two large identical alternators each rated 1.25 MVA with 0.8 P.F. (lagging) supply a load in parallel. The governor setting on the prime-mover of machine No. 1 drops from 50 Hz on no-load to 48 Hz on full load. The second machine (No. 2) drops from 50 to 47 Hz. How will they share a load of 1.5 MW? Machine No. 1 drops 2 Hz for 1000 kW (1250 kVA at 0.8 P.F.), i.e. 0.002 Hz per kW. Machine No. 2 drops 3 Hz for 1000 kW (1250 kVA at 0.8 P.F.), i.e. 0.003 Hz per kW. Let the load on machine No. 1 be X kW so the load on machine No. 2 = (1500 − X) kW

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A.C. Machines – Operation • 253 The frequency must be equal, as they are in parallel. ∴ 50 − 0.002 X = 50 − 0.003(1500 − X) and 0.002 X = 4.5 – 0.003 X or 0.005 X = 4.5 ∴ 0.5X = 4500 or Load on machine No. 1 = 900 kW Load on machine No. 2 = 1500 − 900 = 600 kW. This is a direct solution found by application of the expressions deduced earlier. Thus for machine No. 1, a speed change of 2 Hz in 50 Hz gives a = 4% and, for machine No. 2, a speed change of 3 Hz in 50 Hz gives b = 6% Here the load increase, i.e. from no-load to full load Q = 1500 kW ∴X =

0.06 × 1500 0.06 × 1500 = = × 0.04 + 0.06 0.1

=

d Q − X = 1500 − 900 = 600 kW

k

kVAr LOAD SHARING. The way in which machines run in parallel, sharing the reactive kVA, is largely governed by their relative internal voltages. Voltage regulation characteristics of 2 machines are shown in figure 9.9 plotted against the kVAr load.

Total load

Total ch Voltage

aracteri

stic

Common voltage 2

Load of machine 1

1

Load of machine 2

Leading kVAr

Lagging kVAr

▲ Figure 9.9

As for kW load sharing, the characteristics are plotted in figure 9.10. The positions of the characteristics are determined by the amount of excitation. An increase of this for one machine, e.g. No. 1, raises the curve to L1. Machine No. 1 then takes a larger share of the kVAr load and the busbar voltage raises. Condition L2 shows how machine No. 1 operates at a leading P.F. even though the total load lags. Figure 9.9 shows the machine with the flatter characteristics takes the largest loading share.

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254 • Advanced Electrotechnology

L2

Voltage

1

Leading kVAr

Total load (lagging kVAr)

Voltage

L1 2

Leading kVAr

▲ Figure 9.10

Example 9.5. Two alternators operate in parallel supplying a 1 MW load at 0.8 P.F. (lagging). If the supply of machine No. 1 is adjusted so it is loaded to 600 kW and excitation adjusted so that its P.F. = 0.707 (lagging) (a) find the P.F. at which machine No. 2 operates, (b) if the load and supplies of both machines are unchanged and the excitations altered so the P.F. of machine No. 2 is raised to 0.8 (leading). Find the new P.F. of machine No. 1. Total active load = 1000 kW cos φ = 0.8 (lagging) and sin φ = 0.6 i.e. Total apparent power =

1000 = 1250 kVA 0.8

Total reactive power = −1250 × 0.6 = −750 kVAr (a) Machine No. 1 Active power = 600 kW Apparent power =

600 = 848.7 kVA 0.707

∴ Reactive power = −

600 × 0.707 = −600 kVAr 0.707

Machine No. 2 Active power = 1000 − 600 = 400 kW Reactive power = −750 − (−600) = −150 kVAr

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A.C. Machines – Operation • 255

∴ Apparent power = 4002 + 1502 = 10 4 2 + 1.52 = 102 16 × 2.25 = 10 102 18.25 = 102 × 4.275 = 427.5 kVA cos φ2 =

400 = 0.94 (lagging). 427.5

(b) Even with increased excitation, the active power output of machine No. 2 is still 400 kW, although its P.F. now leads, i.e. there is a positive kVAr condition. ∴ Apparent power =

400 = 500 kVA 0.8

Reactive power = 500 × 0.6 = 300 kVAr Thus for machine No. 1 Reactive power = −750 + 300 = −450 kVAr Apparent power = 102 36 + 20 20.25 25 = 102 56.25 = 10 × 7.5 = 750 kVA ∴ cos φ 1 =

600 = 0.8 (lagging). 750

The Synchronous Motor The basic principles of an alternator’s ‘motoring action’ when connected to busbars have been covered. Here attention is given to a machine when mechanically loaded, i.e. when it takes electrical energy to operate as a motor. This is used to advantage in larger sized machines because the P.F. can be controlled by varying D.C. excitation. When operated in this way it can improve the overall load system P.F., such as in a factory, and when used as a ship’s propulsion motor unity P.F. minimises current so the system copper losses and associated equipment size will not be larger than that required for the appropriate kW rating.

Operating action Consider an alternator run up to speed and synchronised onto the busbars. If the machine delivers no power it is said to ‘float on the bars’. Suppose next the power source is disconnected. The machine draws power from the mains, sufficient to overcome the

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256 • Advanced Electrotechnology no-load losses and the rotor slips back through angle α1 having a phasor diagram as shown in figure 9.11. The busbar voltage V and generated e.m.f. Е1 have a resultant ER 1 ER and the motor draws a current I1 from the supply. I1 = 1 where ZS is the armature’s ZS synchronous impedance. I1 lags behind the applied voltage V by angle φ1 and power input = VI cos φ1 V I2 I1 φ1

ER

φ2

2

ER

1

α2 α1

E2

E1

▲ Figure 9.11

If the motor is loaded, the rotor drop back further to angle α2. The resultant voltage is then ER 2 and the current increases to I2, lagging as before, on V by angle φ2. Power input is now VI2 cos φ2 and α2 adjusts itself until sufficient power is drawn from the supply to manage the load. The phasor diagram is drawn from the busbar point of view and shown in figure 9.12. To the basic voltage phasors are added the voltage drops in the motor due to resistance and synchronous reactance. Here angle α is the ‘angle of retard’. The total mechanical power developed by the motor is E2 I2 cos(φ1 − α2) which to a first approximation is expressed as V2I2 cos φ2. Power available at the shaft is less than this due to the iron, friction and windage losses. An increased load torque causes angle α to increase, meaning a larger current. At the same time φ increases and the P.F. cos φ diminishes. For small φ values I increases at a greater rate than cos φ decreases, so the power developed increases. A value of φ is reached at which I increases at a lower rate than

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A.C. Machines – Operation • 257 V IR

ER =IZS IXS E2

I2 α2 ER

φ2

▲ Figure 9.12

cos φ decreases, so a maximum power condition is obtained. Thus, once the current increase is countered by a P.F. decrease, power starts to fall and, if the load torque required is increased further, the motor will pull out of step and stall. Under normal operating conditions if P is the total mechanical power developed in watts W, V the supply voltage, I the motor current, cos φ the motor P.F. and Ra the effective motor resistance then P = VI cos φ − I2 Ra. Example 9.6. A 500 V, synchronous motor develops 7.5 kW, with 0.9 P.F. (lagging). The effective armature resistance is 0.8 Ω. Iron and friction losses are 500 W and the excitation loss 800 W. Find (a) the motor current, (b) the input power and (c) the overall efficiency (all 1 decimal place). Output power = 7.5 kW = 7500 W Mechanical power developed = 7500 + 500 = 8000 W ∴ 8000 = 500 × I × 0.9 – I2 × 0.8 or 0.8I2 – 450I + 8000 = 0 Whence I = =

450 ± 4502 − 4 × 0.8 × 8000 2 × 0.8 450 ± 4502 − 25 600 16

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258 • Advanced Electrotechnology Giving I = 17.2 A Power input to motor = 500 × 17.2 × 0.9 = 7.74 kW Total power input, inclusive of excitation loss = 7.74 + 0.8 = 8.5 kW Overall efficiency =

7.5 × 100 = 87.6% 8.84

Example 9.7. The input to a 11 kV, three-phase, star-connected synchronous motor is 50 A. The effective synchronous reactance and resistances per phase are 29 and 0.95 Ω respectively. Calculate the power supplied to the motor in kW and the induced e.m.f. for 0.8 P.F. (lagging) (both 3 significant figures). In figure 9.12 it is shown E2 is given by E22

(Vcos − IR )2 + (Vsinφ IX I S )2

Resistance voltage drop per phase = 50 × 0.95 = 47.5 V Reactance voltage drop per phase = 50 × 29 = 1450 V Phase voltage =

11 000 3

= 6352 V

Then E22 = {(6352 × 0.8) − 4 7.5}2 + {(6352 × 0.6) − 1450}2 = 50342 + 23612 = 106(5.0342 + 2.3622) = 106(25.4 + 5.57) = 106 × 30.97 or E2 = 5570 V Thus induced e.m.f. = 3 × 5.57 = 9.635 kV Input power = 3 × 11 000 × 50 × 0.8 × 10 −3 = 762 kW.

Starting Figure 9.13 depicts a polyphase, 2-pole stator winding fed with a polyphase A.C. current. A synchronous rotating magnetic field of constant strength is produced which is assumed to rotate clockwise. Stator poles N1 and S1 occupy the positions shown at a particular instant, the 2-pole rotor being stationary. Under these conditions, the rotor will be urged to move anti-clockwise but a half-cycle later, the stator poles will reverse and S1 will now be at the top and N1 at the bottom. The rotor now tends to move in a

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A.C. Machines – Operation • 259 Rotating magnetic field

N1

N S

S1

▲ Figure 9.13

clockwise manner and, as it cannot respond to an alternating torque, remains at rest and so the motor is not self-starting. If the rotor is run at synchronous speed, the opposite stator and rotor poles ‘lock-in’ and the motor continues to run as considered earlier under the heading of Operating Action. Thus, to operate, the motor must be brought up to speed, by some external means and synchronised onto the supply as was the alternator. There are various practical methods of running a motor up to speed and the following are among the most usual. (1) An A.C. or D.C. pony motor (a small motor that gets a larger motor started) is coupled to the main machine of sufficient power to drive the latter up to slightly above synchronous speed, running up a synchronous motor. In one arrangement the D.C. exciter is used as the pony motor to drive the main machine above synchronous speed. It is then switched off and reconnected as a D.C. generator, to supply the main motor’s field. As the latter slows down and passes through synchronism, paralleling is affected and the load to be driven is then taken up. (2) A synchronous motor can be provided with a cage winding on the rotor to run the machine up to speed as an induction motor. The D.C. field is applied and the rotor pulled into synchronism. This is the method employed in A.C. ship-propulsion propeller motor systems. Since propeller speed is controlled by the main alternator speed, i.e. by the supply frequency, procedure is as follows. Consider a 3 kV system. 1 3000 speed, with or 375 V generated. As induction motor 8 8 1 torque is proportional to the square of supply voltage, a voltage reduction to normal 8 1 of normal. This reduced torque cannot provide much power at produces a torque 64 An alternator runs up to say

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260 • Advanced Electrotechnology the propeller, which may not even turn − thus the need for double excitation. At this 1 stage, the motor is connected to the alternator, and speed raised until near speed, 5 3000 when only = 600 V is generated, double excitation is applied to the alternator 5 field. Thus if the alternator field is rated for 110 V D.C., 220 V will be applied. As field strength is doubled, 2 × 600 = 1200 V is generated, sufficient to run the propeller motor up to speed as an induction motor. When synchronism is approached, the motor D.C. field is applied and the propulsion motor pulled into step to operate as a synchronous motor. From now on, as the alternator speed is raised its excitation is reduced, so by the time normal operating speed value is reached, the alternator field is reduced to the rated value, i.e. 110 V in this example. The induction motor cage winding is made up of copper bars embedded in the rotor pole faces. These damper bars are connected together by short-circuiting rings carried round the rotor circumference.

Induction Motors Induction motors are included under the heading of ‘operation’, as little has been said about their starting and control, and consideration is now given to their operating requirements. As the three-phase induction motor is the A.C. machine invariably used in marine applications, attention is given to usual marine operational practice.

Starting Cage-type motors are the most common and are considered first. Since the torque produced is proportional to applied voltage squared, direct application of line voltage has some advantages. Unlike D.C. motors, which, under standstill conditions, offers low resistance to the line voltage, A.C. motors have enough inherent impedance to limit current to say 4−7 times the full-load current value. Thus a rheostatic starter (which during starting cuts out successively one or more resistors in the rotor circuit), as used for D.C. motors, is unnecessary and A.C. motor-starting methods, for cage machines, takes 3 forms. (1) DIRECT-ON-STARTING. Even though the starting current may be 7 times higher than the full-load value, current will not necessarily burn out a machine, if it is limited to only a short period. This starting method is usual for small motors, up to about 5 kW in shore practice but in marine work, when appropriate alternators are installed, i.e. those arranged to maintain a fairly constant terminal voltage when a large starting

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A.C. Machines – Operation • 261 current is drawn, direct-on starting was developed for motors rated up to 200 kW or more. Starting current depends on motor size and design. The motor must accelerate quickly and current must fall rapidly. The main objection to this starting method is the large voltage drops in the cables which can affect other connected machines. High starting currents produce large heating effects and direct-on starting is not used for repeated operations if a cooling interval is not provided. Starting is effected with a contactor, actuated by a push-button or control switch, connecting the motor to the supply. The contactor usually has a built-in isolator and High Rupture Capacity (H.R.C.) fuses included for short-circuit protection. Over-current protection is provided by an overload unit of 3 series current coils but a thermal trip can be substituted to ensure a cooling period between ‘starts’. Full-voltage working for the closing coil, push-button circuits, etc. is used, but for marine work it is common practice to incorporate an auxiliary 110 V voltage transformer, for control and indicator lamp circuits. (2) STAR-DELTA STARTING. A 2-step starter is used with a motor whose phase windings are normally delta connected. All 6 ends of the phase windings are brought into the starter which is of the ‘2-position’ type. The first position connects the motor phases in 1 1 star so the starting voltage per phase is supply voltage and starting torque is that 3 3 obtained by direct starting. Changeover from star to delta is made by contact arms or segments in the manually operated starter or by contactors in an automatic starter. 1 1 Starting current is reduced since it is phase value or of the corresponding direct 3 3 switching value. A reduced starting current is shown in example 9.8 and is the method’s main advantage. British practice rarely uses the star-delta starting method for marine work due to the interval between change-over; the motor is disconnected from the supply and any induced e.m.f. falls in value and frequency. When supply reconnection is made, transient currents flow which may have peaks larger than those resulting from direct1 on switching. High current conditions can occur and yet torque is of that obtained by 3 direct starting. The method is considered poor for marine work because: (a) if alternators can be designed to cope with the direct-on starting current peaks, the reduced stardelta starting conditions offer little advantage, (b) reduced starting torque limits this starting method mainly to centrifugal pump and fan motors. Example 9.8. The stator winding of a three-phase motor is arranged for star/delta starting. If a motor is rated at 7.5 kW, voltage is 440 V, full-load efficiency 85%, 0.8 P.F. (lagging) and the short-circuit current equals 4 times the full-load current, calculate the current taken from the mains at the instant of starting (2 decimal places).

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262 • Advanced Electrotechnology Full-load output = 7.5 kW = 7500 W Full-load input = 7500 × ∴I =

8823 3 × 440 × 0 8

=

100 = 8823 W = 3 VI cos φ 85

25.06 3

= 14.5 A

The short-circuit current/phase =

588 3

Z ph = Current in start =

Short-circuit current = 4 × 14.5 = 58 A

= 33.5 A 440 = 13.13 Ω 33.5

3 254 = = 19.35 A 13.13 13.13

i.e.

1 of direct-on starting. 3

(3) AUTO-TRANSFORMER STARTING. With a tapped three-phase auto-transformer, any supply voltage fraction can be applied to a motor without changing the motor connections. A reduced voltage results in a certain motor current and the mains current is further reduced by the transformer turns ratio. For example, with a 50% tapping a 1 motor will take half the current of one direct started and the mains current will be 2 1 that of the motor current. So mains current is times the motor current if direct started. 4 Torque is proportional to the square of applied voltage, i.e. a quarter that obtained by direct switching. This starter type is expensive and for marine work is only used for large motors. If switching as described above is used motor connections are broken when the tapping switches are moved and current surges or transients as described for ‘star-delta’ starting occur. To overcome this, methods often use a Korndorfer starter, especially for automatic starters. This starter is used for reduced voltage starting with three-phase auto-transformers, with newer devices providing full automatic operation. Figure 9.14 shows the typical main circuit scheme for such a marine auto-transformer starter type. An electrically controlled switch is used for switching the power circuit, which is similar to a relay, except that it is used with higher current ratings. With operation of the ‘on’ push-button–auxiliary circuits not shown, contactors A and В close first. The motor receives a reduced voltage depending on the transformer tapping used. Time-lagged switches in the auxiliary contactor closing-coil circuits are fitted and after a short period, contactor В opens. The motor now receives an increased voltage with a section of the transformer winding used as a series reactor. After a further time interval, contactor С closes and contactor A opens. Thus the motor is connected to the supply direct and the transformer disconnected. Circuit protection

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A.C. Machines – Operation • 263 C

Motor

A

B

▲ Figure 9.14

for over-current and under-voltage is built into contactor С since it is the main circuit or ‘running’ contactor. Example 9.9. A cage-type induction motor which, if started directly on line, would take a current surge 6.5 times the normal full-load current developing 70% of normal full-load torque, is started with an auto-transformer on a 75% tapping. Calculate the starting current in terms of normal full-load current and torque. With a 75% tapping as current is proportional to the voltage motor current = 0.75 × 6.5 times normal full-load current. Mains current is reduced because of the transformer ratio. 3 3 × × 6 5 times normal full-load current 4 4 or 3.65 times normal full-load current. On a 75% tapping since torque is proportional to 2 70 9 7 ⎛ 3⎞ voltage squared the new torque × times full-load torque = × = 0.394 ⎝ 4 ⎠ 100 16 10 ∴ Starting current, as drawn from the mains =

times the full-load torque. For a wound-rotor induction motor, theory shows this machine may develop maximum torque at starting and during acceleration. Rotor resistance can be used for speed control but resistor units are not usually rated for such a duty. The starter usually comprises a normal three-phase contactor, with the usual protection features to control the stator supply. The rotor resistor controller is interlocked with this contactor to ensure starting can only be attempted with maximum resistance in the rotor circuit. For marine work, if speed control is needed as for some A.C. winch or ‘barring’ motors, resistors are cut out by a drum controller or regulator similar to those used for D.C. machines. Rotor resistors are rated accordingly, although a cagetype 2-speed machine is often favoured.

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264 • Advanced Electrotechnology

Speed and torque control The method by which such control is achieved for the wound-rotor motor has been considered. For a cage-type motor, some control is obtained by rotor design variations. Cage resistance and reactance can be adjusted, and as little has been said about its constructional details, the chapter will end looking at different cage construction features. STATOR. Construction of an induction motor stator for both wound-rotor and cage types is similar to the A.C. generator as it consists of low-loss electrical steel laminations pre-stamped with slots, compressed and secured into a fabricated steel or cast-iron frame to form the body of the machine. Air circulation space is provided between these laminations and the frame, and radial air ducts arranged, if required, by fixing the stampings in stacks, separated by spacers. Stator slots are usually of a semi-enclosed type and, since the induction motor is a small air-gap machine, the bore of the stator is usually turned or ground true. Windings are similar to those described for the alternator, usually of a single layer or concentric type. For relatively large output low-voltage motors, one can use a double layer, 2 bars per slot winding with the following advantages. It is more rigid mechanically and due to the possibility of short-chording, overall machine length can be reduced to save copper. Leakage reactance is also reduced in this way. For medium-sized motors, windings can be either wire or a rectangular bar. If wire coils are used they can be former wound and then roughly shaped, slot portions being untapped. Wires are introduced into the slots one at a time and end sections pulled into shape by hand creating the common ‘Mush’ winding, a type of A.C. armature winding with an odd coil span of long and short conductors with connections made between a long and short conductor only. Coils are made from copper wire covered with abrasion-resisting synthetic enamel and insulated from the core slots. Coils are made for a tight fit in the slots closed by fibre wedges. The stator is impregnated with a high-grade thermosetting resin varnish, to prevent coil movement and is resistant to moisture, salt spray and oil vapour. End windings are taped and the insulation reinforced with suitable fibrous material before varnishing. For large low-voltage motors, when bar windings are used, the conductors of a one-turn coil can be of copper strip, insulated with tape – varnished and baked, or moulded under pressure directly onto the conductors. These can be pushed through slots from one end only, a method to ensure high mechanical strength. Connections are thus all on one side, accessed for testing and inspection.

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A.C. Machines – Operation • 265 ROTOR. A standard cage rotor is designed for optimum running performance and used for applications where normal starting current and torque values are acceptable. Such a rotor develops 120–150% full-load torque with 600–800% full-load current if direct started. Where a smaller starting current and/or higher starting torque is needed, alternative rotor construction exists to give the desired characteristics. Because of the many performance variations it is hard to distinguish between ‘standard’ and ‘hightorque’ motors but appropriate examples are considered under these headings. Typical marine drives for cage motor include centrifugal pumps, fans and hydraulic steeringgear. High starting torque can be specified for a motor but in most cases starting is light and a standard rotor is satisfactory. The core of a cage rotor is built up under pressure from similar materials as is the stator core. A laminated cylindrical core is keyed direct to the shaft for small motors, and to a cast-iron or cast-steel rotor hub or spider, itself keyed to the shaft, for larger machines. The core has slots which may be semi-enclosed or totally enclosed round its outer periphery. Ventilating ducts are arranged to lie opposite corresponding ducts in the stator core along its length. Slots are usually skewed to ensure smooth starting and to avoid magnetic noise. The rotor periphery is accurately machined to ensure a minimum air-gap between rotor and stator, consistent with good design. The shaft is of good quality tensile steel to withstand torsional effects and mechanical shocks which may be transmitted from driven machinery. The rotor is finally dynamically balanced after the cage winding is inserted. STANDARD-CAGE ROTOR. For small motors a winding consists of a die casting of special aluminium alloy. Bars, end-rings and cooling vanes are cast in one piece to avoid joints and to produce a higher starting torque, due to the slightly higher resistance of aluminium compared with copper.

▲ Figure 9.15

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266 • Advanced Electrotechnology Larger machine rotor windings consist of solid copper or bronze bars brazed to end-rings of the same material (figure 9.15). Bars are driven into the slots and the profile and material, as for the end-rings, modified to give desired start and running characteristics. Low cage resistance is a good feature for efficient running and small slip, but disadvantageous for starting. Starting currents are large, P.F.s low and starting torque poor. A standard-cage rotor, with minor modifications, is common since most drives start light, and current can be limited by various starting arrangements, e.g. the star-delta or auto-transformer starter. HIGH-TORQUE CAGE ROTORS. The advantages of wound-rotor motors with external resistance provide improved starting to some extent without losing the robust construction and good running properties of a standard-cage motor. This is achieved by special constructions, and 2 methods are considered. (1) Shaped and Embedded Rotor Bars. By appropriately dimensioned slots and conductors, a high starting characteristic is possible as for large A.C. currents. Conductors are of large cross-section solid copper, so large eddy currents develop and an electromagnetic effect operates, forcing current upwards in the conductors towards the slot-openings. Current density can rise over part of any one conductor, resulting in excessive copper loss. This increases the conductor’s resistance, since a large part of its cross-sectional area is current free. The effect’s size depends on frequency, the higher it is the more pronounced the current localisation. As the slip decreases and rotor frequency falls, the effect disappears, producing a high resistance value on starting which is further increased by conductor shaping (figure 9.16 examples). The apparent increase in cage bar resistance during starting results in a desired starting torque increase. (a)

(b)

▲ Figure 9.16

(2) Double-Cage Rotors. The aim here is to achieve a variable rotor resistance to combine the high starting torque and low starting current provided by a motor more like the inner cage of figure 9.17 with the low normal operating slip and high motor efficiency more like the outer cage of figure 9.17. The solution is provided by having 2 cages with the overall result produced by the sum of the two. Here 2 circumferential

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A.C. Machines – Operation • 267 100 Combined torque

Torque (%)

Outer cage

Inner cage

O

Slip (%)

100

▲ Figure 9.17

rows of holes in the rotor iron constitute the conductor slots. The outer row is spaced from the inner by narrow radial air-gaps linking slot pairs and continues through to the outer rotor surface. The arrangement is shown in figure 9.18 where the radial air-gap ensures the stator flux links with the bottom conductors and does not take the shorter path which would exist if no radial air-gaps were added. Outer cage conductors are of high resistance (bronze) and low reactance while the inner cage is made of high conductivity copper with high reactance.

Pole flux

▲ Figure 9.18

On starting, the rotating magnetic field links mainly with the outer cage conductors as the rotor frequency is the supply frequency and the large reactance of the inner, more highly inductive cage limits current. The outer cage high resistance is thus more operative and the corresponding torque higher. As the motor accelerates, rotor current frequency falls, inner cage reactance decreases and more current flows until it takes over the main part of the driving torque. Figure 9.17

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268 • Advanced Electrotechnology shows a double cage rotor torque/slip curve. Adjusting the separate torques through rotor slot design yields a combined characteristic that can be made to suit any desired condition.

Speed adjustment The cage motor is a single-speed machine and best suited to constant speed drives. It is possible to design a ventilating fan or pump, permitting single-speed operation with no efficiency loss. If single speed is unacceptable, 2 or more speeds can be obtained with a pole-change cage motor, a development of the standard motor providing a convenient and efficient speed changing method at low cost. It is similar to a standard motor but by arrangement of multiple stator windings: 2, 3 or 4 definite speeds are possible. Twospeed motors are provided for marine work and these are considered next. TWO-SPEED MOTORS. These can be built as (1) dual wound or (2) consequent-pole machines. (1) A dual-wound arrangement has been used for many years in marine duties – general service pumps and forced fans, consisting of a standard motor with 2 separate windings, one for each speed. Windings are pitched to divide the stator into 2-pole systems, with double the number of the other. One winding only is connected into the circuit at any time and the usual forms of starting: direct-on, star-delta or auto-transformer used. (2) Consequent-pole machines have only one winding, tapped and connected to appropriate terminals. These motors can only be direct-started or used with autotransformer starters. The arrangements for speed control are effected by suitable switching and the control gear, being more complicated than the dual-wound motor, is not favoured for marine duties so further comment will not be made. Example 9.10. On full-load, a 4-pole, 50 Hz, wound-rotor induction motor operates with 5% slip. Find the resistance inserted into each phase of a rotor circuit to reduce speed to 1200 rev/min, if torque remains constant. The ratio of stator standstill reactance to resistance is 6:1. Since T =

thus T =

KV 2 sR2 R22

s 2) ( sX

2

then T =

CR2 where C is a constant. R22 2 + X2 s s

C C and initially T = 2 2 X s 36R22 0 05 R R + 22 + s R 0 05 R2 2

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A.C. Machines – Operation • 269

Also N2 = N1 − sN1 = N1 (1 − s) and f = So N1 =

PN 120

50 × 120 = 1500 rev/min. 4

Thus 1200 = 1500(1 − s) and

4 = 1 − s so s 5

1 5

0.2

Let mR2 be the new value of rotor resistance per phase Then T =

and

C mR 2 36R22 0 2 + 02 mR2

R2 = 0 05

R22 0 05 mR2 36R22 0 2 = + R2 02 mR2

Thus 204 + 1.8 5m + Whence m =

7.2 or 5m2 − 21.8 m

7.2 = 0

21.8 ± 21.82 − 4 × 5 × 7.2 2×5 475.24 − 144 331.24 = 21.8 ± 10 10 10 331.24 = 21.8 ± = 21.8 8 ± 3.3124 = 2.18 8 1.82 = 4 10 = 21.8 ±

New resistance = mR2 = 4R2 ∴ Additional resistance = 3R2

Practice Examples 9.1 A synchronous motor with a power consumption of 50 kW is connected in parallel with a load of 200 kW having 0.8 P.F. (lagging). If the combined load has 0.9 P.F. (lagging), what is the leading reactive kVA supplied by the motor (1 decimal place) and at what P.F. is it working (3 decimal places)? 9.2 Two 3.3 kV, star-connected alternators when operating in parallel supply the following loads: (a) 800 kW at unity P.F., (b) 600 kW at 0.707 P.F. (lagging). The

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270 • Advanced Electrotechnology current of one machine is 150 A at 0.85 P.F. (lagging). Find the current output (3 significant figures) and the other machine’s P.F. (2 decimal places). 9.3 A ship’s electrical system is supplied by 2 identical three-phase, star-connected alternators operating in parallel. The machines share a total load of 1000 kW at 440 V, 0.8 P.F. (lagging). If the kW loadings of the machines are equal and one machine supplies a 1000 A lagging current, find (a) the current supplied by the second machine (3 significant figures), (b) each machine’s P.F.s (3 decimal places) and (c) the reactive current circulating between two machines (3 significant figures). 9.4 Two alternators A and В operate in parallel. When tested individually, the frequency of machine A falls from 50 to 48.5 Hz when the load is 150 kW, while machine В falls from 50 to 48.5 Hz when the load is 220 kW. If the total load is 200 kW, find the frequency of the paralleled system (2 decimal places), and the load on each machine (2 and 3 significant figures respectively). 9.5 A 15 kW 440 V, three-phase star-connected synchronous motor has an armature of 0.4 Ω effective resistance per phase. It gives full-load output at 0.9 P.F. (leading). If the iron and friction losses are 500 W, find the armature current (1 decimal place). 9.6 A 220 V, single-phase, synchronous motor has a synchronous impedance of 5 Ω and an effective armature resistance of 0.5. Ω. Calculate (a) the minimum armature current (2 decimal places), (b) the generated e.m.f. (3 significant figures) and (c) angle of retard for a total load of 5 kW (in ° and ′), which includes the motor’s iron and friction losses. 9.7 A three-phase, 440 V, 37 kW induction motor has 82% efficiency operating at 0.85 P.F. (lagging). When ‘direct-on’ started the motor takes a current 6 times fullload current and produces a torque = 1.5 times full-load torque. Calculate the current taken from the supply (2 decimal places), and the ratio of starting to fullload torque if the motor is started with an auto-transformer with a 75% tapping. 9.8 Find the ratio of starting to full-load current for a 15 kW, 415 V, three-phase, induction motor with a star-delta starter, given full-load efficiency is 85%, the full-load P.F. is 0.8 (lagging), the short-circuit current is 60 A at 220 V and the magnetising current negligible. 9.9 A 440 V three-phase wound-rotor induction motor has a rotor resistance and standstill reactance of 0.02 and 0.27 Ω per phase respectively. The stator to rotor phase turns ratio is 3:1 with the stator windings connected in delta. If the motor is started with a resistance starter of resistance 0.25 Ω per phase, calculate the current taken by the motor from the supply (a) at starting and (b) under full-load running conditions if the full-load slip is 4%. What would be the current taken

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A.C. Machines – Operation • 271 from the supply if the motor was accidently started with the starting resistance in the ‘run’ position? Neglect the no-load current, resistance and reactance of the stator windings (3 significant figures). 9.10 A 6-pole, three-phase induction motor on full load develops a useful torque of 162.4 Nm and the rotor e.m.f. makes 90 complete cycles per minute. Calculate the power in kW (2 significant figures). If the frictional mechanical torque lost is 13.6 Nm, find the rotor copper loss, the input to the motor in kW (2 decimal places) and the efficiency (1 decimal place). Assume the stator loss is 750 W.

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10

ELECTRONIC EMISSION PROCESSES AND DEVICES Give us the tools to finish the job. Sir Winston Churchill

A knowledge of basic electronics concepts is now essential for all engineers, and students are reminded that in spite of the emergence of semiconductor devices, and the rapid development of associated optoelectronic and spintronic technologies, its impact on power engineering and device applications is still significant. Large solid-state motor starters, controlled rectifiers and static inverters have been introduced into ship enginerooms, as well as smaller but highly sophisticated instrumentation and control apparatus like monitoring and alarm systems, supervisory units and routine data loggers. Because of the complexity of such equipment, specialist knowledge is needed in the event of inevitable breakdowns and malfunctions but electronic diagnostic tools and replacement of diagnosed faulty components by maintenance staff is now routine, especially in modern military equipment, but elsewhere repair is generally performed by an engineer. An introduction to modern electronics is made in this chapter, covering key common components such as the thermionic diode, triode, Cathode Ray Tube (CRT) which is

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Electronic Emission Processes and Devices • 273 now being replaced by Liquid Crystal Display (LCD) and plasma displays which will be discussed fully towards the end of the book in Chapter 14. Nonetheless, understanding the function of the anode, cathode and gate electrodes (triode) is essential to see how the operation of semiconductor technology devices arose and is more than just a review of superseded valve technology. Chapters 11 and 12 respectively will introduce the diode and the transistor, both essential components of solid-state maritime electronics. John Ambrose Fleming (1849–1945) invented the thermionic valve (1904), taking advantage of the particle-like behaviour of electrons. As its name indicates a valve acts like a tap in a water system, which can be turned on and off to permit electricity to flow one way only. The first valves were used in radio circuits detecting radio signals and had 2 elements – a wire and a metal electrode surrounding it, both placed within a vacuum. Lee de Forrest (1873–1961) added a third element, the gate, to the valve producing the triode, allowing greater electrical current control. The triode even provides a small amplification, i.e. making a weak signal stronger, thus fabricating the first crude electronic amplifier. For the first time speech, radio and TV signals could be amplified for the accelerating growth in global communications. However, despite significant improvements valve devices were generally bulky, high on power consumption and somewhat unreliable. It was the invention of the transistor, a semiconductor device with at least three terminals, by Bardeen (the only double Nobel winning physicist), Brattain and Shockley in 1947 that made possible the development of digital electronics and integrated circuits containing millions and millions of ‘invisible’ electrical components.

Electron Emission Electrons associated with the atoms of a metal are normally confined to shells, as described in Volume 6. At a metal surface, ‘free’ electrons in the outer shells may leave the surface due to their increased velocity. However, they are attracted back by the unbalanced electric field created which sets up a potential barrier. If these free electrons can break through the potential barrier, a process termed electron emission results. Electron emission arises from various sources and is described by one of the following processes: (1) thermionic emission, (2) photoelectric emission, (3) secondary emission or (4) cold or field emission. Of these thermionic emission is considered first in some detail due to its relevance to various common electron-related devices operation and laying the principles for solid-state devices. Photoelectric and secondary emission will be discussed with respect to modern image intensifiers for ‘night vision’ while cold emission will not be discussed as significant here.

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274 • Advanced Electrotechnology

Thermionic Emission When a metal is heated, energy is transferred to it and electrons acquire increased random velocities which may permit some to escape the metal surface’s potential barrier, in a similar manner to a rocket needing a certain minimum velocity to escape earth’s gravitational attraction. Such electrons can also be likened to vapour globules given off from a boiling water surface, where unless some means of collecting electrons as they are emitted is found, they will lose their velocities, forming a space charge. Such space charges or electron clouds give rise to a −ve charge which repels further electrons. We can picture electrons leaving a metal surface, repelled by the space charge which has built up, subsequently returning to the metal. To enable electrons to leave a heated metal surface, an additional electrode is placed adjacent to, but insulated from, the metal within a vacuum enclosure. A vacuum is needed as atmospheric pressure prevents electrons moving between –ve cathode and +ve anode. This electrode, the anode, is made +ve with respect to the heated surface, the cathode. This arrangement was first introduced in Volume 6 and is essential for the operation of radio valves, CRTs, mercury-arc rectifiers and discharge lamps. For most modern electronic equipment, thermionic valves have been replaced by semiconductor devices but a basic understanding of the operation of the former is vital in appreciating the functioning of semiconductor devices and other thermionic devices for which no substitute has been found, or in military equipment designed to resist Electro Magnetic Pulses (EMPs). One thermionic example is the CRT used in oscilloscopes, computer video displays or within older TV sets. Here the principles of rectification and amplification were developed in stages with various valve devices and although we now have solid-state components, nonetheless knowledge of the characteristics and limitations of the former assists when the latter are studied further.

The Vacuum Diode A charged plate is placed near the thermionic electron emission source in the valve. One plate is made positive with respect to the electron emitter so electrons are attracted to it and so a space charge does not accumulate. The most basic valve is the vacuum diode, with 2 electrodes sealed in an evacuated glass envelope. One electrode in the shape of a wire is heated by a current flowing through it, resulting in thermionic emission from the wire or filament. The other electrode is a cylinder surrounding the filament, the anode, +ve with respect to the filament or cathode, and attracts emitted electrons.

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Electronic Emission Processes and Devices • 275 Current flows from the anode to the cathode inside the valve as long as the anode is +ve with respect to the cathode. However, if the anode is made −ve with respect to the cathode, emitted electrons are repelled and any current stops. Thus the diode is a unidirectional (one-way) conducting device, i.e. it allows current to flow in one direction only, functioning as a one-way valve. Figures 10.1a and 10.1b show valve construction and its equivalent electrical circuit. (a)

(b)

Cathode

Evacuated glass envelope Anode

Load resistor

Emitted electrons

+ H.T. Battery – – L.T. + battery for heating cathode

Valve pins

▲ Figure 10.1

In practice, the cathode is heated by passing a separate current through the filament electrode or through a heater next to the cathode. Figure 10.2 shows the diode valve’s circuit symbols and its circuit position. Current Directly heated valve

Anode H.T. battery

Cathode L.T. battery

Anode Indirectly heated valve

Cathode

H.T. battery

Heater (energised from separate battery or transformer)

▲ Figure 10.2

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276 • Advanced Electrotechnology For an indirectly heated arrangement, a cathode is composed of a nickel tube sprayed with barium or strontium oxide (materials with a low work function that easily release electrons). The heater wire is looped inside and insulated from it. For A.C. radio sets, a low-voltage heater-transformer energises the heater and any other valves. Figure 10.3 shows the test circuit used to determine diode valve characteristics. Conduction is only possible when the anode is +ve with respect to the cathode and the more +ve the anode, the greater the number of electrons reaching it, i.e. the larger the anode current. However, the anode current cannot increase indefinitely, as there are only a limited number of electrons leaving the cathode every second and will reach a saturation value shown on the characteristic at a point where, when anode voltage is increased beyond a certain value, no increase of anode current is obtained and the curve bends over to the horizontal. To obtain more anode current, cathode temperature must be increased to increase electron emission. Ia (mA)

V

H.T. battery

Anode current Ia

A

4 V on heater

2 V on heater

Anode voltage Va

▲ Figure 10.3

The valve characteristic, i.e. relationship between anode voltage Va and anode current Ia, is obtained by varying the High-Tension (H.T.) voltage applied to the anode, with an adjustable contact on a potentiometer resistor. Plotting Voltmeter and milliammeter readings gives the graph shown. Certain diode factors are derived from the characteristics to help select valves for specific applications. Thus a diode can be used as a rectifier, but its internal resistance in the conducting direction must be known. This resistance value is obtained from the characteristic but it is now important to distinguish between static and dynamic operating conditions. In the circuit (figure 10.3), the valve anode is not connected in series with a load, such as a resistor. The voltage across the anode and cathode is set to a known value during the test, a condition different from that when a valve is loaded or operates dynamically. The second condition is examined and the characteristic considered as (1) static and (2) dynamic.

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Electronic Emission Processes and Devices • 277

Static characteristics Departure from Ohm’s Law in terms of diode behaviour shows that its internal resistance varies with voltage. If a part of the static characteristic, obtained from a test, is examined, it is seen (figure 10.4) that, for any anode current there is a fixed ratio of voltage to current, the anode DC resistance.

P

Anode current Ia

B

O

A Anode voltage Va

▲ Figure 10.4

BP OA or , a value also obtained from the reciprocal PA OB 1 of the slope of the line through the origin, i.e. OB . Point P is the ‘operating point’ and OA the D.C. resistance value relative to this point.

Thus for point P the D.C. resistance=

Although diode operation under A.C. conditions is not considered to any extent, we introduce the term A.C. resistance or slope resistance, whose symbol ra is considered when triode valve operation is described. Assume that the alternator in figure 10.5 generates a sinusoidal voltage of maximum value Em, small compared with the battery H.T. voltage, applied to the diode’s anode. The battery is assumed to offer little resistance to the A.C. voltage and an A.C. current flows superimposed on the D.C. current, so the total current fluctuates between maximum and minimum values. As deduced from figure 10.6, the valve offers a resistance value different to that for D.C. working. If a diode is used in a circuit with a steady D.C. current,

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278 • Advanced Electrotechnology

+

Em

H.T. battery –

Heater supply

▲ Figure 10.5

anode D.C. resistance is used, but if the anode current varies between limits, an A.C. value is used instead. Using the characteristic, note the limits Q and O between which the operating point oscillates. Voltage and current changes are found by drawing triangle XYZ. Hypotenuse ZX is the tangent to the curve at point P. O

Ia

X

P

Q Z

Y

Va

▲ Figure 10.6

⎛ current ⎞ The graph’s slope gives conductance ⎜ but it is more practical to use resistance ⎝ voltage ⎟⎠ or the reciprocal of the slope. Thus the slope’s reciprocal or A.C. resistance value (ra) is given by:

ZY ⎛ voltage ⎞ i.e. . ⎝ current ⎠ XY

So ra = the ratio

δV Small change in anode voltage or ra = a Resulting small cchange in anode current δ Ia

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Electronic Emission Processes and Devices • 279

The sign δ here signifies a small change and in limit is written:

dV Va . dIa

A.C. resistance is less than the D.C. resistance over the working range and example 10.1 shows comparative values. Example 10.1. A diode’s anode current–voltage static characteristic is drawn from the following test values. Deduce the A.C. and D.C. resistance for anode current values of 5 and 10 mA. The A.C. voltage is 2.5 V. Anode current (mA)

0

2

4

6

8

10

Anode voltage (V)

0

30.5

46

58

68

77.5

The Ia/Va characteristic is plotted as shown (figure 10.7).

B

Anode current (mA)

10 8 6 A 4 2 0 10

20

30

40 50 60 Anode voltage (V)

70

80

90

▲ Figure 10.7

Then for the 5 mA value: D.C. resistance: Point A

52.5 5 × 10 −3

5 10

3

Ω or 10.5 kΩ

A.C. resistance: The A.C. voltage of 2.5 V is superimposed on the 52.5 V for a standing current of 5 mA. Draw a small right-angled triangle as shown. Anode voltage varies between 50 and 55 V and the corresponding currents will be 4.6 and 5.5 mA. Or ra =

55 − 50 = 5 55 kΩ (5.5 − 4.6 ) × 10 −3

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280 • Advanced Electrotechnology For the 10 mA value similarly: D.C. resistance: Point B =

77.5 V = 7.75 kΩ 10 mA

A.C. resistance: The A.C. voltage is 2.5 V. For a standing current of 10 mA the corresponding voltages will be 77.5 ± 2.5 = 80 V and 75 V. The currents will be 10.6 and 80 − 75 = 4 55 kΩ. 9.5 mA or ra = (10.5 − 9.5) × 10 −3

Dynamic characteristics – load line A valve is often used with a load which, in simplest form, is a resistor in the anode circuit. Even though the H.T. voltage V is constant, when anode current varies the voltage drop across resistor R will vary and the resulting voltage applied to the anode changes. The valve now operates under dynamic conditions, i.e. with a load and, to investigate the correct circuit conditions, the dynamic characteristic rather than the static characteristic is used. The former is obtained from a test circuit similar to that shown in figure 10.3 except that a load resistor is included and the voltage drop across it is measured at each anode current value.

Stat

ic

A construction method can obtain a dynamic characteristic from a static characteristic. Refer to figure 10.8, let P be a point on the static characteristic. Then Ia = OB and the valve voltage drop equals OA. For the same current, the resistance voltage drop = IaR = VR,

Ia

ic

m

a yn

D

R

P′

P

B VR V

Va

Va O

VR A

Voltage (V)

▲ Figure 10.8

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Electronic Emission Processes and Devices • 281 so the H.T. voltage V = Va + VR and point P obtained on the dynamic characteristic. If the procedure is repeated, the complete characteristic is obtained and then used to find current and voltage values for any H.T. voltage operating condition. There is a separate dynamic characteristic for each value of R and the characteristic approximates to a straight line as load-resistor value increases. LOAD LINE. As shown, the voltage-drop condition in a diode’s anode circuit with a resistor in the circuit is expressed: V = Va + VR or Va = V − VR. As the voltage drop across the resistor is proportional to the anode current, the relationship is a straight-line graph which results from the expression: Va

V − Ia R or Ia =

V Va V Va = − R R R

If the line is superimposed on the static characteristic (figure 10.9), we see a ‘negative slope’, i.e. it slopes in the opposite direction to the valve characteristic because an

30 B

20

ad

Lo e

lin 4 kΩ

Anode current (mA)

25

15

10 ic

nam

Dy

0

ic

at

St

5

10

20

30

40

50

60

70

80

A 90 100 110 120

H.T. voltage (V)

▲ Figure 10.9

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282 • Advanced Electrotechnology increase in Ia results in a decrease in anode voltage. The line cuts the static characteristic to give a point which shows the respective voltage-drop conditions across the valve and resistor for indicated current values. The straight-line graph or ‘load line’ is different for various values of R and is used in the next problem. When the loading condition for one value of R at a specific H.T. voltage is needed, it is not necessary to construct a dynamic characteristic but just to draw the load line. Referring to figure 10.9, point A is obtained by assuming Ia = 0. Under this condition any voltage drop across R is zero and Va is the full H.T. voltage V. Point B is obtained by finding the current which flows if anode voltage is reduced to zero, i.e. if the diode developed a short-circuit between anode and cathode. Under this condition anode V current is given by: Ia = . Example 10.2 shows the full treatment and with figure 10.9, R the two solution methods. Method 1 involves plotting the static characteristic and the load line. Method 2 involves deducing the dynamic characteristic, its values determined from a table. This method is more useful if the operating conditions for various values of H.T. voltage are considered. Example 10.2. Using the test values of example 10.1, deduce the dynamic characteristic for a 4 kΩ load-resistor. Find the circuit current for a H.T. voltage of 110 V and the resistor’s voltage drop. Method 1. If the dynamic characteristic was not required, this method gives the required answer. Consider figure 10.9. Plot the static characteristic and obtain the load line thus: Point A at 110 V, as Ia = 0. Point B =

110 = 27.5 10 -3 = 27.5 mA . Valve resistance is 4 × 10 3

assumed to be zero. The intersection point gives the answer. Circuit current = 9.1 mA. The voltage drop across the resistor = 9.1 × 10−3 × 4 × 103 = 36.4 V. VR = IR = V − Va = 110 − 36.4 = 73.6 V. Reading from the graph: VR = 74 V and Va = 36 V. Method 2. This involves deducing the dynamic characteristic as plotted. Consider one such point with a current of 4 mA, then Va = 46 V and the resistor voltage drop = 4 × 10−3 × 4 × 103 = 16 V. The H.T. voltage = 46 +16 = 62 V, i.e. the value for the dynamic characteristic. The following table is deduced.

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Electronic Emission Processes and Devices • 283

Anode current (mA)

0

2

4

6

8

10

Anode voltage (V)

0

30.5

46

58

68

77.5

Resistor voltage drop (V)

0

8

16

24

32

40

H.T. voltage (V)

0

38.5

62

82

100

117.5

For a H.T. voltage of 110 V, the circuit current is 9.1 mA and the voltage drop across the resistor = 110 − 74 V = 36 V.

The Vacuum Triode We now add a third electrode into the diode assembly and vary its potential. Electron flow is controlled by this electrode which may take the form of an open spiral of wire wound closely round the cathode, an arrangement shown in figure 10.10. This third electrode is called the grid electrode. The circuit symbol for the vacuum triode is also shown. The grid is positioned in the region of the ‘space charge’ and made −ve with respect to the cathode. By altering the grid potential the space charge is modified and the anode current controlled. The principle of a gate electrode is of fundamental consideration to semiconductor device operation, as will be seen in the next chapter.

Grid

Filament or cathode

Anode Directly heated

Indirectly heated

Circuit symbols

▲ Figure 10.10

Consider figure 10.11 which shows the potential gradient between cathode and anode for 3 different grid voltage values. In Curve 1, the grid is only slightly −ve with respect to the cathode. In Curve 2, the −ve grid potential is more negative. In Curve 3, the potential is changed to make the grid very −ve with respect to the cathode. By considering potential gradient, we will deal with electric field intensity, the magnitude of electric force acting on a charge within the field between 2 electrodes. An electron has −ve charge and if emitted from the cathode experiences a force moving it from a point of low to high potential.

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284 • Advanced Electrotechnology + Anode

– Cathode 1

2

3

▲ Figure 10.11

The force magnitude is directly proportional to the potential gradient. If the gradient is +ve, the force is towards the anode and if −ve, the force is towards the cathode. For Curve 1 most of the emitted electrons from the cathode pass the zero potential gradient value and accelerate towards the anode to give a large anode current. For Curve 3, only a few electrons have sufficient velocity to pass the point of zero potential, most return to the cathode. The few electrons which pass the point of zero potential gradient proceed to the anode to give a small anode current. By altering grid potential the anode current is controlled and may be optimised. As the grid is invariably kept at a potential −ve to the cathode, electrons do not reach it and no current is drawn from the grid voltage supply source.

Static characteristics The circuit in figure 10.12 shows how a triode can be tested and anode current controlled by (1) variation of grid voltage and (2) variation of anode voltage. Curves when plotted from the test results divide into 2 static characteristic groups: (1) the Mutual Characteristics, which show the change in anode current Ia for a change in grid

Ia mA A +

Va V Grid – bias battery +

H.T. supply

volts V Vg volts

–

▲ Figure 10.12

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Electronic Emission Processes and Devices • 285 voltage Vg, the anode voltage Va kept constant, and (2) the Anode Characteristics which show the variation of anode current Ia with a change in anode voltage Va, the grid voltage Vg kept constant.

V

00

10

10

5

5

0

0

V

Vg = –5 V

V

Vg = –4

(b) 15 Anode current Ia (mA)

B

V

a

D

=1

50

=2 V

a

V

a

=2

V

50

V

15

Vg = –2

(a)

Vg = 0

MUTUAL OR TRANSFER (Ia/Vg) CHARACTERISTICS. By keeping anode voltage constant, the effect on anode current of a variable grid voltage is investigated by the figure 10.12 circuit. If the results of such tests are plotted a family of curves, as those shown by figure 10.13a, provide information similar to that from the anode characteristics. Manufacturers provide information on particular valve types in the form of such curves allowing 3 coefficients or parameters to be determined, which specify a triode’s circuit behaviour and each parameter, or ‘valve constant,’ represents the relationship between the changes in any 2 variables, the third kept constant.

F

–6

–4

E

–2

Grid potential Vg (volts)

A

100

200

C

300

400

Anode voltage Va (volts)

▲ Figure 10.13

ANODE OR OUTPUT (Ia/Vg) CHARACTERISTICS. By keeping the grid at cathode potential, i.e. Vg = 0, a characteristic curve relating anode current to increasing anode voltage is obtained, as shown in figure 10.13b. If the grid has a small −ve potential and the test repeated, a curve of similar shape is obtained displaced to the right, marked as Vg = −2 V. The displacement is due to the fact that the anode must reach a minimum +ve potential to overcome the repelling force of the grid on emitted electrons. The higher the −ve grid potential, the more the anode characteristic moves to the right. A family of such curves at various −ve grid voltage values is obtained in the manner as shown.

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286 • Advanced Electrotechnology

Valve parameters These are derived from the slopes or change conditions of the graphs and obtained by constructing small triangles as shown. As a straight position of a graph is needed, the sides of a triangle are kept as small as possible and referred to as ‘a small change of’ with the sign δ used as previously. (1) INTERNAL RESISTANCE. As for the diode, the anode characteristic slope, e.g. (figure 10.13b) gives conductance =

AC BC

current . It is more convenient to use valve voltage

AC , other terms are slope or anode resistance, AC internal resistance or A.C. resistance. The symbol ra is used and defined as: ‘Slope or A.C. resistance (ra) is the ratio of the small change of anode voltage to the small change of anode current produced, the grid voltage kept constant’. resistance or the reciprocal of the slope =

Thus ra =

Small change of anode voltage δV or ra = , Va being constant. Resulting small change of aanode current δ Ia

In the limit, mathematically the above is written as: ra =

dV Va . dIa

Triode internal resistance for a particular operating condition is found from either the anode or mutual characteristics and illustrated by the examples which follow. (2) MUTUAL CONDUCTANCE. The influence of the control-grid voltage on anode DF current is determined by the slope of a mutual characteristic curve. For example, EF δ Ia (figure 10.13a) gives the second parameter, , termed the mutual conductance. The δ Va symbol gm is used and defined as: ‘Mutual conductance is the ratio of the small change of anode current to the small change of grid voltage producing it, the anode voltage kept constant’. Thus gm =

δI small change of anode current or gm = a , Va being constant. small change of grid voltage g δ Va

Either anode or mutual characteristic determines gm. Examples show the unit as milliamperes or milliSiemens (mS) but use of per volt or mA/V is common. volts

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Electronic Emission Processes and Devices • 287 (3) AMPLIFICATION FACTOR. This is the third parameter, which represents the maximum theoretical voltage gain obtained from a valve, and is a ratio of 2 voltages without units. The symbol used is μ and defined thus: ‘Amplification factor is the ratio of the change in anode voltage to the change in grid voltage when both have the same anode current change’. Change in anode voltage Change in grid voltage

Thus

μ=

δ Va , Ia being constant. δ Vg

The amplification factor is determined from either the anode or mutual characteristics. A relationship connecting ra, gm and μ is deduced, so the third constant is determined if the other two are known. The examples extend use of the parameters and characteristics.

Parameter relationships The relationship between the 3 valve parameters is now considered. As μ =

δ Va δV δ I equality still exists if written as: μ = a × a δ Vg δ Vg δ Ia

Thus μ =

δ Va δ Ia × giving μ = ra × gm or δ Vg δ Ia

Amplification factor = anode A.C. resistance × mutual conductance. Example 10.3. In a triode, anode current is 5 mA with an anode potential of 220 V and grid potential of −3 V. When anode potential is increased to 260 V, current rises to 7 mA and a change of grid potential to −4 V restores the current to its original value. Determine the valve constants (all 2 significant figures). Here ra =

δ Va ( − ) V 40 × 103 = = = 20 000 Ω or 20 kΩ −3 δ Ia ( − ) × 10 A 2 gm =

δ Ia ( − ) mA = =2m mA/V A/V or 2.0 × 10 -3 A V δ Vg −3 − (− ( )V μ=

δ Va ( − ) V 40 = = = 40 δ Vg −3 − (− ( )V 1

or with the relationship: μ = ra × gm = 20 × 103 × 2 × 10−3 = 40.

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288 • Advanced Electrotechnology Example 10.4. A triode’s static characteristic test values are as follows. Vg = 0 V.

Vg = −3 V.

Then for Va

50

100

150

200

(Volts)

Ia

3

6.3

9.6

13

(milliamperes)

Then for Va

150

200

250

300

(Volts)

Ia

1.4

5

8.5

12.1

(milliamperes)

Plot the anode characteristic and determine: ra, gm and μ (all 2 significant figures).

V –3 g

=

15

V

Anode current (mA)

V

g

=

0V

20

10

5

0

100

200 300 Anode voltage (V)

400

▲ Figure 10.14

Anode characteristics are shown in figure 10.14. With data given in this form, the most convenient solution method is to erect a triangle making contact with an adjacent curve as follows. ra =

δ Va 200280 = = 15 kΩ δ Ia (1325)10 −3 μ=

gm =

δ Ia (1325)10 −3 = = 2.7 mA/V δ Vg 02(23)

δ Va 200 − 80 = = 40. δ Vg 0 − ( −3)

Check μ = ragm = 15 × 103 × 2.667 × 10−3 = 40.

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Electronic Emission Processes and Devices • 289

Ionisation In Volume 6 when ionisation was mentioned, it was stated that if an electron is ‘lost’ from an atom, the latter acquires a +ve charge to become a +ve ion. Ionisation of gas atoms is often accompanied by monochromatic light emission, i.e. a single colour whose wavelength depends on the gas ionised. It is unlike light given off by a tungsten-filament lamp which covers a broad spectrum of visible white light. Gas ionisation is mainly the result of the collision of fast, freely moving electrons with gas atoms. We saw how electrons are emitted from metal surfaces by thermionic emission, and how such electrons travel at high velocity towards an electrode. If the latter is made +ve with respect to the emitter, the higher the p.d. between anode and cathode, the greater the electron velocity. If these collide with gas atoms with velocity sufficient to remove further electrons from the parent gas atoms, ionisation may cause light emission. The extra electrons, subject to the electric field between the electrodes, accelerate towards the anode, colliding with further atoms to maintain ionisation. An electron ‘stream’ makes its way to the anode, with current flowing from anode to cathode. The electronic device, whether a gas-filled valve or discharge tube, depends on ionisation for conduction. When electrons reach the anode, they displace further electrons in the conducting metal of the circuit, which is why a current is drawn from the mains to maintain operation. The +ve gas ions, being heavier than electrons, move slowly to the cathode, combining with electrons around the circuit. The ionised gas current consists of electrons moving from cathode to anode and +ve ions moving from anode to cathode. The +ve ions, moving to the cathode, will neutralise any space charge and, if the voltage across the electrodes is too high, may strike the cathode with velocity sufficient to damage it. Proton mass is some 1850 times bigger than an electron and the heat generated at the cathode by ion bombardment, under normal operation, ensures that the temperature is high enough for electron emission to continue. Let’s now consider the practical applications where electron emission and ionisation feature. The cold-cathode discharge tube was described in Volume 6, Chapter 3 (where electron emission is induced by an electrostatic field and usually from a metal surface into a vacuum). However, many electronic devices, other than semiconductors, rely on thermionic emission. Light given off in most types of discharge is mainly in the Ultra Violet (U.V.) which is invisible, so the tubes or glass envelopes are coated internally with a phosphorescent powder which glows or fluoresces when subject to the U.V. generated through ionisation.

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290 • Advanced Electrotechnology

The hot-cathode discharge lamp (low pressure) Figure 10.15 shows the basic hot-cathode construction. The lamp envelope or tube is long (typically 30 cm to 2.4 m) compared with its diameter (1.5 to 3.8 cm). The tube’s interior is coated with a phosphor, and mounted at each end are small tungsten wire spiral electrodes, usually oxide-coated to enhance electron emission. Guard electrodes may also be fitted. The tube contains small amounts of mercury and argon gas which, under warm working conditions, is some 10−6 bar or 0.1 Pa. Low working pressure allows a lamp to run at a temperature without affecting the phosphor coating. Argon gas helps initiate discharge through the low pressure mercury vapour, and although the white light given off is small, 65% of the input is converted into radiant U.V. energy which acts upon the fluorescent powder to radiate visible light. Supply Supply switch

Choke coil

P.F. correction capacitor Oxide coated filament Tube internally coated

Starter switch

▲ Figure 10.15

At start-up, current passes through both electrodes, heating them, making them suitable for electron emission and reducing the start-up voltage. The current sets up a small potential between the ends of each filament, ionising gas and vapour near the electrodes, assisting the main discharge ‘strike’. When electrodes are hot, the heating current is broken by a starter-switch and a momentary surge voltage (700–1000 V) is set up by a capacitor sufficient to start the main discharge, and then maintained by the normal mains voltage. The coil has two purposes, initially it gives a high-voltage impulse to strike the tube, but then acts as a voltage dropper to maintain the correct p.d. across the tube. ‘Instant-start’ fluorescent lamps are used ashore, but marine systems often have circuits with a starter-switch. Its function is to preheat the lamp electrodes, and apply a striking voltage. When a control-switch is closed, the supply is applied to the circuit and as the starter-switch closes, current flows through the filaments giving sufficient electron emission to start ionisation. At a predetermined

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Electronic Emission Processes and Devices • 291 time, the starter-switch opens automatically to interrupt the main circuit. A collapsing field around the coil induces a high voltage across the tube, causing it to strike. Once the main discharge current flows, the voltage drop across the coil results in the correct p.d. applied across the tube, and a stable running condition results. Lamp efficiency is measured by comparing the lumens out for every watt in. Fluorescent lamp efficiency is typically 40–50 lumens/watt which compared with a tungsten filament lamp is some 14–16 lumens/watt.

The hot-cathode discharge lamp (high pressure) This lamp type is relatively compact and mostly used for outdoor installations on docks, wharfs and for some deck illumination. The commonest hot-cathode discharge lamps are sodium and mercury-vapour, with a sodium lamp illustrated in figure 10.16. When the supply is switched on a 450 V voltage is applied from the secondary winding of a high-reactance transformer connected across the lamp electrodes causing a low discharge through the discharge tube so the neon glows red. Heat produced gradually vaporises the metallic sodium and the sodium vapour ionises to give a bright goldenyellow discharge. A vacuum jacket thermally insulates the discharge tube to improve sodium vaporisation. As the emitted light is monochromatic, a fluorescent coating is not needed. High-pressure mercury-vapour lamps are more often used than sodium lamps to illuminate deck working-spaces and are briefly described in figure 10.16 in conjunction with figure 10.17.

Autotransformer Quartz tube

Electrodes

Sodium Vacuum jacket Supply

▲ Figure 10.16

As a lamp operates at normal mains voltage, no step-up transformer is needed. The inner tube is made of quartz glass to withstand the high operating temperatures and contains mercury with a little neon or argon in an evacuated outer glass envelope

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292 • Advanced Electrotechnology Special 2-pin bayonet or screw cap P.F. correction Choke coil capacitor Quartz tube containing high pressure mercury vapour and argon gas Auxiliary electrode

Resistor Main electrodes Vacuum

▲ Figure 10.17

which may be coated inside with a fluorescent phosphor. There are 2 main electrodes and a starting electrode next to one of them. When the supply is switched on discharge occurs in the inert gas between the main and secondary electrode. The initial discharge ‘triggers’ discharge between the 2 main electrodes by ionising mercury vapour. Once the main discharge strikes, electrodes are kept at their working temperature by bombardment of +ve mercury ions. Having started the lamp, the auxiliary electrode has no further purpose, and current flow through it is minimised by a high ohmic value series resistor. A main circuit stabilising coil also prevents excess build-up of current after the discharge starts. Emitted light is mostly U.V., hence the fluorescent coating. The main electrodes, without a preheating current, are a spiral wire with barium or strontium oxide, which helps increase electron emission when electrodes are heated. The lamp takes several minutes to warm-up, as mercury vaporises and the pressure rises to atmospheric pressure (1 bar). One disadvantage of high working pressures is that a lamp will not restrike immediately after switching off, and must cool down, allowing the pressure to fall, before both discharge can recommence.

The Cathode Ray Oscilloscope One historically important thermionic electron display device is the Cathode Ray Oscilloscope (CRO) developed for a variety of shipborne and land-based applications. In its most basic form, the CRO can measure voltage in magnitude and time. By displaying a luminescent, mobile spot on a fluorescent screen, a deflection trace and/or waveform is both observed and measured. A suitably scaled graticule is often present, and with further amplifiers and transducers, any measurable quantity may be converted into voltage and displayed. Suitable calibration allows the quantity to be measured in

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Electronic Emission Processes and Devices • 293 appropriate units. This device will be described in some detail in Chapter 14; however construction of a CRO or CRT basically consists of 4 distinct parts: (1) A system of electrodes, to produce an electron beam of high velocity and to control the beam intensity. (2) A means of concentrating the beam, by electrodes to sharply focus the spot on screen. (3) A deflecting system, to move a focused beam to any part of the screen. (4) A screen, coated with fluorescent material, emits light when hit by a stream of highvelocity electrons. As with other vacuum electronics devices the whole arrangement is enclosed in an evacuated glass envelope.

Image intensifiers Another very important vacuum electronic device variant in the marine environment is the now popular Image Intensifier. Modern image intensifier systems provide‘Night Vision’ and are used regularly during Bridge Watchkeeping, utilising 2 further electron emitting processes: the Photoelectric effect and Secondary Emission. Modern electro-optic systems incorporating image intensifiers are used by Naval, Coastguard and paramilitary forces in maritime operations. Both thermal and image intensifiers are generically referred to as ‘Night Vision’ but they operate in different ways. Image Intensifiers or Night Vision Goggles (NVGs) amplify low-level visible light and Near Infra Red (NIR) radiation some 20 000–1 million times, but they only amplify light and NIR and so cannot function in total darkness. However, even on relatively dark nights, there is often some light available allowing the device to operate. An image intensifier requires no artificial light source and is often termed a ‘passive’ device. The first proposed imaging tube was fabricated by Holst and De Boer in 1928, leading to the first NIR converter in 1934. During the 1930s the first inverting image intensifier (Generation 0) was developed. The discovery of more effective photocathode materials (caesium-antimonide) led to Generation 1 devices sensitive in the visible (to 650 nm). Generation 2 Micro Channel Plate (MCP) devices used multialkalis (e.g. potassium-caesium-antimonide) and were developed from late Generation 1 systems to provide extended red and reduced blue response. The same technology used to produce optical fibres produced the MCP, a thin glass wafer with an electrode either side with a p.d. of up to 1 kV applied. Generation 3 materials provided higher sensitivity, while recent military Generation 4 developed the first filmless image tubes (which have suffered setbacks due to ion bombardment issues).

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294 • Advanced Electrotechnology

Image intensifier device operation An objective lens projects an image on to a cathode (photocathode) where the Photoelectric effect causes electron emission (figure 10.18). An anode, at about +10 000 V, accelerates electrons to the MCP where electrons are multiplied by Secondary emission, which will be discussed shortly. Many energetic electrons exit the MCP, hit a phosphor screen where light is emitted, producing an image on the screen through the inverse photoelectric effect.

Fibre optic

Fibre optic

Micro Channel Plate

Objective lens

Image

Lens Object Photocathode Focusing anode

Phosphor screen

▲ Figure 10.18 Schematic Diagram of Generation 2 Image Intensifier

Photoelectric Effect Electron emission is explained if light is considered to consist of energy packets or photons. In 1905, Albert Einstein (1879–1955) showed that light could be thought of as discrete quanta, or photons, rather than waves. Based on Max Planck’s theory of black-body radiation, Einstein proposed that photon energy was equal to its frequency multiplied by a constant, later called the Planck’s constant. Einstein states that a photon above a minimum threshold frequency has sufficient energy to eject an electron. This discovery led to the quantum revolution in physics and rightly earned Einstein the Nobel Prize in Physics in 1921. In the Photoelectric Effect electrons are emitted from a metal surface when a metal absorbs energy from electromagnetic radiation of high-frequency radiation, acquiring more energy than the work function (the electron binding energy) of the material, and is ejected. However, if photon energy is too low, electrons cannot escape. The photoelectric effect requires photons with energies above a few electron volts in elements of high atomic number. Electrons emitted in this way are called photoelectrons and they were first observed by Heinrich Hertz (1857–1894) in 1887. Understanding the photoelectric effect led to the conceptual development

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Electronic Emission Processes and Devices • 295 of the quantum nature of light and electrons, influencing ideas such as wave-particle duality. Photon energy (E) is given by E = hf, where ‘h’ is Planck’s constant (6.6 × 10–34 Js) and ‘f’ is the wave frequency. High-frequency (short wavelength) photons carry the most energy, and so if the photon energy is more than the photocathode’s Work Function (φ), a material parameter, it can eject electrons and give them some kinetic energy, i.e. E = hf = φ + (½ mv2)max where ‘m’ is the mass of the electron and ‘v’ is its speed. As photon frequency decreases, so does the emitted photon energy and also its Kinetic Energy (K.E.). Eventually a frequency is reached where an electron is just emitted with zero K.E. This frequency, f0, is the threshold frequency and given by: E0 = hf0 = φ A typical image intensifier gathers low-level visible and NIR light through the objective lens which is focused onto the pixelated surface of the input fibre optic window. Light entering a particular fibre optic stays within the fibre until it reaches the photocathode surface where if it has sufficient energy an electron will be emitted. Emitted electrons are emitted which are accelerated towards the anode but enter the MCP which as discussed produces a shower or avalanche of electrons which exit the pixelated discrete channels of the MCP. These electrons then strike an output phosphor screen to produce visible light (usually green as the human eye is most sensitive to green light).

Secondary Emission Some materials when bombarded by ‘primary’ electrons successfully emit several ‘secondary’ electrons. An MCP usually has several million small channels parallel to each other between the plate surfaces (figure 10.19). Each channel has a diameter of only a few microns and releases several electrons for every electron strike upon it. In this way each channel acts as a separate electron multiplier. When the alternating power supply is reversed the emitted electrons are accelerated and provide further secondary emission through subsequent ‘strikes’. Thus 1 electron rapidly becomes first 3, and then 9, 27, and so on. A typical MCP may be composed of over 5 million individual narrow collimated channels which helps to reduce image blurring due to bright light sources. Modern MCP Image Intensifiers and photomultipliers use secondary emission, which limits the spreading out of a bright light source across a screen, a big problem for early

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296 • Advanced Electrotechnology Power supply – +

Channels Secondary electrons Phosphor screen Primary electron

Output electrons × 10 000 Glass structure

▲ Figure 10.19 Micro Channel Plate

3-stage Cascade Image Intensifiers. It is the benefit of many separate channels that improves the spatial resolution. Individual channel walls have a limited secondary electron supply so if they accept large numbers of ‘primary’ electrons they saturate, but do not affect other nearby channels. Image intensifiers have several advantages, first the photocathode is sensitive to visible and NIR radiation so these Night Vision devices can use the NIR present in the night’s sky. Secondly they are small and lightweight, and being ‘passive’ require only limited battery operation. A green phosphor is used because the human eye is most sensitive to green. However they do have some disadvantages, being reliant upon visible and NIR, they cannot ‘see’ through fog, mist or smoke. They also cannot see without some visible light or NIR present. Nonetheless, they have found wide-spread applications for night observation, driving sights and flying goggles. They can be readily fitted to (TeleVision) TV cameras or low light level Closed Circuit TV (CCTV) systems which makes them ideal for use in dock and wharf surveillance.

Practice Examples 10.1 A diode valve’s anode power dissipation is 336 mW. The anode to cathode voltage is 160 V. Find the anode current (2 significant figures). 10.2 When a 58 V anode voltage is applied to a 2-electrode valve the anode current is 6 mA. Calculate the dissipated anode power (2 significant figures).

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Electronic Emission Processes and Devices • 297 10.3

When the linear part of the Ia/Va characteristic of a diode valve is examined, the slope is 1.64 mA/8.13 V. Find the valve’s A.C. resistance (2 significant figures).

10.4

The Ia/Va characteristic of a diode valve shows an increase of anode voltage from 75 to 129 V and current from 12 to 22 mA. Find the valve A.C. resistance and the D.C. resistance for each voltage condition (2 significant figures).

10.5

The Ia/Va curve of a diode valve is given by: Anode voltage (V) Anode current (mA)

5

10

15

20

25

27.5

0.6

2.0

4.2

7.25

10.5

12

Plot the curve and find the A.C. resistance over the straight line region (2 significant figures). 10.6

A vacuum diode has the following forward characteristics: Anode–cathode voltage (V)

0

10

20

30

40

50

Anode current (mA)

0

25

60

100

145

200

The diode is placed in series with a 300 Ω resistor load and the combination fed from a 60 V D.C. supply, connected to make the diode conduct. Determine the current flowing and the power dissipated in the load resistor (2 significant figures both). 10.7

The following readings were obtained from the linear portions of a triode valve’s static characteristics. Va (volts)

120

120

80

Vg (volts)

–1.3

–3.8

–1.3

10

4

6.2

Ia (mA)

Find the valve’s A.C. resistance, mutual conductance and amplification factor (all 2 significant figures). 10.8

Explain how the photoelectric effect and secondary emission are used in a practical modern MCP Image Intensifier.

10.9

If the work function of a surface of barium on barium oxide is 2.22 eV what frequency will just cause photoelectric electron emission to take place in a vacuum (2 decimal places).

10.10 If an electron is emitted from a photocathode having a work function of 2.1 eV what will be the speed of the electron near the metal surface if the surface is illuminated by radiation of wavelength 200 nm (4 significant places)? Planck’s constant = 6.6 × 10‒34 Js.

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11

SOLID-STATE ELECTRONICS: THE DIODE An amplifier using semiconductors rather than vacuum is in principle possible. William Shockley (1939)

Semiconductors This chapter introduces the semiconductor diode, a device comparable in function to the thermionic diode in the electronic circuits as discussed in Chapter 10. Semiconductor materials have properties allowing their use for many device applications in response to temperature, pressure, stress, light, magnetic and electric fields. As such they are used frequently as sensors and transducers. The term ‘semiconductor’ includes all solids whose electrical conductivity lies between that of metals and insulators. Germanium and silicon, and gallium arsenide are perhaps the most important semiconductor materials of current working device choice, but graphene, a substance made from pure carbon, with atoms arranged in a regular hexagonal pattern similar to graphite but only one atom thick, may provide novel new commercial devices in the near future. However, before investigating the electrical properties of current materials such as germanium and silicon, it is worth recalling the history of semiconductor development in electrical work. The electrical properties of semiconductor materials will then be examined, without which diode and transistor operation will not be understood.

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Solid-State Electronics: The Diode • 299 ‘Solid-state’ devices, as their name might suggest, are not vacuum devices like the thermionic valve, and it is their ability to control conductivity without the need for bulky yet delicate glass bulbs that make solid-state devices so attractive for most practical purposes. The rectifying properties of the diode had been grounded in the fundamental work of the late nineteenth century onwards also providing the advent of radio through the thermionic valve and the rectifying crystal. In 1833, Michael Faraday discovered that silver sulphide had a negative temperature coefficient of resistance. Austrian physicist Braun (1874) investigated the resistance properties of various contacts between metals and their oxides, using a wire as a point contact, and noted that resistance depended on the polarity of the applied voltage. He also observed the rectification effects of selenium. Point contact rectifiers were used for a considerable period, in fact they were used in the early days of wireless up to the 1920s. The rectifying crystal, a natural semiconductor diode, was later replaced by the valve which in time became the triode and used for detection, rectification and amplification. The crystal also became popular as a detector in the early days of broadcasting because of its cheapness and the device was often called the cats-whisker. Use of the crystal lapsed once more as valves became widely available and were adapted for ‘mains’ use, but the crystal was redeveloped eventually for frequency stability use in radar. Various solid-state diodes have now been designed that perform the same required functions of early valve devices such as multi-stage A.C. amplifiers and multi-electrode valves (such as the tetrode and pentode valves) but no recourse to these will be made here as this technology is now largely obsolete and of specialist historical interest only. Simple metal or plate rectifiers, made of new semiconductor layers such as cuprous oxide or selenium bonded between metal electrodes, were introduced into industry between the 1920s and late 1930s even though little was known about their functioning theory at this time. With the advent of World War II, the silicon point-contact detector was developed for radar and other ‘spin-offs’ included the discovery of the properties of germanium and the improved crystal sensitivity arising from ‘doping’ a crystal. By the end of the war, research into semiconductors was well advanced, and the junction diode was demonstrated in 1941. To-date the most remarkable achievements have been in the fields of electronic equipment, power control devices and in miniaturisation of electronics, notably valued in the Apollo Space Programme in order to minimise the platform’s overall launch mass. To understand the action of a semiconductor, diode reference is once again made to electron theory, and the concept of electric current as an electron flow. The fundamental concepts are summarised as an introduction here to basic semiconductor theory.

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300 • Advanced Electrotechnology

Basic theory The atom of a material consists of a nucleus (+ve charge) and planetary electrons (–ve charge). The nucleus has nearly all the mass of the atom and consists of both protons and neutrons with no electrical properties, but as forces of repulsion exist between +ve charged protons, the neutrons have the function of ‘holding’ the nucleus together. An atom contains equal numbers of protons and electrons with charges cancelling to make the atom neutral overall. Atoms of various elements contain different numbers of protons, neutrons and electrons and the greater the number of protons and neutrons in the nucleus, the greater the density of the substance. Electrons may be considered to move in planetary orbits and for the larger atoms, these electrons are arranged in various shells. The atom’s chemical properties may be explained in terms of the number and grouping of the planetary electrons. These ‘quantum shells’ are considered for simplicity to be concentric and 7 in number, distinguished by the sequential letters: K, L, M, N, O, P and Q respectively. The K shell is the closest to the nucleus and the maximum number of electrons possible in each of these shells is 2, 8, 18, 32, 18, 13 and 2 respectively. As germanium is of special interest here it is noted that its 32 electrons are arranged in the K, L, M and N shells thus 2, 8, 18 and 4. Silicon, also of semiconductor diode interest, has 14 electrons arranged in the K, L and M shells thus 2, 8 and 4. A solid in its smallest particle form consists of many crystals joined together and built up from a regular structure of atoms which repeats itself continually to form a lattice. Electrons in the innermost shells of an atom do not play a part in the crystal lattice structure, but those in the outer shells are important, as they decide the atom’s electrical and chemical properties and are known as ‘valence electrons’. The chemical nature of an atom is revealed by the way the atom combines with other atoms to form molecules. Valency is a chemical term and is explained as ‘the property of the atom of an element which enables it to enter into chemical combination with another atom’. Thus when the molecules of a substance are formed from 2 different atoms, this is done by the electrons in the unfilled external quantum shells or valence electrons. The valency of an element is determined by the electron relationship between these 2 outermost shells. Note that 2 shells are involved because a combination of atoms is achieved by the gain or loss of electrons, so it appears that all the outermost shells are complete. The planetary electrons of an atom can thus be divided into 2 classes: (1) valency electrons, those available for linking atom to atom and (2) the core, those which retain their configurations or orbits in all compounds of the element, which are unaffected by the formation of bonds between atoms.

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Solid-State Electronics: The Diode • 301 CO-VALENT BONDING. One way in which an atom may combine with other atoms and bring about a change in the number of electrons in its outer valency shells is by co-valent bonding. A co-valent bond is the sharing of a pair of electrons by 2 atoms, each atom contributing one electron to form a shared pair with a consequent strong bond between them. Consider hydrogen which has one planetary or valence electron. The K shell electrons should be 2. Thus the hydrogen molecule contains 2 hydrogen atoms, with the nuclei linked by the valency electrons to form a pair, as illustrated by figure 11.1a. (a)

Hydrogen nucleus –

– H

H

H –

H – One molecule

Atoms (b) Germanium nucleus Ge –

–

–

–

Ge

Ge

Ge

–

– –

–

Ge

▲ Figure 11.1

Consider germanium whose outermost shell contains 4 electrons. To obtain stability, the atoms of the element build up into a molecular structure giving combinations as shown by figure 11.1b. It is seen that one valence electron of the 4 goes to make up a ‘shared pair’ between 2 adjacent atoms. Shared pairs of electrons can be depicted by

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302 • Advanced Electrotechnology double lines as shown by figure 11.2a or alternatively the crystal structure which leads to a regular arrangement of atoms throughout the lattice (in 3 dimensions), represented by figure 11.2b for germanium.

(a) Ge –

– –

– Ge

Ge

Ge –

– –

– Ge

(b) –

–

–

– Ge –

– Ge –

– Ge –

–

–

–

–

–

–

– Ge –

– Ge –

– Ge –

–

–

–

–

–

–

– Ge –

– Ge –

– Ge –

–

–

–

▲ Figure 11.2

Conduction control Pure germanium and silicon have crystalline structures and their atoms arranged in a crystal lattice are depicted in figure 11.2b, which show these materials to be similar to carbon in its diamond form. The arrangement is frequently called a ‘diamond lattice’ and it is seen that there are no free electrons except for any freed by application of heat or light, this effect, termed ionisation, is considered later. The conductivity of pure germanium, e.g. is poor except when heated, but is improved by injecting an impurity.

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Solid-State Electronics: The Diode • 303 For semiconductors conduction is a function of (a) temperature, and (b) impurity content, which is very important, and an increase of both (a) and (b) which increases conductivity will now be considered. INTRINSIC CONDUCTIVITY. Near –273°C (Absolute Zero) pure germanium has no free electrons as all the valence electrons are firmly attached to their respective atoms. If the crystal is supplied with energy, such as heat or light, bound electrons can then absorb energy and escape from the atom. Thus electrons leave the atoms and move freely in the crystal, increasing conductivity as more energy is supplied. The places from which the electrons leave are known as holes. If a crystal is subjected to a potential gradient, a current will flow, with the electrons and holes acting as charge carriers. Holes are regarded as +ve charge carriers moving towards the negative terminal (or cathode) while electrons, being –ve charge carriers, move towards the positive terminal (or anode). Thus although electrons move in the opposite direction to conventional current flow, hole movement is in the same direction as conventional current flow. In summary, intrinsic conductivity is associated with the movement of equal numbers of oppositely charged carriers. IMPURITY (EXTRINSIC) CONDUCTIVITY. As already stated, the conductivity of a pure semiconductor material can be controlled by the addition of a known impurity such as antimony or indium. The process is known as ‘doping’ and gives a result which can be considered under one of the two following headings. N-type germanium. When a pentavalent atom (an atom with 5 valence electrons), such as antimony or arsenic, is introduced into a crystal lattice of pure germanium or silicon, one impurity atom electron is free to become a –ve charge carrier as shown in figure 11.3b. The pentavalent atom, when bonding with germanium, only uses 4 valence electrons and so one impurity atom electron is not held by covalent bonding. This ‘excess’ electron has sufficient energy to migrate through the lattice structure as a charge carrier. The impurity atom is considered as a donor of electrons or an N-type impurity since it provides an electron. The adulterated or doped germanium is called N-type germanium. P-type germanium. If a trivalent impurity atom (an atom with 3 valence electrons) such as indium or aluminium is introduced into the crystal lattice, bonding is imperfect in that a ‘hole’ is left in the lattice. The term +ve hole is given to a deficiency created in the bonding arrangement, since the valence electrons of the impurity atom can only pair up with 3 of the electrons in its adjacent 4 germanium atoms. A hole exerts an attractive force on any electron which has been liberated from a germanium atom by ionisation and will ‘capture’ it, as it happens to pass near the hole. Thus, as an electron moves from the orbit of 1 germanium atom to fill a deficiency in another, it leaves a similar ‘hole’ in its original orbit. The hole will appear to move in the opposite direction to the

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304 • Advanced Electrotechnology (b)

(a) – –

– –

–

–

–

–

– –

–

–

– Pentavalent antimony – – atom

– –

–

– –

– –

–

–

–

–

–

– –

–

–

– –

–

–

– –

–

–

–

–

–

Trivalent indium atom –

–

– – – – –

– –

–

–

– –

–

–

Hole +

– –

– –

–

Free electron –

–

–

–

–

– –

–

– –

▲ Figure 11.3

electron and as the process is continuous throughout the germanium it constitutes an effective flow of +ve charge carriers in the direction of conventional current flow. The arrangement is shown in figure 11.3b. A ‘hole’ has the characteristics of a +ve charge and can be regarded as a free, mobile, +ve charge. Trivalent impurity atoms are called acceptors, as they accept electrons from covalent bonds and in so doing create holes or +ve charge carriers, which are also termed P-type impurities and the doped germanium, containing such an acceptor impurity, is called P-type germanium. Silicon will behave in a similar way if doped with either pentavalent or trivalent impurity atoms. IONISATION. Electrons can be liberated by the action of heat, light or other radiation, through ionisation. The total charge in a particular material is always zero, since for every charge carrier produced by ‘doping’ or ionisation, there is an incomplete atom or ion carrying a charge of opposite sign remaining. However, some dissociation or breakdown of the covalent bonds will occur for both P- and N-type materials, even at room temperature, and so a few holes may occur in N-type material along with an equal number of extra free electrons. Free electrons will predominate in N-type material and are called the majority carriers, while holes are known as the minority carriers. Similarly for P-type material, holes will be the majority carriers and any free electrons are similarly the minority carriers.

The P-N Junction The junction diode consists in its simplest form as a section of P-type material and N-type material grown together to form a continuous crystal. At the junction electrons

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Solid-State Electronics: The Diode • 305 from the N-type region diffuse across into the P-type region to combine with the holes neutralising them. Similarly some holes from the P-type region diffuse across the junction into the N-type region to combine with the electrons. If the overall current I is made up of two components, a forward current IF and a reverse current IR, we may say that: I = IF − IR = 0. This action results in a narrow depletion layer on either side of the junction and due to both sides losing some of their majority carriers, the P-side will exhibit a slightly –ve charge while the N-side will exhibit a slightly +ve charge. The +ve charge of the N-side relative to the P-side tends to attract electrons so migration ceases and does not continue indefinitely. Similarly the −vе charge of the P-side relative to the N-side attracts holes back also to limit migration. The final charge distribution results in a potential barrier set up across the junction which is likened to a battery connected in the sense as shown (figure 11.4). The potential barrier is normally referred to as the Junction barrier. –

+

P type –

+

–

+

N type –

– – +

– – – + – + – – – – – + + + – – – – – + +

– – – – – – – – –

+ + + + + + + + +

+ + + + + + + + +

+

+ +

+

Donor atoms +vely charged

–

Acceptor atoms –vely charged

– Electrons + Holes

–

+

+

– +

–

+

IF IR

+

–

–

–

–

–

+

–

I = IF – IR = 0

Potential distribution Across barrier

▲ Figure 11.4

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306 • Advanced Electrotechnology In summary for the arrangement shown, few charges can cross the barrier unless they have an energy level high enough to overcome the junction barrier’s potential. The +ve charge of the N-type will however attract the minority carrier electrons from the P-type side and the –ve charge will attract the minority carrier +ve holes from the P-type side. The resulting current I = IF – IR will be zero and since there is no external voltage applied the junction will be in equilibrium with charges and potentials as shown. Semiconductor diode behaviour will be investigated for the condition of an external electromotive force or applied electric potential.

The junction diode FORWARD BIAS (GOOD CONDUCTION). If a battery is now connected across the junction diode with opposite polarity to that of the barrier, the barrier potential is reduced, the effective resistance decreases and a current of several milliamperes is able to flow. A forward voltage of about 0.2 V can produce 4 to 5 mA. The barrier potential having been reduced, the majority carriers in the P-type material are reinforced by the +ve potential and the forward current increases, overcoming the reverse current. The forward current is thus considerably larger than the reverse current or I = IF – IR. The conditions being discussed are shown by figure 11.5. +

–

–

+

+

– – – – – –

–

+ + + + + + IF

IR

I = IF – IR (large value)

Potential distribution

▲ Figure 11.5

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Solid-State Electronics: The Diode • 307 REVERSE BIAS (VERY POOR CONDUCTION). If the external battery is now reversed, i.e. connected across the P-N junction so as to assist or add to the existing barrier potential, the latter will be increased. As shown in figure 11.6 the action increases the effective resistance of the junction considered to ‘reverse’ or to be ‘back biased’ by the battery. The reverse current is of the order of a few microamperes (millionths of an ampere) due to electrons released by ionisation, i.e. ‘intrinsic conductivity’. For this reverse-bias connection, the minority carriers are reinforced and the reverse current, although small, dominates, remaining substantially constant over a wide range of applied reverse potentials.

– – – – – –

–

+

–

+

– – – – – –

+

+ + + + + + + + + + + + +

I = IR (small constant value) IF = 0

Potential distribution

▲ Figure 11.6

At a large reverse voltage a breakdown region occurs where the reverse-biased P-N junction will conduct, this value being known as the Peak Inverse Voltage (P.I.V.) which may be very large. There are two causes of this sudden conduction: the avalanche effect and the zener effect. The avalanche effect occurs when minority carriers, travelling at high velocity through the junction’s barrier layer, ‘displace’ electrons from the parent atoms which, in turn, displace further electrons from their atoms. The zener effect, however, is caused by the potential difference (or electric field) applied across the

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308 • Advanced Electrotechnology narrow barrier layer becoming large enough to force electrons away from their parent atoms. Breakdown is not necessarily permanent and if the voltage is reduced below the P.I.V. the reverse-biased diode will operate as previously discussed. However, if the applied potential greatly exceeds the P.I.V. irreparable breakdown of the device may occur. When a diode is used as a one-way conduction device it is vital to pay attention to the diode’s maximum circuit voltage and the P.I.V. rating for this reason just as one needs to consider the current rating of a fuse. An explanation of what is happening in the device is possible if we consider one side of the junction to have only holes as charge carriers and the other side to have only electrons. The arrangement is shown in figure 11.7. If a potential is applied so the P-type end is +ve with respect to the N-type end, then a current due to this potential consists of either electrons moving right to left or holes moving left to right. Since the N-type side contains free electrons and the P-type side contains holes, current will readily flow across the junction. If the applied voltage is reversed, the current direction would occur either by electrons moving from left to right (and there are no electrons available), or by holes moving from right to left (similarly there are no holes available) and so no current can flow. + – + + + – + + + – + + + – P Yes

– + – – – – – – N

+ + + – + + + – + + + – P No

– – – – – – N

▲ Figure 11.7

DIODE CHARACTERISTIC. If a test circuit is arranged, as in figure 11.8a, and the results of applied voltage plotted against diode current for increasing current values, the semiconductor junction is observed to act like a thermionic diode when the latter conducts. The general shape of the current/voltage characteristic is shown by figure 11.8, it resembles the vacuum diode, but it is noted that when the applied voltage is reversed, a minute reverse current will continue to flow. This reverse current is due to the release of electrons by ionisation, i.e. thermal agitation which gives the P-type side a few electrons and the N-type side a few +ve holes. If the reverse voltage becomes too large, there is a breakdown of the covalent bond structure and the current will increase very rapidly. For a semiconductor diode, the electrode to be connected to the +ve terminal of the supply for forward conduction is clearly marked for this reason and must be connected correctly. As for a diode valve the terms anode and cathode are used accordingly.

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Solid-State Electronics: The Diode • 309 (a) A

+

V

– (b) mA

20

10

100

50

0.5

–V

1.0 V+

50

µA

100

▲ Figure 11.8

The characteristic shown is typical of a small semiconductor diode. Note the change of current scale for the reverse graph, which shows the limit where the reverse current saturation value is broken down, where the ‘zener effect’ occurs, and the semiconductor becomes a conductor. Unlike the vacuum diode the semiconductor diode reverse operation is exploited for various circuits and the zener diode, as it is called, is used for voltage regulation and stabilisation, for meter protection, cut-off, limiting and in clipping circuits and will be considered later.

Rectifier operation Since the junction diode requires no heater supply, it is easily introduced directly into circuit arrangements to allow both half-wave and full-wave rectification. The characteristic shows the device is suitable for a rectifier and if an alternating voltage of

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310 • Advanced Electrotechnology about 1 V (peak to peak) is applied to a P-N junction, the potential barrier will alternatively increase and decrease over time to allow rectification. The reverse current is considered negligible. Silicon is used in preference to germanium for most large power rectifiers as it can carry large currents and operate at higher temperatures and its reverse current is also lower than that for germanium with similar forward current values. The advantages of semiconductor rectifiers over earlier ‘metal’ rectifiers such as selenium lie mostly in their smaller size, longer life and improved regulation and efficiency. The latter features are the result of the low forward resistance and voltage drop and the fact that fewer elements are required in series to handle a given voltage. Example 11.1. The following values refer to a germanium diode. Forward current (mA)

0

0.1

0.18 0.22

0.4

0.6

0.8

1.1 1.6

2.3 3.45

Forward voltage (mV)

100

140

160

180

200

220

240 260 280 300

Reverse current (μA)

–0.2

–0.3

–0.4

–0.4

Reverse voltage (mV)

-200 –400 –600 –800

320

Plot the anode characteristics for the diode and determine from it (a) the ‘forward’ D.C. resistance when the current is 3 mA and (b) the forward anode voltage when the D.C. resistance is 200 Ω. (a) From the graph of figure 11.9 the forward D.C. resistance =

OB 315 × 10 −3 = = 105 Ω OA 3 10 −3

(b) Construct the 200 Ω D.C. resistance lines as shown. Current = 1 mA. The applied voltage will be 1 × 200 = 200 mV. Plot this point P. Draw the voltage drop line to cut the curve. Hence the forward anode voltage will be 272 mV as illustrated. STATIC AND DYNAMIC OPERATION. It is seen from the Ia/Va curve that, for the thermionic diode there was a definite voltage to current ratio for any anode current value, and it is applied here to the solid-state semiconductor junction diode. This is the anode D.C. resistance, determined for any point on the characteristic by dividing the anode-voltage value by the corresponding anode-current value. Operation of the semiconductor diode for A.C. conditions, as when used as an RF demodulator, will not be considered here. The slope or A.C. resistance value for any working range can be determined, as for the vacuum diode, by the ratio: Small change in anode voltage Resulting small change in anode current

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Solid-State Electronics: The Diode • 311

Forward (µA)

4

3

A

Current

2

P

1

Reverse (mV)

0 Reverse (µA)

500

100

200

Voltage

B 300 272

Forward (mV)

▲ Figure 11.9

When used as a rectifier, the solid-state semiconductor diode is loaded or operated dynamically. It is apparent that a dynamic characteristic will be obtained either by direct testing or deduction in a manner similar to that used for the encapsulated diode valve as discussed in Chapter 10. As a separate dynamic characteristic is required for each value of R, the load resistance using a ‘load line’ is more appropriate for solving some problems, as illustrated by the following example. Example 11.2. The characteristic of a germanium diode is shown by figure 11.10. If the value of load resistance = 100 Ω and the average value of applied voltage = 2 V, find the average value of the terminal voltage (2 decimal places) and the corresponding load current (4 significant figures). The characteristic is plotted and the load line drawn for 100 Ω thus: With no current flow the full 2 V is applied across the diode and point A obtained. With a diode D.C. resistance of zero, the maximum current which could possibly flow will be 2 = 20 mA . Thus point B is obtained. The point of intersection between the load line 100 and characteristic shows that the terminal voltage will be about 2 – 0.48 V = 1.52 V and the load current will be 15.25 mA.

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312 • Advanced Electrotechnology

20

B

15.25 mA

Current (mA)

15

10

5

Terminal voltage 0

0.48 V 0.5

P

1.52 V 1.0

1.5

2.0

A

Voltage (V)

▲ Figure 11.10

Rectifier Circuits Most electronic circuits operated from a D.C. supply will likely obtain their voltage supply from batteries while at sea. However, as batteries discharge, continuity of voltage supply cannot be assured and reliability of the equipment falls. Rectifier circuits, using the unidirectional conduction properties of junction diodes, are often incorporated in electronic units which are to be connected to A.C. supplies. The rectified output can thus be either half or full-wave arrangements, which are now considered in turn.

Half-wave When a diode is connected in series with an A.C. supply and resistive load, current is seen to flow in only one direction. Each alternate A.C. half cycle is thus ‘blocked’ and a

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Solid-State Electronics: The Diode • 313 series of positive ‘pulses’ occurs as shown in the output waveform (figure 11.11). A small reverse current will flow but for most practical purposes this may be ignored. If a lowvoltage D.C. output is required from a mains A.C. supply, a transformer with a suitable step-down ratio is incorporated with the rectifier circuit connected to the secondary output.

+ A.C. supply

D.C. load

V –

Input waveform

Output waveform

▲ Figure 11.11

Full-wave (1) BI-PHASE CIRCUIT. With the arrangement shown in figure 11.12 using 2 diodes and a centrally tapped transformer, the full A.C. waveform will conduct in a positive direction. During the alternate half cycles each diode will conduct in turn producing an output as a series of unidirectional pulses over each half-cycle. D1 V A.C.

–

+

D1 D2 D1 D2 D1

D.C. load Input waveform

D2

Output waveform

▲ Figure 11.12

(2) BRIDGE CIRCUIT. The full-wave bridge rectifier circuit illustrated in figure 11.13 eliminates the need for a centrally tapped transformer but requires 4 diodes. During each half cycle alternate pairs of diodes conduct once again producing a series of unidirectional pulses. The alternating nature of the unidirectional output to the load is undesirable for most electronic circuits. To reduce this problem, smoothing circuits are generally incorporated. The simplest method connects a capacitor in parallel with the D.C. load (figure 11.14). However, the capacitor must have a high capacitance value and is usually of the electrolytic type as discussed in Volume 6.

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314 • Advanced Electrotechnology

A.C.

+ D.C. load

Input waveform

Output waveform

–

▲ Figure 11.13

+ A.C. supply

D.C. load

C –

▲ Figure 11.14

During the half cycle in which a diode conducts, capacitor C will charge up to the peak value of the D.C. load voltage. During the next half cycle, when the diode is not conducting, the capacitor discharges through the load. Mention was made in Volume 6, Chapter 8, that the charge stored by a capacitor depends upon both the voltage and the capacitance values. Hence, for a given D.C. voltage, if a much larger capacitor is used, more charge can be stored to maintain the load voltage. Figure 11.15 shows the output waveform for such an arrangement. Capacitor discharging

V

Capacitor charging Output voltage

Unsmoothed output

Time

▲ Figure 11.15

The output voltage has a definite ripple, the ripple voltage, but the ripple is reduced with the use of increasingly higher capacitance values. A capacitor used in this way is sometimes referred to as a storage or reservoir capacitor. The time taken to fully charge a capacitor is still very small and decreases as the capacitor value increases. Thus for practical purposes, it is assumed to charge up instantaneously. In a half-wave rectifier circuit the capacitor may be assumed to discharge in the time taken for a complete

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Solid-State Electronics: The Diode • 315 1⎞ ⎛ cycle t = . Similarly with a full-wave rectifier the circuit discharges in only half this ⎝ f⎠ 1 ⎛ ⎞ as shown (figure 11.16). time t = ⎝ 2f ⎠

V Ripple voltage

Time 1 t=— 2f

▲ Figure 11.16

Example 11.3. A full-wave rectifier with capacitor smoothing, operating from a 50 Hz A.C. supply, produces a D.C. output of maximum value 60 V. The capacitor has a value of 2000 μF. Find the peak to peak value of the ripple voltage when delivering 250 mA to a resistive load (2 decimal places). Quantity of electricity stored during charge Q = CVr Quantity of electricity given out during discharge Q = It, i.e. CVr = It I = Discharge current = 250 mA t = Discharge time = Ripple voltage Vr =

1 1 = = 0.01 01 s 2f 2 × 50

250 × 10 −3 × 0 01 ∴ Ripple voltage = 1.25 V 2000 × 10 −6

In many cases this ripple does not actually affect the load operation but in some electronic circuits its presence is undesirable and should be eliminated. Connecting a capacitor across the load will affect the P.I.V. developed across the diode, because when the diode is in a non-conducting mode, the voltage developed across it is the peak value of the supply PLUS the peak value of the capacitor voltage.

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316 • Advanced Electrotechnology Example 11.4. A 240/20 V transformer feeds a half-wave rectifier circuit which has capacitor smoothing. Calculate the maximum output voltage and the diode’s minimum P.I.V. rating. Note. A.C. voltages are expressed in r.m.s. values: i.e. VRMS

Vm =

Vm

Vm

2

VRMS = 2 × 20

Output voltage Vm = 28.28 V P.I.V. = Peak value of supply + Peak value on capacitor (which just before discharge when the diode is not conducting is the peak value of the D.C. load) P.I.V. = 28.28 + 28.28 volts = 56.56 V Diode P.I.V. rating = 56.56 V minimum

Filter Circuit The storage or reservoir capacitor may not completely eliminate ripple and, if complete ripple suppression is needed, another capacitor and an iron cored inductor should be connected as shown in figure 11.17. L + Rectified unsmoothed D.C.

C1

Steady D.C. output

C2 –

▲ Figure 11.17

Capacitor C1 acts as a smoothing capacitor, which produces a unidirectional potential difference across it with a definite ripple. Inductor L and capacitor C2 act as a filter for the circuit, separating out the steady D.C. voltage from the ripple A.C. component of voltage across C1. This occurs because the inductor L offers greater impedance to the A.C. component but little resistance to the D.C. component, so most of the unwanted ripple voltage is developed across it. As the capacitor resistance is very high the D.C. component is established between its plates. A small D.C. voltage drop will occur across the inductor L due to its resistance.

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Solid-State Electronics: The Diode • 317

Voltage Doubler Circuit Another circuit using the storage charge capacity of a capacitor together with semiconductor diodes is the voltage doubler circuit, an arrangement used in circuits requiring a high voltage, but with low D.C. current from a lower voltage A.C. supply. The circuit requires 2 diodes and 2 capacitors connected as shown in figures 11.18a and 11.18b. (b)

(a)

+

– 2Vp

V1

V2

C1

C2

+

– 2Vp

▲ Figure 11.18

During the positive half cycle of A.C. voltage, C1 charges up to the peak value of the A.C. voltage Vp. During the negative half cycle C2 charges and C1 discharges. However, the voltage now developed across the load comprises both the discharge voltage of C1 and the voltage V2 (i.e. VL = V2 + VC1 = 2Vp). During the next positive half cycle C2 discharges and C1 charges, so the D.C. load voltage this time comprises the discharge voltage of C2 and voltage V1 (i.e. VL = V1 + VC2 = 2Vp).

Stabilised power supplies The rectifier circuit D.C. output, even though it may be smoothed with a capacitor, may vary due to peak value variations of input voltage Vm, which are often encountered as the A.C. mains voltage changes due to supply variations or fluctuations. Similarly as the D.C. load is increased the increased current causes the output voltage to fall due to voltage drops in the circuit. These effects are undesirable in electronic control and monitoring devices and can be effectively reduced by a stabilising circuit which incorporates a zener diode as a voltage reference to provide a steady output voltage.

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318 • Advanced Electrotechnology

The zener diode It has already been shown that a P-N junction diode will break down at a definite voltage when reverse biased. This breakdown, or zener effect, occurs at a definite voltage determined by the manufacturer. A junction diode continuously operated with reverse bias is called a zener diode.

▲ Figure 11.19 Zener diode symbol

Below the breakdown voltage or zener voltage level the diode resistance is very high, hence the reverse current will be extremely small, and is usually ignored. However, beyond the breakdown voltage the resistance falls to a low value and a rapid increase of current will occur. The reverse-biased diode will conduct above the fixed zener voltage as shown in figure 11.20. Beyond this breakdown point the voltage developed across the diode is the zener voltage VZ plus the voltage due to the resistance voltage drop across the diode.

V

Slope of characteristic is resistance Rz of the zener

Breakdown or zener voltage

0

IzRz Vz Voltage across the zener diode i.e. VO = Vz+ IzRz

Iz

▲ Figure 11.20

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Solid-State Electronics: The Diode • 319

Series Stabilisation To produce a stabilised voltage the zener diode may be connected across the load as shown in figure 11.21. The zener diode voltage rating is then determined by the D.C. voltage across the load. Rs

Vi unstabilised input voltage

IL

I Iz

RL

VO = Vz + Iz Rz

▲ Figure 11.21

Under all conditions when the zener diode is conducting VI = VO + IRS. The zener diode conducts when the voltage VZ is reached. At this position it will draw extra current from the unstabilised supply, and increase the voltage drop across RS, bringing the output voltage near to the VZ value depending upon the actual zener resistance and current. As VI rises, IZ increases and the series resistance voltage drop will increase keeping VO constant. As the load current changes the current IZ will change inversely but the voltage across the series resistor will also stay nearly constant, so the load voltage stays virtually constant. However, this only occurs if the maximum load current is not exceeded, as beyond this value the voltage drop across the series resistor RS will be excessive and the output voltage will fall. Example 11.5. A zener diode has a breakdown value of 9.1 V and, beyond this voltage, will have a resistance of 15 Ω. It is connected in a stabilising circuit, with a series resistor of 500 Ω to an unstabilised D.C. supply voltage of 30 V. When the load current is 30 mA calculate the zener diode current, the load voltage and the power dissipated in the diode, refer to figure 11.21 (all 3 significant figures). VO = VZ + IZ × RZ VO = 9.1 + 15 × IZ VI = I × RS + VO 30 = 500 (0.03 + IZ) + 9.1 + 15 × IZ

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320 • Advanced Electrotechnology 30 = 15 + 500 × IZ + 9.1 + 15 × IZ Iz =

30 − 24.1 A 515

Diode current IZ = 11.5 mA VO = 9.1 + (15 × 11.46 × 10–3) Load voltage VO = 9.27 V PZ = VO × IZ = 9.272 × 11.46 × 10–3 W Power dissipated in diode = 106 mW USE AS A VOLTAGE STABILISER. The circuit (figure 11.22) shows one arrangement commonly used to produce a constant output voltage VO for a varying load current IO or varying input/supply voltage VI. These conditions are considered in turn. II RS IO

VI

VZ

VO IZ

▲ Figure 11.22

(1) If the load is varied so that IO rises then II will tend to rise with a consequent larger RS voltage drop, VZ will fall but a small decrease of VZ causes a large decrease in IZ and compensates for the increased IO. II will remain constant as will the RS voltage drop and the output voltage VO. If the load current falls, II will tend to fall reducing the voltage drop across RS. A small increase of VZ will cause a large increase of IZ which counters the effect of a reduced IO. The RS voltage drop is fairly constant and so VO remains fairly constant. When no load current is taken, maximum load current plus IZ(min) flows through the zener diode, while on full load the zener conducts IZ(min).

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Solid-State Electronics: The Diode • 321 (2) If the supply voltage varies, VI rises and VO will rise. VZ will rise causing a large increase of IZ. The RS voltage drop will get larger and compensates for the rise of supply voltage to keep VO largely constant. Conversely if the supply voltage VI falls, VO and VZ will fall, causing a large decrease of IZ resulting in a smaller RS voltage drop offsetting the fall of VI, so VO will tend towards minimising change. VZ will remain constant provided IZ(min) flows continuously. VOLTAGE REGULATION. The rise in terminal voltage when a load is removed is termed voltage regulation, expressed as a ‘per unit’ or a ‘percentage’. This rise in voltage is caused by the load current IO passing through the zener when the load circuit is opened. The Per unit voltage regulation = IO × RZ and the Percentage voltage regulation =

I o Rz × 100 Vo

STABILITY. The following assumptions can be made if VO is kept constant by zener action. V V If VI is fairly constant then II is constant and given by II = I O × 100. If VI rises by 1 volt RS 1 then II rises by amperes. The current rise is diverted through the zener, with the IZ RS R 1 rise equalling the increase of II and the voltage rise VO = IZ × RZ = × RZ = z . RS RS Stability is defined by ‘the output voltage when the input voltage rises by 1 volt’. This R stability value is obtained from the ratio z . RS

Practice Examples 11.1

The following data represent the forward characteristics of a P-N junction diode. I (mA) V (volts)

0.25

0.55

1.1

2.5

1

2

3

4

Estimate (a) the forward D.C. resistance when the current is 1.5 mA and (b) the voltage and current for a forward D.C. resistance of 1.75 kΩ. 11.2

On test, a full-wave, silicon semiconductor rectifier is found to give a constant forward voltage drop of 0.7 V, independent of current. Calculate the power dissipated in the rectifier diode for a D.C. current of 5 A. If the rectifier dissipates

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322 • Advanced Electrotechnology heat, at the rate of 0.1 J/s/°C rise in temperature, find the maximum D.C. output current rating if the temperature rise is to be limited to 70°C. 11.3

The following values refer to a germanium diode.

Forward current (mA)

0

0.1

0.18

0.22

0.4

0.6

0.8

1.1

1.6

2.3 3.45

Forward voltage (mV)

100

140

160

180

200 220

240

260

280 300 320

Reverse current (μA)

–0.2

–0.3

–0.4

–0.4

Reverse voltage (mV)

–200 –400 –600

–800

Plot the anode characteristics for the diode and determine from it (a) the ‘forward’ D.C. resistance when the current is 3.45 mA (2 decimal places). 11.4

A half-wave rectifier with capacitor smoothing is fed from a 230 V, 50 Hz supply via a 12:1 step-down transformer. The volt drop across the diode when conducting is 0.75 V. Calculate the peak value of the D.C. output voltage and the P.I.V. rating of the diode.

11.5

An electronic circuit requires 350 mA D.C. from a half-wave rectifier circuit which incorporates a 1800 μF capacitor for smoothing. If the supply voltage is 50 V, 60 Hz, calculate (a) the peak value of the D.C. output voltage and (b) the peak to peak value of the output ripple voltage.

11.6

A D.C. load is to be supplied from a 250/40 V, 50 Hz transformer and bridge rectifier circuit with a 2000 μF capacitor for smoothing. Determine (a) the P.I.V. Rating of each diode and (b) the peak to peak value of the ripple voltage when supplying 250 mA.

11.7

An A.C. supply of 50 V is connected to a voltage doubler circuit as illustrated in figure 11.18. Calculate the peak voltage across the load resistor assuming the diodes to have negligible forward voltage drop.

11.8

The output from a rectifier circuit with capacitor smoothing is 25 V ± 20%. A zener diode series stabilising circuit is connected to this supply to provide a stabilised output from which currents up to 50 mA will be drawn. The zener diode has a breakdown value of 9.7 V and for stability it must carry a minimum current of 1 mA. The diode resistance above 9.7 V is 12 Ω. Determine (a) a suitable value of series resistance and (b) the stabilised voltage.

11.9

A zener diode has a breakdown value of 8.2 V and, beyond this voltage, will have a resistance of 12 Ω. It is connected in a stabilising circuit, with a series resistor of 400 Ω to an unstabilised D.C. supply voltage of 40 V. When the load current is 40 mA calculate the zener diode current, the load voltage and the power dissipated in the diode (all 3 significant figures).

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Solid-State Electronics: The Diode • 323 11.10 Figure 11.23 illustrates a rectifier circuit incorporating capacitor smoothing and zener stabilisation. The zener diode has a breakdown value of 24 V and requires a minimum current of 1 mA for stable operation. For values of voltage above 24 V the slope resistance is 14 Ω. For a stabilised D.C. load current of 60 mA calculate (a) the peak to peak ripple voltage measured across the capacitor, (b) the required P.I.V. rating of the diodes, (c) the minimum value of RS for stable operation and (d) the power rating of the zener diode. Rs + 250 V 50 Hz

40 V Load –

▲ Figure 11.23

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12

SOLID-STATE ELECTRONICS: THE TRANSISTOR The transistor is the information engine of the digital age. Chris Lavers

The Transistor This chapter continues our discussion of solid-state electronics by introducing the transistor. The transistor is a semiconductor device used to amplify and switch electronics signals and electrical power and has quite simply revolutionised electronics. It is composed of a semiconductor material with at least 3 terminals, for connection to an external circuit. A current or voltage applied to a pair of the transistor’s terminals changes the current flowing through another pair of terminals acting as a controllable electronic switch. Because the controlled output power can be significantly higher than the controlling input power, a transistor can be viewed as an electronic amplifier of a signal. Today, transistors are available as separate devices, or more commonly as one of millions of similar components within integrated circuits. By the end of World War II, research into semiconductors had advanced rapidly during the short period of the war years, with the first junction diode demonstrated in 1941 and the first transistor made at the Bell Laboratories, New York, Christmas Eve 1947, by

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Solid-State Electronics: The Transistor • 325 J. Bardeen and W. Brattain, part of a team working under Dr W. Shockley. The first public demonstration of a transistor took place in June 1948 on a point-contact transistor with an output power greater than the input, substantiating Shockley’s early peace time theories. This was followed by the junction transistor which he also designed and worked in accordance with his theoretical expectations. The next key step forward was the development of the junction-alloy transistor which made the point-contact transistor obsolete by 1956. Physical electronic theory had by then progressed to the stage where a range of different semiconductor junction possibilities could be fabricated. Shockley proposed a 4-layer P-N-P-N device and a different class of uniplanar devices based on the ‘field-effect’ principle. In fact a wide range of solid-state devices are now available, covering various ‘families’ of such devices, but the chief among these is still the junction transistor, the subject of this chapter. This family is split into 2 sub-groups covering P-N-P and N-P-N types. Another family covers the field-effect devices which can also be divided into 2 groups – the Junction Field-Effect Transistor (JFET or FET) and Insulated-Gate Field-Effect Transistor (IGFET) more commonly known as a Metal Oxide Semiconductor FET (MOSFET) and refers to its construction from metal layers (the gate), the oxide (the insulator) and a semiconductor. Further families cover 3-junction devices like the thyristor, diac and triac, while developments in diffusion techniques, metal oxide processing and nanoscale photolithographic etching have created the integrated circuit with transistors, resistors and capacitors formed better during manufacture. Study of these now numerous devices includes in no particular order: Schottky, Avalanche, Darlington transistors, carbon nanotube FET, organic FET and FETS used to sense the environment, and the Single Electron Transistor (SET) which would require in-depth discussion, and so we return to our discussion of the junction transistor.

The junction transistor This device consists of a ‘sandwich’ of 3 layers, being either of P-type material, i.e. rich in +ve charge carriers (holes) or N-type material, i.e. rich in electrons. Adulterated or doped germanium was the first material used but it has since been replaced by silicon which results in lower device leakage currents and has the ability to withstand higher temperatures (180°C) when compared with germanium (75°C). It also has a higher voltage rating. Silicon transistors generally do not operate above 1 kV, while SiC can operate well above 10 kV. Although the first junction transistors were of Ge and Si construction advances in microwave devices employ the semiconductor material Gallium Arsenide (GaAs) and the semiconductor alloy Silicon Germanium (SiGe). GaAs has the highest electron mobility making it useful for high-frequency applications.

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326 • Advanced Electrotechnology Modern Bipolar Junction Transistors (BJT) are of a planar-epitaxial type. Bipolar means that current is conducted by the 2 charge carriers: electrons and holes. Planar-epitaxial refers to its uniformly flat-layered construction. Although the junction transistor is treated in terms of the simple diagram shown (figure 12.1), practical assembly is more in accord with figure 12.2. Figure 12.2a shows a section of an early junction transistor. Figure 12.2b shows a section of the planar-epitaxial transistor where N-type silicon disc is first oxidised, a window photolithographically etched, cleaned chemically and then an impurity, e.g. boron, fused into the window to form a P-type silicon region and an N-P-N transistor. The window is closed by oxidation and a smaller window etched in. Chemical cleaning is repeated and phosphorus is fused into the P-region to convert it into N-type. The window is then closed by oxidation and the disc cut into thin sections to provide millions of transistor chips. Connections are bonded by aluminium fused into the silicon and each chip enclosed inside a can. A P-N-P transistor on the other hand is made in a similar way except different doping elements are used to reverse the polarities. Impurity fusion is achieved by Chemical Vapour Deposition (C.V.D.) which is accurately controlled and combined with mass production methods so some million transistor chips can be produced from one original 300 mm diameter disc or wafer, with the most recent wafer size of 450 mm now coming into production with Intel and Samsung. However, not all of these transistors may work and various factors will affect the final device yield. (a)

+

–

+ Ib

Ie P

(b)

–

Ie

Ic

N

+

Ic e

c

–

P

+ b

(c)

Collector

–

– –

Vbc

– +

Ib

+ V ec –

1%

Emitter 100%

Ib

Ic

99%

Base

–

Veb +

+

+

Ie Hole or current flow

▲ Figure 12.1

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Solid-State Electronics: The Transistor • 327 (a)

N -type germanium Soldered emitter connection

(b)

Thin base region

Connections Oxide layer

Soldered collector connection

Small indium pellet

Large indium pellet

P-type region alloy junction

Soldered base connection

N -type silicon emitter

P-type silicon base

N -type silicon collector

▲ Figure 12.2

OPERATION. Figure 12.3 shows a basic P-N diode and a basic P-N-P transistor arrangement, the latter consists of a region of N-type material sandwiched by P-type material regions. The N-type material, the base, is rich in electrons but is very thin (0.005–0.05 mm). The outer regions – the sandwich ‘slices’ – the emitter and collector, are both rich in +ve charge carriers or holes. An N-P-N transistor will have the materials reversed. Amplifying action is obtained because the current between the emitter and base influences the current at a higher power level between the collector and base. Before describing this action in detail, we will revise the explanation of junction diode operation by remembering that on one side there are only holes (figure 12.3a), while on the other side there are only electrons. Application of a voltage across the junction, so the P-end is +ve and the N-end is −ve, results in current flow since holes move left to right and electrons move from right to left. Reversal of applied potential results in no current flow since there are no holes free to move right to left or electrons free to move left to right. The junction provides an ‘easy’ path in one direction only and is virtually open circuit in the other direction. With a +ve potential applied to the P-side and a −ve potential on the N-side, the diode is forward biased and current readily passes provided the voltage is above the barrier potential, typically 0.6 V for silicon, and 0.2 V (smaller) for germanium. (a)

+ – n p + + + – – – + ++ – – – – + + – – – + +

(b)

+ –

+

–

n p p + + + –– + + + + + + – + + + –– + ++ + + + + + – + + Emitter Base Collector

▲ Figure 12.3

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328 • Advanced Electrotechnology Consider figure 12.3b. The arrangement shows 2 P-N junctions close together or ‘back to back’ in a single piece of semiconductor. It should be no surprise to the reader that current through one junction influences the current passing through the other junction and the arrangement constitutes a transfer-resistor or transistor (a term created by John R. Pierce), whose action is understood by considering the action of the device in 2 stages. Consider the collector–base junction only. If the collector is made −ve with respect to the base, the arrangement will be that of a diode with reverse bias. The only current is due to minority carriers, i.e. electrons from the P-type collector and holes from the N-type base. Minority carriers are small and so the resulting current is small. Consider next the emitter-base diode. If the emitter is made +ve with respect to the base, the arrangement is that of a forward-biased diode and current will be large, being due largely to majority carriers. For a modern bipolar transistor both P- and N- type materials are heavily doped, so current is conducted by both carrier types. Holes injected into the base from the emitter collect in high concentration at the junction and because of the base thinness diffuse quickly towards the collector junction. If a large −ve potential is applied across the collector-base junction, any holes that reach this barrier are rapidly swept across, reaching the collector and increasing the collector current. Note that not all the injected holes from the emitter reach the collector because some combine with electrons in the base. However, by keeping the base thin – only a few microns – the electrons available for recombination are reduced and most of the holes leaving the emitter reach the collector. For a typical transistor for every 1 mA leaving the emitter, the collector current is about 0.99 mA. Amplification occurs because of the different impedances in which these currents flow. The emitter current is in the direction of ‘easy’ flow, so the impedance may be less than 50 Ω, while the collector current is in the direction of difficult flow and resistance may be above 1 MΩ or more. To make up for electrons lost in the base by recombination with holes, the base requires a small electron flow, i.e. a conventional current flow from the base is needed. Current flow from the base is regarded as a flow of holes, the result of such a flow is reduction of the base positive charge and the greater the flow, the greater the base-collector junction depletion layer is reduced. This is equivalent to reducing junction resistance, so the collector current increases. A base current increase results in a collector current increase. This condition must be kept in mind when transistor operation in various configurations is considered. Some of these configurations are described shortly. The diagrams of figure 12.1 show the P-N-P transistor arrangement, the standard circuit symbol and an illustration symbolising hole flow. The conditions for аn N-P-N transistor are shown in figure 12.4. CURRENT TRANSFER RATIO. All transistors behave like a control unit in that current, in one part of a device, can alter the current in another part of the circuit. Early research showed that the effectiveness of this control depended on the thinness of the base

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Solid-State Electronics: The Transistor • 329 (a)

–

+

–

+

Ib

Ic

Ie N

P

(c)

e

– +

N

Ic

Ie

(b)

c

+ –

b

Ib

Ic

Collector 99% +

+ +

Vcb –

Base + Ib

1%

+

Vce

–

Vbe

–

Emitter 100%

–

–

Ie

Hole or current flow

▲ Figure 12.4

region, which ensured that the base current value is only a small fraction of the total emitter current so the ratio of these currents is largely constant. The ratio value varies quite considerably for different individual transistors of the same type due to the mass production methods used. Consideration of the diagrams (figures 12.1 and 12.4) leads to the deduction that Ie = Ic+ Ib, Ie is the emitter current, Ib the base current and Ic the I collector current. c is used as a measure of the device control effectiveness and given Ie I the term current-transfer ratio and the symbol hF. If α = c as Ic is less than Ie, then α will Ie be less than unity. Here manufacturing improvements have resulted in smaller Ib values leading to a reduction in the difference value between Ic and Ie and α will approach unity. I A more useful ratio is c – symbol β. A simple relationship between α and β is deduced Ib but it is noted that, as for the thermionic valve, characteristic curves are obtained and the linear parts of these give values for α and β. It is useful to consider small changes of current instead of actual current values, so we write: α=

δ Ic δI and β = c . Also since Ie = Ic + Ib δ Ie δ Ib

Then δIe = δIc + δIb and δIb = δIe + δIc Substituting in β =

δ Ic δ Ic = δ Ib δ e δ Ic

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330 • Advanced Electrotechnology

=

δ Ic δ Ib

δ Ie δ Ic α α − = . Thus β = δ Ie δ Ie 1 − α 1− α

With the ‘forward current transfer ratio’ hFB = current transfer ratio’ or hFB =

δ Ic , similarly β is given the term ‘forward δ Ie

δ Ic . δ Ib

These current ratios are defined by the static value of the short-circuit forward current ratio with the symbol hF. Another subscript is added to indicate the configuration, hence the above hFB and hFE. Transistor use shows that any value of α must be specified accurately and this requires a symbol which states within limits, the conditions the value was measured under. To show α was measured in ‘grounded emitter’ mode it is stated α ʹ. To show it was measured at zero frequency (D.C.) it is written α 0′ . If measured at large current values, it is written as αo etc. Use of h with subscripts to define the operation modes gives clearer information but it is noted that other h values are given in manufacturer’s data; e.g. hFE (small-signal gain, i.e. change of Ic for unit change in Ib). Such h hybrid parameters are termed small-signal parameters and related to circuit theory involving special circuits, etc. Symbols α and β are used for basic study, with reference made to the h representation as appropriate. Example 12.1. For a base current of 0.1 mA in a transistor, the corresponding collector current under short-circuit conditions was recorded as 1 mA. What is the transistor’s common base short-circuit current gain? Here δ

b

δ Ic = 1 mA A ∴δδ

e

α=

10 = 0.909 1.1

δIe is deduced from the fact that emitter current = collector current + base current.

Transistor characteristics Transistor performance can be investigated by making suitable tests and from the results and the characteristics of each configuration drawn. These are the (1) Input Characteristic, (2) Output Characteristic and (3) Transfer Characteristic. The last can be deduced from the first two but a test is useful in that an average value for the forward current transfer ratio is obtained directly. In figure 12.5 the emitter arrow indicates the conventional current direction, i.e. the direction in which +ve holes are injected into the base. For N-P-N transistors, the current carriers (collector current) consist of

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Solid-State Electronics: The Transistor • 331 –

(a)

–

(b) e

c

–

(c)

c

b

b OUT

IN

e

OUT IN

IN e

+

e

b

Common emitter H.T. +

+ Common base

b

c

OUT

+ Common collector

H.T. +

c

H.T. +

OUT IN

IN H.T. –

OUT

H.T. –

IN

OUT H.T. –

▲ Figure 12.5

electrons flowing from the N-type emitter into the P-type base. Conventional current (hole) flow is in the opposite direction, hence the arrow reversal on the emitter which points away from the base. Device polarities are reversed with the collector +ve with respect to the base and the emitter –ve biased. N-P-N transistor use has increased, as diode manufacturing techniques have improved unit cost reduction. CIRCUIT CONFIGURATION. Transistor electrodes are of course digital analogies of those found in the thermionic triode, the device whose miniaturisation the transistor was designed to replace. The emitter corresponds to the cathode, the base to the grid and the collector to the anode. Like the valve, the device is used in one of 3 different circuit configurations, and as for the valve, attention will be given to the most commonly used. Figure 12.5 illustrates the 3 circuit configurations and the corresponding triode valve connections given alongside for comparison. The first and most commonly used (a) is the ‘common or grounded emitter’ where the cathode is ‘earthed’, the grid is the input electrode and the anode the output electrode. The second configuration (b) is the ‘common or grounded base’ mode which as for the triode is rarely used. Mode (c) shows the ‘common or grounded collector’, mainly used for impedance matching, e.g. where an amplifier must be coupled to a low-impedance load, as this mode has high input impedance and low output impedance.

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332 • Advanced Electrotechnology The features of the various transistor modes are as follows: (1) Common or grounded emitter. Medium impedance for both input and output circuits. The battery +ve line is common (as the capacitors, see figure 12.15b, connect the emitter to the +ve line for A.C. signals), the emitter is also common to input and output circuits. There is a phase shift of 180° and A.C. current and voltage gains are high. (2) Common or grounded base. Low input impedance to emitter, high output impedance from collector. Capacitors (figure 12.9) allow A.C. signals while blocking D.C. ones. The base is grounded to A.C. and common to input and output signal circuits. The phase shift is zero and A.C. current gain less than 1 with A.C. voltage gain high. (3) Common or grounded collector. This device configuration has a high input impedance but low output impedance. The battery offers negligible impedance to A.C. signals and the input is applied between base and collector and the output taken from the emitter and collector. The corresponding phase shift is zero while A.C. current is high, providing a voltage gain of less than 1. COMMON-BASE CONFIGURATION. The circuit for testing a transistor is connected in the common-base mode as shown in figure 12.6. Ic

Ie +

– Ve

Ie

Ic

Vc

–

+

▲ Figure 12.6

The ‘input characteristic’ is obtained by keeping the base-collector voltage Vc constant and recording the emitter current Ie for various emitter-base voltages Ve. Voltage adjustments are made with a circuit potentiometer and graphs, of the form in figure 12.5, are plotted from test results. Graphs can be plotted as Ie against Ve or more commonly Ve against Ie, which allows input impedance to be determined for all working δI conditions. Thus the slope of the graph in figure 12.7a is e and the input impedance δ Ve δV 1 is the reciprocal of the slope or rin = . Thus input impedance rin = e . The slope of slope δ Ie the graph of figure 12.7b gives the input impedance directly.

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Solid-State Electronics: The Transistor • 333 (a)

(b)

δVe

δIe

Ve

Ie

δVe

δIe

Ve

Ie

▲ Figure 12.7

So Input impedance rin =

δ Ve . δ Ie

The ‘output characteristic’ is obtained by holding Ie constant and obtaining Ic for various base-collector voltages Vc. A family of curves, (figure 12.8b) is found which resemble the thermionic pentode’s Ia/Va curves, as intended from semiconductor manufacture. The key difference is that each curve is for a separate value of emitter current instead of grid voltage, demonstrating that transistors are current-operated devices. (b)

6 mA

4

4 mA

2

2 mA

V

6

8

5

Ie = 8 mA

Ic (mA)

Ic (mA)

V

c

=

8

4.

(a)

δIc 4

0 +1

0

–2

–4 Vc (volts)

–6

–8

δIe

4 Ie (mA)

α=

δIc δIe

8

▲ Figure 12.8

The reason for the shapes of the curves is that the collector will always collect holes if it has a −ve bias. Current Ic is almost independent of applied voltage, because it is a reverse current, and depends only on the minority carriers (+ve holes) which are

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334 • Advanced Electrotechnology available, themselves depending on Ie. The ‘transfer characteristic’ is found by a direct test, holding the base-collector voltage at a fixed value and varying Ie. The effect on Ic is noted, as shown in figure 12.8b, and its slope is a measure of α or hFB – the forward current transfer ratio (current gain), whose value is deduced directly from the output δI characteristic. For this mode α = C for a fixed value of Vc, the change in Ic noted for a δ Ie change in Ie, and current gain deduced. For example (figure 12.8a) for a constant Vc value of −4 V, when the emitter current is raised from 2 to 4 mA (difference 2 mA), the 18 collector current rises from 1.9 to 3.8 mA (1.8 mA difference). Thus α = = 0 9 (approx.). 2 Difficulties arise in achieving the required accuracy for α, when using this method, and a more reliable result is achieved by plotting the transfer characteristic. Amplification. A single-stage amplifier arrangement, connected in common-base configuration, is shown (figure 12.9). Battery polarity is vital and a P-N-P transistor considered.

RE VIN

RL

VOUT

▲ Figure 12.9

Assume a 10 mV supply voltage is applied to the emitter to produce a 1 mA current of which 0.98 mA passes to the collector and 0.02 mA to the base. Transistor supply voltages are often multiples of 1.5 V and even the smallest voltage unit may give too high a current if applied directly to the emitter. RE is included to limit the emitter current to a few milliamperes. By inserting a load resistor RL in the collector circuit, amplifying action is obtained. With no A.C. input signal a steady emitter current flows, limited to a convenient value (e.g. 1 mA) by RE. If the input signal varies by 5 mV to produce an emitter current fluctuation of 0.5 mA, the collector current will vary by about 0.5 mA (some 0.5 × 0.8 = 0.49 mA). If RL = 1000 Ω, the output voltage A.C. fluctuation = 1000 × 0.5 × 10−3 = 0.5 V. Thus a voltage amplification of output to input voltage of about 05 = 0 1× 10 −3 = 100 is obtained. 5 1023

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Solid-State Electronics: The Transistor • 335 Figure 12.10 shows the phase relationship between the input and output voltages; they are in phase which is the opposite of that observed for a valve where the input and output voltages are 180° out of phase. In the common-base mode, the transistor’s full abilities are not exploited, in that control by the base current of the collector current is not used. The configuration gives a current gain of less than 1 and amplification is only achieved by the different impedance levels in the emitter and collector circuits. It is convenient to deal in terms of voltage gain, but true amplification is considered in terms of a power gain and this aspect is important for transistor work. Thus for the circuit of figure 12.9, the forward-biased transistor emitter-base diode is of low impedance, e.g. 15 Ω. If a 0.5 mA current change takes place, the power change = 0.52 × 10−6 ×15 = 3.75 μW. In the collector circuit, a corresponding change of 0.49 mA takes place with an impedance of say 1000 Ω. Thus power change is 0.492 × 10−6 × 103 = 0.492 × 10−3 = 240 0.24 ×10−3 or 240 μW, giving a power gain of = 64 (approx.). 3 75

Voltage (V) and current (mA)

VOUT

Ie Ω Ic 1.5 1.0 0.5 mA VIN Time

▲ Figure 12.10

To achieve this gain, it is vital to match the signal source to the input impedance and the load to the transistor output impedance. In the common-base mode current amplification cannot be obtained as ‘a’ is less than unity. Voltage amplification is possible because the emitter-base junction has a low resistance and if a large load resistance value is used in the collector circuit, a high-voltage gain results. Power gain can be large, as shown, and is mostly used for R.F. circuits. Example 12.2. For a common-base arrangement (figure 12.9) assume RL = 50 kΩ and transistor current gain = 0.95. Find the voltage gain if an input signal causes a 100 μA emitter current change (3 significant figures). Transistor input resistance = 100 Ω.

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336 • Advanced Electrotechnology

Since α =

δ Ic δI then 0.95 95 = c ∴δδ δ Ie δ Ie

μA

c

Input-signal voltage variation = 100 × 10−6 × 100 = 10−2 = 0.01 V Output-voltage variation = 95 × 10−6 × 50 × 103 = 475 × 10−2 = 4.75 V

∴Voltage gain

4.75 = 475. 0 01

In the circuit of figure 12.9, the emitter resistance RE is shown. An explanation for its presence is useful, in case it is confused with the value of 100 Ω given in the example for the input resistance. Input resistance is a transistor value and does not refer to RE. Tests show voltage applied between the emitter and base causes a normal current flow of about 100 mV. If a fixed voltage is applied, then as a result of temperature variation, emitter current will alter since the effective emitter base resistance decreases with a temperature rise. To overcome this, a resistor RE is included in emitter circuits with a value much greater than the emitter-base resistance so the current will vary little with temperature change. This technique comes under the heading of ‘stabilisation’ and prevents ‘thermal run away’. COMMON-EMITTER CONFIGURATION. Figure 12.11 shows a test circuit for obtaining transistor characteristic curves connected in common-emitter configuration. Potential adjustments for both input and output circuits are made by means of resistance potentiometers. Ic Ib Ic

– Vb +

Ie

– Vc

+

▲ Figure 12.11

One set of characteristics is obtained for constant collector voltages Vc, by varying the base-emitter potentiometer to give a series of different values of Vb, current is Ib noted for each value of Vb. A typical curve for this ‘input characteristic’ is shown in figure 12.12a, plotted as Ib against Vb or, Vb against Ib. The latter is more common – figure 12.12b.

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Solid-State Electronics: The Transistor • 337 (a) 150 145

S

(b) Q

80

93

P

Vb (mV)

Ib (μA)

100

50 35

δVb δIb

T

U

0 0

50

93 100

80

150

Vb (mV)

Ib (μA)

▲ Figure 12.12

Example 12.3. The ‘input characteristic’ of figure 12.13 is of a typical transistor connected in grounded-emitter mode. Find the input impedance when Ib = 40 μА and Vc is kept constant at −4.5 V. Point P is determined on the characteristic and a tangent drawn to determine the slope, giving an input impedance or A.C. resistance.

4.5

Vc

200

V

=–

Emitter-base voltage Vb (mV)

175 150

P

135

δVb

δIb

100

50

0

0

20

40 50 60

100

150

200

Base current Ib (μA)

▲ Figure 12.13

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338 • Advanced Electrotechnology

∴ rm =

δ Vb ( = δ Ib (

)10 −3 40 × 103 = = 1× 103 Ω = 1 kΩ. )10 −6 40

− −

A further set of graphs are termed ‘output characteristics’. The collector current/collector voltage curve is plotted by obtaining corresponding readings for Ic and Vc, while Ib is held constant for each set of readings. Typical curves are shown in figure 12.14. (a)

(b) A

Ic (mA)

8

80 μA

6

40 μA

4 2

Ic (mA)

I b = 120 μ

10

δIc δ Ib δIc α ′= – δIb

0 0

–2

–4

–6

–8

Ib (μA)

Vc (volts)

▲ Figure 12.14

A ‘transfer characteristic’ is obtained by plotting Ic against Ib, with Vc constant. This can be deduced from the output characteristic and is shown in figure 12.14b. Curves obtained from the tests are static characteristics. When a small A.C. signal is applied between base and emitter, the effect is shown on the input characteristic – figure 12.12a. The varying signal causes a small swing of Ib and Vb over part of the curve (shown by PQ). The reciprocal of the slope PQ gives the input A.C. resistance. With a working point of 93 mV voltage input varies to give points P and Q. With US a straight line through PQ, the working points’ A.C. resistance is obtained as the slope’s reciprocal. Thus

UT ( = TS (

− −

) × 10 −3 100 = × 103 = 909 Ω ) × 10 −6 110

Input resistance depends on the working D.C. bias value because the gradient of the curve changes with Vb, with output resistance obtained from figure 12.14a given by the slope of the appropriate curve. Output A.C. resistance will be high and nearly constant for a given Ib value if the working conditions are beyond the curve’s turn-over point. When Vc is small, i.e. below this value, output resistance is much reduced. The reader’s h α attention is drawn to the deduction made earlier: β or hFE = or FB . 1− α 1 − hFB

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Solid-State Electronics: The Transistor • 339 Example 12.4. If the change of collector current is 2 mA when the emitter current (in a common-base circuit) changes by 2.2 mA (collector voltage being constant), calculate the transistor current gain when connected in a common-emitter circuit (1 decimal place). For the common-base mode α = α=

δ Ic , where δIc = 2 mA and δIe = 2.2 mA δ Ie

2 1 α 0 91 0 91 91 = = 0 91 also β = = = . Thus β = = 10.1. 2.2 1.1 1 − α 1 − 0.91 0 09 9

Amplification. The arrangement for a single-stage amplifier connected in commonemitter configuration is shown in figure 12.15. The base and collector are made −ve with respect to the emitter. In figure 12.15b base bias is obtained by a voltage drop along RB as RB forms a bias potentiometer with the base-emitter resistance in series. (a)

(b) RB

VIN

RB

RL

VOUT

VIN

RL

VOUT

▲ Figure 12.15

The common or ground-emitter arrangement often takes advantage of a transistor’s I high current gain. Here c is the current gain as the base current is now the input current. Ib As Ib is very small, the ratio is large with a value of about 50. Figure 12.15 shows how supply voltage is applied between emitter and collector and circuit current is controlled by base current. As stated earlier α′ or β is used for common-emitter ‘current-gain’ but hFE is recommended, i.e. the ‘forward current transfer ratio’, with values between 20 and 100. The transfer characteristic slope (figure 12.14b) gives this value. It is easy to explain the high gain for this mode. Two separate current paths are seen in figure 12.15. For one, a current flows into the emitter and out through the base. For the other, current flows out of the collector and returns via the supply to the emitter. The collector current is less than the emitter current by the very small base current. Alternatively, the base current is the difference between the emitter and collector currents: Ib = Ie − Ic. Any change occurring in Ie or Ic leads to a change in Ib equal to this difference. Thus for a transistor with a common-base gain factor α = 0.98, when Ie is increased by 1 mA,

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340 • Advanced Electrotechnology Ic changes by 0.98 mA and the resulting change in Ib is 0.02 mA. Here a certain base current change is associated with a change in the collector current which is 49 times bigger, and there is a current gain between the circuits. Example 12.5. The ‘output characteristic’ of figure 12.16 is that of a typical transistor mentioned in example 12.3 connected in a grounded-emitter mode. Find the output impedance when Vc = −4.5 V and Ib = −40 mА. Also find the current gain for this point. δI Point P is determined on the −40 mA characteristic and the slope found, Slope = c . δ Vc 10

140 μ

Ib = –

Collector current Ic (mA)

–120

A

μA

–100 μ A –80 μ A

5

–60 μA 3.4 P

2.3 2 2.25 0

–40 μA –20 μA

0.5

2.8

4.5 5 5.5

10

Emitter-collector voltage Vc (V )

▲ Figure 12.16

Since output impedance is the reciprocal of the slope then output impedance r0′ =

δV δI 1 5.5 − 2.8 2.7 × 103 = c = = = 9 × 103 Ω = 9 kΩ kΩ , the current gain α ′ = C . slope δ Ic (2.3 − 2.0 )103 03 δ Ib

For a voltage value of −4.5 V a base current change of (60 − 40) = 20 mA or 20 × 10−6 A is associated with an Ic current change of (3.4 – 2.25) = 1.15 × 10–3 mA. ∴α ′ =

δ IC 1.15 × 10 −3 0.115 = = × 103 = 0.0575 × 103 = 57.5 or current gain = 57.5. δ Ib 20 × 10 −6 2

Example 12.6. A 50 μА transistor base current is shown in the common-emitter amplifier (figure 3.15b), the collector supply voltage = 10 V. Find the single resistor value RB used to provide the bias current. Ignore the base-emitter forward conduction voltage drop.

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Solid-State Electronics: The Transistor • 341 As Ib = 50 μА and the base-emitter voltage is neglected the base is 10 V positive with respect to the −ve line, i.e. the full 10 V is dropped across RB. ∴ RB =

10 = 2 × 105 = 200 kΩ 50 × 10 −6

The amplification action is similar to the common-base arrangement. Here operation is applied to the circuit of figure 12.11. A change of emitter-base signal voltage of 10 mV produces a base current change δIb, e.g. 0.05 mA and because of the high β value for this mode, a large change in current Ic occurs. Suppose this current change δIe = 2.5 mA, then almost all the current appears in the collector current Ic and if it passes through a high load resistor RL, say 1 kΩ, A.C. output voltage change across this resistor will be: δIc × RL =2.5 × 10−3 × 103 = 2.5 V 25 is obtained. From figure 12.17 waveforms, an 10 × 10 −3 inversion of output to input voltage is seen, similar to common triode valve operation where the input and output voltages are 180° out of phase. Transistor inversion occurs because on application of a small alternating voltage signal across the base-emitter junction, the base current fluctuates, with such variations resulting in similar collector current variations. Thus a rise of input voltage produces an increase in collector current. Voltage drop across the load resistor RL increases and the amplifier output voltage falls as a result. Similarly a fall in input voltage produces a fall of collector current and a rise of output voltage. Considering the Power in = 10 mV × 0.05 mA = 0.5 μW, and the Power out = IC3 RL = 2.52 × 10 −6 × 1000 = 6.25 × 10 −3 V = 6250 μW . A voltage amplification of about

VOUT

Voltage and current

Ie Ω Ic

Ib

VIN

Time

▲ Figure 12.17

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342 • Advanced Electrotechnology

Thus power gain =

6250 = 12 500. 05

COMMON-COLLECTOR CONFIGURATION. This arrangement is little used and is given no further attention here.

Load lines Use of a load line, in conjunction with the transistor characteristics of a particular mode, follows valve procedures. Example 12.7 shows how it can be constructed and applied. Example 12.7. Figure 12.18 shows a transistor’s typical common-emitter characteristic. From the information available draw the load line for a collector load resistance = 4 kΩ and supply voltage = 15 V. Assume it is possible to swing a sinusoidal signal over the full range of the load line, state a suitable Ic value and calculate (a) the D.C. power I dissipated in the load resistor, (b) the A.C. power available and (c) the c gain. Estimate Ib the saturating value of Vc (4, 3 and 2 significant figures respectively). Ib = 60 μA

5

Ib= 50 μA 4

AX

40 μA

3.6mA

Ic (mA)

3 30 μA O

2

1.8mA

20 μA

1

10 μA Y 0 μA

0 0 2 0.7 V

4

6

8

10

12

14 B 16

Vc (volts)

▲ Figure 12.18

The load-line is drawn using the expression Vc = Vs − IcRL. Vc is the collector-emitter voltage, VS is the supply voltage, and RL the load resistance. when Vc = 0 , then

Vs 15 = × 10 −3 = 3.75 mA. RL 4

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Solid-State Electronics: The Transistor • 343 Mark Point A on the diagram for Vc = 0 and Ic = 3.75 mA. When Ic = 0 then Vc = VS or the voltage on the collector is 15 V, giving point В on the diagram. Join A and В to give the load line. The saturating value of Vc for this condition is about 0.7 V. The collector current is 3.6 mA (approx.). A suitable value for the quiescent point О will be

36 = 1 8 mA 2

(1) The D.C. power dissipated will be IC2 RL = ( .8 × 10 −3 )2 × 4000 W = 1.82 × 4103 = 12.96 mW (2) The power available = =

(

(

c(max) ( )

− .7)(3.. − ) × 10 8

−

(min)

)( c(max) ( ) −

(min)

)

8 −3

=

14.3 × 3.6 mW 12.87 = = 6.44 mW 8 2

Ic gain is determined for working about the operating point O between 20 and Ib 30 μА base current characteristics

(3) The

We have β or hFE =

2.3 − 1.5 0 8 × 103 = = 80. (30 − 20 )10 − 3 10

Leakage current It is noted from example 12.7 (figure 12.18) that collector current values are shown for the Ib = 0 characteristic, i.e. a current of increasing magnitude flows for rising values of Vc. This is the normal diode junction reverse current and usually its effect can be neglected, especially for silicon transistors. However, leakage current even for silicon transistors can cause concern when operating temperatures exceed 100°С and its effect must be taken into consideration when designing circuits.

The practical amplifier Figure 12.19 shows a practical circuit for the commonest transistor amplifier, in a common-emitter configuration, one battery source with components’ values obtained from transistor static characteristics, and knowledge of a circuit’s current-flow directions. R1 and R2 provide the base-emitter bias by acting as a potential divider. The function of R3 is covered under ‘stabilisation’.

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344 • Advanced Electrotechnology

R1

Ic

I1 + Ib

Ib R2

I1

R3 Ie =Ic + Ib

▲ Figure 12.19

STABILISATION. One of the main difficulties found in transistor circuit operation is a tendency for operating conditions to alter or ‘drift’. Drift is caused by temperature changes and action must be taken to reduce this. As germanium resistance falls as temperature rises, a larger current will flow which, in turn, increases transistor heating – even destroying it in extreme cases. Such destruction is called ‘thermal runaway’ and is minimised by circuit stabilisation methods. One efficient and common method of stabilising the base bias uses a potential divider made up from R1 and R2. As the current I1 in the series resistors is more than the base current, the base potential is held fairly constant in spite of base current variation when A.C. input signals are applied. Use of resistor R3, inserted between the emitter and the battery +ve terminal, was explained in relation to example 12.2 for resistor RE. Its value is much larger than the transistor internal emitter resistance and the current is largely unaffected when transistor temperature changes, so thermal runaway is avoided. Use of silicon, instead of germanium for semiconductor devices, is mainly to raise operating temperature and stabilising circuit techniques are still used for silicon transistors.

The transistor as a switch This device can operate as a highly efficient switch at very high speed (micro-seconds). Furthermore it may be powered with a fraction of the main current it controls. It is highly sensitive and can operate with various transducers to sense tiny physical changes of temperature, pressure, strain and light. For the switching function, the device is used in common-emitter mode with high current gain. As the resistance between collector

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Solid-State Electronics: The Transistor • 345 and emitter is controlled by the base bias, if the base is made +ve with respect to the emitter, the base-emitter junction is reverse biased and will not emit, so no collector current flows. With the base made −ve with respect to the emitter, both a base current and collector current will flow. As the base is made more negative, collector current increases until it is limited by the collector supply voltage and load resistance. Under this condition only a fraction of a volt is dropped across the collector-emitter junction, so full supply voltage will be available for the load. The transistor is thus seen as a switch. With a +ve bias on the base the switch is open and with a −ve potential of suitable magnitude, the switch is closed. PRACTICAL CIRCUIT. A simple circuit is shown in figure 12.20. Current only passes between the emitter and collector when the base is 0.7 to 1 V negative with respect to the emitter. The easiest way to achieve this is shown in figure 12.20 using related subcircuits (a) or (b). (a)

Rx

(b) A

A

A

B

B

TR1

C

C

R1

Relay

RX Water level

B

–

–t°

Supply source TR2 +

▲ Figure 12.20

The arrangement in figure 12.20a is a simple remote control switch where a circuit is closed by a rising water level. The appropriate bias, decided by the value of series resistor RX, is applied and the transistor TR1 turned on. Current then flows via R1 through the emitter-collector junction. The potential at the top of R1 will be about −1 V and TR2 is turned hard ‘on’ causing a relay to operate. With the remote switch open, no bias is applied to the base of TR1 and the transistor is ‘off ’ ensuring that TR2 is ‘off ’. For arrangement in figure 12.20b the bias on the base of TR1 is decided by a potential divider. A thermistor is connected in series with a variable resistor. The potential of point В varies according to the thermistor’s resistance value, which varies with its temperature, i.e. resistance decreases as temperature increases. The potential of В and thus the base of TR1 alters with temperature and control is effected. The series variable resistor RX helps adjustment of the operating point. Another example of a switch would be as a lighting component.

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346 • Advanced Electrotechnology

The thyristor This 4-layer device is termed a Silicon-Controlled Rectifier (SCR), and is mainly used as a semiconductor switch for controlling application of voltage to a load. It is provided with 3 terminals, the anode, cathode and gate or trigger. Figure 12.21 illustrates its general construction and the appropriate symbols. It is seen that 2 types are available, namely N- and P-gates. The latter is almost universally used with the anode made +ve with respect to the cathode. Theory shows how a thyristor conducts in one direction only although bidirectional thyristors or ‘triacs’ are still available. Cathode Gate Hermetically sealed ceramic/metal case

Anode

a

Cathode contact Inert gas N P N P Anode contact Mounting stud

c

g General symbol a

c

g N-gate (anode controlled) a

c

g P-gate (cathode controlled)

▲ Figure 12.21

If the anode of a P-gate thyristor is made −ve with respect to the cathode, the 2 outer diodes of the 4-layer sandwich, see figure 12.22, are reverse biased and conduction does not occur. With the anode made +ve with respect to the cathode, the centre diode only has a reverse bias which prevents conduction if no input is applied to the gate. The gate’s function causes the device to conduct or ‘fire’ when an appropriate potential is applied. Current carriers from the outer layers flow to the centre junction. When this occurs, anode and cathode are considered to be connected so a Short-Circuit (S.C.) occurs across the device. Once the gate has initiated conduction, current carriers from the outer layers maintain the main current even when the gate input is removed. Thus to ‘switch off ’, the current must be interrupted with an external means or circuit condition. This is achieved if an A.C. voltage is controlled as the current value at the end of a half-cycle falls to zero. The applied voltage potential then reverses and a condition of non-conduction is obtained. The thyristor switches off at the end of a +ve half-cycle and will not conduct again until another +ve gate impulse is applied during another +ve half-cycle of mains voltage across the anode-cathode terminals. When operating to control a D.C. circuit, extra circuitry is required. Usually a second thyristor is connected

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Solid-State Electronics: The Transistor • 347 A

A

P1

e

N1

G

c

N2

TR1

P

b G

P2

A

N P

C

Nc P b N e

TR2

G

C

C

▲ Figure 12.22

in parallel with the main device with the anodes capacitively coupled. Triggering the second thyristor switches a load off at the period of capacitor charging. THEORY OF OPERATION. A thyristor is represented by a 4-layer sandwich (figure 12.22). If the centre P-N section is split to give the arrangement shown, we can develop the 2 complementary transistor analogy. TR1 is а P-N-P device and TR2 an N-P-N device. Assume a short positive current pulse to be applied to the base of TR2. With the correct bias on the base of TR2, the transistor is ‘turned on’. The TR2 collector current is drawn from the base of TR1. Since the emitter-base diode of TR1 is forward biased this unit is also ‘turned on’. TR1 collector current now supplies the base of TR2 and gate current is no longer required. Both transistors now continue to conduct as long as the supply polarities to anode and cathode are correct and anode current does not fall below the holding current value.

Forward current I

If a thyristor’s current/voltage characteristic is considered (figure 12.23), further explanation of the holding current can be made. With no gate connection, if an increasing voltage is applied between the anode and cathode, a small but constant

Avalanche reverse breakdown

Holding current

Conducting state Forward voltage drop VBO

3

–V Leakage current

Leakage current (blocking state)

VBO

2

Forward breakover voltage VBO

Forward voltage

1

V+

▲ Figure 12.23

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348 • Advanced Electrotechnology current flows through the device between the terminals. At a certain voltage value, the Breakover Voltage (VBO), current suddenly increases in value, limited only by load conditions. The device has switched itself into a state of low impedance by internal electron-hole rearrangement and stays in this state until current is reduced below the holding value. The thyristor then reverts to its ‘blocking state’. With the gate connected and an electrical bias of the correct relation to the anode applied, it is found that the VBO value can be controlled but, as stated, once the thyristor has ‘fired’ the gate has no further effect. CONTROL METHODS. The load-current value can be controlled by the duration the thyristor conducts and this can be varied in one of 2 ways: (1) Phase-shift Control, or by (2) Burst-fire or Burst-triggering Control. (1) Phase-shift Control. Figure 12.24 illustrates the method. The device is triggered by a short-duration current pulse applied to the gate during each +ve half-cycle of the supply voltage to the anode, cathode and load circuit. The instant of pulse application or triggering can be altered in relation to the start of a supply voltage half-cycle so the conduction time for the half-wave can be controlled and the average current value through the load is varied. A triggering circuit usually involves a variable potential divider and capacitance circuit with variable time constant. The voltage value built up across the capacitor initiates conductance through a transistor which then discharges through a resistor, thus applying a pulse to the thyristor’s gate. Figure 12.24 shows how this control method can trigger a thyristor at an instant when the supply voltage is high and a large current can be switched suddenly. This condition can give rise to radio interference and suitable suppression arrangements are needed. (2) Burst-fire Control. Figure 12.25 shows the method. The thyristor is switched on for several consecutive positive half-cycles. Conduction is thus for one or more half-waves

Load current

+

Conduction period

0

Time

Current

– + Gate pulse 0

Time

▲ Figure 12.24

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Solid-State Electronics: The Transistor • 349 of load current with periods of non-conduction interposed. The method is suitable for large loads where thermal or mechanical inertia ensures that the effect can be smoothed out to equate to that of a smaller constant (averaged) current. A gate pulse is applied by a timing circuit at the instant just before the start of a +ve half-cycle of the main circuit voltage and as the latter rises from zero, load current rises in a sinusoidal manner and falls similarly. This occurs for each conducting half-cycle and radio interference will be minimised. The number of conduction cycles can be varied by the duration of pulsing the gate but the heating at the gate electrode must be taken into account by the circuit designer. Since the gate supplies current during the pulsing period, this source of added temperature rise can be reduced using a H.F. supply for the trigger circuit.

Current

+

Conduction period

0 Time – +

Current

Gate pulse

0

Time

▲ Figure 12.25

Practice Examples 12.1

A transistor connected in a common-emitter circuit shows changes in emitter and collector currents of 1.0 and 0.98 mA, respectively. What mA change in base current produces these changes (2 decimal places), and what is the current gain of the transistor (2 significant figures)?

12.2

In a certain transistor, a change in emitter current of 1 mA produced a change in base current of 0.1 mA. Calculate the common-emitter and common-base short-circuit current gains (both 1 decimal place).

12.3

A certain transistor has a common-emitter short-circuit current gain α′ or β = 50 or β = 50. Calculate the common-base short-circuit current gain (2 decimal places).

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350 • Advanced Electrotechnology 12.4

For a common-emitter transistor, the transfer characteristic is given by the values: Ib (mA)

20

40

60

80

100

120

Ic (mA)

1.4

2.5

3.6

4.7

5.8

6.9

Find the current gain (2 significant figures). 12.5

From the output characteristic shown in figure 12.16, deduce the transfer characteristic, and hence find the current gain for the transistor when connected in the common or grounded-emitter mode (1 decimal place).

12.6

The data given in the table refers to a P-N-P transistor in the common-base configuration. Draw the collector-current/collector-voltage characteristic for the various values of emitter current and calculate the resistance of the transistor in kΩ (3 significant figures). Collector voltage Vc (volts)

Collector Current Ic (mA) Emitter current Ie = 0 mA

Emitter current Ie = 2 mA

Emitter current Ie =4 mA

Emitter current Ie = 6 mA

Emitter current Ie =8 mA

–5

0

–1.9

–3.7

–5.7

–7.6

–30

–0.1

–2.0

–3.8

–5.8

–7.7

–55

–0.2

–2.1

–3.9

–5.9

–7.8

12.7

From the data given for example 12.6, plot the collector current/emitter-current or transfer characteristic for Vc = –30 V and hence calculate the current gain α (2 decimal places). From this value deduce the current gain (β) for the commonemitter mode (2 significant figures).

12.8

The data given in the table refers to a P-N-P transistor in the common-emitter configuration. Collector voltage Vc (volts)

Collector current Ic (mA) Base current Ib = −20 mA

Base current Ib = −40 mA

Base current Ib = −60 mA

Base current Ib = −80 mA

−3

−0.91

−1.6

−2.3

−3.0

−5

−0.93

−1.7

−2.5

−3.25

−7

−0.97

−1.85

−2.7

−3.55

−9

−1.0

−2.05

–3.0

−4.05

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Solid-State Electronics: The Transistor • 351 Plot the collector current/collector voltage characteristic for base currents of −20, −40, −60 and −80 μA and using these determine (a) the output resistance of the transistor for the Ib = −60 μA condition, in kΩ (1 significant figure) and (b) the current gain when the collector voltage is −6 V (2 significant figures). 12.9

From figure 12.26, calculate the value of the bias resistor to give a standing base current of 40 μA in kΩ (3 significant figures). The supply voltage is 9 V and for the transistor VBE = −0.2 V and VCE = −5 V. – RL

RB

c b e +

▲ Figure 12.26

12.10 Figure 12.27 is given for a typical common-emitter amplifier. If the current through the emitter resistor is 0.5 mA, determine the battery voltage (1 decimal place). Assume a base-emitter voltage drop of 0.1 V. –

47 kΩ

4.7 kΩ

4.7 kΩ 1.2 kΩ +

▲ Figure 12.27

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13

ELECTRONIC AIDS TO NAVIGATION We have always been taught that navigation is the result of civilization, but modern archeology has demonstrated very clearly that this is not so. Thor Heyerdahl

Previous chapters have looked at both the theory and application of A.C. and D.C. machines and transformers, before introducing solid-state semiconductor devices. In these last two chapters we will introduce recent developments in electronic aids for navigation and modern display devices which present relevant navigational awareness information to the maritime user, concluding this present volume of Advanced Electrotechnology for Marine Engineers. Probably the most important non-human sensor to date for Officers on civilian and naval vessels is the use of modern radar systems, discussed in a little detail in Volume 14 Stealth Warship Technology in the chapter titled ‘Radar’. Prior to the introduction of electronic and inertial navigation, navigation was a hazardous affair relying upon a combination of piloting, dead reckoning, experience and celestial navigation to cross the unknown oceans of the world. Radar gave a sudden step change in capabilities, but pulse delay range radar shows only a contact without providing a target image, unlike thermal imagers and Near Infra Red (NIR) Image Intensifier ‘night vision’ devices. Laser Ranging systems, using the same operational principle as radar ranging, now routinely maintain safe distances during Replenishment at Sea (RAS) and both basic theory of laser operation and ship deployment of such a RAS system will be discussed. We will briefly examine the history of radar and marine radar applications before discussing the Automatic Radar Plotting Aid (ARPA) which allied with the use of Electronic Chart Display and Information Systems (ECDIS) in the civil domain offer a sophisticated

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Electronic Aids to Navigation • 353 layered approach to charting, mirrored by the use of War-fighting ECDIS (WECDIS) in the military sector to provide further war-fighting layers in such a database approach. Further vessel positional information is provided by use of now common place Global Positioning Systems (GPS) and Automatic Identification of Ships (AIS), as well as Emergency Personal Indicating Radio Beacons (EPIRB). Radio navigation related aids arising from the creation of wide area interference patterns, based on the use of multiple transmitter systems such as DECCA, OMEGA and LORAN-C, will be introduced, but electronic navigation aids are more fully discussed in Volume 15, Electronics, Navigational Aids and Radio Theory for Electrotechnology Officers. Several generations of display devices now exist side by side from the bulky older Cathode Ray Tube (CRT) to the modern Liquid Crystal Display (LCD), as well as Plasma, Organic Light Emitting Diode (OLED) displays, as well as Electroluminescent and holographic displays which are still some way off from shipborne deployment but offer exciting possibilities for future for marine navigation applications and will be discussed in our final chapter.

Radar Radar, the first sensor mentioned, was secretly developed by several nations as a matter of urgency both immediately before and during World War II. The term RADAR was first coined in 1940 by the U.S. Navy as an acronym for ‘RAdio Detection And Ranging’ being synonymous with the British system developed at the same time termed ‘Radio Aid for Direction finding And Ranging’. There are many varied modern uses of radar, including commercial marine and naval radars to locate other ships, hazards and landmarks, ocean surveillance systems, Air Traffic Control (ATC), as well as space and land-based systems and even ground-penetrating radar. Maritime development of radar arose from the work of a German inventor Christian Hülsmeyer who was the first scientist to use radio waves to detect distant metallic maritime objects. In 1904 he demonstrated the ability to detect a ship in dense fog, and patented this device although he was unable to provide valuable ranging information. A little later in 1922 Hoyt Taylor and Leo Young, researchers for the U.S. Navy, with a transmitter and a receiver on opposite sides of the Potomac River discovered that a ship passing through the beam caused the received signal to fade in and out as the ship interrupted the direct path between transmitter and receiver. Taylor suggested that this could be used to detect ships in low visibility, but the Navy was slow to take this work forward. Nonetheless, prior to World War II, researchers notably in the United

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354 • Advanced Electrotechnology States, Nazi Germany, the United Kingdom, Japan, Holland and the Soviet Union developed several technologies that led to the inception of radar. In 1934, following systematic studies on the microwave generating magnetron, the research branch of the Compagnie Generale de Telegraphie Sans Fil (CSF) headed by Maurice Ponte developed an obstacle locating radio apparatus, part of which was installed on the Normandie liner in 1935 with the first radar pulsed system demonstrated in December 1934 by American Robert Page working at the U.S. Naval Research Laboratory. In the 1930s with the rearmament of Nazi Germany and the significant expansion of the German Luftwaffe the British government became alarmed at these developments and recognised the urgent requirement for an effective aircraft warning system to be developed quickly and deployed in the imminent likelihood of war in Europe. Driven by fears of such a German Luftwaffe attack on mainland Britain Robert Watson-Watt was approached in January 1935 as to whether radio might be used to detect aircraft approaching the shores of England. He wrote a memorandum to the Air Defence Subcommittee of the Committee of Imperial Defence to promote developments of radio Direction Finding (DF). From this a concentrated radar development programme began with the first prototype radio DF station (latterly called the Chain Home Command system) built at Bawdsey Manor near Felixstowe, Suffolk, under the control of the Air Ministry in 1937 for the Royal Air Force (RAF) operating at a relatively low frequency of 22 MHz. His team’s work resulted in the design and installation of the aircraft detection and tracking Chain Home Command System along the entire South and East coasts of England just in time for the outbreak of World War II in 1939. The system provided vital advanced information that helped the RAF win the Battle of Britain. Under good atmospheric conditions it could detect propeller-driven aircraft as far away as 150 km at 3000 m elevation providing sufficient time for successful interception.

Marine radar applications The information provided by radar includes both range and bearing, providing precise location information for any object along the sea surface, but insufficient for an aircraft or missile at an unknown elevation, thus the requirement for some sort of height finding device. Although the first use of radar was intentionally for military purposes – to locate air, ground and sea targets, through technology transfer – it has now migrated into similar civilian applications for ship and aircraft movements. Ship radars are used to measure the bearing and distance of ships to prevent collision with other ships, to navigate safely and to fix their position at sea when within range of shore or other fixed references such as light houses and islands. In port or harbour, vessel traffic service radar systems monitor and regulate ship movements in busy sea lanes.

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Electronic Aids to Navigation • 355 In essence a radar system has a transmitter that emits a focused beam of radio waves in a specific direction. When these come into contact with an object some of the waves may be reflected or scattered in different directions. Radar signals are reflected very well by materials of high electrical conductivity, such as metals, seawater and wet land. It is the radar signals that reflect back towards a transmitter that most radar use. In addition if an object or ‘target’ moves either towards or away from the radar transmitter there will be a small change in the returned radio wave frequency, caused by the Doppler Effect which can help discriminate against unwanted background returns. Radar receivers are usually but not always located in the same place as the transmitter and are termed monostatic radar. Radar with a separately located receiver are known as bistatic radars, and where multiple receivers are used the term multistatic radar is used. For the case of marine radar we will only discuss monostatic radar and so bistatic and multistatic systems will not be discussed further. The reflected radar signal detected by even a relatively large ship’s search radar is usually very weak, but can be amplified significantly using sophisticated electronic amplifiers and digital signal processing methods and with modern display systems now provide a surprisingly high degree of accuracy. However, it must be pointed out that there are many things which do not show up on radar displays and the information provided by the radar on its own provides at best an incomplete tactical picture and situational awareness of the navigational and possible war-fighting scene. Radar is without doubt regarded as an essential component for safety at sea both near and far from shore. A ship’s officer needs to be able to manoeuvre his ship within feet in the worst of conditions and to navigate ‘blind’. One student I taught returning to Brunei, during periods of the worst forest fires in that region, spent the best part of a month navigating blind using instrumentation and without visible contact. In commercial ships radar is fully integrated into a suite of marine instruments including chart plotting aids, sonar, two-way VHF radio (and other bands) and emergency locators such as EPIRB. Use of radar is an essential aid to avoiding collision and is used to get the most accurate information of ship movements and to assess the risk of collision in terms of bearing, distance, Closest Point of Approach (CPA) and to display other ships in its vicinity. Marine radar systems provide useful and highly accurate radar navigation information for navigation officer on-board ships, and ship position can be fixed by the bearing and distance information of land targets on the radar screen. Shipborne radar will typically have performance adjustment controls for brightness and contrast, gain, swept gain, methods to reduce sea and rain clutter, minimise interference and noise reduction. Common controls consist of range scales, bearing and distance cursors, and range markers. As radar relies on its own transmissions it is available 24/7 and in almost every weather condition. However, radar use from a military perspective is very much

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356 • Advanced Electrotechnology a double-edged sword, as radar operation can reveal its presence to another vessel in a conflict. Radio and the visible part of the spectrum provide two ‘windows’ of good atmospheric transmission where the weak absorption of radio waves by the atmosphere enables radar to detect targets at relatively long ranges, ranges at which other electromagnetic wavelengths, such as visible light and NIR, are much too strongly attenuated. Weather phenomena such as fog, clouds, rain, snow, sleet and volcanic ash that block visible light are often transparent to radio waves. From an aircraft perspective volcanic ash is a deadly hazard as it does not show up well at usual radar frequencies. Certain radio frequencies that are absorbed or scattered by water vapour, rain drops or atmospheric gases such as oxygen are generally avoided when designing radars, except when their detection is intended, i.e. for weather radar. Radar waves scatter in many different ways depending on the relationship between the size of the radio wave and a target’s size and shape. If the wavelength is much smaller than the target’s size, waves will reflect in a similar way to light reflected from a mirror. If however the wavelength is much bigger than the target size it may not be visible at all due to poor reflection. Low-frequency radar also depends strongly on resonances for detection, and cannot provide direct target identification. Short-wave radio waves reflect from corners in a way similar to optical reflection from smooth glass, so most reflective targets for short wavelengths have 90° angles between the reflective surfaces. A corner reflector consisting of 2 or 3 flat surface meeting as within a box has a structure which reflects waves directly back to the source over a broad angle range. They are commonly used as radar reflectors to make otherwise difficult to detect vessels, easier to ‘see’. Consequently corner reflectors on boats make them more detectable to avoid collisions or helping during search and rescue. For military reasons, platforms intending to avoid detection are designed to remove such surfaces and edges in anticipated threat directions, leading to the unusual shapes of modern stealth ships and aircraft and use of sophisticated metamaterials. These precautions do not completely eliminate reflection because of diffraction, especially at the longer wavelengths. Half wavelength long wires or strips of conducting material, such as chaff, are very reflective but do not direct the scattered energy just back to the source. The extent to which an object reflects or scatters radio waves is called its Radar Cross Section (RCS), a measure of the detected signal compared with a reference target. The basic principle of radar, or more strictly echo location, is quite simple. Visible light, with which we are all familiar, is simply part of a wider family of waves which form the electromagnetic spectrum covering waves such as X-rays, gamma rays and ultraviolet radiation which all have more energy than visible light, and also waves of longer

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Electronic Aids to Navigation • 357 wavelength such as Infra Red (I.R.) heat, radar and Very Low Frequency (VLF) waves used to communicate with submerged submarines with energies less than visible light. If these radar frequency electromagnetic waves are sent out as short duration directional pulses from a ship’s navigation radar they may strike an object with a flat surface, so that some of the wave energy transmitted by the radar is reflected back to the radar receiver (figure 13.1). Radar transmitter Outgoing pulse

Returning echo

▲ Figure 13.1

If the elapsed time t from transmission of the short radar pulse to the time the echo is received is measured electronically knowledge of the wave speed allows the contact range to be calculated accurately. Since electromagnetic waves travel close to the speed of light in vacuum, and only a little less in our planet’s thin atmosphere, with a value of c = 3 × 108 ms−1, the distance they travel may be given as follows: distance = ct. Hence contact range R, which is half the total distance, is given by: R = ct/2. If the elapsed recorded time from pulse transmission to reception is 1 ms the radar range from the transmitted to the reflecting target will be given by: R = 3 × 108 × 1 × 10−6 / 2 = m. This relationship is used in pulse radars to measure contact range, and the method is commonly referred to as ‘pulse delay ranging’. Other important radar parameters include the transmitted radar frequency f, the Pulse Repetition Frequency or PRF (the number of pulses transmitted per second), the pulse duration τ and the Antenna Rotation Rate (ARR) measured in Revolutions Per Minute.

Maximum Detection Range (MDR) and Radar Cross Section (RCS) An individual search radar’s Maximum Detection Range (MDR) depends on several factors such as transmitted power, a ship’s reflecting properties, the antenna or aerial size and the receiver’s sensitivity. The shipborne MDR is given by simplifying the standard radar range equation, to provide the maximum range a radar will detect a chosen ship (or target) of a given size.

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358 • Advanced Electrotechnology

MDR = 4

PGσ t ot Aefff ( 4π )2 Smin

Here P is the average transmitted power, G the antenna gain, i.e. the ship’s RCS (a measure of the contact size seen by the radar beam), tot is the time the radar contact is illuminated by the rotating radar beam, Aeff is the effective size of the receive antenna and Smin is the minimum signal energy required for detection by the receiver. Changes to any of these parameters will alter the MDR. Generally wave loss increases with increasing frequency and so the MDR falls. Clearly a ship’s search radar has control of all but the elusive RCS of the vessels around it and the environmental conditions. Not all ship platforms, from a navigation or search radar perspective, may have good radar reflecting surfaces. This may arise due to the materials from which a ship is made, e.g. Glass Reinforced Plastic (GRP) is a weak radar reflector compared with metal, and the ship surfaces may be inclined at an angle to the vertical so much of the reflected radar energy is deflected away from the search or navigation radar system and therefore undetected. However, as mentioned earlier, ship geometry can be put to advantage in the form of corner reflectors, dihedral and trihedral surfaces (where 2 or 3 surfaces respectively meet together at 90°) to generate significant radar returns, which are used to great effect to enhance radar returns and thus the detectability of commercial vessels, passenger ferries and privately owned yachts, enhancing their visibility in busy shipping lanes. Pulse forming network

Master oscillator

Modulator

Radar

Power amplifier

Duplexer

Signal processor

Receiver

Display

▲ Figure 13.2 Radar block diagram

A monostatic pulsed radar system (figure 13.2) will include the following components: An Oscillator which creates a continuous lower power electromagnetic wave. The frequency of the output low-amplitude signal is designed to work about a certain frequency, but may vary within certain limits.

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Electronic Aids to Navigation • 359 A Modulator which converts a single continuous wave of energy into a series of regularly spaced, identical pulses of fixed duration. It is basically controlled by discharge of a bank of capacitors. A Power amplifier which boosts the modulator output signal power to increase radar range. A Duplexer which allows the same antenna to both transmit and receive detected signals, in turn. This unit separates the sensitive receiver system from the highpower transmitter as it would be damaged if exposed to the high power of the outgoing pulses. A Receiver when reflected target echoes are detected by the antenna they are very weak. Before signals can be analysed, they must be amplified before signal processing. TRANSMITTED POWER. This is usually measured in the range of 10–100 kW, and is the power transmitted from the radar antenna. Long-range search radars may have a transmitted power in the region of a MW. Compare this with a microwave oven rated perhaps at only 800 W. Marine radar designers and operators usually want to know the average power produced by the radar over a period of time and this is directly affected by 3 factors: peak power Pp, pulse duration τ and PRF, as Pav pp × τ × PRF . PEAK POWER. For radar peak power is the maximum pulse power, but the average power is much lower as the radar power is zero during the receive period (which should be much greater than the time the radar transmits). Increasing peak power increases average power up to practical limits where a radar may overheat or sparking could occur. Secondly, increasing the time each pulse transmits increases the radar’s average power, however this means the radar listens for less time. Finally increasing the rate the radar transmitter can send out pulses, as the PRF can vary from a few hundred to over ten thousand pulses per second for certain maritime military radar. Increasing the PRF also increases the radar’s average power but also decreases the time a radar listens as well, introducing earlier radar resetting pulses.

Antennas Small radar antennas are especially vulnerable to wind and precipitation. As well as physical damage, precipitation causes changes to an antenna’s electrical properties so to protect high-frequency antennas, they are often enclosed in radomes made of GRP designed for strength and transparent at the transmission frequencies. The size and

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360 • Advanced Electrotechnology shape of a radar antenna is important, and their size is determined by the wavelengths used. The larger the wavelength the larger an antenna should be. However, the size that can be used may be limited by factors such as the space available. The simplest radar antenna is a dish reflector (as in figure 13.2) which acts as a focusing mirror for radiation and is similar to a torch’s focusing reflector and is usually of parabolic shape. Marine radar also has the ability to distinguish 2 or more targets along the same bearing direction. Signal pulses occupy a certain length of space in proportion of the speed it travels at and the amount of time the pulse is transmitted. As radar pulses travel at the speed of light, even a pulse duration of a few microseconds will have a pulse of considerable length. Thus if 2 contacts lie on the same bearing close to each other it is possible for the return signals from the targets to overlap and the radar incorrectly detects only a single contact (figure 13.3).

Outgoing pulse Radar transmitter Returning overlapping echoes

▲ Figure 13.3 Radar resolution and multiple target discrimination

The formula Range Resolution = ct/2 is extremely helpful here. If vessels are separated by more than the range resolution the vessels will be resolved or discriminated separately, but if vessels are less than this distance apart they will not. It also means the shorter the pulse duration, the better the range resolution of the radar system. A similar problem exists in angular resolution, the ability to distinguish 2 contacts close together at the same range, separated by a small difference in their bearing from the radar. This is critical to correctly deter the number of contacts such as concentrations of fishing vessels. Radar can only separately resolve 2 contacts provided the vessels’ angular separation is more than the radar beamwidth. However, once a conventional radar is designed and installed there is little flexibility to alter the size of the beam unlike the case of range resolution just discussed. The formula for radar beamwidth is useful 60 λ to consider here. Beamwidth α = where λ is the wavelength, and D is the aperture D size of the transmitting antenna, e.g. if the wavelength is 1 m and the aperture size is 60 × 1 = 20°. 5 m the Beamwidth in degrees will be α = 5

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Electronic Aids to Navigation • 361

Signal Processing When an echo signal is received it is often ‘mixed up’ with signals from other radars, background noise and clutter. A signal processor needs to filter and clean up the incoming signal converting them into a suitable display format.

Clutter Radar returns are produced from nearly all types of surfaces when illuminated by radar. Therefore in competition with the ship’s radar returns there are many sources of unwanted echoes from our radar transmissions. Clutter may include sea returns, returns from cliffs and weather itself. The definition of clutter depends on the radar function, e.g. weather is not clutter in a weather detecting radar. The two main types of clutter observed on a Plan Position Indicator (PPI) display are Surface Clutter and Volume Clutter. Surface clutter is composed of sea returns associated with the movement of waves. As wind speed and sea state increases, so does the amount of sea clutter which may be appreciable from close range. Short-range sea clutter can be a major problem at short range on a radar display but may be eliminated by correct use of Swept Gain or Sensitivity Time Control (STC) which reduces the radar amplifier’s sensitivity at close range. Volume clutter covers all types of precipitation and typical volume clutter depends on the wave frequency, droplet diameter and precipitation rate. Rain gives larger echoes than snow or hail. Volume clutter requires use of a Differentiator circuit, which looks at the Fast Time Constant (FTC) response at the leading edge of a rain storm and is often referred to simply as ‘Anti Rain’.

Noise Noise is a random source of energy which can either be man-made or of natural origins. Noise is most apparent in regions with low signal levels, such as the weak received echo-signal in a radar receiver. Noise is characterised by its statistical properties and if it contains frequencies of equal amplitudes it is often referred to as white noise. Noise occurs in almost all electronic devices and results from a variety of effects. Man-made noise sources include other radar and communications systems, and electronic circuits.

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362 • Advanced Electrotechnology Natural occurring noise sources include thunder storms and solar events. The effects of noise can be mitigated by the comparison of several consecutive pulses which may have been reflected from the same marine platform, and uses important relationships in not only frequency and amplitude but phase as well.

Environmental effects The condition of the atmosphere in terms of pressure, density, humidity, particulates, etc. has a significant bearing on the transmission of radar signals. Although there is extensive use of conventional rotating swept radar systems modern electronically scanned array radars are now becoming more widespread. In this case a radar beam is steered electronically and is produced from a 2D array of emitters. In some cases, power is supplied to the masthead through waveguide trunking up the mast itself (a passive array) or from active microwave emitters mounted on the masthead itself (an active array). These radar will not be discussed further here.

Lasers Lasers for marine applications The laser is a device that emits light through a process of optical amplification based on the stimulated emission of electromagnetic radiation. The term laser comes from the acronym for Light Amplification by Stimulated Emission of Radiation (LASER). Modern lasers can operate right across the electromagnetic spectrum and have extremely good spatial coherence in that the output being a diffraction limited narrow beam can be focused into a very tiny spot, achieving a high irradiance and can also achieve high power levels even at considerable range. A laser typically consists of a gain medium, a mechanism to supply energy to it and something to provide optical feedback. The gain medium is a material with properties that allow it to amplify light by stimulated emission. Light of a specific wavelength that passes through the gain medium is amplified. For the gain medium to amplify light, it must be supplied with energy, through the process of pumping. Energy is typically supplied as an electrical current, or as light at a different wavelength. Pump light may be provided by a flash lamp or another laser.

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Electronic Aids to Navigation • 363 The most common type of lasers uses feedback from an optical cavity with a pair of mirrors at each end of the gain medium. Electromagnetic radiation bounces back and forth between the mirrors, passing through the gain medium and is amplified on each transit. Typically one of the two mirrors, the output mirror is partially transparent so some of the light escapes through this mirror.

Stimulated emission Quantum mechanical effects cause individual electrons to take on discrete positions in orbits so that electrons are found in specific energy levels of an atom, two of which are shown in figure 13.4. Before emission Excited level

After emission

E2

hν

hν Incident photon

ΔE

E1 Atom in excited state

hν

Ground level

Atom in ground state

E2 – E1 = ΔE = hν

▲ Figure 13.4 Laser stimulated emission

When an electron absorbs energy from a light photon it receives that incident quantum of energy. Transitions are only permitted between discrete energy levels such as the two shown in figure 13.4. When an electron is excited from a lower to a higher energy level it will stay there for a short interval of time. An excited state may decay to a lower energy state which is unoccupied, after a particular time constant or relaxation time characterising that transition. When such an electron decays without external influence to emit a photon it is called spontaneous emission. The phase associated with the emitted photon is random. A material with many atoms in such an excited state may result in radiation which is spectrally limited, centred about one specific wavelength, but individual photons have no common phase relationship and move in random directions. However, an external electromagnetic field at a frequency associated with a transition can affect the atom’s quantum mechanical state. As the electron in the atom makes a transition between 2 states it enters a transition state which has a dipole field, and which oscillates at a characteristic frequency so the probability of the atom entering

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364 • Advanced Electrotechnology the transition state increases. The rate of transition between the 2 stationary states is enhanced above that due to spontaneous emission. A transition from the higher to a lower energy state will now produce an added photon of the same frequency and phase through the process of stimulated emission.

Laser safety Laser products are usually labelled with a safety class number which identifies how dangerous a laser is. Class 1 is inherently safe, usually because the light is contained in an enclosure, e.g. in CD players or those used in supermarkets. Class 2 is safe during normal use, as the human eye blink reflex will prevent damage. Class 3A involves a small risk of eye damage within the time of the blink reflex. Staring into such a beam for several seconds is quite likely to damage the retina. Class 3B can cause immediate eye damage on exposure. Class 4 lasers are the most dangerous, as they can burn skin, and in some cases even indirectly scattered light has been known to cause eye and/or skin damage.

Laser range finding and maintaining ships’ proximity in RAS A laser range finder is a device which uses a laser beam to determine the distance to an object in the same way as a microwave radar system does for navigation or search radars. At optical frequencies the wavelength is very short and so very precise distance measurements can be made. Some systems with very short and well-shaped laser pulses can range an object to just a few millimetres. Range finders provide an exact distance which is ideal for military and civilian engineering applications. Laser Range finders are generally considered to be eye safe, meeting the Class 1 safety requirements. Several laser systems can maintain an exact distance between vessels necessary for RAS operations. One such system, Laser Atlanta SPROX (Ship PROXimity) @TM, is a recognised ‘laser distance measuring kit’ ensuring ships’ crews maintain a safe distance during underway replenishment and RAS, and is approved by the U.S. Chief of Naval Operations and by NATO for its accuracy and ease of use. The SPROX system measures and clearly displays ship distances in a fraction of a second with great accuracy on an easy-to-read waterproof ElectroMagnetic Interference (EMI)-protected LED display. An integrated Central Processor Unit (CPU) automatically analyses and

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Electronic Aids to Navigation • 365 stores recorded readings. The portable equipment easily withstands rugged outdoor use and lowers the risk of dangerous manoeuvres while at sea, ideal for sea operations in all weather conditions, and can measure ranges of up to 5000 m to an accuracy of ± 0.5 m.

Automatic Radar Plotting Aid (ARPA) Next in our discussion is the Automatic Radar Plotting Aid (ARPA). Radar with ARPA fitted can create tracks using radar contacts. The system can calculate tracked objects’ courses, speeds and Closest Points of Approach (CPA), thereby establishing if a danger of collision with another ship or landmass exists. ARPA development began following the sinking of the Italian liner SS Andrea Doria which collided in dense fog off the East coast of the United States. The first commercially available ARPA was installed on the cargo liner MV Taimyr in 1969 manufactured by Norcontrol, now a part of the naval manufacturer Kongsberg Maritime. ARPA-enabled radars are now available for most small yachts as well. The International Maritime Organisation (IMO) sets certain standards concerning the International Convention for the Safety Of Life At Sea (SOLAS) requirements regarding the carrying of suitable automated radar plotting aids. The primary function of ARPA is to improve the standard of collision avoidance at sea, to reduce the workload of observers by enabling them to automatically obtain information so they can perform as well with multiple targets as they can by manually plotting a single target. A typical ARPA creates a presentation of the current situations and uses computer technology to predict future situations. ARPA assesses the risk of collision, and enables operators to see proposed movements of their own ship. ARPA generally provides the following characteristics: true or relative motion radar presentation, automatic target acquisition with manual acquisition, digital read-out of acquired targets to provide course, speed, range bearing and CPA, the ability to display collision assessment information directly on to the Radar’s Plan Position Indication (PPI) display, the ability to perform trial manoeuvres including course, speed and combined course speed changes and automatic stabilisation for navigation purposes. ARPA processes radar information more rapidly than conventional radar but are still subject to the same limitations. ARPA data are only as accurate as the data that come from its data inputs. It should be noted that similar electronic tools exist for Air Traffic Control in aviation as well.

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366 • Advanced Electrotechnology

Electronic Chart Displays Electronic Chart Display and Information System (ECDIS) Electronic Chart Display and Information Systems (ECDIS) are computer-based navigation information systems that comply with IMO regulations which can be used authoritatively as an authenticated alternative to paper nautical charts and are now quickly becoming established as the tool of preference. An ECDIS system displays the information from Electronic Navigational Charts (ENC) or Digital Nautical Charts (DNC) and integrates position information of real vessels from position, heading and speed through water reference systems and other navigational sensors such as AIS. ECDIS provides continuous position and navigational safety information, and can provide audible and visual alarms when a vessel is in proximity to dangerous navigational hazards. The two commonest types of electronic chart data are ENC and raster charts. ENC charts are vector charts that conform to the requirements for ECDIS chart databases with standard content, structure and format, and are issued for use with ECDIS on the authority of the government-authorised hydrographic office, conforming to the International Hydrographic Organisation’s (IHO) specifications. ENCs contain all the chart information required for safe navigation but may contain supplementary information in addition to that contained on paper charts. Systems using ENC charts can be programmed to give warning of impending danger in relation to the vessel’s current position and future predicted movement. Raster Navigational charts are raster charts that also conform to IHO specifications and are produced by converting paper charts to digital images by scanner. Images are similar to digital camera pictures which may be ‘zoomed in’ for more detailed information as can ENC.

War-fighting ECDIS WECDIS systems contain all the layers of information maintained within an ECDIS system but provide additional layers of information for naval War-fighting ECDIS (WECDIS) which add to civilian-based information layers. This subject will not be discussed any further here.

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Electronic Aids to Navigation • 367

Radio Navigation Aids Interference-based navigation-related sensors Navigation is the process of planning and controlling vessel movements from one location to another. All navigational techniques involve locating a navigator’s position compared to known locations, and several important navigation-related sensors help navigators achieve this. Radar navigation uses radar to determine distance from or bearing of objects whose position is known, and is primarily used within radar range of land. Radio navigation on the other hand uses radio waves to determine position by either radio DF systems or hyperbolic systems, such as LORAN-C, and may be used out of sight of land. LORAN-C is similar in principle to Omega, and Decca which was discontinued due to the advent of Global Positioning Systems (GPS). Satellite navigation uses artificial earth satellite systems such as GPS to determine position and is used worldwide. I will focus briefly on the use of LORAN-C as it arises from combinations of 2 source interference. Such systems unlike active radar and laser-based sensors operate in the mode of passive radio navigation. Vessels need not emit any signals to obtain positional information.

LORAN-C LORAN (LOng RAnge Navigation) is a terrestrial navigation system using Low-Frequency (LF) radio transmitters. Before satellite-based GPS systems, it was the primary location method used in marine applications. The United States, Canada and Japan still uses LORAN. Russia uses a similar system called CHAYKA. The current version, LORAN-C, is an ocean navigation aid, requiring a special receiver. A Master transmitter and Slave transmitter emit pulses at the same frequency, and is an example of 2 source interference. Measuring the difference between the arrival times of the two pulses determines on which hyperbolic Line of Crossed LOPs from the 2 master-slave pairs, and provides the observer at point P with a hyperbolic fix (figure 13.5). The LOP is the locus of all points in space where the observed time differences between arrival of signals from the 2 paired stations are constant, forming a hyperbola. LORAN-C is thus described as a

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368 • Advanced Electrotechnology

P

M

X

Y LOP 1 LOP 2

▲ Figure 13.5 Two source interference

hyperbolic radio navigation system. A LORAN-C chain consists of a Master transmitter with 2, 3 or 4 Slaves designated W, X, Y and Z. LF transmitters operate at 100 kHz and emit groups of typically microsecond duration pulses at a specified interval, the Group Repetition Interval (GRI), unique to each chain and selected to avoid interference with other chains. Slave station transmission is delayed with respect to the Master, the secondary coding delay, ensuring pulse groups are always received in sequence (M, W, X, Y, Z). The master transmits 9 pulses while slaves transmit 8, with the master’s 9th pulse used for identification. LORAN was an American development of the British GEE radio navigation system, used in World War II, with a range of about 400 miles. In LORAN, one station is kept constant in each application, the Master, paired separately with 2 other Slave stations. Given 2 secondary stations, the Time Difference (TD) between the Master and first secondary identifies one curve, and the TD between the Master and second secondary identifies another curve. The intersection of lines determines a geographic point in relation to the position of the 3 stations. These curves are referred to as Time Delay or TD lines. Essentially we measure the distance we have travelled by counting the number of constructive to destructive half wavelengths moved. Modern LORAN receivers display Latitude and Longitude instead of signal TDs, with high accuracy.

Limitations of LORAN LORAN suffers from the effects of weather and also from ionospheric effects near sunset and sunrise. The most accurate signal is the Ground wave that follows earth’s surface along a water path (best surface conductivity). At night indirect Skywaves, taking paths refracted back to the earth’s surface by the ionosphere, are a problem as multiple signals arrive from different paths. The ionosphere’s response to sunset and

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Electronic Aids to Navigation • 369 sunrise accounts for significant disturbances and errors during these periods. Magnetic storms can also seriously effects LORAN-C performance (in the range 90 to 100 kHz), as it is an RF-based system. Due to such interference, the very phenomenon LORAN is based upon, and further propagation loss suffered by LF signals from land features and man-made structures, LORAN signal accuracy degrades significantly in land. With the growth in GPS-based systems it is possible LORAN will also be replaced as was Omega, although LORAN uses a strong signal which is difficult to jam, and also acts as a complementary independent system of electronic navigation in the unwanted advent of satellite system failure. Omega was itself the first global radio navigation system for aircraft. It enabled aircraft and ships to determine their position by receiving VLF radio signals in the range 10 to 14 kHz, and was transmitted by a small network of fixed terrestrial radio beacons until it was shut down in 1997.

Global Positioning Systems These are satellite-based systems designed to accurately plot the user’s location anywhere on or above the world’s surface any time of day or night and in all weather conditions. These systems all work by using accurate timed radio signals using a standard based on the Caesium atomic clock, to pinpoint their range from a number of satellites at known orbital positions. Using various satellite ranges, projected geometric spheres can be intersected to determine a user’s position, in a similar manner to radar Direction Finding. To be able to measure the time taken for the signal to reach us, the satellite and the receiver codes must first be synchronised. Each satellite transmits a unique digital code, which is synchronised with the code on the GPS receiver. The code contains information about the position of the satellite and the time the code was transmitted (figure 13.6). Time difference between identical parts of GPS code

Signal from satellite Signal from ground receiver

▲ Figure 13.6 Synchronisation of satellite and ground receiver codes

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370 • Advanced Electrotechnology It is then possible to work out when the signal left the satellite, as the same code is used by the receiver and the satellite. It is then necessary to look at the incoming code from the satellite and see how long ago the receiver generated the same code, and then to work out the exact time the coded signal left the satellite. The signal from one satellite determines the receiver to be somewhere on a sphere, a set radius from the satellite. A second subsequent measurement narrows the receiver’s position to be narrowed down to the intersection of 2 spheres. A third measurement narrows this down further to just 2 points in space (figure 13.7). In practice 3 measurements are usually enough to determine a position accurately, if the receiver height is known. Of the 2 possible solutions from the software algorithm one can usually discard one point because it will usually be a ridiculous answer, either very far out in space or moving at exceptionally high speed. For an accurate position the receiver clock should be synchronised with the various satellites’ clocks, which requires a fourth satellite to be detected. With an inaccurate receiver clock, the fourth measurement will not intersect with the other 3 previously found. So the receiver works out a single clock correction factor that it subtracts from its timing measurement to make all the signals intersect at a single given point. Two possible positions, at the intersection of 3 spheres corresponding to the 3 time delays found.

▲ Figure 13.7 Finding surface position with GPS

The U.S. GPS NAVSTAR project was the first developed in 1973 to overcome the limitations of previous navigation systems. GPS consists of 3 major segments: a space segment, a control segment and a user segment (figure 13.8). GPS users are only really concerned with the visual representation of the latter component, and of course, its accuracy.

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Electronic Aids to Navigation • 371 Space segment

Control segment

User segment GPS receiver

▲ Figure 13.8 Segments of the GPS system

The space segment is composed of the orbiting GPS satellite in 6 orbital planes with 4 satellites in each. The 6 orbital planes have approximately 55° inclination (tilted relative to Earth’s equator) and are separated by 60° right ascension of the ascending node (the angle along the equator from a reference point to the orbits’ intersection). Orbits are arranged so that at least 6 satellites are always kept within line of sight from virtually anywhere on the earth’s surface (figure 13.9). Currently there are about 30 satellites in the U.S. Navstar GPS constellation orbiting at an altitude of about 20 200 km.

▲ Figure 13.9 GPS satellite constellation

Each satellite continuously broadcasts a navigation message on the L1 Coarse Acquisition (C/A) frequency and L2 Precise (P/Y) frequency as well as further L3, L4 and L5 frequencies. All satellites broadcast at the same frequencies. Signals are encoded

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372 • Advanced Electrotechnology using Code Division Multiple Access (CDMA) allowing messages from individual satellites to be distinguished from each other based on unique digital encodings from each satellite that the receiver must be aware of. Two distinct types of CDMA encodings are used: the C/A code which is accessible by the general public, and a P(Y) code, which is encrypted so only the U.S. military can access it. All satellites broadcast at the same 2 frequencies L1 (1.57542 GHz) and L2 (1.2276 GHz). The satellite network uses a CDMA spread-spectrum technique where a low bit rate message data are encoded with a high rate Pseudo Random Number (PRN) digital sequence that is different for each satellite. The receiver must know the unique PRN code for each satellite to reconstruct the actual message data. The C/A code, for civilian use, transmits data at 1.023 M code bits, or chips, per second, while the P(Y) code, for U.S. military users, transmits at 10.23 M chips per second. The L1 carrier is modulated by both the C/A and P(Y) codes, while the L2 carrier is only modulated by the P(Y) code. The P(Y) code can be encrypted as a socalled P(Y) code that is only available to military equipment with a proper authorised decryption key which prevents unauthorised users accessing the accurate positional information. Both the C/A and P(Y) codes impart the precise time of day to the user, although relativistic corrections are needed due to orbital eccentricity (2% eccentricity gives rise to an 8.6 ns delay).

NAVSTAR The first operational GPS was the U.S. Navigation Satellite Timing and Ranging GP system or NAVSTAR GPS. The NAVSTAR GPS is currently managed by the U.S. Department of Defence (DoD) who launched the first generation of test satellites (Block I) in 1978. The first operational satellites (Block II and Block IIA) were launched from 1989 onwards and now the next generation of satellites (Block IIR) is being launched.

GLONASS The second of the world-wide GPS systems is the Russian Ex-Soviet Global Navigation Satellite System or GLONASS. It is operated by the Russian Federation Government and it uses many of the same principles as its U.S. counterpart. The first satellites were launched by the Soviet Union in 1982 and the system became fully operational in 1991. The system was completed in 1995 but could not be maintained by the Russian Federation and fell into a state of disrepair for a period of time with only 8 satellites still operational in 2003. The Russian government has now fully restored the GLONASS system to operational status which was also opened up to civilian users in 2007, with further third-generation satellites to be launched into the system over time.

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Electronic Aids to Navigation • 373

GALILEO The European Union (EU) GALILEO project was formed from the need to have a separate system independent from the United States, to allow the European Union to stand alone on its own if ever needed. It also stemmed from the need for a more accurate horizontal and vertical positioning for civilian engineering and aviation requirements not readily available from the other existing networks. The system will be interoperable with NAVSTAR GPS and GLONASS, thereby increasing constellation visibility and positional accuracy with the use of a greater number of satellites. GALILEO will achieve better coverage at high latitudes (Northern Europe) and also make up part of the Global Maritime Distress and Safety Service (GMDSS) and exceptionally high accuracy is achieved using signals from additional strategically located ground stations (differential GPS). One company TOPCON now provides receivers which can detect GPS NAVSTAR, GLONASS and GALILEO to provide unprecedented accuracy from over 80 separate satellites. The system was developed from 2001 to 2005 with the deployment of the satellites due to take place from 2007 which was realised with several satellite launches in 2013. The European Union and its partners of the European Space Agency have signed agreements with non-European countries, most notably China, although China has its own plans to extend its own current Beidou capability (limited to Asia and the West Pacific) to create COMPASS, the People’s Republic of China Global System, planned to be operational by 2020. India intends its own regional navigational system, to be operational starting from 2014 covering India and the Northern Indian Ocean, the IRNSS system. Satellite systems provide considerable position-fixing advantages: continuous 24-hour coverage (day and night), worldwide coverage, 3D positioning, weather independence, a common reference system (WGS84) and very fast data processing.

Global Maritime Distress Safety System (GMDSS) The GMDSS is a maritime communications system for all vessels. It is used for emergencies and can be used for vessel to vessel, vessel to shore and shore to vessel routine communications. Commercial vessels over 300 gross tonnage and certain smaller vessels, including some fishing boats, must use GMDSS equipment. Most offshore yacht races now require yachts to be GMDSS enabled. GMDSS is composed of several components: radio Digital Selective Calling (DSC), Navtex for weather and navigation information, Search And Rescue (SART) Transponders (Radar transponders)

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374 • Advanced Electrotechnology and Emergency Position Indicating Radio Beacons (EPIRBs). Even if private leisure crafts do not have all 3 individual components VHF DSC should be considered the very minimum requirement. Equipment should comply with international specifications as in Europe in accordance with European Union Radio and Telecommunications terminal Equipment (R&TTE) directives. DSC provides the ability to raise a Distress Alert sending out a boat’s identity (MMSI), its position and the type of distress. The DSC alert will be heard immediately by all DSC-equipped vessels and shore stations within range, VHF provides typically 30 to 50 nm range. The distress message will be repeated automatically after a set number of minutes until acknowledged either by a Coastguard rescue coordinator centre or ship within radio range, as well as safety broadcasts, routine calls and group calls. All big ships and almost all European Coastguards are fully equipped for DSC. GMDSS is a world-wide system, which can be used anywhere. Being a digital data exchange network it can also provide marine and port communications, satellite and radio telephone services and a degree of aerial photography and colour diagrams. GMDSS is a fully automatic digital system and as such it can avoid potentially life-threatening language problems that may otherwise arise!

Automatic Identification System (AIS) The Automatic Identification System (AIS) is an autonomous and continuous broadcast system operating in the VHF maritime mobile band. It exchanges information such as vessel identification, position, course, speed, etc. and can also be applied to Aids to Navigation. The International Convention for the Safety Of Life At Sea (SOLAS), 1974, requires AIS to be fitted aboard all ships of 300 gross tonnage and upwards engaged on international voyages, cargo ships of 500 gross tonnage and upwards not engaged on international voyages and all passenger ships irrespective of size. The requirement became effective for all ships on 31 December 2004, and it is estimated that over 76 193 ships now have AIS fitted (October 2013). Regulations require that AIS shall provide information including the ship’s identity, type, position, course, speed, navigational status and other safety-related information automatically to appropriately equipped shore stations, other ships and aircraft. AIS needs to automatically receive such information from similarly fitted ships, monitor and track ships and to exchange data with shore-based facilities. AIS will display different classes of vessels with different symbology, from passenger vessels to large tankers. AIS data are available on the internet, e.g. www.marinetraffic.com.

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Electronic Aids to Navigation • 375 AIS transponders automatically broadcast information via VHF transmitter. The information originates from the ship’s navigational sensors, typically its Global Navigation Satellite System (GNSS) receiver and gyrocompass. Other information such as the vessel name and VHF call sign are programmed when installing the equipment and is also transmitted regularly. The signals are received by AIS transponders fitted on other ships or land-based systems, such as Vessel Traffic Services (VTS) systems. The received information can be displayed on a screen or chart plotter, showing the other vessels’ positions. There are 2 main classes of AIS unit recognised by the IMO: Class A, mandated for use on SOLAS vessels and Class B, a low-power lower cost derivative for leisure and non-SOLAS markets. AIS units can transmit and receive 26 different message types. Data are sent every 2 to 10 seconds depending on the vessel’s size and speed while underway, or every 3 minutes if the vessel is at anchor. AIS will state the vessel’s Maritime Mobile Service Identity (MMSI), navigation status, rate of turn, speed over the ground, position, course over ground, true heading and time stamp. Additional information such as ship type, cargo, ship dimensions, draught and destination ETA is broadcast less frequently on up to 6 minute intervals. AIS was developed in the 1990s as a high-volume, short-range identification and tracking network, and at that time, it was not anticipated that it would be detected from space. However, AIS systems have now been demonstrated in space by various companies and by the European Space Agency, e.g. in 2010 the Norwegian AISSat 1 satellite demonstrated an AIS system with a receiver made by Kongsberg Seatex. The big challenge for AIS satellite operators is the ability to receive a large number of AIS messages simultaneously from a satellite’s large reception footprint. The addition of satellite-based Class A and B messages could enable truly global AIS coverage, but are likely to augment the existing terrestrial-based system. MARITIME SECURITY. AIS enables authorities to identify specific vessels and their activity within or near a nations’ Exclusive Economic Zone (EEZ). When AIS data are fused with existing radar systems, authorities can differentiate between differently classed vessels more easily. AIS data can be automatically processed to create normalised activity patterns for individual vessels which when breached create an alert, thus highlighting potential threats for more efficient use of security assessment. Real-time AIS position data are freely available on the internet through variously privately operated Geographic Information Systems (GIS). Satellite AIS, unlike binoculars, do not need line of sight and clear visibility to operate and so are not restricted by fog or where a land mass obstructs the view.

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376 • Advanced Electrotechnology

Emergency Beacons Distress radio beacons, also known variously as emergency beacons, Personal Locator Beacons or PLBs, Emergency Locator Transmitters (ELTs), or Emergency Position Indicating Radio Beacons (EPIRBs), are tracking transmitters which can aid in the detection and location of sinking vessels, aircraft and people in distress. They are radio beacons that interface with worldwide services of Cospas-Sarsat, the recognised international satellite system for Search And Rescue (SAR). When manually activated, or automatically activated upon immersion, such beacons send out a distress signal. The distress signal is monitored worldwide and the location of the distress is then detected by various non-geostationary satellites, and then located by a combination of GPS and Doppler triangulation. The main purpose of a distress radio beacon is to help rescuers find survivors within the so-called golden day (the first 24 hours following a traumatic event) during which most survivors are usually saved. Since the inception of Cospas-Sarsat in 1982, distress beacons have helped rescue over 28 000 people in some 7000 distress situations. In 2010 alone, the system helped rescue 2388 persons in 641 distress situations. Most PLBs are brightly coloured and waterproof. PLBs vary in size from cigarette-packet to paperback book size and weigh between 0.2 and 1 kg. EPIRBs and ELTs are generally larger, but will need to fit into a cube of some 30 cm a side, weighing between 2 and 5 kg. They can be purchased from marine suppliers, and hiking supply stores. The units have a life of 10 years, and operate across the range of −40 to 40°C, and will transmit for 24 to 48 hours.

How a typical beacon operates All systems work something like this: a beacon is activated by a sinking vessel, or manually by survivors. The beacon’s transmission is picked up by one or more satellites. The satellite transmits the beacon’s signal to its ground control station. The ground station processes the signals and forwards the data, including the rough location, to a national Coastguard authority, e.g. Falmouth U.K. The national authority forwards the data to the appropriate rescuing authority. The rescuing authority uses its own receiving equipment to locate the beacon and makes the rescue or recovery. Once the satellite data are in, it takes less than a minute to forward the data to any signatory nation. There are several systems in use, with beacons of varying expense, different types of satellites and varying performance. Even the oldest systems provide an immense improvement in safety, compared to not having a beacon at all.

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Activation There are 2 ways to activate a beacon: manually or automatically. Automatic EPIRBs are water activated, while automatic ELTs are impact activated. Some EPIRBs also deploy; this means that they must physically leave their mounting bracket on the outside of the vessel (usually by going into water). For a marine EPIRB to start signal transmission it must come out of its bracket (or ‘deploy’). This can happen either manually – where someone has to physically remove it from its bracket, or automatically, where water pressure causes a hydrostatic release unit to release the EPIRB from the bracket. If it does not come out of the bracket it will not activate. There is a magnet in the bracket which operates a reed safety switch in the EPIRB, which prevents accidental activation in case the unit gets wet from rain or shipped seas (which is quite likely in rough weather). Once deployed, EPIRBs can be activated, depending on the circumstances, manually (by flicking a switch) or automatically (as water comes into contact with the unit’s ‘sea-switch’). Modern EPIRBs provide both activation and deployment methods and are labelled ‘Manual and Automatic Deployment and Activation.’

Hydrostatic Release Unit A Hydrostatic Release Unit (HRU) is a pressure-activated mechanism designed to automatically deploy when certain conditions are met. In the marine environment this occurs when submerged to a maximum depth of 4 m. The pressure of the water against a diaphragm within the sealed casing causes a plastic pin to be cut thereby releasing the containment bracket casing, allowing the EPIRB to float freely. Some common characteristics of HRUs are as follows: Water pressure sensitive at depths not to exceed 4 m or less than 2 m. They are used for single use only, require replacement if activated and cannot be serviced, only replaced. They are waterproof and sealed against moisture and tampering and labelled with an expiry date, which will be 2 years from the month of installation. Before fitting EPIRB will undergo tests for corrosion resistance, temperature, submergence and Manual Release, strength tests and further performance tests.

GPS beacon operation The most modern 406 MHz beacons with GPS track to a precision of 100 m in 70% of the world nearest to the equator, and send a serial number so the responsible authority can

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378 • Advanced Electrotechnology look up phone numbers to notify the registrator (e.g. next-of-kin) in about 4 minutes. The GPS system permits stationary, wide-view geosynchronous communications satellites to enhance the Doppler position received by Low Earth Orbit (LEO) satellites. EPIRB beacons with built-in GPS are usually called GPS Position-Indicating Radio Beacons (GPIRBs). Rescue will not begin until a Doppler track is available. COSPAS-SARSAT specifications state that a beacon location is not ‘resolved’ unless at least 2 Doppler tracks match or a Doppler track confirms an encoded (GPS) track.

High-precision registered beacon operation An intermediate technology 406 MHz beacon (now largely replaced by GPS-enabled units) has worldwide coverage and locates to within 2 km (12.5 km2 search area), notifying family and rescuers within 2 hours (46 min average), and has a serial number to look up family phone numbers, etc. This can take up to 2 hours because it uses non-Geostationary weather satellites to locate the beacon. To help locate a beacon, the beacon’s frequency is controlled to 2 parts per billion, with a power level of only 5 W. Both the above types of beacons usually include an auxiliary 25 mW beacon at 121.5 MHz to guide SAR aircraft. When considering emergency systems and GPS systems in general power is a vital matter. Even though GPs information should be rightly considered as only one possible choice of an aid to navigation, it is increasingly being viewed as the primary source for positional information, and as such is utterly dependent upon the system’s network of accurate satellite clocks. The problem is that with these satellites operating in orbits of tens of thousands of kilometres above earth’s surface the typically 100 W signal transmitted by a GPS satellite and the even smaller signal from an activated emergency location device can easily spread itself out over an area equivalent to an appreciable part of the earth’s surface, so that by the time it is detected it is both extremely weak and vulnerable to accidental and malicious interference, as well as deliberate jamming from even a relatively low-power transmitter.

Non-GPS Doppler location The Cospas-Sarsat system uses Doppler processing with terrestrial Local Unit Terminals (LUTs) detecting non-geostationary satellites monitoring the Doppler frequency shift recorded by Low Earth Orbit (LEO) and Medium Earth Orbit (MEO) SAR satellites as they pass over an activated distress beacon transmitting at a fixed frequency. Range and bearing are measured from the rate of change of the recorded frequency, which

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Electronic Aids to Navigation • 379 will vary according to the satellite’s path in space and the earth’s rotation to enable the activated beacon’s position to be triangulated. A faster change in the Doppler frequency indicates the beacon is closer to the satellite’s ground track as the Doppler shift’s magnitude depends on the relative motion along the direct line between the transmitting source of waves (the beacon) and the satellite (the detecting observer). The more accurate the beacon the more precisely it can be located, saving search time and potentially lives. A modern 406 MHz frequency beacon is now manufactured to an accuracy of 2 parts per billion, and yielding a search area of just 2 km2, compared to older beacons (accurate to 50 parts per million yielding 200 km2 of search area). To increase useful power levels, and handle multiple simultaneous beacons, modern 406 MHz beacons transmit in bursts, and then remain silent for about 50 seconds.

Satellites The Russian Federation developed the original different types of SARSAT system using several types of satellite, which substantially reduced the programme’s cost by including some weather satellites set in ‘ball of yarn’ orbits, inclined at 99°. The longest period that all satellites can be out of line-of-sight of a beacon is now reduced to about 2 hours. The first Sarsat satellite constellation was launched in the early 1970s by the Soviet Union, the United States, Canada and France. Some geosynchronous satellites now also have beacon receivers. Since 2003 there have been 4 such GEOstationary SAR satellites (GEOSAR) launched, now covering over 80% of the earth’s surface and located above the equator. Consequently GEOSAR satellites do not cover the polar caps, but as they see the Earth as a whole, they will see a beacon immediately, but having no motion (geostationary means they are fixed relative to the earth), they cannot provide a Doppler frequency shift to locate it. However, if the beacon transmits GPS data, a geosynchronous satellite can give a very rapid response.

Search and rescue response Emergency beacons currently operate at a frequency of 406 MHz and transmit a unique 15, 22 or 30 character serial number called a Hex Code. When the beacon is purchased the Hex Code should be registered with the relevant national (or international) authority. Registration provides SAR agencies with vital information such as phone numbers to call in the event of a disaster, the vessel’s description or person in the case of a PLB, the vessel’s home port and any further information that might be useful to SAR agencies.

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380 • Advanced Electrotechnology Registration information also allows SAR agencies to start a successful rescue more quickly. For example, if a shipboard telephone number listed in the registration is unreachable, it can be assumed that a real distress event is taking place. Conversely, the information provides a quick and easy way for the agencies to check and eliminate any false alarms (potentially sparing the beacon’s owner a significant false alert fine!). An unregistered 406 MHz beacon carries limited information: just the manufacturer and the beacon’s serial number, and in some cases, the MMSI. Despite the clear benefits of beacon registration, even an unregistered beacon is better than no beacon at all, because the Hex Code received from a 406 MHz beacon confirms the signal as a real SAR alert. Beacons operating on 121.5 and 243.0 MHz just transmit an anonymous siren tone, and carry no information to SAR agencies. Such beacons now rely solely on the terrestrial or aeronautical monitoring of frequencies, and are therefore of limited maritime value. In summary, various classes of location device have been used in recent decades with satellite-enabled automatic identification systems and alert detection abilities demonstrating the greatest utility. It is also expected that existing systems such as AIS will expand their capabilities to handle increasing numbers of vessels and increased data traffic as the volume of global maritime traffic increases. As the number of ocean and littoral marine traffic increases the vessels and passengers likely to benefit in a distress situation from use of such digital electronic navigations aids and GMDSS distress alert systems is rising. All marine users would be sensible to purchase, install and register up-to-date SOLAS-compliant equipment on their vessels for the undesired possibility of disaster at sea.

Practice Examples 13.1

13.2

A radar contact is detected using pulse delay ranging. If the elapsed recorded time from pulse transmission to reception is 0.2 ms, and given that c = 3 × 108 ms−1, find the radar range of the target (1 decimal place). PGσ t ot Aefff Considering the formula for Maximum Detection Range MDR = 4 ( 4π )2 Smin what will be the increase in MDR if the transmitted power is increased by a factor of 16? Is this worthwhile in practice and could this have any consequences for a military vessel?

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Electronic Aids to Navigation • 381

13.3

13.4

PGσ t ot Aefff if the transmitted power ( 4π )2 Smin is increased by 60% and the Gain of the antenna increased by 30% what will be the decrease in effective area caused by rotation of the radar antenna for the MDR to remain unchanged (2 significant figures)?

Considering the formula for MDR, MDR = 4

60 λ where λ is the wavelength, D and D is the aperture size of the transmitting antenna, find the horizontal beamwidth of a navigation radar with a 2.4 m wide aperture which operates at a frequency of 9.5 GHz (2 decimal places). Using the equation for radar beamwidth, α =

13.5

If a long-range search radar has a peak power Pp of 1.1 MW, a pulse duration τ = 60 ms and a PRF = 700 pulses per second, use the formula Pav Pp × τ × PRF to find the average power of the search radar (3 significant figures).

13.6

Explain how a GPS system works.

13.7

Use a web-based AIS application to track your vessel over a period of several days.

13.8

Compare the NAVSTAR GPS and GLONASS satellite systems.

13.9

State how a GMDSS system operates.

13.10 Explain the operation of an EPIRB system.

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14

DISPLAY DEVICES There must be a beginning of any great matter, but the continuing unto the end until it be thoroughly finished yields the true glory. Sir Francis Drake

At the heart of all the marine engineering systems discussed in both Reeds Electrotechnology volumes lies the issue of being able to monitor system performance, and for the human operator to adjust the system parameters as required to achieve the desired system output levels, e.g. power. To do this the system must have some output transducer which provides the required decision-making information in a form a human being can understand. This information is usually provided in visual form on some sort of ‘display’ or monitor. Clearly it is important for the Master, Deck Officers or Ratings to have good eyesight in order to view the display correctly (i.e. there are no problems with visual fields, colour vision, night blindness, double vision, etc.) as laid down in STCW minimum in-service eyesight standards. However, it is equally important that the displays provide the required standard necessary to present both data and images to the operator unambiguously. Over the past few decades there has been a significant growth in the diversity of new display systems building upon the first widespread use of Cathode Ray Tube (CRT) and Oscilloscope displays. In this chapter we will look at several common display systems that can now be found on merchant and naval vessels, starting with the CRT, considering some of their advantages and disadvantages.

Cathode Ray Tube The CRT is a vacuum tube containing one or more electron guns (electron emitters) and a fluorescent screen which accelerates and deflects an electron beam onto a screen to view images which can represent maritime radar targets, electrical waveforms and pictures for TV and computer monitors. A CRT uses an evacuated glass envelope which

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Display Devices • 383 is large and heavy, and being made of glass is relatively fragile. In the interest of safety, the screen face is usually made of thick shatter-resistant lead glass which also blocks X-ray emissions. CRTs have now largely been superseded by new display technologies such as: LCD, plasma and OLED displays which have lower manufacturing costs, power consumption, weight and bulk. In TV and computer monitors, the front of the tube is scanned repetitively in a fixed pattern or raster scan. In modern CRT monitors and TVs, beams are bent by magnetic deflection, a varying magnetic field is generated by coils and driven by electronic circuits around the tube’s neck, although electrostatic deflection is often used in oscilloscopes. Most displays are now based on Liquid Crystal Display (LCD) technology with only a few specialist applications and older marine fits using CRTs, although market position was not long ago reversed. The basic CRT structure has a heated Cathode electrode, a Grid electrode through which electrons can pass (negative with respect to the Cathode) and an Anode electrode (positive with respect to the Cathode) mounted in a vacuum, figure 14.1. CRT deflection coils

Focusing electrode

Cathode Thermionic heater

Beam spot

Phosphor screen Vacuum Electron beam Anode with conductive coating

Grid electrode

▲ Figure 14.1

When a cathode is heated some of its electrons gain enough energy to escape the electrode surface. The electric field between the anode and cathode draws them towards the positive anode while the electric field between the grid and cathode opposes that between the anode and cathode, so the difference between them controls the number of electrons that reach the anode. Making the grid more negative reduces the number of electrons reaching the anode. In a CRT the anode is arranged to allow electrons to pass through it and strike a phosphor-coated screen to produce visible light. Grid voltage controls the brightness of light produced. A focusing system often has an electrostatic lens which helps narrow the electron beam to a point where it strikes the phosphor.

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384 • Advanced Electrotechnology Two common deflection types are used: (1) Electrostatic. Plates on either side of the beam have voltages applied between them, deflecting the beam towards the more positive plate (used in all screen displays). (2) Electromagnetic. Coils on either side of the tube produce a magnetic field across it, deflecting the electron beam. TV and large screen displays use this deflection method.

Production of displays To measure time a reference is provided on a CRT which causes the spot to travel across the screen at a constant speed. The signal that causes this is called a time base. Deflection to the right B

Time

A

C

D

Deflection to the left

▲ Figure 14.2 A typical time base waveform for TV and marine radar ‘A type’ displays

A to B shows a steadily increasing deflection which moves the spot from the left-hand edge of the screen A to the right-hand side B, writing a line across the screen in the process. B to C shows a rapid return of the spot back to the left-hand edge of the screen, usually occurring while the spot is switched off electronically. C to D shows the waiting period before D at which point the trace is redrawn. The vertical distance CB controls the width of the trace, the time interval AC controls the spot’s speed across the tube, and the range scale, and the time interval AD controls the frequency with which the trace is repeated, the Pulse Repetition Frequency (PRF). RASTER SCAN. This type of scan is used in various marine computer, radar, sonar, TV and thermal imaging displays. The display consists of many horizontal lines spaced down the screen. The higher the display resolution the greater the number of lines.

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Display Devices • 385 Lines are produced using 2 time-bases. The X-time base is a high-frequency time base which writes lines across the screen. The Y-time base is a lower frequency time base causing the spot to move steadily down, writing each horizontal line underneath the previous one. Information is displayed by modulating the spot intensity and this is achieved by altering grid potential. In TV the modulation produces a continuous variation along each line. A colour CRT TV picture is produced with 3 electron beams and a single deflection system. Electron beams illuminate 3 phosphors: red, green and blue, and the screen colours are combinations of these. A raster scan causes the 3 beams to scan the screen together and by modulating the 3 electron beams different colours are produced along the trace. The brightness, colour and illumination persistence depend on the CRT screen phosphor types used. Phosphors are available with persistences ranging from microseconds to seconds. If you look at a screen closely you will see the 3 coloured phosphors, arranged in stripes or clusters called ‘triads’. A shadow mask uses a metal plate with tiny holes, so an electron beam only illuminates the correct phosphor on the tube face; holes are tapered so electrons that strike the inside of a hole are reflected back, if they are not absorbed, instead of bouncing through a hole to strike a wrong coloured spot on the screen. Computer display, producing graphics and data use a dot-matrix display. A horizontal line is broken into a number of points, each of which can be individually addressed by computer. By illuminating selected points, characters are drawn on the screen. The first cathode rays experiments are credited to J. J. Thomson, an English physicist with the earliest CRT invented by German physicist Braun in 1897. It was a cold-cathode diode, a modification of the Crookes tube with a phosphor-coated screen. The first CRT to use a hot cathode (easy to emit many electrons) was developed by Johnson and Weinhart with the first commercially made electronic CRT TV manufactured by Telefunken in Germany in 1934.

Oscilloscope CRTs In oscilloscope CRTs, electrostatic deflection is used, rather than the magnetic deflection used in TV and larger CRTs because the magnetic coils’ inductive reactance limits the instrument’s frequency response. A beam is deflected horizontally by applying an electric field between a pair of plates to its left and right, and vertically by applying an electric field to plates above and below. If the shadow mask becomes magnetised, its magnetic field will deflect electron beams passing through it, causing colour distortion as the beams bend through the mask holes to hit ‘wrong’ phosphors and not those they were intended to strike; e.g. some electrons from a green beam may hit red phosphors, giving red parts of the image a

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386 • Advanced Electrotechnology wrong ‘tint’. This effect will be localised to a specific screen area if the magnetisation of the shadow mask is localised so it is important the shadow mask is demagnetised. Most colour CRT displays, i.e. TV and computer monitors, have a built-in degaussing circuit, the main component of which is a degaussing coil round the edge of the CRT face to prevent magnetisation. On powering-up a CRT display, the degaussing circuit produces a brief A.C. current through the degaussing coil which decays in strength to zero over a few seconds, producing a decaying A.C. magnetic field from the coil. This field is usually large enough to remove any magnetisation. In cases of strong magnetisation where the internal degaussing field is not sufficient, the shadow mask may be degaussed externally with a portable degausser. An excessively strong magnetic field, whether A.C. or constant, may mechanically bend the shadow mask, causing permanent colour display distortion which looks similar to a magnetising effect. To repeat degaussing, the CRT display is switched off and left for several seconds, allowing the circuit to cool to ambient temperature; switching the display off and immediately back on results in a weak degaussing cycle.

CRT advantages CRTs have high contrast ratios (over 15 000:1), and excellent colour and have no input lag. There is no ghosting and ‘smearing’ artefacts during fast motion due to sub-millisecond response time. CRTs have near zero colour, saturation, contrast or brightness distortion and provide a wide viewing angle range. They can be used or stored in extremely hot and cold temperature conditions without harm to the system.

CRT disadvantages Although the backbone display technology for decades, CRT-based computer monitors and TVs are now largely a redundant technology. The demand for CRT screens dropped significantly after 2000. Rapid advances and falling prices of LCD flat panel technology have been a key factor in the demise of competing display technologies such as CRT, which despite recent advances have remained relatively heavy and large, taking up a lot of space compared with other display technologies. Even a 50 cm display can weigh more than 10 kg. CRT screens require large vacuum tubes compared with flat panel displays for a given screen size, and are impractical beyond 102 cm. CRT disadvantages became decisive in the light of rapid LCD and plasma flat-panel technological advancements which allow them to exceed 102 cm besides being thin and wallmountable, 2 key features increasingly demanded by consumers. CRTs have high power consumption, typically consuming up to 10 × the power an identical sized LCD monitor

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Display Devices • 387 consumes, depending on the type of backlight LCD screen, and brightness. A great deal of heat can be emitted while in operation, due to the high power consumption and they can suffer screen burn-out. Image edges are blurred and distorted compared with the sharp image of a flat panel pixel array as text, computer-generated diagrams, line drawings and other images composed mostly of sharp edges, require very sharp pixel definition. CRTs are still popular in specific market niches and provide better colour fidelity, contrast and better off-axis viewing (wider viewing angle) than many flat screen LCDs. CRTs emit a small amount of X-ray radiation from the electron beam’s bombardment of the shadow mask/aperture and phosphors but the amount of radiation leaving the monitor is considered harmless. Older colour and monochrome CRTs contain toxic substances, such as cadmium, in the phosphors. Many countries treat CRTs as toxic waste and prohibit disposal in landfills or by incineration. In Europe, disposal of CRT televisions and monitors is covered by a relevant EU Directive. As electronic waste, CRTs are considered one of the hardest types to recycle having high concentration of lead and phosphors, both of which are needed for the CRT display. Under some circumstances, signals radiated from the electron guns can be captured remotely to show what was on the CRT. Special maritime TEMPEST shielding can mitigate against this. Such potentially exploitable signals, however, occurs with other display technologies and electronics in general. CRTs are affected by earth’s magnetic field and are factory preset biased for operation in north or south hemispheres, or equatorial regions, but may require ‘trimming’ if crossing between hemispheres on the world’s oceans, as ‘crossing the bar’ is not an uncommon event. Adjustment requires considerable technical skill, besides the required safety precautions associated with opening the CRT display housing. CRTs are sensitive to magnetic interference, which causes images to shimmer (e.g. if a transformer or other electro-magnetic source is close to the screen) or colours change (if unshielded speakers are also too near).

Liquid Crystal Displays Liquid Crystal Displays (LCDs) are now available in a wide range of devices from simple digital watches to mobile phones, laptops and high-resolution flat panel TVs. There has been an incredible pace of development in LCD technology over the past 100 years with possibilities for future sensing and communications applications. They are available in a wider range of screen sizes than CRT and plasma displays, and as they do not use phosphors, the screen surface will not burn out. LCDs are more energy efficient than CRTs as they require no cathode heating and LCDs are often made up of segments filled

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388 • Advanced Electrotechnology with liquid crystals and arrayed in front of a light source or reflector to produce both colour and monochrome images.

A brief history of liquid crystals Just over 125 years ago on 14 March 1888, Austrian botanist and chemist Friedrich Reinitzer noted that the organic material cholesteryl benzoate showed unusual behaviour on melting. Initially, it became a cloudy liquid on heating to 145°C and then clear at 187.5°C. Subsequent studies on the part of eminent scientists of his day – Lehmann, Vorlander and Friedel – established that the intermediate phases of this and similar compounds represented new states of matter which Lehmann named liquid crystals. As a general rule a substance in a liquid crystal or mesomorphic state is strongly anisotropic (dissimilar in different directions) in some or all of its properties and exhibits a high level of fluidity comparable with an isotropic liquid. The existence of liquid crystal behaviour among inorganic substances is uncommon, compared with about 1 in 200 of organic compounds. Several thousand organic compounds exhibit liquid crystalline behaviour with further synthesis adding to this number daily. For Liquid Crystal (L.C.) behaviour to occur it is essential the molecules are highly geometrically anisotropic, usually long rod-like molecules, although they need not be rod-like, in fact doughnut- or disk-shaped molecules will do. Depending on the molecular geometry the liquid may pass through one or more phases on heating before it ends up as an isotropic liquid where all its properties are the same in every direction. Transitions to these intermediate states are brought about by either thermal heating processes, or solvents. The second of these occur when an L.C. is dissolved in a suitable solvent, most often water, and the L.C. phase formed over a specific solvent concentration range. These so-called lyotropic L.C.s are generally derived from long helical rod-type structures or amphiphilic compounds, (viruses are typical long rod shaped molecules, e.g. the Tobacco Mosaic Virus (T.M.V.) which has large molecular dimensions: 300 Å long by 200 Å wide). This class is characterised by having a section of the molecule hydrophobic (water hating) and a section hydrophilic (water loving). As the concentration of water solvent is increased the organic molecules form various packing structures from a simple layered (lamellar) to a cubic and finally a hexagonally close packed phase (figure 14.3). Lyotropic L.C. phases are of fundamental importance in biological systems and occur abundantly in nature, including the human body. Phospholipids are amphiphilic molecules which form the basis of cell membranes, which mixed with a small amount of water stack into multilayered or smectic liquid crystals. Amphiphilic molecules

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Display Devices • 389

Water

Water

Cubic

Lamellar

Water

Hexagonal close pack

▲ Figure 14.3

organise themselves into bilayers separated by water so the head groups lie at the lipid–water interface. A familiar example is soap (sodium dodecyl sulphate) in water. In the lamellar phase, water is sandwiched between the polar heads of adjacent layers, while the hydrocarbon tails are in a non-polar environment. In the cubic phase layers bend to form spherical units. In the hexagonal phase the layers are rolled up into tight cylinders. However, the L.C.s of the greatest present benefit to the display industry are the thermotropic types. This phase commonly used in flat panel, low-power display devices is the nematic L.C. which has long range order of the molecular axis (figure 14.4). There was little interest in commercial applications in the display area until the development of the first L.C. electro-optic device using ‘dynamic scattering’ noted in 1968 by Heilmeier. Competition and progress were then swift leading to the Twisted Nematic (T.N.) display device which have made LCDs widespread in everyday life. Schadt and Helfrich described what is now known as the T.N. effect, where an electric field applied perpendicular to the plates of an L.C. cell realigns in the bulk to a uniform nematic with the long molecular axes lying perpendicular to the plates (homeotropic).

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390 • Advanced Electrotechnology Surface alignment direction

Smectic layering

SC phase SA phase Nematic phase

▲ Figure 14.4

In the T.N. state if cell thickness is much greater than the incident wavelength the twisted structure guides light through the cell. But under application of a D.C. voltage the uniform homeotropic nematic no longer guides light through the sample as waveguide conditions change too fast, giving high contrast ratios between light and dark states (figure 14.5).

Observer Polarised light

Backlit Liquid crystal off Random light

Vertical polariser

Common electrode

Segment electrode

Horizontal polariser

No light emerges

Liquid crystal on

▲ Figure 14.5 TN cell

The simplest practical cell is the T.N. cell. Light of one polarisation passes through the first polaroid (polariser) and then a second polaroid (analyser). If the second polaroid is twisted by 90º the half twist structure will transmit polarised light. Voltages do not need to be high and are seldom above about 12 V as it is the electric field which is responsible for molecular reorientation (E = V/d where d is the cell thickness). Since typical liquid crystal cells in the nematic phase are about 10–20 μm thick, and in the more ordered smectic phases less than 5 μm, field strengths are typically about 1 MV/m.

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Display Devices • 391

Liquid crystal phases Following the method proposed by Friedel liquid crystals are classified into 3 types: nematic, cholesteric and smectic. NEMATICS. Friedel coined the word ‘nematic’ from the Greek nema, meaning a thread, and refers to certain thread-like defects (or disclination lines) which are commonly observed. A schematic representation of the order in a nematic phase is shown in figure 14.4 and a real optical texture of a typical nematic L.C. phase is shown in figure 14.6.

▲ Figure 14.6 Nematic texture

In a nematic phase molecules are on average aligned so their axes are parallel so that a preferred direction is imposed on the system (figure 14.4). The optic axis or director coincides with the defined direction. For most nematics there is rotational (equal) symmetry about the long molecular axis (cigar-like) so the phase is identical in all its properties at right angles to its long axis (uniaxial). The absence of long range positional order determines the nematic phase’s fluid character, typically with a viscosity of 0.1 Poise, 10 times more than pure water at room temperature. A nematic phase can occur either with molecules that have mirror symmetry, or with an equal mixture or right- and left-handed cholesteric or chiral nematic phases. CHOLESTERICS. The Cholesteric or Chiral Nematic (N*) phase in addition to long-range orientational order has a gradual twisted spatial variation of the director leading to a helical structure. The cholesteric phase is represented in figure 14.7. If a series of planes perpendicular to the helical axis is considered, orientational order exists in each plane as in the nematic phase. However, local molecular alignment is rotated slightly between adjacent planes. The most striking feature of cholesterics is their strong optical activity

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392 • Advanced Electrotechnology

▲ Figure 14.7 Cholesteric phase

and selective light reflectivity. Optical activity and selective light reflectivity are due to the twisted structure and inherently related to the temperature-dependent twist pitch. SMECTICS. Smectics comes from the Greek smegma, meaning soap, with the 2 simplest commercialised smectic phases being the Smectic A and C* phases. Structurally all smectics are layered, with a well-defined layer spacing, measured by X-ray diffraction. The first X-ray evidence to support well-defined layer spacing was obtained by E. Friedel, the son of G. Friedel who discovered the Smectic A phase. Smectics are more ordered than nematics and have lower symmetry than the nematic and cholesteric phases. A Smectic A phase is also shown schematically in figure 14.4. Inside each layer there is no long-range order. Smectic C* phase. In this more complicated phase the long molecular axis is on average aligned with all the other molecules in the layer but tilted with respect to the layer normal by an angle ‘Θ’ (figure 14.4). However, the presence of a permanent molecular dipole attached to a molecule makes an L.C. ferroelectric, which means the molecule can reorientate under an applied electric field to change the crystal symmetry. All L.C. structures exist in various textures, the optical patterns observed in thin layers with a polarising microscope. These textures are normally complicated by line defects at which an abrupt change in molecular orientation, or ‘disclination’ occurs. For device applications it is vital the L.C. phase which forms must be uniform over large areas, if not all the display. Once formed L.C. defects have a bad habit of either growing or staying firmly put after temperatures cycling, although controlled temperature cycling can help ‘iron’ out some defects.

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Display Devices • 393 A chevron in the layering. As temperature reduces from the Smectic A to Smectic C* phase the layers start to shrink. As the layers are firmly anchored at the aligning surface, to maintain the fixed spacing of the smectic A phase (the density wave layer spacing), the Smectic C* layers must tilt as a response to the ‘shrinking’ which occurs. Something similar is seen when you take a stiff piece of upright card and try to bend it by pushing inwards from both ends. Initially it bows or ‘bananas’ out slightly and can then suddenly kink (or chevron) near the middle of the card. The stiffer the card the greater the energy required to deform the card. Similarly in the L.C. layering, when sufficient energy is supplied the layer suddenly kinks and the kink deformation which forms, just as in the card, is ‘locked’ into the cell leaving a ‘memory’ if you like, even after temperature cycling. This layer shrinking and layer tilting forms a clear ‘Chevron’ in the layers (figure 14.8) and was verified experimentally by X-ray scattering in the late 1980s and by optical probing techniques [1–2].

▲ Figure 14.8 Chevron in SC* layering

Various frustrated liquid crystal modellers came up with many fruitless and fruitful models to explain what was going on, including the ‘banana’ model which for several years seemed a more plausible model than the idea of a single ‘kink’ found near the middle of a thin L.C. cell. To further confuse things the chevron in the layers caused molecules to give a large fixed optical ‘twist’ out over most of the cell, irrespective of the L.C. molecules anchoring at the surface.

Switching speeds But why all the interest in Ferroelectric Liquid Crystal (F.L.C.) displays anyway? Two reasons initially, speed and memory (or bistability). For the various effects which LCDs employ the ‘switch on’ ton and ‘switch off ’ toff times are often quoted at 90% and 10% saturated response respectively. High-speed devices rely on electron movement, while thermal and molecular processes are unfortunately slower, but how slow? LCDs are in the slower category because the process involves realignment of molecules. In the ferroelectric phases the realignment process is given by coupling the ferroelectric dipole Ps to the Electric field E so the 1/τ switching time is proportional to Ps × E. Unfortunately this was not the full story, as although scientists started to synthesise

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394 • Advanced Electrotechnology new Smectic C* liquid crystals, it was later shown fast potential switching speeds were offset by the presence of a dielectric term –εE2 giving rise to a minimum in switching time. Switching times typically are milliseconds rather than hoped for microseconds but still offer many potential applications, although unsuitable in current form for fast digital data rate communications applications. A second problem involves a cell which sits in one of 2 optical states optically equivalent to a digital ‘on’ or ‘off ’ until application of a voltage of opposite polarity. This potentially reduces problems encountered by large area nematic LCDs where the time between pulsing a voltage across individual display elements or ‘pixels’ becomes large due to the physical display size. Instead of a clear dark/light contrast partial reorientation of molecules occurs so most of the time elements give an average ‘grey’ intensity as molecules ‘relax’ back to their resting state. Under application of a D.C. biased voltage the equal likelihood (or degenerate bistability) of these 2 states is broken, so dipoles in both regions try to align in the same direction parallel to an applied electric field. Reversing the applied electric field polarity causes both regions to switch into the other possible uniform state. Two types of defects are often observed in display cells, designated ‘lightning’ (figure 14.9a) and ‘hairpin’ (figure 14.9b) defects for obvious reasons. These defects are

Lightning defect

Hairpin defect >>>*