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English Pages [49] Year 2004
SP.764, Practical Electronics
Dr. James A. Bales
Lecture 1: Course Introduction
Topics: 1) Resistors, and the voltage divider. 2) Voltage divider as a power supply. (Make 3 volts out of 5 volts with voltage divider.) Know when it is a good idea and when it is not. 3) Get intuitive sense of voltage divider. 4) Simple rules of thumb. 5) Diodes: Diodes are one of the fundamental building blocks of electrical circuits. Most common diode is the LED (Light Emitting Diode). Demos:
Red Light Emitting Diode (LED) The LED lens gathers the light and focuses it into a beam. If an LED is introduced in water, light disperses in every direction. Since the index of refraction of plastic is similar to that of water, the LED plastic casing doesn’t act as a lens anymore.
120LED array.
The 120LED array shown above has an intense cyan color with a broad angle of light. About 2530 watts are being put into the apparatus. (The red LED was about a 10th of a watt.) Some LED arrays may appear to be more efficient (while consuming less power) than others. However, this is because a lens is used to concentrate the light, as opposed to the 120LED array shown above that was designed to disperse light.
Theory: 1) Voltage dividers a. Review b. As Voltage Source
2) Diodes
a. Ideal Diode b. Our Model Diode c. Real Diodes d. Driving LEDs e. Diodes as Voltage Source
+ V IN R1
VOUT
I↓ R2
SP.764 Practical Electronics Dr. James A. Bales
⎛ R2 ∆V1 ∆V2 =I= → VOUT = V IN ⎜⎜ R1 R2 ⎝ R1 + R 2
⎞ ⎟⎟ ⎠
Lecture 1
Page 2 of 5
Always analyze extreme cases. 1) What if R1 goes to infinity? Output voltage is zero. 2) What if R1 goes to 0? Output voltage is equal to the input voltage. 3) What happens if R1=R2? Output voltage is half the input voltage. 4) What if R2 = 2R1? Twothirds of input voltage. 5) R2 = 9R1? Ninetenths of input voltage Note: A) VOUT is closest to the voltage at the other end of the lowervalued resistor. B) R1 ≈ R2, VOUT ≈ ½ VIN.
Kirchkoff’s Voltage Law (KVL)
+ 9V Symbol for a battery
R1 VOUT
9V
+
R1
VOUT
⇔
R2
R2

KVL: The sum of voltage drops going around a closed loop path is zero.
10Ω ⎛ ⎞ VOUT = 9V ⎜ ⎟ ⎝ 80Ω + 10Ω ⎠
80 Ω
VOUT = 1V
9V 10 Ω
= 1V
∑ ∆V = (9V − 1V ) + (1V − 0V ) + (0V − 9V ) loop
SP.764 Practical Electronics Dr. James A. Bales
Lecture 1
Page 3 of 5
Kirchkoff’s Current Law (KCL)
I1 ↓
← I2
I1 + I 2 + I 3 = 0
I3 ↑
As long as currents are given a direction, the total sum of the currents equals zero. Diodes The “ideal” diode only allows current to flow in one direction.
Anode (+)
Allows current to flow
No current flows
Cathode ()
An easy to way to tell difference between the anode and the cathode is that electrons come from the cathode (i.e. CRT tube in a TV – cathode rays emitted hit the TV screen). Plot of I vs. V for Diode and Resistor
Resistor slope = 1/R Ideal Diode
Our Model of a Diode
VFORWARD
SP.764 Practical Electronics Dr. James A. Bales
Lecture 1
Page 4 of 5
Example:
+ 9V
R = 700Ω
I1 ↓
VOUT LED
I1 ↓
VFORWARD = 2V
Questions: 1) What is I1? Current that goes in, must go out. Well, VOUT = 2V (the LED holds the voltage at that value)
∆V resistor = 9V − 2V
= 7V
I=
∆V 7V 1
= = A R 700Ω 100
= 0.01 A = 10 mA
SP.764 Practical Electronics Dr. James A. Bales
Lecture 1
Page 5 of 5
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SP.764, Practical Electronics Dr. James A. Bales Lecture 2: Switches, Rectifiers and Generators
Topics: 1) Homework Review
2) Switches
3) Bridge Rectifiers
4) AC vs. DC
5) Function Generators and Oscilloscopes
Homework Review: Homework 1: Voltage dividers
+ 20V 1K
∆V1 = IR1 = 20V − VOUT VOUT = 18V
9K
∆V 2 = IR 2 = VOUT − 0V
∆V1 = IR1
1K
I=
VOUT 9K
∆V2 = IR2
− 20V
⎛ R2 VOUT = V IN ⎜⎜ ⎝ R1 + R 2
I=
V ∆V1 0 − VOUT = = − OUT R1 R1 R1 ∆V2 VOUT − (− 20V ) = R2 R2
⎞ ⎟⎟ ⎠
⎛ 1k ⎞ = 20V ⎜ ⎟ = 18V ⎝ 1k + 9k ⎠
− −
VOUT VOUT 20V = + R1 R2 R2 ⎛ 1 20V 1 ⎞ ⎟⎟ = VOUT ⎜⎜ + R R1 R 2 ⎠ ⎝ 1 ⎛ R1 ⎞ ⎟⎟ VOUT = −20V ⎜⎜ R + R 2 ⎠ ⎝ 1
= −2V
Homework 2: Diodes
+ 5V
R1 VF = 2V
R2 VF = 4V
R1 =
(5V − 2V ) = 150Ω
R2 =
(5V − 4V ) = 50Ω
20mA
20mA
There are two diodes that will be used in class: •
Zener diodes: If you put a negative voltage across a Zener diode, it has a second turnon point. The slope of the IV curve of the second turnon point is even more abrupt than the slope of the IV curve of the first turnon point. You can tail the breakdown voltage from a couple of volts to hundreds of volts. A Zener diode is used in reverse bias to clamp and hold the voltage. The name Zener diode comes from a Physics term called the Zener effect, which is not even related to the Zener diode at all. However, the Zener diode does exhibit the Avalanche effect.
•
Silicon diodes: These diodes do not emit light. Have same functionality as other diodes. Their forward voltages are really small (~0.60.7 volts).
Diodes don’t have resistance. If the voltage is below a diode’s forward voltage, then the element looks like an open circuit. If the voltage is below the forward voltage, the element behaves as a short circuit.
SP.764, Practical Electronics Dr. James A. Bales
Lecture 2
Page 2 of 4
+ 5V
I1 ↓
330 Ω
VF = 2V
I3 ↓
I1 =
RL
(5V − 2V ) = 9mA 330Ω
I 1 = I 2 + I 3 = 9mA
I2 ↓
If RL is such that VOUT = 1V, there is no current flow through the diode. The diode conducts (turns on) when VOUT hits RL.
+ 5V R L > 220Ω
RL = 220Ω
2V RL
RL < 220Ω
One gets VOUT = 2V when RL = 220 ohms. Now, try RL = 330 ohms. The voltage divider predicts VOUT = 2.5V. BUT the diode clamps at 2.0V! The diode will steal any necessary current to stay at 2V. The bigger RL is, the more current passes through the diode.
SP.764, Practical Electronics Dr. James A. Bales
Lecture 2
Page 3 of 4
AC vs. DC One can implement a simple DC to AC converter with a switch configuration like the following:
+
A

With switch to one position:

B
+
9V
VA  VB = 9V With switch on the other position: VB  VA = 9V
SP.764, Practical Electronics Dr. James A. Bales
− 9V
Lecture 2
Page 4 of 4
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SP.764, Practical Electronics
Dr. James A. Bales
Lecture 3: Capacitors
Topics: 1) Review Homework 2) Capacitors 
Time varying systems

Water analogy

What capacitors are

How they work

Constitutive systems

RC circuits
Homework Review:
2.0V Func Gen
1.5V
A
B RL
1.5V
2.0V
For “positive” part of cycle: VA = 3V, VB = 2V For “negative” part of cycle: VA = 1.5V, VB = 3.5V
V +5
t
−5
VA 3
− 1.5
t
VB
2 t
− 3.5
2
1
t
SP.764, Practical Electronics Dr. James A. Bales
Lecture 3
Page 2 of 5
+5 4
V
t
−3 −5
VA
3
− 1.5
t
VB
2
− 1.5
− 3.5
t
V A − VB 2
1
t
SP.764, Practical Electronics Dr. James A. Bales
Lecture 3
Page 3 of 5
Capacitors: Water Analogy
Voltage Ù Pressure
Current Ù Volumetric Flow Rate
A capacitor is a bucket for charge! In the fluid system, voltage is equivalent to pressure. Current maps to the volumetric flow rate of water. Looking inside a pipe, one would say water is flowing at a rate of gallons per second. The water analogy is used because the easiest way to think about this is that the capacitor is a bucket for charge. That is, the capacitor can store charge and can release charge as needed.
Symbol for a capacitor
Close switch!
VO
C
I (t)
VO
++++ C −−−−
The symbol for a capacitor is two metal plates (two pieces of metals separated by air. If a battery is added with voltage V0 and a switch, then current will flow if the switch is closed. How is I related to q? 
For a resistor, I = V/R. Since I is the flow rate of charge, then I = dq/dt.

For a capacitor, the amount of charge “q” on a plate obeys: q = CV. Therefore when the switch is closed, current starts flowing until q = CV0. Resistor
I=
SP.764, Practical Electronics Dr. James A. Bales
V dq = R dt
Capacitor
I=
dq dV =C dt dt
Lecture 3
Page 4 of 5
R
Position 2
Position 1
C
VO
The switch starts at position 1, and at time t = 0, the switch goes instantaneously to position 2. Question: What is VCAP as a function of time?
VCAP
For t ≤ 0, VCAP ? For t ≤ 0, q = 0, so VCAP = 0.
V0 t
At t = 0, flip switch:
q(t) = CVCAP (t)
∆VR
∆V across resistor is ∆V(t = 0) = V0
V0
I (t = 0) =
V0
R
⎛ VCAP (t) = V0 ⎜⎜1− e ⎝
t − RC
t
⎞
⎟
⎟ ⎠
I
t
I (t) =
V0 − RC e R
RC = time constant
SP.764, Practical Electronics Dr. James A. Bales
V0
R
t
Lecture 3
Page 5 of 5
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SP.764, Practical Electronics
Dr. James A. Bales
Lecture 4: RC Circuits
Topics: 1) RC circuit review
2) Equations for RC
3) Time Response of RC circuits
4) RC Filters
Question: What factors contribute to the capacitance of a capacitor?
Area A Material has a dielectric constant
C =ε
ε
A d
d q = CV
If d becomes smaller, the capacitance goes up. Therefore, the capacitor can store more charge.
++++++++ Electric Field
E
E=
_________
VCAP d
For a given voltage, if one makes the separation d smaller, the electric field E increases. Why is this a problem? If the electric field gets too large, the insulator breaks down. A small spark will jump between plates and short the capacitor, which will not work anymore.
In summary, if V gets too large or d gets too small, the dielectric material will “breakdown” and destroy the capacitor. In some cases the capacitor “pops”, or may not act properly anymore. A capacitor is specified by: 
Capacitance value

Maximum voltage across capacitor

Polarized or nonpolarized.
What is polarized? 
Polarized capacitors are marked with “+” lead and “” lead.

If the positive and negative leads are connected the wrong way, the capacitor can blow up.
If one is working with high voltages, then one would want to keep electric field small by putting plates apart, but one would loose some capacitance. RC circuit review:
I (t) “A” “B”
R C
VO
Move switch from “A” to “B” at time t = 0. t − ⎛ ⎞ RC ⎟ ⎜ VCAP (t) = V0 ⎜1− e ⎟ ⎝ ⎠
t
,
V − I (t) = 0 e RC R
RC = time constant [OHM][FARAD]=[SEC]
SP.764, Practical Electronics Dr. James A. Bales
Lecture 4
Page 2 of 6
The current I(t) starts at peak value V0/R, and then drops off and dies out exponentially. The voltage across the capacitor VCAP(t) starts at 0, goes up rapidly, and asymptotically approaches V0. Time Response of RC circuits: RC circuits can do interesting things. Consider the following example: Problem: A square wave that goes between +/Vo with period T is applied to the following RC circuit.
R
V0 C
t
− V0 T
Question: Complete the following graphs. Plot VCAP(t) for: 1. RC ≈ T
VCAP V0
2. RC > T
− V0
t
Hint: e1 = 1/3
Solution:
Assume that the square wave signal has been “ON” for a long period of time. At least
one knows VCAP(t) will range between +V0 and –V0 (but will never reach those
voltages).
SP.764, Practical Electronics Dr. James A. Bales
Lecture 4
Page 3 of 6
VCAP (t) + VO
t
− VO + VO
RC ≈ T
t
− VO + VO
RC > T t
− VO
SP.764, Practical Electronics Dr. James A. Bales
Lecture 4
Page 4 of 6
RC Filters: Swing analogy If one pushes a person sitting in a swing slowly, the person swings smoothly, so the swing works fine. However, if one starts pushing too many times per second, the swing barely moves at all. The performance of the swinging motion depends on the frequency at which the person is pushing the swing. Frequency is just the inverse of the period T,
f =
1 T
When RC > T Æ f is “HIGH”
(Signal gets through)
(Signal is blocked)
IF RC 0 VBE 0.6V
(i.e. the baseemitter junction looks like a diode) Analysis: Assume I B
> 0, then
VBE = 0.6V = VB − VE
0
⇒ VB = 0.6V Current through 100Ω resistor is:
I=
∆V 5V − 0.6V = = 0.044 A = 44mA R 100Ω
SP.764, Practical Electronics Dr. James A. Bales
Lecture 5
Page 5 of 6
Then,
I C = β I B = 4.4 A
!!
Lamp limits
IC
IC
Load is saturated
IC
Transistor limits
IC
Slope = β
IB VCE 5V VCE ≈ 0.2V (saturation)
IB
Pros: 
Small and handy

Cheap

FAST!
Cons: 
No Electrical Isolation

Cannot work with High Voltages & Currents
SP.764, Practical Electronics Dr. James A. Bales
Lecture 5
Page 6 of 6
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SP.764, Practical Electronics
Dr. James A. Bales
Lecture 6: Op Amps
Topics: 1) Op Amps as Amplifiers

Example

“Golden Rules” of their Use

Analyze example

More Examples
2) Op Amp as a Comparator Op Amps: 
Op amps are semiconductor devices.

Op amps are mostly optimized to have highspeed or lownoise features.

Even though they are remarkably sophisticated circuits, they are really
easy to work with.

If you connect the power supply incorrectly, the op amp might blow up.
Comparator: 
Remarkably useful: Feed two voltages, and comparator compares them.

The comparator is heart of digital electronics because of its ability to decide whether a voltage is higher or lower than a threshold value.
Question 1: What is the output going to be?
1mA
10kΩ
1mA 1kΩ
+1V
A

VOUT
+
∆V = IR = 1mA ×10kΩ = 10V
VA = 0
⇒ VOUT = −10V Question 2: If instead of 1V, the input is half a volt. 
You would get half the output.
Question 3: What about double? 
Tricky! The op amp saturates at 15V, so output cannot go beyond 15V.
Op amps are usually run off bipolar supplies (+/15V).
Op Amps:
+V
(V+ ,V− )

2 Inputs:

Noninverting Inverting
Input Input
1 Output: ( −V < VOUT < +V )

2 Voltage Supplies:

( +V , −V )
+ V
Golden Rules: 1) The OpAmp Inputs draw no current. 2) With Negative Feedback, the inputs are at the same voltage (i.e., V+=V)
SP.764, Practical Electronics Dr. James A. Bales
Lecture 6
Page 2 of 5
Now, one can generalize the first circuit:
IF I IN
RF
VIN

VOUT
RI I IN
+
V −V = IN − RI
Golden Rule #2 tells us
Golden Rule #1 requires
V− = 0 ⇒ I IN =
I F = I IN VIN RI
⇒ (V− − VOUT ) = I F RF 0
⇒ −VOUT =
VIN RF RI
⎛R ⎞ ⇒ VOUT = − ⎜ F ⎟VIN ⎝ RI ⎠ “Negative Feedback” means there is a current path from VOUT to the “” input. More examples:
VOUT = f (VIN ) Unity Gain Buffer +V 
VOUT VIN
+ V
SP.764, Practical Electronics Dr. James A. Bales
VIN = VOUT
Lecture 6
Page 3 of 5
R4
+V 
R3
VOUT
R1 VIN
+ V
R2 ⎛ R2 ⎞ V− = V+ = VIN ⎜ ⎟ ⎝ R1 + R2 ⎠ VOUT − V− V− = R4 R3
R4 V− R3
⎛ R ⎞ R3 VOUT = V− ⎜ 1+ 3 ⎟ R4 R4 ⎠ ⎝
⎛ R2 ⎞ ⎛ R4 + R3 ⎞ R3 VOUT = VIN ⎜ ⎟⎜ ⎟ R4 ⎝ R1 + R2 ⎠ ⎝ R4 ⎠
VOUT =
SP.764, Practical Electronics Dr. James A. Bales
R2 R3
⎛ R3 + R4 ⎞ ⎜ ⎟ VIN ⎝ R1 + R2 ⎠
Lecture 6
Page 4 of 5
Assignment: 
Work out the following circuits before next Lab.

For the following circuits,

Find VOUT.
VOUT = f (V A , VB )
R2
Circuit A
R1 VA

VOUT
VB
+
R3 R4
Circuit B
RF
RA VA 
VOUT
VB RB
SP.764, Practical Electronics Dr. James A. Bales
+
Lecture 6
Page 5 of 5
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SP.764, Practical Electronics Dr. James A. Bales Lecture 7: FlipFlops and 555 Timer Circuit
Topics: 1) Comparator Review
2) Flip Flops
3) 555 as oscillator
4) 555 as “oneshot”
Comparator Review: V
V−
V+
VOUT
+
If
VOUT is:
V+ > V−
+V (i.e. maximum)
V+ < V−
−V (i.e. minimum)
+V Comparators 
Feedback loop is not used.

Decides if one voltage is greater than the other.

Takes analog voltages and convert them into a series of bits.

Binary representation of 4 digits give you 16 values (4bit converter).

Circuit above is a 1bit converter: o
“0” or “1” output depending which voltage is greater than the other.
Flip Flops: RS Flip Flop
R
Q
S
Q
R ≡ Reset S ≡ Set
Two Values TRUE
“1”
Hi Voltage
FALSE
“0”
Lo Voltage
INPUTS For some circuits:
We use:
R
S
Lo
Lo
OUTPUTS
Q
Q
Holds last value
Hi ≡ 5V
Hi ≡ +V
Lo
Hi
Hi
Lo
Lo ≡ 0V
Lo ≡ −V
Hi
Lo
Lo
Hi
Hi
Hi
Not Allowed!
Once can force the output Q to be “HI” by setting S to “HI”. Similarly, one can force the Q output to “LO” by resetting R to “LO”. If one drives both R and S to “HI”, there is no guarantee about the output’s state. 555 as Oscillator:
V0 8
6
+
5k IN
5k
R
Q
S
Q
OUT 3
+
5k 
2 7
1
SP.764, Practical Electronics Dr. James A. Bales
Lecture 7
Page 2 of 4
Voltage @ Pin 2 & 6 ( V2−6 )
Output of
Output of
Output
CR
CS
Q
Q
Transistor @ Pin 7
< 1 V0 3
Lo
Hi
Hi
Lo
OFF
1 V 2 V0 3
Lo
Lo
Stay
Stay
Stay
Hi
Lo
Lo
Hi
ON
Output
555 as Oscillator:
V0 8
V0 6
+
5k
RA

5k
RB
R
Q
S
Q
OUT 3
+
5k 
2 7
C
1
SP.764, Practical Electronics Dr. James A. Bales
Lecture 7
Page 3 of 4
Assume there is no charge in the capacitor at start. Because VCAP is at 0V and it connects to pins 2 and 6, the input is at 0V at time = 0. When the circuit is powered up, the capacitor starts charging. When the VCAP reaches 2/3 V0, the transistor turns on and grounds pin 7. Therefore, the capacitor starts to discharge through RB until VCAP reaches 1/3 V0, at which point the transistor turns off and the capacitor starts to charge up again.
V @ pins 2&6
V0
2 V 3 0 1 V 3 0 t Transistor
ON
OFF
t
Q V0
t SP.764, Practical Electronics Dr. James A. Bales
Lecture 7
Page 4 of 4
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SP.764, Practical Electronics
Dr. James A. Bales
Lecture 8: Final Project – Fan Controller
Topics: 1) Final Project Introduction
2) Background Information
Final Project Introduction: 
The final project will evolve from the 555 timer circuit.

The project will consist of building a fan controller.

Involves giving the system a reference signal from a thermistor. (Therefore,
one is expected to know how to read thermistor values.)

Use the 555 timer circuit as a oneshot to catch any input signal and produce a fixedwidth pulse output signal.

The thermistor driving a 555 timer will give a variable frequency signal.
Background information: 
PWM

How motors work.
o
Constant current
o
Variable voltage
o
Behave like a LP filters
o
Require lots of current
Voltage 1V 2V 3V
Speed 2500 rpm 3500 rpm 4500 rpm
Current 0.5A 0.5A 0.5A
Transistor multiplies current
IN
+
555
PMW
Fan motor needs high current
Controlling the fan: 
Measure temperature o

Thermometer
Measure revolutions per minute (RPM) o
Phototransistor
Displaying temperature values:
+5V
(
V 25D C
)
+5V
37 D C
27 D C
Connect the input of a series of opamps in such a way that there is a common voltage (VTHERM) reaching the inputs of all op amps. Choose resistor values to create reference voltages that represent different temperature values. Since the 555 is better at sinking current rather than sourcing it, connect LEDs as shown above.
SP.764, Practical Electronics Dr. James A. Bales
Lecture 8
Page 2 of 3
+5V
R1
8
4
7
R2
3
555 6
Out
2 1
2/3
1/3
5
C1
555PIN −2 VTHERM +5V
555 PIN − 2 VTHERM
t VOUT
t
Since VTHERM is a variable voltage, use a comparator to control the average voltage delivered to the motor. If VTHERM increases (implying higher temperature), the output of the comparator will have a higher duty cycle and the motor will spin faster. If VTHERM decreases, the duty cycle of the output decreases and the motor spins slower.
SP.764, Practical Electronics Dr. James A. Bales
Lecture 8
Page 3 of 3
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SP.764, Practical Electronics Dr. James A. Bales Lecture 9: Final Project Concepts
Topics: 1) Concepts 
Generate concepts

Explore concepts

Select concepts
2) Plan & Begin Design Concepts: I.
Temperaturecontrolled fan, w/ readout a. Sense temperature i. Thermistor b. Display temperature i. LEDs ii. Comparator c. Control speed of motor w/ PWM i. Transistor ii. 555
II. “Color Organ” a. Pick up sound i. Microphone ii. Amplifier iii. Filter b. Sense volume (Extract envelope) i. Rectifier ii. Lowpass Filter c. Sense freq. band & volume in that band d. Light LED bar
III. Sunflower a. Sense light i. Photocells b. Compare intensity i. Comparator c. Actuator i. Solenoids ii. Motor d. Motor needs to stop! (Implement “deadband”)
SP.764, Practical Electronics Dr. James A. Bales
Lecture 9
Page 2 of 2
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