Part
1
Electronic Principles
Eighth Edition
Chapter 1
Introduction
SELF-TEST
1.
2.
3.
4.
5.
6.
a
c
a
b
d
d
7.
8.
9.
10.
11.
12.
b
c
b
a
a
a
13.
14.
15.
16.
17.
18.
c
d
b
b
a
b
19.
20.
21.
22.
23.
b
c
b
b
c
us more insight into how changes in load resistance affect the
load voltage.
12. It is usually easy to measure open-circuit voltage and shortedload current. By using a load resistor and measuring voltage
under load, it is easy to calculate the Thevenin or Norton
resistance.
PROBLEMS
1-1.
JOB INTERVIEW QUESTIONS
Note: The text and illustrations cover many of the job interview
questions in detail. An answer is given to job interview questions
only when the text has insufficient information.
2. It depends on how accurate your calculations need to be. If
an accuracy of 1 percent is adequate, you should include the
source resistance whenever it is greater than 1 percent of the
load resistance.
5. Measure the open-load voltage to get the Thevenin voltage
VTH. To get the Thevenin resistance, reduce all sources to
zero and measure the resistance between the AB terminals to
get RTH. If this is not possible, measure the voltage VL across
a load resistor and calculate the load current IL. Then divide
VTH – VL by IL to get RTH.
6. The advantage of a 50 Ω voltage source over a 600 Ω voltage
source is the ability to be a stiff voltage source to a lower
value resistance load. The load must be 100 greater than the
internal resistance in order for the voltage source to be considered stiff.
7. The expression cold-cranking amperes refers to the amount
of current a car battery can deliver in freezing weather
when it is needed most. What limits actual current is the
Thevenin resistance caused by chemical and physical
parameters inside the battery, not to mention the quality of the
connections outside.
8. It means that the load resistance is not large compared to the
Thevenin resistance, so that a large load current exists.
9. Ideal. Because troubles usually produce large changes in
voltage and current, so that the ideal approximation is adequate for most troubles.
10. You should infer nothing from a reading that is only
5 percent from the ideal value. Actual circuit troubles
will usually cause large changes in circuit voltages. Small
changes can result from component variations that are still
within the allowable tolerance.
11. Either may be able to simplify the analysis, save time when
calculating load current for several load resistances, and give
Given:
V = 12 V
RS = 0.1 Ω
Solution:
RL = 100RS
RL = 100(0.1 Ω)
RL = 10 Ω
Answer: The voltage source will appear stiff for values of
load resistance of ≥10 Ω.
1-2.
Given:
RLmin = 270 Ω
RLmax = 100 kΩ
Solution:
RS < 0.01 RL (Eq. 1-1)
RS < 0.01(270 Ω)
RS < 2.7 Ω
Answer: The largest internal resistance the source can
have is 2.7 Ω.
1-3.
Given: RS = 50 Ω
Solution:
RL = 100RS
RL = 100(50 Ω)
RL = 5 kΩ
Answer: The function generator will appear stiff for
values of load resistance of ≥5 kΩ.
1-4.
Given: RS = 0.04 Ω
Solution:
RL = 100RS
RL = 100(0.04 Ω)
RL = 4 Ω
Answer: The car battery will appear stiff for values of
load resistance of ≥ 4 Ω.
1-1
“Copyright © McGraw-Hill Education. Permission required for reproduction or display.”
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1-5.
Given:
RS = 0.05 Ω
I=2A
Solution:
RL = 0.01RS (Eq. 1-4)
RL = 0.01(250 kΩ)
RL = 2.5 kΩ
Solution:
V = IR (Ohm’s law)
V = (2 A)(0.05 Ω)
V = 0.1 V
Answer: The voltage drop across the internal resistance
is 0.1 V.
1-6.
Given:
V=9V
RS = 0.4 Ω
Solution:
I = V/R (Ohm’s law)
I = (9 V)/(0.4 Ω)
I = 22.5 A
IL = IT [(RS)/(RS + RL)] (Current divider formula)
IL = 5 mA [(250 kΩ)/(250 kΩ + 10 kΩ)]
IL = 4.80 mA
Answer: The load current is 4.80 mA, and, no, the current
source is not stiff since the load resistance is not less than
or equal to 2.5 kΩ.
1-12. Solution:
VTH = VR2
VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
VR2 = 36 V[(3 kΩ)/(6 kΩ + 3 kΩ)]
VR2 = 12 V
RTH = [R1R2/R1 + R2] (Parallel resistance formula)
RTH = [(6 kΩ)(3 kΩ)/(6 kΩ + 3 kΩ)]
RTH = 2 kΩ
Answer: The load current is 22.5 A.
1-7.
Given:
IS = 10 mA
RS = 10 MΩ
Answer: The Thevenin voltage is 12 V, and the Thevenin
resistance is 2 kΩ.
Solution:
RL = 0.01 RS
RL = 0.01(10 MΩ)
RL = 100 kΩ
Answer: The current source will appear stiff for load
resistance of ≤100 kΩ.
1-8.
R1
3 kV
R2
6 kV
R1
3 kV
R2
36 V
Given:
RLmin = 270 Ω
RLmax = 100 kΩ
Solution:
RS > 100 RL (Eq. 1-3)
RS > 100(100 kΩ)
RS > 10 MΩ
6 kV
36 V
VTH
Answer: The internal resistance of the source is greater
than 10 MΩ.
1-9.
Given: RS = 100 kΩ
Solution:
RL = 0.01RS (Eq. 1-4)
RL = 0.01(100 kΩ)
RL = 1 kΩ
Answer: The maximum load resistance for the current
source to appear stiff is 1 kΩ.
1-10. Given:
IS = 20 mA
RS = 200 kΩ
RL = 0 Ω
Solution:
RL= 0.01RS
RL= 0.01(200 kΩ)
RL= 2 kΩ
Answer: Since 0 Ω is less than the maximum load resistance of 2 kΩ, the current source appea rs stiff; thus the
current is 20 mA.
1-11. Given:
I = 5 mA
RS = 250 kΩ
RL = 10 kΩ
RTH
(a) Circuit for finding VTH in Prob. 1-12. (b) Circuit for
finding RTH in Prob. 1-12.
1-13. Given:
VTH = 12 V
RTH = 2 kΩ
Solution:
I = V/R (Ohm’s law)
I = VTH/(RTH + RL)
I0Ω = 12 V/(2 kΩ + 0 Ω) = 6 mA
I1kΩ = 12 V/(2 kΩ + 1 kΩ) = 4 mA
I2kΩ = 12 V/(2 kΩ + 2 kΩ) = 3 mA
I3kΩ = 12 V/(2 kΩ + 3 kΩ) = 2.4 mA
I4kΩ = 12 V/(2 kΩ + 4 kΩ) = 2 mA
I5kΩ = 12 V/(2 kΩ + 5 kΩ) = 1.7 mA
I6kΩ = 12 V/(2 kΩ + 6 kΩ) = 1.5 mA
Answers: 0 Ω 6 mA; 1 kΩ, 4 mA; 2 kΩ, 3mA; 3 kΩ,
2.4 mA; 4 kΩ, 2 mA; 5 kΩ, 1.7 mA; 6 kΩ, 1.5 mA.
1-2
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Solution:
RN = RTH (Eq. 1-10)
RTH = 10 kΩ
RTH
VTH
IN = VTH/RTH (Eq. 1-12)
VTH = INRN
VTH = (10 mA)(10 kΩ)
VTH = 100 V
RL
Thevenin equivalent circuit for Prob. 1-13.
1-14. Given:
VS = 18 V
R1 = 6 kΩ
R2 = 3 kΩ
Answer: RTH = 10 kΩ, and VTH = 100 V
RTH
VTH
Solution:
VTH = VR2
VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
VR2 = 18 V[(3 kΩ)/(6 kΩ + 3 kΩ)]
VR2 = 6 V
RTH = [(R1 × R2)/(R1 + R2)] (Parallel resistance formula)
RTH = [(6 kΩ × 3 kΩ)/(6 kΩ + 3 kΩ)]
RTH = 2 kΩ
100 V
Thevenin circuit for Prob. 1-17.
1-18. Given (from Prob. 1-12):
VTH = 12 V
RTH = 2 kΩ
Answer: The Thevenin voltage decreases to 6 V, and the
Thevenin resistance is unchanged.
Solution:
RN = RTH (Eq. 1-10)
RN = 2 kΩ
1-15. Given:
VS = 36 V
R1 = 12 kΩ
R2 = 6 kΩ
Solution:
VTH = VR2
VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
VR2 = 36 V[(6 kΩ)/(12 kΩ + 6 kΩ)]
VR2 = 12 V
RTH = [(R1R2)/(R1 + R2)] (Parallel resistance formula)
RTH = [(12 kΩ)(6 kΩ)/(12 kΩ + 6 kΩ)]
RTH = 4 kΩ
Answer: The Thevenin voltage is unchanged, and the
Thevenin resistance doubles.
1-16. Given:
VTH = 12 V
RTH = 3 kΩ
Solution:
RN = RTH
RN = 3 kΩ
Answer: IN = 4 mA, and RN = 3 kΩ
4 mA
IN = VTH/RTH (Eq. 1-12)
IN = 12 V/2 kΩ
IN = 6 mA
Answer: RN = 2 kΩ, and IN = 6 mA
IN
RN
6 mA
2 kV
Norton circuit for Prob. 1-18.
1-19. Shorted, which would cause load resistor to be connected
across the voltage source seeing all of the voltage.
1-20. a. R1 is open, preventing any of the voltage from reaching
the load resistor. b. R2 is shorted, making its voltage drop
zero. Since the load resistor is in parallel with R2, its voltage drop would also be zero.
1-21. The battery or interconnecting wiring.
IN = VTH/RTH
IN = 12 V/3 kΩ
IN = 4 mA
IN
10 kV
RN
3 kV
Norton circuit for Prob. 1-16.
1-17. Given:
IN = 10 mA
RN = 10 kΩ
1-22. RTH = 2 kΩ
Solution:
RMeter = 100RTH
RMeter = 100(2 kΩ)
RMeter = 200 kΩ
Answer: The meter will not load down the circuit if the
meter impedance is ≥ 200 kΩ.
CRITICAL THINKING
1-23. Given:
VS = 12 V
IS = 150 A
Solution:
RS = (VS)/(IS)
RS = (12 V)/(150 A)
RS = 80 mΩ
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Answer: If an ideal 12 V voltage source is shorted and
provides 150 A, the internal resistance is 80 mΩ.
1-24. Given:
VS = 10 V
VL = 9 V
RL = 75 Ω
Solution:
VS = VRS + VL (Kirchhoff’s law)
VRS = VS – VL
VRS = 10 V – 9 V
VRS = 1 V
IRS = IL = VL/RL (Ohm’s law)
IRS = 9 V/75 Ω
IRS = 120 mA
RS = VRS/IRS (Ohm’s law)
RS = 8.33 Ω
RS < 0.01 RL (Eq. 1-1)
8.33 Ω < 0.01(75 Ω)
8.33 Ω ≮ 0.75 Ω
Answer: a. The internal resistance (RS) is 8.33 Ω. b. The
source is not stiff since RS ≮ 0.01 RL.
1-25. Answer: Disconnect the resistor and measure the voltage.
1-26. Answer: Disconnect the load resistor, turn the internal
voltage and current sources to zero, and measure the
resistance.
1-27. Answer: Thevenin’s theorem makes it much easier to
solve problems where there could be many values of a
resistor.
1-28. Answer: To find the Thevenin voltage, disconnect the load
resistor and measure the voltage. To find the Thevenin
resistance, disconnect the battery and the load resistor,
short the battery terminals, and measure the resistance at
the load terminals.
1-29. Given:
RL = 1 kΩ
I = 1 mA
Solution:
RS > 100RL
RS > 100(1 kΩ)
RL > 100 kΩ
V = IR
V = (1 mA)(100 kΩ)
V = 100 V
Answer: A 100 V battery in series with a 100 kΩ resistor.
1-30. Given:
VS = 30 V
VL = 15 V
RTH < 2 kΩ
Solution: Assume a value for one of the resistors. Since
the Thevenin resistance is limited to 2 kΩ, pick a value
less than 2 kΩ. Assume R2 = 1 kΩ.
VL = VS[R2/(R1 + R2)] (Voltage divider formula)
R1 = [(VS)(R2)/VL] – R2
R1 = [(30 V)(1 kΩ)/(15 V)] – 1 kΩ
R1 = 1 kΩ
RTH = (R1R2/R1 + R2)
RTH = [(1 kΩ)(1 kΩ)]/(1 kΩ + 1 kΩ)
RTH = 500 Ω
Answer: The value for R1 and R2 is 1 kΩ. Another possible solution is R1 = R2 = 4 kΩ. Note: The criteria will
be satisfied for any resistance value up to 4 kΩ and when
both resistors are the same value.
1-31. Given:
VS = 30 V
VL = 10 V
RL > 1 MΩ
RS < 0.01RL (since the voltage source must be stiff)
(Eq. 1-1)
Solution:
RS < 0.01RL
RS < 0.01(1 MΩ)
RS < 10 kΩ
Since the Thevenin equivalent resistance would be the
series resistance, RTH < 10 kΩ.
Assume a value for one of the resistors. Since the
Thevenin resistance is limited to 1 kΩ, pick a value less
than 10 kΩ. Assume R2 = 5 kΩ.
VL = VS[R2/(R1 + R2)] (Voltage divider formula)
R1 = [(VS)(R2)/VL] – R2
R1 = [(30 V)(5 kΩ)/(10 V)] – 5 kΩ
R1 = 10 kΩ
RTH = R1R2/(R1 + R2)
RTH = [(10 kΩ)(5 kΩ)]/(10 kΩ + 5 kΩ)
RTH = 3.33 kΩ
Since RTH is one-third of 10 kΩ, we can use R1 and R2
values that are three times larger.
Answer:
R1 = 30 kΩ
R2 = 15 kΩ
Note: The criteria will be satisfied as long as R1 is twice
R2 and R2 is not greater than 15 kΩ.
1-32. Answer: First, measure the voltage across the terminals.
This is the Thevenin voltage. Next, connect the ammeter to the battery terminals—measure the current. Next,
use the values above to find the total resistance. Finally,
subtract the internal resistance of the ammeter from this
result. This is the Thevenin resistance.
1-33. Answer: First, measure the voltage across the terminals.
This is the Thevenin voltage. Next, connect a resistor
across the terminals. Next, measure the voltage across
the resistor. Then, calculate the current through the load
resistor. Then, subtract the load voltage from the Thevenin voltage. Then, divide the difference voltage by the
current. The result is the Thevenin resistance.
1-34. Solution: Thevenize the circuit. There should be a
Thevenin voltage of 0.148 V and a resistance of 6 kΩ.
IL = VTH/(RTH + RL)
IL = 0.148 V/(6 kΩ + 0)
IL = 24.7 μA
IL = 0.148 V/(6 kΩ + 1 kΩ)
IL = 21.1 μA
IL = 0.148 V/(6 kΩ + 2 kΩ)
IL = 18.5 μA
IL = 0.148 V/(6 kΩ + 3 kΩ)
IL = 16.4 μA
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IL = 0.148 V/(6 kΩ + 4 kΩ)
IL = 14.8 μA
IL = 0.148 V/(6 kΩ + 5 kΩ)
IL = 13.5 μA
IL = 0.148 V/(6 kΩ + 6 kΩ)
IL = 12.3 μA
Answer: 0, IL = 24.7 μA; 1 kΩ, IL = 21.1 μA; 2 kΩ, IL =
18.5 μA; 3 kΩ, IL = 16.4 μA; 4 kΩ, IL = 14.8 μA; 5 kΩ,
IL = 13.5 μA; 6 kΩ, IL = 12.3 μA.
1-35. Trouble:
1: R1 shorted
2: R1 open or R2 shorted
3: R3 open
4: R3 shorted
5: R2 open or open at point C
6: R4 open or open at point D
7: Open at point E
8: R4 shorted
2-4.
500,000 free electrons
2-5.
a. 5 mA
b. 5 mA
c. 5 mA
2-6.
a.
b.
c.
d.
e.
2-7.
Given:
Barrier potential at 25°C is 0.7 V
Tmin = 25°C
Tmin = 75°C
Solution:
ΔV = (–2 mV/°C) ΔT (Eq. 2-4)
ΔV = (–2 mV/°C)(0°C – 25°C)
ΔV = 50 mV
Vnew = Vold + ΔV
Vnew = 0.7 V + 0.05 V
Vnew = 0.75 V
1-36. R1 shorted
1-37. R2 open
ΔV = (–2 mV/°C) ΔT (Eq. 2-4)
ΔV = (–2 mV/°C)(75°C – 25°C)
ΔV = –100 mV
1-38. No supply voltage
1-39. R4 open
Vnew = Vold + ΔV
Vnew = 0.7 V – 0.1 V
Vnew = 0.6 V
1-40. R2 shorted
Chapter 2 Semiconductors
Answer: The barrier potential is 0.75 V at 0°C and
0.6 V at 75°C.
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
d
a
b
b
d
c
b
b
c
a
c
c
b
b
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
a
b
d
d
a
a
d
a
a
a
d
b
b
a
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
d
c
a
a
b
a
b
c
c
a
b
a
b
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
b
b
c
a
c
d
a
a
d
c
b
d
b
JOB INTERVIEW QUESTIONS
9. Holes do not flow in a conductor. Conductors allow current
flow by virtue of their single outer-shell electron, which is
loosely held. When holes reach the end of a semiconductor,
they are filled by the conductor’s outer-shell electrons entering at that point.
11. Because the recombination at the junction allows holes and
free electrons to flow continuously through the diode.
PROBLEMS
2-1.
–2
2-2.
–3
2-3.
a.
b.
c.
d.
p-type
n-type
p-type
n-type
p-type
2-8.
Given:
IS = 10 nA at 25°C
Tmin = 0°C – 75°C
Tmax = 75°C
Solution:
IS(new) = 2(ΔT/10)IS(old) (Eq. 2-5)
IS(new) = 2[(0°C – 25°C)/10]10 nA
IS(new) = 1.77 nA
IS(new) = 2(ΔT/10) IS(old) (Eq. 2-5)
IS(new) = 2[(75°C – 25°C)/10)] 10 nA
IS(new) = 320 nA
Answer: The saturation current is 1.77 nA at 0°C and
320 nA at 75°C.
2-9.
Given:
ISL = 10 nA with a reverse voltage of 10 V
New reverse voltage = 100 V
Solution:
RSL = VR/ISL
RSL = 10 V/10 nA
RSL = 1000 MΩ
ISL = VR/RSL
ISL = 100 V/1000 MΩ
ISL = 100 nA
Answer: 100 nA.
Semiconductor
Conductor
Semiconductor
Conductor
2-10. Answer: Saturation current is 0.53 μA, and surfaceleakage current is 4.47 μA at 25°C.
2-11. Reduce the saturation current, and minimize the RC time
constants.
2-12. R1 = 25 Ω
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2-13. R1 open
3-3.
2-14. D1 shorted
2-15. D1 open
2-16. V1 = 0 V
Solution: Since the diodes are in series, the current
through each is the same.
Answer: 400 mA
Chapter 3 Diode Theory
3-4.
SELF-TEST
1.
2.
3.
4.
5.
6.
b
b
c
d
a
b
7.
8.
9.
10.
11.
12.
c
c
a
a
b
b
13.
14.
15.
16.
17.
a
d
a
c
b
18.
19.
20.
21.
22.
b
a
b
a
c
4. If you have a data sheet, look up the maximum current rating and the breakdown voltage. Then, check the schematic
diagram to see whether the ratings are adequate. If they are,
check the circuit wiring.
7. Measure the voltage across a resistor in series with the diode.
Then, divide the voltage by the resistance.
8. With the power off, check the back-to-front ratio of the diode
with an ohmmeter or use the diode test function on a DMM.
If it is high, the diode is OK. If it is not high, disconnect one
end of the diode and recheck the back-to-front ratio. If the
ratio is now high, the diode is probably OK. If you are still
suspicious of the diode for any reason, the ultimate test is to
replace it with a known good one.
10. Connect a diode in series between the alternator and the
battery for the recreational vehicle. The diode arrow points
from the alternator to the RV battery. This way, the alternator can charge the vehicle battery. When the engine is
off, the diode is open, preventing the RV battery from
discharging.
11. Use a voltmeter or oscilloscope for a diode in the circuit. Use
an ohmmeter, DMM or curve tracer when the diode is out of
the circuit.
IL = VL/RL (Ohm’s law)
IL = 20 V/1 kΩ
IL = 20 mA
PL = (IL)(VL)
PL = (20 mA)(20 V)
PL = 400 mW
PD = (ID)(VD)
PD = (20 mA)(0 V)
PD = 0 mW
PT = PD + PL
PT = 0 mW + 400 mW
PT = 400 mW
Answer:
IL = 20 mA
VL = 20 V
PL = 400 mW
PD = 0 mW
PT = 400 mW
3-5.
PROBLEMS
Since it is a series circuit, the current flowing through the
diode is the same as the current through the resistor.
Answer: 27.27 mA
3-2.
Given:
VD = 0.7 V
ID = 100 mA
Solution:
P = VI
P = (0.7 V)(100 mA)
P = 70 mW
Answer: 70 mW
Given:
VS = 20 V
VD = 0 V
RL = 2 kΩ
Solution:
IL = VL/RL (Ohm’s law)
IL = 20 V/2 kΩ
IL = 10 mA
Given:
R = 220 Ω
V=6V
Solution:
I = V/R
I = 6 V/220 Ω
I = 27.27 mA
Given:
VS = 20 V
VD = 0 V
RL = 1 kΩ
Solution:
VS = VD + VL (Kirchhoff’s law)
20 V = 0 V + VL
VL = 20 V
JOB INTERVIEW QUESTIONS
3-1.
Given:
VD1 = 0.75 V
VD2 = 0.8 V
ID1 = 400 mA
Answer: 10 mA
3-6.
Given:
VS = 12 V
VD = 0 V
RL = 470 Ω
Solution:
VS = VD + VL (Kirchhoff’s law)
12 V = 0 V + VL
VL = 12 V
IL = VL/RL (Ohm’s law)
IL = 12 V/470 Ω
IL = 25.5 mA
PL = (VL)(IL)
PL = (12 V) (25.5 mA)
PL = 306 mW
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PD = (VD)(ID)
PD = (0 V) (25.5 mA)
PD = 0 mW
PT = PD + PL
PT = 0 mW + 306 mW
PT = 306 mW
Answer:
VL = 12 V
IL = 25.5 mA
PL = 306 mW
PD = 0 mW
PT = 306 mW
3-7.
Given:
VS = 12 V
VD = 0 V
RL = 940 Ω
Solution:
IL = VL/RL (Ohm’s law)
IL = 12 V/940 Ω
IL = 12.77 mA
Answer: 12.77 mA
3-8.
Given:
VS = 12 V
RL = 470 Ω
Solution: The diode would be reversed-biased and acting
as an open. Thus the current would be zero and the voltage would be source voltage.
Answer:
VD = 12 V
ID = 0 mA
3-9.
Given:
VS = 20 V
VD = 0.7 V
RL = 1 kΩ
Solution:
VS = VD + VL (Kirchhoff’s law)
20 V = 0.7 V + VL
VL = 19.3 V
IL = VL/RL (Ohm’s law)
IL = 19.3 V/1 kΩ
IL = 19.3 mA
PL = (IL)(VL)
PL = (19.3 mA)(19.3 V)
PL = 372 mW
PD = (ID)(VD)
PD = (19.3 mA)(0.7 V)
PD = 13.5 mW
PT = PD + PL
PT = 13.5 mW + 372 mW
PT = 386 mW
Answer:
IL = 19.3 mA
VL = 19.3 V
PL = 372 mW
PD = 13.5 mW
PT = 386 mW
3-10. Given:
VS = 20 V
VD = 0.7 V
RL = 2 kΩ
Solution:
IL = VL/RL (Ohm’s law)
IL = 19.3 V/2 kΩ
IL = 9.65 mA
Answer: 9.65 mA
3-11. Given:
VS = 12 V
VD = 0.7 V
RL = 470 Ω
Solution:
VS = VD + VL (Kirchhoff’s law)
12 V = 0.7 V + VL
VL = 11.3 V
IL = VL/RL (Ohm’s law)
IL = 11.3 V/470 Ω
IL = 24 mA
PL = (VL)(IL)
PL = (11.3 V)(24 mA)
PL = 271.2 mW
PD = (VD)(ID)
PD = (0.7 V)(24 mA)
PD = 29.2 mW
PT = PD + PL
PT = 29.2 mW + 271.2 mW
PT = 300.4 mW
Answer:
VL = 11.3 V
IL = 24 mA
PL = 271.2 mW
PD = 29.2 mW
PT = 300.4 mW
3-12. Given:
VS = 12 V
VD = 0.7 V
RL = 940 Ω
Solution:
VS = VD + VL (Kirchhoff’s law)
12 V = 0.7 V + VL
VL = 11.3 V
IL = VL/RL (Ohm’s law)
IL = 11.3 V/940 Ω
IL = 12 mA
Answer: 12 mA
3-13. Given:
VS = 12 V
RL = 470 Ω
Solution: The diode would be reversed-biased and acting
as an open. Thus the current would be zero and the voltage would be source voltage.
Answer:
VD = 12 V
ID = 0 mA
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Note for Probs. 3-14 to 3-18: Since the bulk resistance
of a 1N4001 is 0.23 Ω and meets the criteria of Eq. (3-6)
(rB < 0.01 RTH), it can be ignored for these calculations.
3-14. Given:
VS = 20 V
VD = 0.7 V
RL = 1 kΩ
Solution:
VS = VD + VL (Kirchhoff’s law)
20 V = 0.7 V + VL
VL = 19.3 V
IL = VL /RL (Ohm’s law)
IL = 19.3 V/1 kΩ
IL = 19.3 mA
Answer:
VL = 11.3 V
IL = 24 mA
PL = 271.2 mW
PD = 29.2 mW
PT = 300.4 mW
3-17. Given:
VS = 12 V
VD = 0.7 V
RL = 940 Ω
Solution:
VS = VD + VL (Kirchhoff’s law)
12 V = 0.7 V + VL
VL = 11.3 V
PL = (IL)(VL)
PL = (19.3 mA)(19.3 V)
PL = 372 mW
IL = VL/RL (Ohm’s law)
IL = 11.3 V/940 Ω
IL = 12 mA
PD = (ID)(VD)
PD = (19.3 mA)(0.7 V)
PD = 13.5 mW
Answer: 12 mA
PT = PD + PL
PT = 13.5 mW + 372 mW
PT = 386 mW
Answer:
IL = 19.3 mA
VL = 19.3 V
PL = 372 mW
PD = 13.4 mW
PT = 386 mW
3-15. Given:
VS = 20 V
VD = 0.7 V
RL = 2 kΩ
Solution:
IL = VL/RL (Ohm’s law)
IL = 19.3 V/2 kΩ
IL = 9.65 mA
Answer: 9.65 mA
3-16. Given:
VS = 12 V
VD = 0.7 V
RL = 470 Ω
Solution:
VS = VD + VL (Kirchhoff’s law)
12 V = 0.7 V + VL
VL = 11.3 V
IL = VL/RL (Ohm’s law)
IL = 11.3 V/470 Ω
IL = 24 mA
PL = (VL)(IL)
PL = (11.3 V)(24 mA)
PL = 271.2 mW
PD = (VD)(ID)
PD = (0.7 V)(24 mA)
PD = 29.2 mW
PT = PD + PL
PT = 29.2 mW + 271.2 mW
PT = 300.4 mW
3-18. Given:
VS = 12 V
RL = 470 Ω
Solution: The diode would be reversed-based and acting
as an open. Thus the current would be zero, and the voltage would be source voltage.
Answer:
VD = 12 V
ID = 0 mA
3-19. Open
3-20. The diode voltage will be 5 V, and it should burn open the
diode.
3-21. The diode is shorted, or the resistor is open.
3-22. The voltage of 3 V at the junction of R1 and R2 is normal if it is a voltage divider with nothing in parallel
with R2. So, the problem is in the parallel branch. A
reading of 0 V at the diode resistor junction indicates
either a shorted resistor (not likely) or an open diode.
A solder bridge could cause the resistor to appear to be
shorted.
3-23. A reverse diode test reading of 1.8 V indicates a leaky
diode.
3-24. 1N4004
3-25. Cathode band. The arrow points toward the band.
3-26. The temperature limit is 175°C, and the temperature of
boiling water is 100°C. Therefore, the temperature of the
boiling water is less than the maximum temperature and
the diode will not be destroyed.
CRITICAL THINKING
3-27. Given:
1N914: forward 10 mA at 1 V; reverse 25 nA at 20 V
1N4001: forward 1 A at 1.1 V; reverse 10 μA at 50 V
1N1185: forward 10 A at 0.95 V; reverse 4.6 mA at
100 V
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Solution:
1N914 forward:
R = V/I (Ohm’s law)
R = 1 V/10 mA
R = 100 Ω
V = VR + VD (Kirchhoff’s law)
V = 1.25 V + 0.7 V
V = 1.95 V
1N914 reverse:
R = V/I (Ohm’s law)
R = 20 V/25 nA
R = 800 MΩ
VR1 = VS – V (Kirchhoff’s law)
VR1 = 12 V – 1.95 V
VR1 = 10.05 V
This is the voltage at the junction of R1 and R2. Next find
the voltage drop across R1.
I = V/R (Ohm’s law)
I = 10.05 V/30 kΩ
I = 335 μA
1N4001 forward:
R = V/I (Ohm’s law)
R = 1.1 V/1 A
R = 1.1 Ω
Now that the current through R1 is known, this is the total
current for the parallel branches. The next step is to find
the current through R2.
1N4001 reverse:
R = V/I (Ohm’s law)
R = 50 V/10 μA
R = 5 MΩ
I2 = I1 – ID (Kirchhoff’s law)
I2 = 335 μA – 0.25 mA
I2 = 85 μA
1N1185 forward:
R = V/I (Ohm’s law)
R = 0.95 V/10 A
R = 0.095 Ω
1N1185 reverse:
R = V/I (Ohm’s law)
R = 100 V/4.6 mA
R = 21.7 kΩ
Answer:
1N914:
forward R = 100 Ω
reverse R = 800 MΩ
The next step is to use the voltage and current to calculate
the resistance.
R2 = V/I2 (Ohm’s law)
R2 = 1.95 V/85 μA
R2 = 23 kΩ
Answer: R2 = 23 kΩ
3-30. Given:
500 mA at 1 V
0 mA at 0.7 V
Solution:
rB = (V2 – V1)(I2 – I1) (Eq. 3-7)
rB = (1 V – 0.7 V)/(500 mA – 0 mA)
rB = 600 mΩ
1N4001:
forward R = 1.1 Ω
reverse R = 5 MΩ
1N1185:
forward R = 0.095 Ω
reverse R = 21.7 kΩ
3-28. Given:
VS = 5 V
VD = 0.7 V
ID = 20 mA
Answer: rB = 600 mΩ
3-31.
1.
2.
3.
4.
5.
Solution:
VR = VS – VD (Kirchhoff’s law)
VR = 5 V – 0.7 V
VR = 4.3 V
6.
R = V/I (Ohm’s law)
R = 4.3 V/20 mA
R = 215 Ω
8.
Answer: R = 215 Ω
3-29. Given:
VD = 0.7 V
ID = 10 mA
R1 = 30 kΩ
R3 = 5 kΩ
Solution: Find the voltage required on the parallel branch
to achieve a diode current of 0.25 mA.
VR = IR3 (Ohm’s law)
VR = (0.25 mA)(5 kΩ)
VR = 1.25 V
7.
9.
10.
11.
12.
IR = ISL + IS
5 μA = ISL + IS(old)
ISL = 5 μA – IS(old)
100 μA = ISL + IS(new)
IS(new) = 2(∆T/10) IS(old) (Eq. 2-6)
Substitute formulas 2 and 5 into formula 4.
100 μA = 5 μA –IS(old) + 2(∆T/10)IS(old)
Put in the temperature values.
100 μA = 5 μA –IS(old) + 2[(100ºC – 25ºC)/10]IS(old)
Move the 5 μA to the left side, and simplify the
exponent of 2.
95 μA = – IS(old) + 27.5 IS(old)
Combine like terms.
95 μA = (27.5– 1)IS(old)
95 μA = (180.02) IS(old)
Solve for the variable.
IS(old) = 95 μA/(180.02)
IS(old) = 0.53 μA
Using formula 3:
13. ISL = 5 μA – IS(old)
14. ISL = 5 μA – 0.53 μA
15. ISL = 4.47 μA
Answer: The surface-leakage current is 4.47 μA at 25°C.
3-32. Given:
R1 = 30 kΩ
R2 = 10 kΩ
R3 = 5 kΩ
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This condition will not occur if the diode is normal. It can
be either opened or shorted. If it is shorted, the resistance
would be 0 Ω. If it is open, it would be the resistance of
the resistors.
Solution: The circuit would have R1 and R2 in parallel,
and the parallel resistance in series with R3.
R = [(R1)(R2)]/(R1 + R2) (Parallel resistance formula)
R = [(30 kΩ)(10 kΩ)]/(30 kΩ + 10 kΩ)
R = 7.5 kΩ
RT = 5 kΩ + 7.5 kΩ
RT = 12.5 kΩ
Answer: The resistance would be 12.5 kΩ if the diode is
open and 0 Ω if the diode is shorted.
9. When you need a high dc output voltage from the power
supply, but a step-up transformer is neither available nor
practical in the design.
11. Because a transformer with a high turns ratio produces a few
thousand volts, which means more insulation and expense.
13. There is probably a short in the circuit that caused excessive
current through the resistor. You have to look at the schematic
diagram and test the different components and wiring to try
to locate the real trouble.
PROBLEMS
4-1.
3-33. During normal operation, the 15-V power supply is supplying power to the load. The left diode is forward-biased
and allows the 15-V power supply to supply current
to the load. The right diode is reversed-based because
15 V is applied to the cathode and only 12 V is applied to
the anode. This blocks the 12-V battery. Once the 15-V
power supply is lost, the right diode is no longer reversedbiased, and the 12-V battery can supply current to the
load. The left diode will become reverse-biased, preventing any current from going into the 15-V power supply.
Since the average and the dc values are the same:
Vdc = 0.318 VP (Eq. 4-2)
Vdc = 0.318 (70.7 V)
Vdc = 22.5 V
Answer: The peak voltage is 70.7 V, the average voltage
is 22.5 V, and the dc voltage is 22.5 V.
3-34. D1 is shorted
4-2.
3-35. D1 is open
3-36. Power supply has failed and is 0 V
3-37. R3 is shorted
3-38. D1 is reverse biased
3-39. 1N4001 silicon rectifier diode
3-41. Cathodes
Answer: The peak voltage is –21.2 V, the average voltage
is –6.74 V, and the dc voltage is –6.74 V.
3-42. Normally reverse biased
4-3.
3-43. Normally reverse biased
Chapter 4 Diode Circuits
SELF-TEST
b
a
b
c
c
b
b
8.
9.
10.
11.
12.
13.
c
c
d
b
b
c
Given: Vin = 15 V ac
Solution:
VP = 1.414 Vrms
VP = 1.414 (15 V ac)
VP = –21.2 V
Vp(out) = Vp(in) (Eq. 4-1)
Vp(out) = –21.2 V
Since the average and the dc values are the same:
Vdc = 0.318 Vp (Eq. 4-2)
Vdc = 0.318 (–21.2 V)
Vdc = –6.74 V
3-40. Yes, the 1N4002 has the same forward current rating and
a higher reverse breakdown rating.
1.
2.
3.
4.
5.
6.
7.
Given: Vin = 50 V ac
Solution:
VP = 1.414 Vrms
VP = 1.414 (50 V ac)
VP = 70.7 V
Vp(out) = Vp(in) (Eq. 4-1)
Vp(out) = 70.7 V
14.
15.
16.
17.
18.
19.
a
b
a
d
c
c
20.
21.
22.
23.
24.
25.
c
a
b
a
c
c
JOB INTERVIEW QUESTIONS
7. The LC type is preferable when tighter regulation is required
and (or) power cannot be wasted. Examples include transmitters, lab test equipment, and military gear when cost is not of
primary concern. The LC filter ideally dissipates no power.
The less costly RC filter consumes power in the resistor.
8. A full-wave rectifier is made up of two back-to-back halfwave rectifiers.
Given: Vin = 50 V ac
Solution:
VP = 1.414 Vrms
VP = 1.414 (50 V ac)
VP = 70.7 V
VP(out) = VP(in) – 0.7 V (Eq. 4-4)
VP(out) = 70.0 V
Since the average and the dc values are the same:
Vdc = 0.318 V (Eq. 4-2)
Vdc = 0.318 (70.0 V)
Vdc = 22.3 V
Answer: The peak voltage is 70.0 V, the average voltage
is 22.3 V, and the dc voltage is 22.3 V.
4-4.
Given: Vin = 15 V ac
Solution:
VP = 1.414 Vrms
VP = 1.414 (15 V ac)
VP = –21.2 V
VP(out) = Vp(in) – 0.7 V
Vp(out) = –20.5 V
(Eq. 4-4)
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VP = 1.414 Vrms
VP = 1.414 (15 V ac)
VP = 21.21 V
Since the average and the DC values are the same:
Vdc = 0.318 V (Eq. 4-2)
Vdc = 0.318 (–20.5 V)
Vdc = –6.52 V
Answer: The peak voltage is –20.5 V, the average voltage
is –6.52 V, and the dc voltage is –6.52 V.
Vp(out) = Vp(in) – 0.7 V (Eq. 4-4)
Vp(out) = 20.51 V
Vdc = 0.318 VP (Eq. 4-2)
Vdc = 0.318 (20.51 V)
Vdc = 6.52 V
Answer: The peak voltage is 20.51 V, and the dc voltage
is 6.52 V.
Output waveform for Probs. 4-1 and 4-3. Waveform is
negative for Probs. 4-2 and 4-4.
4-5.
Given:
Turns ratio = N1/N2 = 6:1 = 6
V1 = 120 Vrms
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 Vrms/6
V2 = 20 Vrms
VP = (1.414) (Vrms)
VP = (1.414) (20 Vrms)
VP = 28.28 VP
Answer: The secondary voltage is 20 Vrms or 28.28 VP.
4-6.
Given:
Turns ratio = N1/N2 =1:12 = 0.083333
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/0.083333
V2 = 1440 V ac
VP = 1.414 Vrms
VP = 1.414 (1440 V ac)
VP = 2036.16 V
Answer: The secondary rms voltage is 1440 V ac, and the
peak voltage is 2036.16 V.
4-7.
Given:
Turns ratio = N1/N2 = 8:1 = 8
V1 = 120 V ac (rms)
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/8
V2 = 15 V ac
VP = 1.414 Vrms
VP = 1.414 (15 V ac)
VP = 21.21 V
Vp(out) = Vp(in) (Eq. 4-1)
Vp(out) = 21.21 V
Vdc = 0.318 VP (Eq. 4-2)
Vdc = 0.318 (21.21 V)
Vdc = 6.74 V
Answer: The peak voltage is 21.21 V, and the dc voltage
is 6.74 V.
4-8.
4-9.
Given:
Turns ratio = N1/N2 = 4:1 = 4
V1 = 120 Vrms
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 Vrms/4
V2 = 30 Vrms
Since it is a center-tapped transformer, each half of the
secondary is half of the total secondary voltage.
Vupper = ½ V2
Vupper = ½ (30 Vrms)
Vupper = 15 Vrms
Vlower = ½ VP
Vlower = ½(30 Vrms)
Vlower = 15 Vrms
VP = (1.414) (Vrms)
VP = (1.414) (15 Vrms)
VP = 21.21 VP
Answer: Each half of the secondary has an rms voltage
of 15 V and a peak voltage of 21.21 V.
4-10. Given:
Turns ratio = N1/N2 = 7:1 = 7
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/7
V2 = 17.14 V ac
VP = 1.414 Vrms
VP = 1.414 (17.14 V ac)
VP = 24.24 V
VP(in) = 0.5 VP
VP(in) = 0.5(24.24 V)
VP(in) = 12.12 V
Vp(out) = Vp(in) (Eq. 4-1)
Vp(out) = 12.12 V
Since the average and the dc values are the same:
Vdc = 0.636 Vp (Eq. 4-6)
Vdc = 0.636 (12.12 V)
Vdc = 7.71 V
Answer: The peak output voltage is 12.12 V, and the dc
and average values are 7.71 V.
Given:
Turns ratio = N1/N2 = 8:1 = 8
V1 = 120 V ac (rms)
4-11. Given:
Turns ratio = N1/N2 = 7:1 = 7
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/8
V2 = 15 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/7
V2 = 17.14 V ac
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VP = 1.414 Vrms
VP = 1.414 (17.14 V ac)
VP = 24.24 V
VP(in) = 0.5 VP
VP(in) = 0.5(24.24 V)
VP(in) = 12.12 V
Vp(out) = Vp(in) – 0.7 V (Eq. 4-4)
Vp(out) = 11.42 V
Since the average and the dc values are the same:
Vdc = 0.636 VP (Eq. 4-6)
Vdc = 0.636 (11.42 V)
Vdc = 7.26 V
Answer: The peak output voltage is 11.42 V, and the dc
and average values are 7.26 V.
4-12. Given:
Turns ratio = N1/N2 = 8:1 = 8
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/8
V2 = 15 V ac
VP(in) = 1.414 Vrms
VP(in) = 1.414 (15 V ac)
VP(in) = 21.21 V
Vp(out) = Vp(in) (Eq. 4-1)
Vp(out) = 21.21 V
Since the average and the dc values are the same:
Vdc = 0.636 Vp (Eq. 4-6)
Vdc = 0.636 (21.21 V)
Vdc = 13.49 V
Answer: The peak output voltage is 21.21 V, and the dc
and average values are 13.49 V.
4-13. Given:
Turns ratio = N1/N2 = 8:1 = 8
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/8
V2 = 15 V ac
VP(in) = 1.414 Vrms
VP(in) = 1.414 (15 V ac)
VP(in) = 21.21 V
Vp(out) = Vp(in) – 1.4 V (Eq. 4-8)
Vp(out) = 19.81 V
Since the average and the dc values are the same:
Vdc = 0.636 Vp (Eq. 4-6)
Vdc = 0.636 (19.81 V)
Vdc = 12.60 V
Answer: The peak output voltage is 19.81 V, and the dc
and average values are 12.60 V.
Output waveform for Probs. 4-10 to 4-13.
4-14. Given:
Turns ratio = N1/N2 = 8:1 = 8
V1(max) = 125 V ac
V1(min) = 102 V ac
Solution:
V2(max) = V1(max)/(N1/N2) (Eq. 4-5)
V2(max) = 125 V ac/8
V2(max) = 15.63 V ac
V2(min) = V1(min)/(N1/N2) (Eq. 4-5)
V2(min) = 105 V ac/8
V2(min) = 13.13 V ac
VP(in)max = 1.414 V2(max)
VP(in)max = 1.414 (15.63 V ac)
VP(in)max = 22.10 V
VP(in)min = 1.414 V2(min)
VP(in)min = 1.414 (13.13 V ac)
VP(in)min = 18.57 V
Vp(out)max = Vp(in)max (Eq. 4-1)
Vp(out)max = 22.10 V
Vp(out)min = Vp(in)min (Eq. 4-1)
Vp(out)min = 18.57 V
Vdc(max) = 0.636 Vp(out)max (Eq. 4-6)
Vdc = 0.636 (22.10 V)
Vdc = 14.06 V
Vdc(min) = 0.636 Vp(out)min (Eq. 4-6)
Vdc = 0.636 (18.57 V)
Vdc = 11.81 V
Answer: The maximum dc output voltage is 14.06 V, and
the minimum is 11.81 V.
4-15. Given:
Vin = 20 V
XL = 1 kΩ
XC = 25 Ω
Solution:
Vout = (XC/XL)Vin (Eq. 4-9)
Vout = (25 Ω/1 Ω)(20 V)
Vout = 500 mV
Answer: The ripple voltage would be 500 mV.
4-16. Given:
Vin = 14 V
XL = 2 kΩ
XC = 50 Ω
Solution:
Vout = (XC/XL)Vin (Eq. 4-9)
Vout = (50 Ω/2 kΩ)(14 V)
Vout = 350 mV
Answer: The ripple voltage would be 350 mV.
4-17. Given:
Turns ratio = N1/N2 = 8:1 = 8
V1 = 120 V ac
RL = 10 kΩ
C = 47 μF
fin = 60 Hz
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/8
V2 = 15 V ac
VP = 1.414 V2
VP = 1.414 (15 V ac)
VP = 21.2 V (This is the dc output voltage due to the
capacitor input filter.)
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I = V/R (Ohm’s law)
I = 21.2 V/10 kΩ
I = 2.12 mA
fout = fin (Eq. 4-3)
fout = 60 Hz
VR = I/(fC) (Eq. 4-10)
VR = (2.12 mA)/[(60 Hz)(47 μF)]
VR = 752 mV
Answer: The dc output voltage is 21.2 V with a 752 mVp-p
ripple.
Output waveform for Prob. 4-17.
4-18. Given:
Turns ratio = N1/N2 = 7:1 = 7
V1 = 120 V ac
RL = 2.2 kΩ
C = 68 μF
fin = 60 Hz
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/7
V2 = 17.14 V ac
VP = 1.414 Vrms
VP = 1.414 (17.14 V ac)
VP = 24.24 V
VP(in) = 0.5 VP
VP(in) = 0.5(24.24 V)
VP(in) = 12.12 V
Vp(out) = Vp(in) (Eq. 4-1)
Vp(out) = 12.12 V (This is the dc output voltage due to the
capacitor input filter.)
I = V/R (Ohm’s law)
I = 12.12 V/2.2 kΩ
I = 5.51 mA
fout = 2fin (Eq. 4-7)
fout = 2(60 Hz)
fout = 120 Hz
VR = I/(fC) (Eq. 4-10)
VR = (5.51 mA)/[(120 Hz)(68 μF)]
VR = 675 mV
Answer: The dc output voltage is 12.12 V, with a
675 mVp-p ripple.
4-19. Answer:
VR = I/(fC) (Eq. 4-10)
If the capacitance is cut in half, the denominator is cut in
half and the ripple voltage will double.
4-20. Answer:
VR = I/(fC) (Eq. 4-10)
If the resistance is reduced to 500 Ω, the current increases
by a factor of 20; thus the numerator is increased by a
factor of 20 and the ripple voltage goes up by a factor
of 20.
4-21. Given:
Turns ratio = N1/N2 = 9:1 = 9
V1 = 120 V ac
RL = 1 kΩ
C = 470 μF
fin = 60 Hz
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/9
V2 = 13.33 V ac
VP = 1.414 Vrms
VP = 1.414 (13.33 V ac)
VP = 18.85 V
Vp(out) = Vp (Eq. 4-1)
Vp(out) = 18.85 V (This is the dc output voltage due to the
capacitor input filter.)
I = V/R (Ohm’s law)
I = 18.85 V/1 kΩ
I = 18.85 mA
fout = 2fin (Eq. 4-7)
fout = 2(60 Hz)
fout = 120 Hz
VR = I/(fC) (Eq. 4-10)
VR = (18.85 mA)/[(120 Hz)(470 μF)]
VR = 334 mV
Answer: The dc output voltage is 18.85 V, with a
334 mVp-p ripple.
Output waveform for Probs. 4-18 and 4-21.
4-22. Given:
Turns ratio = N1/N2 = 9:1 = 9
V1 = 105 V ac
RL = 1 kΩ
C = 470 μF
fin = 60 Hz
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 105 V ac/9
V2 = 11.67 V ac
VP = 1.414 Vrms
VP = 1.414 (11.67 V ac)
VP = 16.50 V
Vp(out) = VP (Eq. 4-1)
Vp(out) = 16.50 V (This is the dc output voltage due to the
capacitor input filter.)
4-23. Given: Vp = 18.85 Vp from Prob. 4-21
Solution:
PIV = VP (Eq. 4-13)
PIV = 18.85 V
Answer: The peak inverse voltage is 18.85 V.
4-24. Given:
Turns ratio = N1/N2 = 3:1 = 3
V1 = 120 Vrms
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 Vrms/3
V2 = 40 Vrms
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VP = (1.414) (Vrms)
VP = (1.414) (40 Vrms)
VP = 56.56 VP
PIV = VP (Eq. 4-13)
PIV = 56.56 V
Answer: The peak inverse voltage is 56.56 V.
4-25. Solution:
From the information in Section 4-8.
a. Secondary output is 12.6 V ac.
VP = 1.414 Vrms
VP = 1.414 (12.6 V ac)
VP = 17.8 V
b. Vdc = 17.8 V
c. I = V/R (Ohm’s law)
Idc = 17.8 V ac/1 kΩ
Idc = 17.8 mA
Rated current is 1.5 A.
Answer: The peak output voltage is 17.8 V, and the
dc output voltage is 17.8 V. It is not operating at
rated current, and thus the secondary voltage will be
higher.
4-26. Given:
Assume Pin = Pout
Vdc = 17.8 V from Prob. 4-25
Idc = 17.8 mA from Prob. 4-25
Solution:
Pout = Idc Vdc
Pout = (17.8 mA)(17.8 V)
Pout = 317 mW
Pin = 317 mW
Pin = V1Ipri
Ipri = Pin/V1
Ipri = 317 mW/120 V
Ipri = 2.64 mA
4-29. Given: Vp(out) = 18.85 V from Prob. 4-21
Solution: Without the filter capacitor to maintain the voltage at peak, the dc voltage is calculated the same way it
would be done if the filter was not there.
Vdc = 0.636 VP
Vdc = 0.636(18.85 V)
Vdc = 11.99 V
Answer: The dc voltage is 11.99 V.
4-30. Answer: With one diode open, one path for current flow
is unavailable. The output will look similar to a halfwave rectifier with a capacitor input filter. The dc voltage
should not change much from the original 18.85 V, but
the ripple will increase to approximately double because
the frequency drops from 120 to 60 Hz.
4-31. Answer: Since an electrolytic capacitor is polaritysensitive, if it is put in backward, it will be destroyed and
the power supply will act as if it did not have a filter.
4-32. Answer: VP will remain the same, DC output equals VP,
Vripple = 0 V.
4-33. Answer: Since this is a positive clipper, the maximum
positive will be the diode’s forward voltage, and all the
negative will be passed through. Maximum positive is
0.7 V, and maximum negative is –50 V.
Output waveform for Prob. 4-33.
4-34. Answer: Since this is a negative clipper, the maximum
negative will be the diode’s forward voltage, and all the
positive will be passed through. The maximum positive is
24 V, and the maximum negative is –0.7 V.
Answer: The primary current would be 2.64 mA.
4-27. Given:
VDC = 21.2 V from Prob. 4-17
VDC = 12.12 V from Prob. 4-18
Solution:
Fig. 4-40(a)
Idiode = V/R
Idiode = (2.12 V)/(10 kΩ)
Idiode = 212 μA
Fig. 4-40(b)
I = V/R
I = (12.12 V)/(2.2 kΩ)
I = 5.5 mA
Idiode = 0.5 I
Idiode = (0.5)/(5.5 mA)
Idiode = 2.75 mA
Answer: The average diode current in Fig. 4-40(a) is
212 μA and the current in Fig. 4-40(b) is 2.75 mA.
4-28. Given: Idc = 18.85 mA from Prob. 4-21
Solution:
Idiode = (0.5)Idc
Idiode = (0.5)(18.85 mA)
Idiode = 9.43 mA
Output waveform for Prob. 4-34.
4-35. Answer: The limit in either direction is two diode voltage
drops. Maximum positive is 1.4 V, and maximum negative is –1.4 V.
4-36. Given:
DC voltage 15 V
R1 = 1 kΩ
R2 = 6.8 kΩ
Solution:
Voltage at the cathode is found by using the voltage
divider formula.
Vbias = [R1/(R1 + R2)]Vdc (Eq. 4-18)
Vbias = [1 kΩ/(1 kΩ + 6.8 kΩ)]15 V
Vbias = 1.92 V
The clipping voltage is the voltage at the cathode and the
diode voltage drop.
Vclip = 1.92 V + 0.7 V
Vclip = 2.62 V
Answer: Since it is a positive clipper, the positive voltage
is limited to 2.62 V and the negative to –20 V.
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Output waveform for Prob. 4-36.
4-37. Answer: The output will always be limited to 2.62 V.
4-38. Answer: Since this is a positive clamper, the maximum
negative voltage will be –0.7 V and the maximum positive will be 29.3 V.
VP = 1.414 Vrms
VP = 1.414 (600 V ac)
VP = 848.4 V
Since it is a tripler, the output is 3VP.
Vout = 3VP
Vout = 3 (848.4 V)
Vout = 2545.2 V
Answer: The output voltage will be 2545.2 V.
4-43. Given:
Turns ratio = N1/N2 = 1:7 = 0.143
V1 = 120 V ac
Output waveform for Prob. 4-38.
4-39. Answer: Since this is a negative clamper, the maximum
positive voltage will be 0.7 V and the maximum negative
will be –59.3 V.
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/0.143
V2 = 839.2 V ac
VP = 1.414 Vrms
VP = 1.414 (839.2 V ac)
VP = 1186.6 V
Since it is a quadrupler, the output is 4VP.
Vout = 4VP
Vout = 4(1186.6 V)
Vout = 4746.4 V
Output waveform for Prob. 4-39.
4-40. Answer: The output will be 2VP or Vp-p, which is 40 V. If
the second approximation is used, the maximum for the
clamp will be 39.3 V instead of 40 V, and since there is
also a diode voltage drop, the output would be 38.6 V.
Output waveform for Prob. 4-40.
4-41. Given:
Turns ratio = N1/N2 = 1:10 = 0.1
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/0.1
V2 = 1200 V ac
VP = 1.414 Vrms
VP = 1.414 (1200 V ac)
VP = 1696.8 V
Since it is a doubler, the output is 2VP.
Vout = 2VP
Vout = 2 (1696.8 V)
Vout = 3393.6 V
Answer: The output voltage will be 3393.6 V.
4-42. Given:
Turns ratio = N1/N2 = 1:5 = 0.2
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/0.2
V2 = 600 V ac
Answer: The output voltage will be 4746.4 V.
CRITICAL THINKING
4-44. Answer: If one of the diodes shorts, it will provide a low
resistance path to either blow a fuse or damage the other
diodes.
4-45. Given:
Turns ratio = N1/N2 = 8:1 = 8
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/8
V2 = 15 V ac
VP = 1.414 Vrms
VP = 1.414 (15 V ac)
VP = 21.21 V
Since each resistor is in the same current path and both
have the same value, they equally divide the voltage.
Since they both have a capacitor input filter, they divide
the peak voltage.
Answer: Each power supply has 10.6 V, but the load
connected to the right side of the bridge is a positive
10.6 V and the load connected to the left side is a
negative 10.6 V.
4-46. Given:
VP = 21.21 VP from Prob. 4-1
R = 4.7 Ω
Solution: The maximum surge current would be all of the
peak voltage dropped across the resistor.
I = V/R (Ohm’s law)
I = 21.21 V/4.7 Ω
I = 4.51 A
Answer: The maximum surge current will be 4.51 A.
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4-47. Answer: The signal is a sine wave, and thus the shape of
the curve is a function of sine. The formula for the instantaneous voltage at any point on the curve is V = Vp sin θ.
Using this formula, calculate the values for each point on
the curve, add all 180 of the 1° points together and divide
by 180.
4-48. Given:
Turns ratio = N1/N2 = 8:1 = 8
V1 = 120 V ac
Solution:
V2 = V1/(N1/N2) (Eq. 4-5)
V2 = 120 V ac/8
V2 = 15 V ac
VP = 1.414 Vrms
VP = 1.414 (15 V ac)
VP = 21.21 V
With the switch in the shown position, it is a bridge rectifier with a capacitor input filter. Thus the output voltage
would be 21.21 V.
With the switch in the other position, it is a full-wave
rectifier with a capacitor input filter. Since it is a centertapped transformer, the peak voltage is half.
VP = 10.6 VP
The output would be 10.6 V.
Answer: With the switch in the shown position, 21.21 V;
with the switch in the other position, 10.6 V.
4-49. Answer: Both capacitors will charge to approximately
56 mV with opposite polarities. Vout will equal 56 mV –
56 mV. Vout will equal zero volts.
4-50. Fault 1—Since the load voltage is 0.636 of the peak voltage, the capacitor input filter is not doing its job; thus
the capacitor is bad.
Fault 2—Since the load voltage dropped a little and the
ripple doubled, one of the diodes is open; this causes
the frequence of the ripple to drop to half, which in
turn causes the ripple to double.
Fault 3—Since V1 is zero, the fuse must be blown. Since
the load resistance is zero, the load resistor is shorted.
This caused the excessive current in the secondary,
which fed back to the primary and blew the fuse.
Fault 4—Since V2 is good and all other voltages are bad,
the transformer and fuse are good. R and C are good:
thus either all four diodes opened (not likely) or there
is an open in the ground circuit.
Fault 5—Since V1 is zero, the fuse must be blown.
Fault 6—The load resistor is open. No current is drawn,
and thus there is no ripple.
Fault 7—Since V1 is good and V2 is bad, the transformer
is the problem.
Fault 8—Since V1 is zero, the fuse must be blown. Since
the capacitor reads zero, the capacitor is shorted. This
caused the excessive current in the secondary, which
fed back to the primary and blew the fuse.
Fault 9—Since the load voltage is 0.636 of the peak voltage, the capacitor input filter is not doing its job and
thus the capacitor is bad.
4-51. C1 is open
4-52. Full-wave bridge is open
4-53. C1 is shorted
4-54. V1 failed
4-55. XMFR secondary winding is open
4-56. Full-wave bridge rectifier
4-57. 15 Vac
4-58. Full-wave rectifier
4-59. Approximately 8.1 V
4-60. 14 V
Chapter 5 Special-Purpose Diodes
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
d
b
b
a
a
c
c
a
9.
10.
11.
12.
13.
14.
15.
16.
c
b
c
a
b
d
d
a
17.
18.
19.
20.
21.
22.
23.
24.
c
c
b
b
a
c
c
c
25.
26.
27.
28.
29.
30.
31.
32.
b
d
a
c
b
b
d
a
JOB INTERVIEW QUESTIONS
3. The zener regulation is dropping out of regulation during
worst-case conditions of low line voltage and high load
current.
4. The LED is connected backward, or the LED current is
excessive either because the series resistor is too small or the
driving voltage is too high.
5. The basic idea is that a varactor is a voltage-controlled
capacitance. By using a varactor as part of an LC tank
circuit, we can control the resonant frequency with a dc
voltage.
6. To provide a high degree of electrical isolation between input
and output circuits.
7. The cathode lead is shorter than the anode lead. Also, the flat
side of the dome package is the cathode.
PROBLEMS
5-1.
Given:
VS = 24 V
VZ = 15 V
RS = 470 Ω
Solution:
IS = IZ = (VS – VZ)/RS (Eq. 5-3)
IS = IZ = (24 V – 15 V)/(470 Ω)
IS = IZ = 19.1 mA
Answer: The zener current is 19.1 mA.
5-2.
Given:
VS = 40 V
VZ = 15 V
RS = 470 Ω
Solution:
IS = IZ = (VS – VZ)/RS (Eq. 5-3)
IZ = (40 V – 15 V)/470 Ω
IZ = 53.2 mA
Answer: The maximum zener current is 53.2 mA.
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5-3.
Given:
VS = 24 V
VZ = 15 V
RS = 470 Ω ± 5%
RS(max) = 493.5 Ω
RS(min) = 446.5 Ω
IZ = IS – IL (Eq. 5-6, Kirchhoff’s current law)
IZ = 20.16 mA – 9.52 mA
IZ = 10.64 mA
Answer: The maximum zener current is 10.64 mA.
5-7.
Solution:
IS = IZ(max) = (VS – VZ)/RS(min) (Eq. 5-3)
IS = IZ = (24 V – 15 V)/(446.5 Ω)
IS = IZ = 20.16 mA
Solution: Maximum current will occur at maximum
voltage.
IS = (VS – VZ)/RS (Eq. 5-3)
IS = (40 V – 15 V)/470 Ω
IS = 53.2 mA
Answer: The maximum zener current is 20.16 mA.
5-4.
Given:
VS = 24 V
VZ = 15 V
RS = 470 Ω
RL = 1.5 kΩ
IL = VL/RL (Eq. 5-5, Ohm’s law)
IL = 15 V/1.5 kΩ
IL = 10 mA
Solution:
VL = [RL/(RS+RL)]VS (Voltage divider formula)
VL = [1.5 kΩ/(470 Ω + 1.5 kΩ)]24 V
VL = 18.27 V
Answer: The load voltage is 18.27 V.
5-5.
IZ = IS – IL (Eq. 5.6, Kirchhoff’s current law)
IZ = 53.2 mA – 10 mA
IZ = 43.2 mA
Answer: The maximum zener current is 43.2 mA.
5-8.
Given:
VS = 24 V
VZ = 15 V
RS = 470 Ω
RL = 1.5 kΩ
IS = (VS – VZ)/RS (Eq. 5-3)
IS = (24 V – 12 V)/470 Ω
IS = 25.5 mA
IL = VL/RL (Eq. 5-5, Ohm’s law)
IL = 15 V/1.5 kΩ
IL = 10 mA
IL = VL/RL (Eq. 5-5, Ohm’s law)
IL = 12 V/1.5 kΩ
IL = 8 mA
IZ = IS – IL (Eq. 5-6, Kirchhoff’s current law)
IZ = 19.15 mA – 10 mA
IZ = 9.15 mA
IZ = IS – IL (Eq. 5-6, Kirchhoff’s current law)
IZ = 25.5 mA – 8 mA
IZ = 17.5 mA
Answer: The series current is 19.15 mA, the zener current
is 9.15 mA, and the load current is 10 mA.
Given:
VS = 24 V
VZ = 15 V
RS = 470 Ω ± 5%
RS(max) = 493.5 Ω
RS(min) = 446.5 Ω
RL = 1.5 kΩ
RL(max) = 1.575 kΩ
RL(min) = 1.425 kΩ
Solution: Looking at Eq. (5-6), the maximum zener current would occur at a maximum series current and a minimum load current. To achieve these conditions, the series
resistance would have to be minimum and the load resistance would have to be maximum.
IS = (VS – VZ)/RS(min) (Eq. 5-3)
IS = (24 V – 15 V)/446.5 Ω
IS = 20.16 mA
IL = VL/RL(max) (Eq. 5-5. Ohm’s law)
IL = 15 V/1.575 kΩ
IL = 9.52 mA
Given:
VS = 24 V
VZ = 12 V
RS = 470 Ω
RL = 1.5 kΩ
Solution:
VL = VZ = 12 V
Solution:
IS = (VS – VZ)/RS (Eq. 5-3)
IS = (24 V – 15 V)/470 Ω
IS = 19.15 mA
5-6.
Given:
VS = 24 V to 40 V
VZ = 15 V
RS = 470 Ω
Answer: The load voltage is 12 V and the zener current
is 17.5 mA.
5-9.
Given:
VS = 20 V
VZ = 12 V
RS = 330 Ω
RL = 1 kΩ
Solution:
VL = VZ = 12 V
IS = (VS – VZ)/RS (Eq. 5-3)
IS = (20 V – 12 V)/330 Ω
IS = 24.24 mA
IL = VL/RL (Eq. 5-5, Ohm’s law)
IL = 12 V/1 kΩ
IL = 12 mA
IZ = IS – IL (Eq. 5-6, Kirchhoff’s current law)
IZ = 24.24 mA – 12 mA
IZ = 12.24 mA
Answer: The load voltage is 12 V, and the zener current
is 12.24 mA.
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RS
330 V
20 V
12 V
RL
1 kV
Zener regulator for Prob. 5-9.
5-10. Given:
RS = 470 Ω
RZ = 1 4Ω
VR(in) = 1 Vp-p
Solution:
R
VR(out) = __
​  RZS ​VR(in) (Eq. 5-8)
VR(out) = (14 Ω /470 Ω )/1 Vp-p
VR(out) = 29.8 mVp-p
Answer: The ripple voltage across the load resistor is
29.8 mVp-p.
5-11. Given:
VS = 21.5 to 25 V
RS = 470 Ω
RZ = 14 Ω
VZ = 15 V
Solution:
IS = (VS – VZ)/RS (Eq. 5-3)
IS = (25 V – 15 V)/470 Ω
IS = 21.28 mA
IL = VL/RL (Eq. 5-5, Ohm’s law)
IL = 15 V/1.5 kΩ
IL = 10 mA
IZ = IS – IL (Eq. 5-6, Kirchhoff’s current law)
IZ = 21.28 mA – 10 mA
IZ = 11.28 mA
ΔVL = IZRZ (Eq. 5-7)
ΔVL = (11.28 mA)(14 Ω )
ΔVL = 157.9 mV
IS = (VS – VZ)/RS (Eq. 5-3)
IS = (21.5 V – 15 V)/470 Ω
IS = 13.83 mA
IL = VL/RL (Eq. 5-5, Ohm’s law)
IL = 15 V/1.5 kΩ
IL = 10 mA
IZ = IS – IL (Eq. 5-6, Kirchhoff’s current law)
IZ = 13.83 mA – 10 mA
IZ = 3.83 mA
ΔVL= IZRZ (Eq. 5-7)
ΔVL = (3.83 mA)(14 Ω)
ΔVL = 53.6 mV
Answer: The load voltage changes from 15.054 V when
the supply is 21.5 V, to 15.158 V when the supply is
25 V.
5-12. Given:
VS = 24 V
RS = 470 Ω
RL = 1.5 kΩ
VZ = 15 V
Solution: The regulation is lost once the load voltage
drops below 15 V.
VL = [(RL)/(RS + RL)]VS (Voltage divider formula)
VS = VL/[RL/(RS + RL)]
VS = 15 V[(1.5 kΩ)/(470 Ω + 1.5 kΩ)]
VS = 19.7 V
Answer: The regulation will be lost when the source voltage drops below 19.7 V.
5-13. Given:
VS = 20 to 26 V
RS = 470 Ω
RL = 500 to 1.5 kΩ
VZ = 15 V
Solution: The regulation is lost once the load voltage
drops below 15 V.
IL = VL/RL (Eq. 5-5, Ohm’s law)
IL = 15 V/1.5 kΩ
RS(max) = [(VS(min) /VZ ) – 1]RL(min) (Eq. 5-9)
RS(max) = [(20 V/15 V) – 1]500 Ω
RS(max) = 167 Ω
Answer: The regulator will fail since the series resistor
is greater than the maximum series resistance. For this
regulator to work properly, the series resistor should be
167 Ω or less.
5-14. Given:
VS = 18 to 25 V
RS = 470 Ω
IL = 1 to 25 mA
VZ = 15 V
Solution:
V
– VZ
RS(max) = ______
​  S(min)
IL(max) ​ (Eq. 5-10)
RS(max) = (18 V – 15V)/25 mA
RS(max) = 120 Ω
Answer: Yes, the regulator will fail since the series resistance is greater than the maximum series resistance. For
this regulator to work properly, the series resistor should
be 120 Ω or less.
5-15. Given:
VS = 24 V
RS = 470 Ω
VZ = 15 V
Solution:
RS(max) = [(VS(min) /VZ) – 1]RL(min) (Eq. 5-9)
RL(min) = RS(max) /[(VS(min) /VZ) – 1]
RL(min) = 470 Ω /[(24 V/15 V) – 1]
RL(min) = 783 Ω
Answer: The minimum load resistance is 783 Ω .
5-16. Given:
VZ = 10 V
IZ = 20 mA
Solution:
PZ = VZIZ (Eq. 5-11)
PZ = (10 V)(20 mA)
PZ = 0.2 W
Answer: The power dissipation is 0.2 W.
5-17. Given:
VZ = 20 V
IZ = 5 mA
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Solution:
PZ = VZIZ (Eq. 5-11)
PZ = (20 V)(5 mA)
PZ = 0.1 W
Answer: The power dissipation is 0.1 W.
5-18. Given:
VS = 24 V
RS = 470 Ω
RL = 1.5 kΩ
VZ = 15 V
IS = 19.15 mA (from Prob. 5-5)
IL = 10 mA (from Prob. 5-5)
VZ = 9.15 mA (from Prob. 5-5)
Solution:
P = I2R
PS = (19.15 mA)2(470Ω)
PS = 172.4 mW
P = I2R
PL = (10 mA)2(1.5 kΩ)
PL = 150 mW
PZ = VI
PZ = (15 V)(9.15 mA)
PZ = 137.3 mW
Answer: The power dissipation of the series resistor is
172.4 mW. The power dissipation of the load resistor
is 150 mW. The power dissipation of the zener diode is
137.3 mW.
5-19. Given: VZ = 15 V ± 5%. The tolerance is determined from
the data sheet.
Solution:
(15 V)(0.05) = 0.75 V
15 V + 0.75 V = 15.75 V
15 V – 0.75 V = 14.25 V
Answer: The minimum voltage of 14.25 V and the maximum voltage is 15.75 V.
5-20. Given:
T = 100°C
Solution:
100°C – 50°C = 50°C
Derating factor 6.67 mW/°C
Answer: P = 667 mW
5-21. Given:
VS = 24 V
RS = 470 Ω
RL = 1.5 kΩ
RZ = 15 V
Solutions:
a. With the diode in parallel with the load, the load resistor is also effectively shorted and the output voltage
would be 0 V.
b. With the diode open, the load resistor and the series
resistor form a voltage divider:
VL = [RL/(RS + RL)]VS (Voltage divider formula)
VL = [1.5 kΩ/(470 Ω + 1.5 kΩ)]24 V
VL = 18.27 V
c. With the series resistor open, no voltage reaches the
load; thus the output voltage would be 0 V.
d. The voltage drop across a short is 0 V.
Answers:
a. 0 V
b. 18.27 V
c. 0 V
d. 0 V
5-22. Answer: From the previous problem, the only trouble that
caused this symptom is an open zener diode.
5-23. Answer: Check the series resistor. If it is shorted, it could
damage the diode. If it had been operating correctly, the
output voltage should have been 18.3 V.
5-24. Answers:
a. If the V130LA2 is open, it will remove the over voltage protection and the LED will remain lit.
b. If the ground is opened, there is no path for current
and thus the LED will not be lit.
c. If the filter capacitor is open, the voltage will have
more ripple but the LED should remain lit.
d. If the filter capacitor is shorted, the voltage across all
devices in parallel with it will be zero; thus the LED
will not be lit.
e. If the 1N5314 is open, it will have no effect on the
LED.
f. If the 1N5314 is shorted, the voltage across all devices
in parallel with it will be zero; thus the LED will not
be lit.
5-25. Given:
VS = 15 V
VD = 2 V
RS = 2.2 kΩ
Solution:
IS = (VS – VD)/RS (Eq. 5-13)
IS = (15 V – 2 V)/2.2 kΩ
IS = 5.91 mA
Answer: The diode current is 5.91 mA.
5-26. Given:
VS = 40 V
VD = 2 V
RS = 2.2 kΩ
Solution:
IS = (VS – VD)/RS (Eq. 5-13)
IS = (40 V – 2 V)/2.2 kΩ
IS = 17.27 mA
Answer: The diode current is 17.27 mA.
5-27. Given:
VS = 15 V
VD = 2 V
RS = 1 kΩ
Solution:
IS = (VS – VD)/RS (Eq. 5-13)
IS = (15 V – 2 V)/1 kΩ
IS = 13 mA
Answer: The diode current is 13 mA.
5-28. Answer: From Prob. 5-27, the resistor value will be 1 kΩ.
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Solution:
RS(max) = [(VS(min) – VZ)/IL(max)] (Eq. 5-10)
RS(max) = [(20 V – 6.8 V)/30 mA]
RS(max) = 440 Ω
CRITICAL THINKING
5-29. Given:
VS = 24 V
RS = 470 Ω
RZ = 14 Ω
VZ = 15 V
RS(min) = [(VS – VZ)/IZM]
RS(min) = [(20 V – 6.8 V)/147 mA]
RS(min) = 90 Ω
Solution:
IS = (VS – VZ)/RS (Eq. 5-3)
IS = (24 – 15)/470 Ω
IS = 19.15 mA
Answer: Any similar design as long as the zener voltage
is 6.8 V and the series resistance is less than 440 Ω, to
provide the desired maximum output current, and greater
than 90 Ω, if a 1N4736A is used to prevent overcurrent if
it becomes unloaded. The load resistance does not need
to be specified because, as a power supply, the load resistance can vary. The only load parameter that is necessary
is maximum current, and it is given.
IL = VL/RL (Eq. 5-5, Ohm’s law)
IL = 15 V/1.5 kΩ
IL = 10 mA
IZ = IS – IL (Eq. 5-6, Kirchhoff’s current law)
IZ = 19.15 mA – 10 mA
IZ = 9.15 mA
ΔVL = IZRZ (Eq. 5-7)
ΔVL = (9.15 mA)(14 Ω )
ΔVL = 128.1 mV
VL = 15.128 V or approximately 15.13 V
Answer: The load voltage would be 15.13 V.
5-30. Given:
VS = 24 V
RS = 470 Ω
RZ = 14 Ω
VZ = 15 V
RL = 1 kΩ to 10 kΩ
IS = 19.15 mA (from Prob. 5-29)
Solution:
IL(max) = VL/RL(min) (Eq. 5-5, Ohm’s law)
IL(max) = 15 V/1 kΩ
IL(max) = 15 mA
IL(min) = VL/RL(max) (Eq. 5-5, Ohm’s law)
IL(min) = 15 V/10 kΩ
IL(min) = 1.5 mA
IZ(min) = IS – IL(max) (Eq. 5-6, Kirchhoff’s current law)
IZ(min) = 19.15 mA – 15 mA
IZ(min) = 4.15 mA
IZ(max) = IS – IL(min) (Eq. 5-6, Kirchhoff’s current law)
IZ(max) = 19.15 mA – 1.5 mA
IZ(max) = 17.65 mA
ΔVL(min) = IZ(min)RZ
ΔVL(min) = (4.15 mA)(14 Ω )
ΔVL(min) = 58.1 mV
ΔVL(max) = IZ(max)RZ
ΔVL(max) = (17.65 mA)(14 Ω )
ΔVL(max) = 247.1 mV
VL(min) = 15.058 V
VL(max) = 15.247 V
Answer: The minimum load voltage would be 15.06 V
and the maximum voltage would be 15.25 V.
5-31. Given:
VS = 20 V
VZ = 6.8 V
VL = 6.8 V
IL = 30 mA
440 V
20 V
6.8 V
Load
Zener regulator for Prob. 5-31.
5-32. Given: VLED = 1.5 to 2 V
ILED = 20 mA
VS = 5 V
Imax = 140 mA
Solution:
RS = [(VS – VLED(min))/ILED]
RS = [(5 V – 1.5 V)/20 mA]
RS = 175 Ω
Answer: Same as Fig. 5-20 with resistor values of 175 Ω,
which limits each branch to a maximum of 20 mA and a
total of 140 mA.
5-33. Given:
VLine = 115 V ac ± 10%
VSec = 12.6 V ac
R2 = 560 Ω ± 5%
RZ = 7 Ω
VZ = 5.1 V ± 5%
Solution: To find the maximum zener current, the maximum secondary voltage, the minimum zener voltage,
and the minimum resistance of R2 must be found. If the
line voltage varies by 10 percent, the secondary voltage
should also vary by 10 percent.
VSec(max) = VSec + VSec (10%)
VSec(max) = 12.6 V ac + 12.6 V ac (10%)
VSec(max) = 13.86 V ac
VP = 1.414 V ac
VP = 1.414 (13.86 V ac)
VP = 19.6 V
VZ = 5.1 V ± 5%
VZ = 5.1 V – [(5.1 V) (5%)]
VZ = 4.85 V
R2 = 560 Ω ± 5%
R2 = 560 Ω – [(560 Ω ) (5%)]
R2 = 532 Ω
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The circuit can be visualized as a series circuit with a
19.6 V power supply, a 532 Ω R2, a 7 Ω RZ, and a 4.85 V
zener diode.
IS = (VS – VZ)/(RS + RZ) (Eq. 5-13)
IS = (19.6 V – 4.85 V)/(532 Ω + 7 Ω )
IS = 27.37 mA
Answer: The maximum diode current is 27.37 mA.
5-34. Given:
VSec = 12.6 V ac
VD = 0.7 V
I1N5314 = 4.7 mA
ILED = 15.6 mA
IZ = 21.7 mA
C = 1000 μF ± 20%
fin = 60 Hz
Solution: The dc load current is the sum of all of the
loads.
I = I1N5314 + ILED + IZ
I = 4.7 mA + 15.6 mA + 21.7 mA
I = 42 mA
fout = 2fin (Eq. 4-7)
fout = 2(60 Hz)
fout = 120 Hz
likely cause of that is overcurrent. The only device
that could short and cause the zener to burn open is RS.
5-37. Troubles:
5. Open at A. Since all the voltages are zero, the power
must not be getting to the circuit.
6. Open RL, an open between B and C, or an open
between RL and ground. To solve this problem, the
second approximation must be used. With the load
resistor operating normally, only part of the total current flows through the zener, which causes the 0.3-V
increase from its nominal voltage. But when the load
resistor opens, all the total current flows through the
diode, causing the voltage drop across the internal
resistance to increase to 0.5 V.
7. Open at E. Since the voltages at B, C, and D are
14.2 V, which is the voltage that would be present if
the circuit were just a voltage divider with no zener
diode, suspect something in the diode circuit. Since
the diode reads OK, that only leaves an open in the
return path.
8. The zener is shorted or a short from B, C, or D to
ground. Since the voltages at B, C, and D are 0, and A
is 18 V, this could be caused by an open RS or a short
from B, C, or D to ground. Since the diode reads 0 Ω ,
it confirms that the fault is a short.
The minimum capacitance will give the maximum ripple.
C = 1000 μF ± 20%
C = 1000 μF – 1000 μF (20%)
C = 800 μF
5-38. RS is open
VR = I/(fC) (Eq. 4-10)
VR = 42 mA/(120 Hz)(800 μF)
VR = 0.438 V
5-41. RL is shorted
Answer: The maximum ripple voltage will be 0.438 V.
5-35. Given:
VS = 6 V ac
VD = 0.25 V
Solution:
VP(in) = 1.414 Vrms
VP(in) = 1.414 (6 V ac)
VP(in) = 8.48 V
VP(out) = VP(in) – 0.5 V (Eq. 4-8; the 1.4 was changed
to reflect using Schottky diodes)
VP(out) = 7.98 V
Answer: The voltage at the filter capacitor is 7.98 V.
5-36. Troubles:
1. Open RS, since there is voltage at A and no voltage at
B; also could be a short from B or C to ground.
2. Open between B and D or an open at E. Since the
voltages at B and C are 14.2 V, which is the voltage
that would be present if the circuit were just a
voltage divider with no zener diode, suspect something
in the diode circuit. Since the diode is good, it is either
an open between B and D or an open at E.
3. The zener is open. Since the voltages at B and C are
14.2 V, which is the voltage that would be present if
the circuit were just a voltage divider with no zener
diode, suspect something in the diode circuit. Since
the diode reads an open, it is bad.
4. RS shorted, which caused the zener to open. With all
the voltages at 18 V, the problem could be an open in
the return path. But the zener is open, and the most
5-39. Power supply failed (0 V)
5-40. Zener diode open
5-42. Zener diode is forward biased
5-43. Common anode
5-44. Biased off
5-45. 25 mA
5-46. 200 Ω
5-47. 470 Ω, 1/2 watt
Chapter 6 BJT Fundamentals
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
9.
b
a
c
a
b
b
b
b
d
10.
11.
12.
13.
14.
15.
16.
17.
18.
b
b
d
b
a
a
b
b
d
19.
20.
21.
22.
23.
24.
25.
26.
27.
c
a
a
a
b
d
d
c
c
28.
29.
30.
31.
32.
33.
34.
35.
c
b
d
c
b
d
c
b
JOB INTERVIEW QUESTIONS
6. A transistor or semiconductor curve tracer.
7. Since there is almost zero power dissipation at saturation and
cutoff, I would expect that the maximum power dissipation is
in the middle of the load line.
10. Common emitter.
14. An increase in temperature almost always increases the
current gain.
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PROBLEMS
6-1.
Solution:
The minimum resistance will yield the maximum
current.
RB = 470 kΩ ± 5%
RB = 470 kΩ – 470 kΩ(5%)
RB = 446.5 kΩ
Given:
IE = 10 mA
IC = 9.95 mA
Solution:
IE = IC + IB (Eq. 6-1)
IB = IE – IC
IB = 10 mA – 9.95 mA
IB = 0.05 mA
Answer: The base current is 0.05 mA.
6-2.
Given:
IC = 10 mA
IB = 0.1 mA
IB = [(VBB – VBE)/RB] (Eq. 6-6)
IB = [(10 V – 0.7 V)/446.5 kΩ]
IB = 20.83 μA
Answer: The base current is 20.83 μA.
6-8.
Solution:
βdc = IC/IB
βdc = 10 mA/0.1 mA
βdc = 100
Solution:
VCE = VCC – ICRC
VCE = 20 V – (6 mA)(1.5 kΩ)
VCE = 11 V
Answer: The current gain is 100.
6-3.
Given:
IB = 30 μA
βdc = 150
Solution:
IC = βdcIB
IC = 150(30 μA)
IC = 4.5 mA
Answer: The collector current is 4.5 mA.
6-4.
Given:
IC = 100 mA
βdc = 65
Solution:
IB = IC/βdc (Eq. 6-5)
IB = 100 mA/65
IB = 1.54 mA
IE = IB + IC
IE = 1.54 mA + 100 mA
Answer: The emitter current is 101.54 mA.
6-5.
Given:
VBB = 10 V
RB = 470 kΩ
VBE = 0.7 V
Solution:
IB = [(VBB – VBE)/RB] (Eq. 6-6)
IB = [(10 V – 0.7 V)/470 kΩ]
IB = 19.8 μA
Answer: The base current is 19.8 μA.
6-6.
6-7.
Answer: The base current is unaffected by the current
gain since 9.3 V/470 kΩ always equals 19.8 μA. The
current gain will affect the collector current in this
circuit.
Given:
VBB = 10 V
RB = 470 kΩ ± 5%
VBE = 0.7 V
Given:
IC = 6 mA
RC = 1.5 kΩ
VCC = 20 V
Answer: The collector to emitter voltage is 11 V.
6-9.
Given:
IC = 100 mA
VCE = 3.5 V
Solution:
PD = VCEIC (Eq. 6-8)
PD = (3.5 V)(100 mA)
PD = 350 mW
Answer: The power dissipation is 350 mW.
6-10. Given:
VBB = 10 V
RB = 470 kΩ
VBE = 0.7 V (second approximation)
VBE = 0 V (ideal)
RC = 820 Ω
VCC = 10 V
βdc = 200
Solution:
Ideal
IB = [(VBB – VBE)/RB] (Eq. 6-6)
IB = [(10 V – 0 V)/470 kΩ]
IB = 21.28 μA
IC = βdcIB
IC = 200(21.28 μA)
IC = 4.26 mA
VCE = VCC – ICRC
VCE = 10 V – (4.26 mA)(820 Ω)
VCE = 6.5 V
PD = VCEIC
PD = (6.5 V)(4.26 mA)
PD = 27.69 mW
2nd Approximation
IB = [(VBB – VBE)/RB] (Eq. 6-6)
IB = [(10V – 0.7 V)/470 kΩ]
IB = 19.8 μA
IC = βdcIB
IC = 200(19.8 μA)
IC = 3.96 mA
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VCE = VCC – ICRC
VCE = 10 V – (3.96 mA)(820 Ω )
VCE = 6.75 V
PD = VCEIC
PD = (6.75 V)(3.96 mA)
PD = 26.73 mW
Answer: The ideal collector-emitter voltage is 6.5 V and
power dissipation is 27.69 mW. The second approximation collector-emitter voltage is 6.75 V and the power
dissipation is 26.73 mW.
6-11. Given:
VBB = 5 V
RB = 330 kΩ
VBE = 0.7 V (second approximation)
VBE = 0 V (ideal)
RC = 1.2 kΩ
VCC = 15 V
βdc = 150
Solution:
Ideal
IB = [(VBB – VBE)/RB] (Eq. 6-6)
IB = [(5 V – 0 V)/330 kΩ]
IB = 15.15 μA
IC = βdcIB
IC = 150(15.15 μA)
IC = 2.27 mA
VCE = VCC – ICRC
VCE = 15 V – (2.27 mA)(1.2 kΩ)
VCE = 12.28 V
PD = VCEIC
PD = (12.28 V)(2.27 mA)
PD = 27.88 mW
2nd Approximation
IB = [(VBB – VBE)/RB] (Eq. 6-6)
IB = [(5 V – 0.7 V)/330 kΩ]
IB = 13.3 μA
IC = βdcIB
IC = 150(13.03 μA)
IC = 1.96 mA
VCE = VCC – ICRC
VCE = 15 V – (1.96 mA)(1.2 kΩ)
VCE = 12.65 V
PD = VCEIC
PD = (12.65 V)(1.96 mA)
PD = 24.79 mW
Answer: The ideal collector-emitter voltage is 12.28 V
and power dissipation is 27.88 mW. The second approximation collector-emitter voltage is 12.65 V, and power
dissipation is 24.79 mW.
6-12. Given:
VBB = 12 V
RB = 680 kΩ
VBE = 0.7 V (second approximation)
VBE = 0 (ideal)
RC = 1.5 kΩ
VCC = 12 V
βdc = 175
Solution:
IB = [(VBB – VBE)/RB] (Eq. 6-6)
IB = [(12 V – 0)/680 kΩ]
IB = 17.6 μA (ideal)
IB = [(12 V – 0.7 V)/680 kΩ]
IB = 16.6 μA (second approximation)
IC = βdcIB (Eq. 6-4)
IC = 175(17.6 μA)
IC = 3.08 mA (ideal)
IC = 175(16.6 μA)
IC = 2.91 mA (second approximation)
VCE = VCC – ICRC (Eq. 6-7)
VCE = 12 V – (3.08 mA)(1.5 kΩ)
VCE = 7.38 V (ideal)
VCE = 12 V – (2.91 mA)(1.5 kΩ)
VCE = 7.64 V (second approximation)
PD = VCEIC (Eq. 6-8)
PD = (7.38 V)(3.08 mA)
PD = 22.73 mW (ideal)
PD = (7.64 V)(2.91 mA)
PD = 22.23 mW (second approximation)
Answer: The ideal collector-emitter voltage is 7.38 V,
and power dissipation is 22.73 mW. The second approximation collector-emitter voltage is 7.64 V, and power
dissipation is 22.23 mW.
6-13. Answer: From the maximum ratings section, –55
to +150°C.
6-14. Answer: From the on characteristics section, 70.
6-15. Given:
PD(max) = 1 W
IC = 120 mA
VCE = 10 V
Solution:
PD = VCEIC (Eq. 6-8)
PD = (10 V)(120 mA)
PD = 1.2 W
Answer: The power dissipation has exceeded the maximum rating, and the transistor’s power rating is damaged
and possibly destroyed.
6-16. Given:
PD = 625 mW
Temperature = 65°C
Solution:
ΔT = 65°C – 25°C
ΔT = 40°C
ΔP = ΔT (derating factor)
ΔP = 40°C(2.8 mW/°C)
ΔP = 112 mW
PD(max) = 350 mW – 112 mW
PD(max) = 238 mW
Answer: The transistor is operating outside of its limits;
the power rating is affected.
6-17. Answer: β = 30
6-18. Answer: β = 85
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6-19. Given:
VCC = 20 V
VBB = 10 V
RB = 1 MΩ
RC = 3.3 kΩ
Solution:
IC(sat) = VCC/RC
(Eq. 6-11)
IC(sat) = 5 V/470 Ω
IC(sat) = 10.64 mA
VCE(cutoff) = VCC
VCE(cutoff) = 5 V
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 20 V/3.3 kΩ
IC(sat) = 6.06 mA
VCE(cutoff) = VCC (Eq. 6-12)
VCE(cutoff) = 20 V
Answer: The collector current at saturation is 6.06 mA,
and the collector-emitter voltage at cutoff is 20 V. The
load line would connect these points.
Answer: The collector current at saturation is 10.64 mA,
and the collector-emitter voltage at cutoff is 5 V. The
load line would connect these points.
6-24. Given:
VCC = 10 V
VBB = 5 V
RB = 680 kΩ
RC = 470 Ω
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 10 V/470 Ω
IC(sat) = 21.28 mA
6-20. Given:
VCC = 25 V
VBB = 10 V
RB = 1 MΩ
RC = 3.3 kΩ
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 25 V/3.3 kΩ
IC(sat) = 7.58 mA
Answer: The load line moves futher away from the origin.
6-21. Given:
VCC = 20 V
VBB = 10 V
RB = 1 MΩ
RC = 4.7 kΩ
VCE(cutoff) = VCC
VCE(cutoff) = 10 V
6-25. Given:
VCC = 5 V
VBB = 5 V
RB = 680 kΩ
RC = 1 kΩ
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 5 V/1 kΩ
IC(sat) = 5 mA
VCE(cutoff) = VCC
VCE(cutoff) = 5 V
VCE(cutoff) = VCC
VCE(cutoff) = 20 V
6-22. Given:
VCC = 20 V
VBB = 10 V
RB = 500 k
RC = 3.3 kΩ
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 20 V/3.3 kΩ
IC(sat) = 6.06 mA
VCE(cutoff) =VCC
VCE(cutoff) = 20 V
Answer: The load line does not change.
6-23. Given:
VCC = 5 V
VBB = 5 V
RB = 680 kΩ
RC = 470 Ω
(Eq. 6-12)
Answer: Load line moved farther from the origin on the
graph.
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 20 V/4.7 kΩ
IC(sat) = 4.25 mA
Answer: The left side of the load line would move down
while the right side remains at the same point.
(Eq. 6-12)
(Eq. 6-12)
Answer: The left side of the load line will decrease by
half, and the right will not move.
Ic
(b)
(a)
(c)
VCE
Load lines for (a) Prob. 6-23, (b) Prob. 6-24, and (c)
Prob. 6-25
6-26. Given:
VCC = 20 V
VBB = 10 V
RB = 1 MΩ
RC = 3.3 kΩ
β = 200
Solution:
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (10 V − 0.7 V)/1 MΩ
IB = 9.3 μA
IC = β IB
(Eq. 6-3)
IC = 200(9.3 μA)
IC = 1.86 mA
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VCE = VCC – ICRC
(Eq. 6-15)
VCE = 20 V – (1.86 mA)(3.3 kΩ)
VCE = 13.86 V
RB = 680 kΩ
RC = 470 Ω
β =150
Answer: The voltage between the collector and ground
is 13.86 V.
Solution:
IB = (VBB − VBE)/RB
(Eq. 6-13)
IB = (5 V − 0.7 V)/680 kΩ
IB = 6.32 μA
6-27. Given:
VCC = 20 V
VBB =10 V
RB =1 MΩ
RC = 3.3 kΩ
βdc(min) = 25
βdc(max) = 300
IB = 9.3 μA (from the previous problem)
Solution:
IC(min) = βdc(min)IB
(Eq. 6-3)
IC(min) = 25(9.3 μA)
IC(min) = 232.5 μA
IC(max) = βdc(max)IB
(Eq. 6-3)
IC(max) = 300(9.3 μA)
IC(max) = 2.79 mA
VCE(min) = VCC − IC(max)RC
(Eq. 6-15)
VCE(min) = 20 V − (2.79 mA)(3.3 kΩ)
VCE(min) = 10.79 V
VCE(max) = VCC − IC(min)RC
(Eq. 6-15)
VCE(max) = 20 V − (232.5 μA)(3.3 kΩ )
VCE(max) = 19.23 V
Answer: The maximum collector to ground voltage is
19.23 V, and the minimum is 10.79 V.
6-28. Given:
VCC = 20 V ± 10%
VBB = 10 V ± 10%
RB = 1 MΩ ± 5%
RC = 3.3 kΩ ± 5%
βdc(min) = 50
βdc(max) = 150
Solution:
IB(min) = (VBB(min) – VBE)/RB(max) (Eq. 6-13)
IB(min) = (9 V – 0.7 V)/1.05 MΩ
IB(min) = 7.90 μA
IB(max) = (VBB(max) – VBE)/RB(min) (Eq. 6-13)
IB(max) = (11 V – 0.7 V)/0.95 MΩ
IB(max) = 10.84 μA
IC(min) = βdc(min)IB(min) (Eq. 6-3)
IC(min) = 50(7.90 μA)
IC(min) = 395 μA
IC(max) = βdc(max)IB(max) (Eq. 6-3)
IC(max) = 150(10.84 μA)
IC(max) =1.63 mA
VCE(min) =VCC(max) – IC(max)RC(max) (Eq. 6-15)
VCE(min) =18 V – (1.63 mA)(3.47 kΩ)
VCE(min) =12.34 V
VCE(max) =VCC(max) – IC(min)RC(min) (Eq. 6-15)
VCE(max) =22 V – (395 μA)(3.14 kΩ)
VCE(max) = 20.76 V
Answer: The maximum collector to ground voltage is
20.76 V, and the minimum is 12.34 V.
6-29. Given:
VCC = 5V
VBB = 10 V
IC = βIB
(Eq. 6-3)
IC = 150(6.32 μA)
IC = 948 μA
VCE = VCC − ICRC
(Eq. 6-15)
VCE = 5 V − (948 μA)(470 Ω)
VCE = 4.55 V
Answer: The voltage between the collector and ground
is 4.55 V.
6-30. Given:
VCC = 5 V
VBB =10 V
RB = 680 Ω
RC = 470 Ω
β(min) = 100
β(max) = 300
IB = 6.32 μA (from the previous problem)
Solution:
IC(min) = β(min) IB
(Eq. 6-3)
IC(min) =100(6.32 μA)
IC(min) = 632 μA
IC(max) = β(max) IB
(Eq. 6-3)
IC (max) = 300(6.32 μA)
IC (max) = 1.90 mA
VCE(min) = VCC − IC(max) RC
(Eq. 6-15)
VCE(min) = 5 V − (1.9 mA)(470 Ω)
VCE(min) = 4.1 V
VCE (max) = VCC − IC (min) RC
(Eq. 6-15)
VCE(max) = 5 V − (6.32 μA)(470 Ω)
VCE(max) = 4.99 V
Answer: The maximum collector to ground voltage is
4.99 V, and the minimum is 4.1 V.
6-31. Given:
VCC = 5 V ± 10%
VBB = 5 V ±10%
RB = 680 kΩ ± 5%
RC = 470 Ω ± 5%
βdc(min) = 50
βdc(max) = 150
Solution:
IB(min) = (VBB(min) − VBE)/RB(max) (Eq. 6-13)
IB(min) = (4.5 V − 0.7 V)/714 kΩ
IB(min) = 5.32 μA
IB(max) = (VBB(max) −VBE)/RB(min) (Eq. 6-13)
IB(max) = (5.5 V − 0.7 V)/646 kΩ
IB(max) = 7.43 μA
IC(min) = βdc(min)IB(min) (Eq. 6-3)
IC(min) = 50(5.32 μA)
IC(min) = 266 μA
IC(max) = βdc(max)IB(max) (Eq. 6-3)
IC(max) = 150(7.43 μA)
IC(max) = 1.11 mA
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VCE(min) = VCC (max) − IC(max)RC(max) (Eq. 6-15)
VCE(min) = 4.5 V − (1.11 mA)(493.5 Ω)
VCE(min) = 3.95 V
IC = βdcIB (Eq. 6-3)
IC = 50(9.3 μA)
IC = 465 μA
VCE(max) = VCC(max) − IC(min)RC(min) (Eq. 6-15)
VCE(max) = 5.5 V − (266 μA)(446.5 Ω)
VCE(max) = 5.38 V
Answer: The transistor is not in saturation because the
calculated collector current is less than the saturation
current.
Answer: The maximum collector to ground voltage is
5.38 V, and the minimum is 3.95 V.
6-32a. Given:
VCC = 20 V
VBB = 10 V
RB = 33 kΩ
RC = 3.3 kΩ
hFE = βdc = 100
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 20 V/3.3 kΩ
IC(sat) = 6.06 mA
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (10 − 0.7)/33 kΩ
IB = 281.8 μA
IC = βdcIB (Eq. 6-3)
IC = 100(281.8 μA)
IC = 28.18 mA
Answer: The transistor is in saturation because the calculated collector current is greater than the saturation
current.
6-32b. Given:
VCC = 20 V
VBB = 5 V
RB = 1 MΩ
RC = 3.3 kΩ
hFE = βdc = 100
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 20 V/3.3 kΩ
IC(sat) = 6.06 mA
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (5 − 0.7)/1 MΩ
IB = 4.3 μA
IC = βdcIB (Eq. 6-3)
IC = 200(4.3 μA)
IC = 860 μA
Answer: The transistor is not in saturation because the
calculated collector current is less than the saturation
current.
6-32c. Given:
VCC = 20 V
VBB = 10 V
RB = 1 MΩ
RC = 10 kΩ
hFE = βdc = 50
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 20 V/10 kΩ
IC(sat) = 2 mA
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (10 − 0.7)/1 MΩ
IB = 9.3 μA
6-32d. Given:
VCC = 10 V
VBB = 10 V
RB = 1 MΩ
RC = 3.3 kΩ
hFE = βdc = 100
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 10 V/3.3 kΩ
IC(sat) = 3.03 mA
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (10 − 0.7)/1 MΩ
IB = 9.3 μA
IC = βdcIB (Eq. 6-3)
IC = 100(9.3 μA)
IC = 930 μA
Answer: The transistor is not in saturation because the calculated collector current is less than the saturation current.
6-33a. Given:
VCC = 5
VBB = 5 V
RB = 51 kΩ
RC = 470 Ω
hFE = βdc = 100
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 5 V/470 Ω
IC(sat) = 10.64 mA
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (5 − 0.7)/51 kΩ
IB = 84.3 μA
IC = βdcIB (Eq. 6-3)
IC = 100(84.3 μA)
IC = 8.43 mA
Answer: The transistor is not in saturation because the
calculated current is less than the saturation current.
6-33b. Given:
VCC = 5 V
VBB = 10 V
RB = 680 kΩ
RC = 470 Ω
hFE = βdc = 500
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 5 V/470 Ω
IC(sat) = 10.64 mA
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (10 − 0.7)/680 kΩ
IB = 13.68 μA
IC = βdcIB (Eq. 6-3)
IC = 500(13.68 μA)
IC = 6.84 mA
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Answer: The transistor is not in saturation because the
calculated collector current is less than the saturation
current.
6-33c. Given:
VCC = 5 V
VBB = 5 V
RB = 680 kΩ
RC = 10 kΩ
hFE = βdc = 100
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 5 V/10 kΩ
IC(sat) = 0.5 mA
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (5 − 0.7)/680 kΩ
IB = 6.32 μA
IC = βdcIB (Eq. 6-3)
IC = 100(6.32 μA)
IC = 632 μA
Answer: The transistor is in saturation because the calculated collector is greater than the saturation current.
6-33d. Given:
VCC = 10 V
VBB = 5 V
RB = 680 kΩ
RC = 470Ω
hFE = βdc = 100
Solution:
IC(sat) = VCC/RC (Eq. 6-11)
IC(sat) = 10 V/470 Ω
IC(sat) = 21.28 mA
IB = (VBB − VBE)/RB (Eq. 6-13)
IB = (5 − 0.7)/680 kΩ
IB = 6.32 μA
IC = βdcIB (Eq. 6-3)
IC = 100(6.32 μA)
IC = 632 μA
Answer: The transistor is not in saturation because the
calculated current is less than the saturation current.
6-34. Answer: With the switch open, the collector voltage is
5 V, and with the switch closed, the collector voltage
is 0 V.
6-35. a. Increase: With the base resistor shorted, the baseemitter junction will have excessive current and will
open, stopping all conduction. Thus source voltage is
read from collector to emitter.
b. Increase: With the base resistor open, the transistor
goes into cutoff and source voltage is read from collector to emitter.
c. Increase: With the collector resistor shorted, it is the
only thing in the circuit with the source and will read
source voltage at all times.
d. Decrease: There will be no voltage present at the collector or emitter.
e. Increase: With the base supply gone, the transistor
goes into cutoff and source voltage is read from collector to emitter.
f. Decrease: There will be no voltage present at the collector or emitter.
CRITICAL THINKING
6-36. Given: βdc = 200
Solution:
αdc = IC /IE (Eq. 6-2)
IC = αdcIE
βdc = IC /IB (Eq. 6-3)
IE = IC + IB (Eq. 6-1)
Substitute Eq. (6-2) for IC: βdc = αdcIE /IB
Substitute Eq. (6-1) for IE: βdc = αdc(IC + IB)/IB
Distribute αdc:
βdc = (αdcIC + αdcIB)/IB
βdc = αdcIC /IB + αdcIB /IB
Simplify: βdc = αdcIC /IB + αdc
Substitute Eq. (6-3) for IC /IB: βdc = αdcβdc + αdc
Factor out the αdc: βdc = αdc( βdc + 1)
Solve for αdc:
αdc = βdc/(βdc + 1)
αdc = 200/(200 + 1)
αdc = 0.995
Answer: The αdc is 0.995.
6-37. Given: αdc = 0.994
Solution: From the previous solution:
βdc = αdcβdc + αdc
βdc – αdcβdc = αdc
Factor out the βdc: βdc(1 – αdc) = αdc
Solve for βdc:
βdc = αdc/(1 – αdc)
βdc = 0.994/(1 – 0.994)
βdc = 165.67
Answer: The βdc is 165.67.
6-38. Given:
VBB = 5 V
VCC = 15 V
hFE = βdc = 120
IC = 10 mA
VCE = 7.5 V
VBE = 0.7 V
Solution:
βdc = IC /IB (Eq. 6-3)
IB = IC/βdc
IB = 10 mA/120
IB = 83.33 μA
IB = (VBB – VBE)RB (Eq. 6-6)
RB = (VBB – VBE)/IB
RB = (5 V – 0.7 V)/83.33 μA
RB = 51.6 kΩ
VCE = VCC – ICRC (Eq. 6-7)
RC = (VCC – VCE)/IC
RC = (15 V – 7.5 V)/10 mA
RC = 750 Ω
Answer: The base resistor needs to be 51.6 Ω, and the
collector resistor needs to be 750 Ω. Note: These may not
be standard values, so it may take more than one resistor,
or a potentiometer may be used.
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VBE = 0.7 V (second approximation)
RC = 820 Ω
VCC = 10 V
βdc = 200
6-39. Given:
VBB = 10 V
VCE = 6.7 V
VBE = 0.7 V (second approximation)
VBE = 0 V (ideal)
RC = 820 Ω
VCC = 10 V
βdc = 200
Solution:
Ideal
VCC = VRC + VCE
VRC = VCC – VCE
VRC = 10 V – 6.7 V
VRC = 3.3 V
IC = VRC/RC
IC = 3.3 V/820 Ω
IC = 4 mA
IB = IC/βdc
IB = 4 mA/200
IB = 20.1 μA
IB = [(VBB – VBE)/RB]
RB = [(VBB – VBE)/IB]
RB = [(10 V – 0 V)/20.1 μA]
RB = 497.5 kΩ
2nd Approximation
VCC = VRC + VCE
VRC = VCC – VCE
VRC = 10 V – 6.7 V
VRC = 3.3 V
IC = VRC/RC
IC = 3.3 V/820 Ω
IC = 4 mA
IB = IC/βdc
IB = 4 mA/200
IB = 20.1 μA
IB = [(VBB – VBE)/RB]
RB = [(VBB – VBE)/IB]
RB = [(10 V – 0.7 V)/20.1 μA]
RB = 462.69 kΩ
Answer: (Ideal) RB = 497.5 kΩ, (2nd Approximation)
RB = 462.69 kΩ.
6-40. Given:
PD = 350 mW @25°C
T = 50°C
VCE = 10 V
Solution:
ΔT = 50°C – 25°C
ΔT = 25°C
ΔP = ΔT (derating factor)
ΔP = 25°C (2.8 mW/°C)
ΔP = 70 mW
PD(max) = 350 mW – 70 mW
PD(max) = 280 mW
PD = VCEIC
IC = PD/VCE
IC = 280 mW/10 V
IC = 28 mA
Answer: The maximum collector current is 28 mA.
6-41. Given:
VBB = 10 V
RB = 470 kΩ
Solution:
IB = [(VBB – VBE)/RB] (Eq. 6-6)
IB = [(10 V – 0.7 V)/470 kΩ]
IB = 19.8 μA
IC = βdcIB
IC = 200(19.8 μA)
IC = 3.96 mA
Answer: The LED current is 3.96 mA.
6-42. Answer: VCE(sat) = 0.3 V
6-43. No collector supply
6-44. RC is shorted
6-45. RB is shorted
6-46. No base supply
6-47. RB is open
6-48. Base bias; common emitter
6-49. Synch input; clock output
6-50. 5 mA
6-51. Saturated
6-52. +5 V
Chapter 7 BJT Biasing
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
b
b
d
b
a
a
c
a
c
d
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
a
a
d
b
b
b
a
c
a
b
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
a
c
c
c
a
b
a
d
a
c
31.
32.
33.
34.
35.
36.
37.
38.
39.
a
d
b
b
c
b
c
a
d
JOB INTERVIEW QUESTIONS
2. The collector current changes only slightly, if at all.
4. Emitter-feedback bias and collector-feedback bias. They
were developed in an attempt to stabilize the Q point against
transistor replacement and temperature changes.
6. No. Saturation and cutoff.
7. Changes in current gain will change the collector current.
The base resistors should be made smaller to satisfy the condition described in the text.
10. The circuit will be highly sensitive to changes in current gain.
PROBLEMS
7-1.
Given:
VBB = 2.5 V
VCC = 20 V
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RC = 10 kΩ
RE = 1.8 kΩ
VBE = 0.7 V
RC = 910 Ω
RE = 180
VBE = 0.7 V
Solution:
VE = VBB − VBE (Eq. 7-1)
VE = 2.5 V − 0.7 V
VE = 1.8 V
Solution:
VE = VBB − VBE (Eq. 7-1)
VE = 2 V − 0.7 V
VE = 1.3 V
IE = VE/RE (Ohm’s law)
IE = 1.8 V/1.8 kΩ
IE = 1 mA
IE ≈ IC
IE = VE/RE (Ohm’s law)
IE = 1.3 V/180 Ω
IE = 7.22 mA
IE ≈ IC
VC = VCC − ICRC (Kirchhoff’s law)
VC = 20 V − (1 mA)(10 kΩ)
VC = 10 V
VC = VCC − ICRC (Kirchhoff’s law)
VC = 10 V − (7.22 mA)(910 Ω)
VC = 3.43 V
Answer: The collector voltage is 10 V, and the emitter
voltage is 1.8 V.
7-2.
Answer: The collector voltage is 3.43 V.
7-5.
Given:
VBB = 2.5 V
VCC = 20 V
RC = 10 kΩ
RE = 3.6 kΩ
VBE = 0.7 V
Solution:
VE = VBB − VBE (Eq. 7-1)
VE = 2.3 V − 0.7 V
VE = 1.6 V
Solution:
VE = VBB − VBE (Eq. 7-1)
VE = 2.5 V − 0.7 V
VE = 1.8 V
IE = VE /RE (Ohm’s law)
IE = 1.6 V/360 Ω
IE = 4.44 mA
IE ≈ IC
IE = VE /RE (Ohm’s law)
IE = 1.8 V/3.6 kΩ
IE = 0.5 mA
IE ≈ IC
VC = VCC − ICRC (Kirchhoff’s law)
VC = 10 V − (4.44 mA)(910 Ω)
VC = 5.96 V
VC = VCC − ICRC (Kirchhoff’s law)
VC = 20 V − (0.5 mA)(10 kΩ)
VC = 15 V
VCE = VC −VE (Eq. 7-2)
VCE = 5.96 V − 1.6 V
VCE = 4.36 V
VCE = VC − VE (Eq. 7-2)
VCE = 15 V − 1.8 V
VCE = 13.2 V
Answer: The collector-emitter voltage is 13.2 V.
7-3.
Given:
VBB = 2.5 V
VCC = 15 V
RC = 10 kΩ
RE = 1.8 kΩ
VBE = 0.7 V
Solution:
VE = VBB − VBE (Eq. 7-1)
VE = 2.5 V −0.7 V
VE = 1.8 V
IE = VE/RE (Ohm’s law)
IE = 1.8 V/1.8 kΩ
IE = 1 mA
IE ≈ IC
VC = VCC − ICRC (Kirchhoff’s law)
VC = 15 V − (1 mA)(10 kΩ)
VC = 5 V
Answer: The collector voltage is 5 V.
7-4.
Given:
VBB = 2 V
VCC = 10 V
Given:
VBB = 2.3 V
VCC = 10 V
RC = 910 Ω
RE = 360 Ω
VBE = 0.7 V
Answer: The collector-emitter voltage is 4.36 V.
7-6.
Given:
VBB = 1.8 V
VCC = 15 V
RC = 910 Ω
RE = 180 Ω
VBE = 0.7 V
Solution:
VE = VBB −VBE (Eq. 7-1)
VE = 1.8 V −0.7 V
VE = 1.1 V
IE = VE/RE (Ohm’s law)
IE = 1.1 V/180 Ω
IE = 6.11 mA
IE ≈ IC
VC = VCC − ICRC (Kirchhoff’s law)
VC = 15 V − (6.11 mA)(910 Ω)
VC = 9.44 V
VCE = VC − VE
(Eq. 7-2)
VCE = 9.44 V − 1.1 V
VCE = 8.34 V
Answer: The collector-emitter voltage is 8.34 V.
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7-7.
Given:
VCC = 5 V
VBB = 2 V
RE = 100 Ω
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]25 V
VBB = 4.51 V
Solution:
VE = VBB −VBE
(Eq. 7-1)
VE = 2 V − 0.7 V
VE = 1.3 V
VE = VBB – VBE (Eq. 7-5)
VE = 4.51 V – 0.7 V
VE = 3.81 V
IE = VE /RE (Ohm’s law)
IE = 1.3 V/100 Ω
IE = 13 mA
IE ≈ IC = ID
Answer: The diode current is 13 mA.
7-8.
Given:
VCC = 5 V
VBB = 1.8 V
RE = 100 Ω
VD ≈ 2 V
Solution:
VE = VBB − VBE
(Eq. 7-1)
VE = 1.8 V− 0.7 V
VE = 1.1 V
IE = VE /RE (Ohm’s law)
IE = 1.1 V/100 Ω
IE = 11 mA
IE ≈ IC = ID
VC = VCC −VD
VC = 5 V −2 V
VC = 3 V
Answer: The diode current is 11 mA, and the collector
voltage is 3 V.
7-9.
Answer: RC could be shorted; the transistor could be open
collector-emitter; RB could be open, keeping the transistor in cutoff; open in the base circuit; open in the emitter
circuit.
7-10. Answer: With the ground open, the base would read VBB
and the collector would read VCC because source voltage
is read above an open.
7-11. Answer: Shorted transistor; RB value very low; VBB too
high.
7-12. Answer: RC could be shorted; the transistor could be open
collector-emitter; RB could be open, keeping the transistor in cutoff; RE could be open; open in the base circuit;
open in the emitter circuit.
7-13. Answer: With the emitter resistor open, the base would
read VBB and the collector would read VCC because source
voltage is read above an open.
7-14. Answer: Shorted transistor collector-emitter because the
emitter voltage should be 1.1 V; open collector resistor;
loss of VCC.
7-15. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
VCC = 25 V
VBE = 0.7 V
IE = VE/RE (Eq. 7-6)
IE = 3.81 V/1 kΩ
IE = 3.81 mA
IC ≈ IE (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-8)
VC = 25 V – (3.81 mA)(3.6 kΩ)
VC = 11.28 V
Answer: The emitter voltage is 3.81 V, and the collector
voltage is 11.28V.
7-16. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 2.7 kΩ
RE = 1 kΩ
VCC = 15 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V
VBB = 2.7 V
VE = VBB – VBE (Eq. 7-5)
VE = 2.7 V – 0.7 V
VE = 2.0 V
IE = VE/R3 (Eq. 7-6)
IE = 2.0 V/1 kΩ
IE = 2 mA
IC ≈ IE (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-8)
VC = 15 V – (2 mA)(2.7 kΩ)
VC = 9.59 V
Answer: The emitter voltage is 2.0 V, and the collector
voltage is 9.59 V.
7-17. Given:
R1 = 330 kΩ
R2 = 100 kΩ
RC = 150 kΩ
RE = 51 kΩ
VCC = 10 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [100 kΩ/(330 kΩ + 100 kΩ)]10 V
VBB = 2.33 V
VE = VBB – VBE (Eq. 7-5)
VE = 2.33 V – 0.7 V
VE = 1.63 V
IE = VE/RE (Eq. 7-6)
IE = 1.63 V/51 kΩ
IE = 31.96 μA
IC ≈ IE (Eq. 7-7)
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VC = VCC – ICRC (Eq. 7-8)
VC = 10 V – (31.96 μA)(150 kΩ)
VC = 5.21 V
VC(min) = VCC – IC(max)RC(max) (Eq. 7-8)
VC(min) = 10 V – (37.36 μA)(157.5 kΩ)
VC(min) = 4.12 V
Answer: The emitter voltage is 1.63 V, and the collector
voltage is 5.21 V.
Answer: The lowest collector voltage is 4.12 V, and the
highest collector voltage is 6.14 V.
7-18. Given:
R1 = 150 Ω
R2 = 33 Ω
RC = 39 Ω
RE = 10 Ω
VCC = 12 V
VBE = 0.7 V
7-20. Given:
R1 = 150 Ω
R2 = 33 Ω
RC = 39 Ω
RE = 10 Ω
VCC = 12 V ± 10%
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [33 Ω/(150 Ω + 33 Ω)]12 V
VBB = 2.16 V
Solution:
VBB(max) = [R2/(R1 + R2)]VCC(max) (Eq. 7-4)
VBB(max) = [33 Ω/(150 Ω + 33 Ω)]13.2 V
VBB(max) = 2.38 V
VE = VBB – VBE (Eq. 7-5)
VE = 2.16 V – 0.7 V
VE = 1.46 V
VE(max) = VBB(max) – VBE (Eq. 7-5)
VE(max) = 2.38 V – 0.7 V
VE(max) = 1.68 V
IE = VE/RE (Eq. 7-6)
IE = 1.46 V/10 Ω = 146 mA
IE(max) = VE(max)/RE (Eq. 7-6)
IE(max) = 1.68 V/10 Ω
IE(max) = 168 mA
IC ≈ IE (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-8)
VC = 12 V – (146 mA)(39 Ω)
VC = 6.3 V
Answer: The emitter voltage is 1.46 V. The collector
voltage is 6.3 V.
7-19. Given:
R1 = 330 kΩ ± 5%
R2 = 100 kΩ ± 5%
RC = 150 kΩ ± 5%
RE = 51 kΩ ± 5%
VCC = 10 V
VBE = 0.7 V
Solution:
VBB(max) = [R2(max)/(R1(min) + R2(max))]VCC (Eq. 7-4)
VBB(max) = [105 kΩ/(313.5 kΩ + 105 kΩ)]10 V
VBB(max) = 2.51 V
VBB(min) = [R2(min)/(R1(max) + R2(min))]VCC (Eq. 7-4)
VBB(min) = [95 kΩ/(346.5 kΩ + 95 kΩ)]10 V
VBB(min) = 2.15 V
VE(max) = VBB(max) – VBB (Eq. 7-5)
VE(max) = 2.51 V – 0.7 V
VE(max) = 1.81 V
VE(min) = VBB(min) – VBE (Eq. 7-5)
VE(min) = 2.15 V – 0.7 V
VE(min) = 1.45 V
IE(max) = VE(max)/RE(min) (Eq. 7-6)
IE(max) = 1.81 V/48.45 kΩ
IE(max) = 37.36 μA
IE(min) = VE(min)/RE(max) (Eq. 7-6)
IE(min) = 1.45 V/53.55 kΩ
IE(min) = 27.08 μA
IC ≈ IE (Eq. 7-7)
VC(max) = VCC – IC(min)RC(min) (Eq. 7-8)
VC(max) = 10 V – (27.08 μA)(142.5 kΩ)
VC(max) = 6.14 V
VBB(min) = [R2/(R1 + R2)]VCC(min) (Eq. 7-4)
VBB(min) = [33 Ω/(150 Ω + 33 Ω)]10.8 V
VBB(min) = 1.95 V
VE(min) = VBB(min) – VBE (Eq. 7-5)
VE(min) = 1.95 V – 0.7 V
VE(min) = 1.25 V
IE(min) = VE(min)/RE (Eq. 7-6)
IE(min) = 1.25 V/10 Ω
IE(min) = 125 mA
IC ≈ IE (Eq. 7-7)
VC(max) = VCC(max) – IC(min)RC (Eq. 7-8)
VC(max) = 13.2 V – (125 mA)(39 Ω)
VC(max) = 8.33 V
VC(min) = VCC(min) – IC(max)RC (Eq. 7-8)
VC(min) = 10.8 V – (168 mA)(39 Ω)
VC(min) = 4.25 V
Answer: The lowest collector voltage is 4.25 V and the
highest collector voltage is 8.33 V.
7-21. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
VCC = 25 V
VBE = 0.7 V
VBB = 4.51 V (from Prob. 7-15)
VE = 3.81 V (from Prob. 7-15)
IE = IC = 3.81 mA (from Prob. 7-15)
VC = 11.28 V (from Prob. 7-15)
Solution:
VCE = VC – VE (Eq. 7-9)
VCE = 11.28 V – 3.81 V
VCE = 7.47 V
Answer: The Q point is IC = 3.81 mA, and VCE = 7.47 V.
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IC ≈ IE (Eq. 7-7)
7-22. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 2.7 kΩ
RE = 1 kΩ
VCC = 15 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V
VBB = 2.7 V
VE = VBB – VBE (Eq. 7-5)
VE = 2.7 V – 0.7 V
VE = 2.0 V
IE = VE/RE (Eq. 7-6)
IE = 2.0 V/1 kΩ
IE = 2 mA
VC = VCC – ICRC (Eq. 7-8)
VC = 12 V – (146 mA)(39 Ω)
VC = 6.3 V
VCE = VC – VE (Eq. 7-9)
VCE = 6.3 V – 1.46 V
VCE = 4.85 V
Answer: The Q point is IC = 146 mA, and VCE = 4.85 V.
7-25. Given:
R1 = 330 kΩ ± 5%
R2 = 100 kΩ ± 5%
RC = 150 kΩ ± 5%
RE = 51 kΩ ± 5%
VCC = 10 V
VBE = 0.7 V
Solution:
VBB(max) = [R2(max)/(R1(min) + R2(max))]VCC (Eq. 7-4)
VBB(max) = [105 kΩ/(313.5 kΩ + 105 kΩ)]10 V
VBB(max) = 2.51 V
IC ≈ IE (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-8)
VC = 15 V – (2 mA)(2.7 kΩ)
VC = 9.59 V
VBB(min) = [R2(min)/(R1(max) + R2(min))]VCC (Eq. 7-4)
VBB(min) = [95 kΩ/(346.5 kΩ + 95 kΩ)]10 V
VBB(min) = 2.15 V
VCE = VC – VE (Eq. 7-9)
VCE = 9.59 V – 2.0 V
VCE = 7.59 V
VE(max) = VBB(max) – VBE (Eq. 7-5)
VE(max) = 2.51 V – 0.7 V
VE(max) = 1.81 V
Answer: The Q point is IC = 2 mA, and VCE = 7.59 V.
VE(min) = VBB(min) – VBE (Eq. 7-5)
VE(min) = 2.15 V – 0.7 V
VE(min) = 1.45 V
7-23. Given:
R1 = 330 kΩ
R2 = 100 kΩ
RC = 150 kΩ
RE = 51 kΩ
VCC = 10 V
VBE = 0.7 V
VBB = 2.33 V (from Prob. 7-17)
VE = 1.63 V (from Prob. 7-17)
IE = IC = 31.96 μA (from Prob. 7-17)
VC = 5.21 V (from Prob. 7-17)
Solution:
VCE = VC – VE (Eq. 7-9)
VCE = 5.21 V – 1.63 V
VCE = 3.58 V
Answer: The Q point is IC = 31.96 μA, and VCE = 3.58 V.
7-24. Given:
R1 = 150 Ω
R2 = 33 Ω
RC = 39 Ω
RE = 10 Ω
VCC = 12 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [33 Ω/(150 Ω + 33 Ω)]12 V
VBB = 2.16 V
IE(max) = VE(max)/RE(min) (Eq. 7-6)
IE(max) = 1.81 V/48.45 kΩ
IE(max) = 37.36 μA
IE(min) = VE(min)/RE(max) (Eq. 7-6)
IE(min) = 1.45 V/53.55 kΩ
IE(min) = 27.08 μA
IC ≈ IE (Eq. 7-7)
Answer: The lowest collector current is 27.08 μA, and the
highest collector current is 37.36 μA.
7-26.
Given:
R1 = 150 Ω
R2 = 33 Ω
RC = 39 Ω
RE = 10 Ω
VCC = 12 V ± 10%
VBE = 0.7 V
Solution:
VBB(max) = [R2/(R1 + R2)]VCC(max) (Eq. 7-4)
VBB(max) = [33 Ω/(150 Ω + 33 Ω)]13.2 V
VBB(max) = 2.38 V
VE(max) = VBB(max) – VBE (Eq. 7-5)
VE(max) = 2.38 V – 0.7 V
VE(max) = 1.68 V
VE = VBB – VBE (Eq. 7-5)
VE = 2.16 V – 0.7 V
VE = 1.46 V
IE(max) = VE(max)/RE (Eq. 7-6)
IE(max) = 1.68 V/10 Ω
IE(max) = 168 mA
IE = VE/RE (Eq. 7-6)
IE = 1.46 V/10 Ω
IE = 146 mA
VBB(min) = [R2/(R1 + R2)]VCC(min) (Eq. 7-4)
VBB(min) = [33 Ω/(150 Ω + 33 Ω)]10.8 V
VBB(min) = 1.95 V
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VE(min) = VBB(min) – VBE VE(min) = 1.95 V – 0.7 V
VE(min) = 1.25 V
(Eq. 7-5)
IE(min) = VE(min)/RE (Eq. 7-6)
IE(min) = 1.25 V/10 Ω
IE(min) = 125 mA
Answer: The lowest collector current 125 mA, and the
highest collector current is 168 mA.
7-27. Given:
RB = 10 kΩ
RC = 4.7 kΩ
RE = 10 kΩ
VCC = 12 V
VEE = –12 V
Solution:
IE = (–0.7 V – VEE)/RE
IE = [–0.7 V – (–12 V)]/10 kΩ
IE = 1.13 mA
VC = VCC – ICRC
VC = 12 V – (1.13 mA)(4.7 kΩ)
VC = 6.69 V
Answer: The emitter current is 1.13 mA, and the collector
voltage is 6.69 V.
7-28. Given:
RB = 20 kΩ
RC = 9.4 kΩ
RE = 20 kΩ
VCC = 12 V
VEE = –12 V
Solution:
IE = (–0.7 V – VEE)RE
IE = [–0.7 V –(–12 V)]/20 kΩ
IE = 565 μA
VC = VCC – ICRC
VC = 12 V – (565 μA)(9.4 kΩ)
VC = 6.69 V
Answer: The emitter current is 565 μA, and the collector
voltage is 6.69 V.
7-29. Given:
RB = 10 kΩ ± 5%
RC = 4.7 kΩ ± 5%
RE = 10 kΩ ± 5%
VCC = 12 V
VEE = –12 V
Solution:
IE(max) = (–0.7 V – VEE)/RE(min) (Eq. 7-11)
IE(max) = [–0.7 V – (–12 V)]/9.5 kΩ
IE(max) = 1.19 mA
VC(max) = VCC – IC(min)RC(min) (Eq. 7-12)
VC(max) = 12 V – (1.08 mA)(4465 Ω)
VC(max) = 7.18 V
IE(min) = (–0.7 V – VEE)/RE(max) (Eq. 7-11)
IE(min) = [–0.7 V – (–12 V)]/10.5 kΩ
IE(min) = 1.08 mA
VC(min) = VCC – IC(max)RC(max) (Eq. 7-12)
VC(min) = 12 V – (1.19 mA)(4935 Ω)
VC(min) = 6.13 V
Answer: The maximum collector voltage is 7.18 V. The
minimum collector voltage is 6.13 V.
7-30. a. I ncrease: If R1 increases, VB decreases, VE decreases,
IE decreases, IC decreases, the voltage drop across RC
decreases, and VC increases.
b. Increase: If R2 decreases, VB decreases, VE decreases,
IE decreases, IC decreases, the voltage drop across RC
decreases, and VC increases.
c. Increase: RE increases, IE decreases, IC decreases, the
voltage drop across RC decreases, and VC increases.
d. Increases: RC decreases, the voltage drop across RC
decreases, and VC increases.
e. Increases: If VCC increases and the voltage drop across
RC does not change, VC increases.
f. Remain the same: βdc does not affect IC. Therefore the
voltage drop across RC does not change, nor does VC.
7-31. a. D
ecreases: If R1 increases, VB increases, VE increases,
IE decreases, IC decreases, the voltage drop across the
collector resistor decreases, and VC decreases.
b. Increases: If R2 increases, VB decreases, VE decreases,
IE increases, IC increases, the voltage drop across the
collector resistor increases, and VC increases.
c. Decreases: RE increases, IE decreases, IC decreases, the
voltage drop across the collector resistor decreases,
and VC decreases.
d. Increase: IC remains the same, RC increases, the voltage drop across the collector resistor increases, and VC
increases.
e. Increase: Since VBE does not increase in proportion
to the increase in voltage supply, as do VB and VEE,
the voltage drop across the emitter resistor increases,
causing IE to increase. This causes the voltage drop
across the collector resistor to increase and VC to
increase.
f. Remain the same: βdc does not affect IC. Therefore the
voltage drop across RC does not change, nor does VC.
7-32. a. T
he approximate collector voltage is 12 V when R1 is
open due to no collector current.
b. The approximate collector voltage is 2.93 V when
R2 is open, the transistor is in saturation. CEB can be
approximated as a short.
c. The approximate collector voltage is 12 V when RE is
open due to no collector current.
d. The approximate collector voltage is 0.39 V when RC
is open. The collector current is zero, therefore the
base current is equal to the emitter current. The circuit
becomes a voltage divider of 150 Ω and 33 Ω driving
10 Ω through the base-emitter diode. Thevenize the
base voltage divider to get a VTH = 2.16 V and a RTH =
27 V Ω. This Thevenin circuit has a load of 10 Ω and
a diode. Now solve for a current of 39.57 mA, which
leads to an emitter voltage of 395 mV.
e. The approximate collector voltage is 12 V when the
collector-emitter is open due to no collector current.
7-33. a. I f R1 is open, the base voltage increases to 10 V and
the transistor cuts off. Therefore, the collector voltage
is zero.
b. If R2 is open, the transistor goes into saturation, similar to the preceding problem. Again, you can approximate the saturated transistor as a CEB short; that is,
all three terminals shorted. Then, 10 kΩ is in parallel with 3.6 kΩ, which is 2.65 kΩ. This is in series
with 1 kΩ and 10 V. The series current is 10 V divided
by 3.65 kΩ, or 2.74 mA. Multiply by 2.65 kΩ to get
7.26 V, the approximate value of collector voltage.
c. With RE open, there is no collector current and the collector voltage is zero.
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d. With RC open, the transistor has no collector current.
Similar to the preceding problem, the circuit becomes
a voltage divider driving the emitter resistor through
the base-emitter diode. The Thevenin voltage and
resistance facing the base-emitter diode are 1.8 V and
1.8 kΩ. The current through the emitter resistor is
(1.8 V – 0.7 V) divided by (1.8 kΩ + 1 kΩ), or
0.393 mA. Multiply by 1 kΩ to get 0.393 V for the
voltage across the emitter resistor. Subtract this from
10 V to gel 9.6 V at the emitter node. Subtract 0.7 V to
get 8.9 V at the base node. Add 0.7 V to get the voltage at the collector node. The final answer is therefore
9.4 V at the collector when RC is open. If you don’t
believe it, build the circuit and measure the collector
voltage with the collector resistor open.
e. When the collector-emitter terminals are open, there
is no collector current and the collector voltage is
zero.
7-34. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RC = 3.6 kΩ
VEE = 10 V
VBE = 0.7 V
Solution:
V2 = [R2/(R1 + R2)]VEE
V2 = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V
V2 = 1.8 V
VRE = V2 – 0.7 V
VRE = 1.8 V – 0.7 V
VRE = 1.1 V
IE = VRE/RE
IE = 1.1 V/1 kΩ
IE = 1.1 mA
IC ≈ IE (Eq. 7-7)
VC = ICRC
VC = (1.1 mA)(3.6 kΩ)
VC = 3.96 V
Answer: The collector voltage is 3.96 V.
7-35. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RC = 3.6 kΩ
VEE = 10 V
VBE = 0.7 V
V2 = 1.8 V (from Prob. 7-34)
VRE = 1.1 V (from Prob. 7-34)
IE = 1.1 mA (from Prob. 7-34)
VC = 3.96 V (from Prob. 7-34)
Solution:
VCE = VCC – VC – VRE
VCE = 10 V – 3.96 V – 1.1 V
VCE = 4.94 V
Answer: The collector-emitter voltage is –4.94 V since
the collector is less positive than the emitter.
7-36. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RC = 3.6 kΩ
VEE = 10 V
VBE = 0.7 V
V2 = 1.8 V (from Prob. 7-34)
VRE = 1.1 V (from Prob. 7-34)
IE = 1.1 mA (from Prob. 7-34)
VC = 3.96 V (from Prob. 7-34)
Solution: Because of the voltage divider, there will always
be a 1.1 V drop across RE, and at saturation VCE = 0 V.
This leaves 8.9 V across RC at saturation.
IC = 8.9 V/RC
IC = 8.9 V/3.6 kΩ
IC = 2.47 mA
At cutoff, the maximum possible voltage across VCE is
8.9 V.
Answer: The saturation current is 2.47 mA, and the
collector-emitter cutoff voltage is 8.9 V.
7-37. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RC = 3.6 kΩ
VCC = –10 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)] 10 V
VBB = –1.8 V
VE = V2 + 0.7 V
VE = –1.8 V + 0.7 V
VE = –1.1 V
IE = VE/RE
IE = 1.1 V/1 kΩ
IE = 1.1 mA
IC ≈ IE (Eq. 7-7)
VC = VCC + ICRC
VC = –10 V + (1.1 mA)(3.6 kΩ)
VC = –6.04 V
Answer: The collector voltage is –6.04 V, and the emitter
voltage is –1.1 V.
CRITICAL THINKING
7-38. The circuit is no longer considered stiff or independent
of Beta. The base current is not small as compared to the
voltage divider current.
7-39. The maximum power dissipation of the 2N3904 is
625 mW. The transistor is dissipating 705 mW. The transistor will probably overheat and fail.
7-40. As long as the voltmeter has a high enough input resistance, it should read approximately 4.83 V.
7-41. Increase the power supply value, short R1.
7-42. Connect an ammeter between the power supply and the
circuit. Measure VR1 and VC, then calculate and add their
respective currents.
7-43. Given: (for Q1):
R1 = 1.8 kΩ
R2 = 300 Ω
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RE = 240 Ω
RC = 1 kΩ
VCC = 15 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [300 Ω/(1.8 kΩ + 300 Ω)]15 V
VBB = 2.14 V
VE = VBB – 0.7 V (Eq. 7-5)
VE = 2.14 V – 0.7 V
VE = 1.44 V
IE = VE/RE (Eq. 7-6)
IE = 1.44 V/240 Ω
IE = 6 mA
IC ≈ IE (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-8)
VC = 15 V – (6 mA)(1 kΩ)
VC = 9.0 V
Given (for Q2):
R1 = 910 Ω
R2 = 150 Ω
RE = 120 Ω
RC = 510 Ω
VCC = 15 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [150 Ω/(910 Ω + 150 Ω)]15 V
VBB = 2.12 V
VE = VBB – 0.7 V (Eq. 7-5)
VE = 2.12 V – 0.7 V
VE = 1.42 V
IE = VE/RE (Eq. 7-6)
IE = 1.42 V/120 Ω
IE = 11.83 mA
IC ≈ IE (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-8)
VC = 15 V – (11.83 mA)(510 Ω)
VC = 8.97 V
Given (for Q3):
R1 = 1 kΩ
R2 = 180 Ω
RE = 150 Ω
RC = 620 Ω
VCC = 15 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [180 Ω/(1 kΩ + 180 Ω)]15 V
VBB = 2.29 V
VE = VBB – 0.7 V (Eq. 7-5)
VE = 2.29 V – 0.7 V
VE = 1.59 V
IE = VE/RE (Eq. 7-6)
IE = 1.59 V/150 Ω
IE = 10.6 mA
IC ≈ IE (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-8)
VC = 15 V – (10.6 mA)(620 Ω)
VC = 8.43 V
Answer: The collector voltage for Q1 is 9.0 V, for Q2 is
8.97 V, and for Q3 is 8.43 V.
7-44. Given:
R1 = 10 kΩ
RE = 1 kΩ
RC = 8.2 kΩ
VCC = 20 V
VD = 0.7 V
Solution:
VBB = 3(VD)
VBB = 3(0.7 V)
VBB = 2.1 V
VE = VBB – 0.7 V (Eq. 7-5)
VE = 2.1 V – 0.7 V
VE = 1.4 V
IE = VE/RE (Eq. 7-6)
IE = 1.4 V/1 kΩ
IE = 1.4 mA
IC ≈ IE (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-8)
VC = 20 V – (1.4 mA)(8.2 kΩ)
VC = 8.52 V
Answer: The emitter current is 1.4 mA, and the collector
voltage is 8.52 V.
7-45. Given:
VBB(1) = 2 V
RE(1) = 200 Ω
RC(1) = 1 kΩ
RE(2) = 1 kΩ
VCC = 16 V
Solution:
VE(1) = VBB(1) – 0.7 V (Eq. 7-5)
VE(1) = 2.0 V – 0.7 V
VE(1) = 1.3 V
IE(1) = VE/RE (Eq. 7-6)
IE(1) = 1.3 V/200 Ω
IE(1) = 6.5 mA
IC ≈ IE (Eq. 7-7)
VC(1) = VCC – ICRC (Eq. 7-8)
VC(1) = 16 V – (6.5 mA)(1 kΩ)
VC(1) = 9.5 V
VC(1) = VBB(2)
VE(2) = VBB(2) – 0.7 V (Eq. 7-5)
VE(2) = 9.5 V – 0.7 V
VE(2) = 8.8 V
Answer: The output voltage is 8.8 V.
7-46. Given:
R1 = 620 Ω
R2 = 680 Ω
RE = 200 Ω
VEE = 12 V
VBE = 0.7 V
Solution:
V2 = [R2/(R1 + R2)]VEE
V2 = [680 Ω/(620 Ω + 680 Ω)]12 V
V2 = 6.28 V
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VRE = V2 – 0.7 V
VRE = 6.28 V – 0.7 V
VRE = 5.58 V
IE = VRE/RE (Eq. 7-6)
IE = 5.58 V/200 Ω
IE = 27.9 mA
ILED ≈ IE
Answer: The LED current is 27.9 mA.
7-47. Given:
R1 = 620 Ω
RE = 200 Ω
VEE = 12 V
VBE = 0.7 V
VZ = 6.2 V
Solution:
VRE = VZ – 0.7 V
VRE = 6.2 V – 0.7 V
VRE = 5.5 V
there must be an open below it. If the transistor is open, VB
would be 0 V; therefore the trouble is an open RE.
Trouble 8: R2 is shorted.
7-54. Answer:
Trouble 9: Since the base voltage is 1.1 V, it appears that
the voltage divider is working but not properly. The emitter voltage is 0.7 V less than the base, so the emitterbase junction is working. If RC is open, the meter would
complete the circuit and give a low voltage reading. The
trouble is an open RC.
Trouble 10: This is very similar to trouble 9 except that
the collector voltage is 10 V. Since source voltage is
read above an open, the trouble is an open collector-base
junction.
7-55. Answer:
Trouble 11: Since all the voltages are 0 V, the power
supply is not working.
Trouble 12: With the emitter voltage at 0 V and the base
voltage at 1.83 V, the emitter-base diode of the transistor
is open.
IE = VRE/RE (Eq. 7-6)
IE = 5.5 V/200 Ω
IE = 27.5 mA
7-56. R2 is shorted
ILED ≈ IE
7-57. RC is shorted
Answer: The LED current is 27.5 mA.
7-58. RE is shorted
7-48. Given:
RE = 51 kΩ
R1 = 3.3R2; this ratio is necessary to prevent moving the
Q point. Assume βdc = 100
Solution:
R1 || R2 < 0.01 βdcRE (Eq. 7-12)
R1 || R2 = 0.01(100)(51 kΩ)
R1 || R2 = 51 kΩ
Since R2 is the smaller of the two resistors, make it 51 kΩ.
Then the parallel resistance will not be higher than 51 kΩ,
which satisfies the requirement.
R1 = 3.3R2
R1 = 3.3(51 kΩ)
R1 = 168.3 kΩ
Answer: R1 maximum of 168.3 kΩ, R2 maximum of
51 kΩ, and the ratio between them 3.3:1.
7-49. Answer: With VB at 10 V and R2 is good, the trouble is R1
shorted.
7-50. Answer: Since VB is 0.7 V and VE is 0 V, the trouble is RE
is shorted.
7-51. Answer:
Trouble 3: Since VC is 10 V and VE is 1.1 V, the transistor is good. Therefore the trouble is RC, which is shorted.
Trouble 4: Since all the voltages are the same, the trouble
is that all the transistor terminals are shorted together.
7-52. Answer:
Trouble 5: Since VB is 0 V, it is either R1 open or R2
shorted. R2 is OK, so the trouble is R1 open.
Trouble 6: R2 is open.
7-53. Answer:
Trouble 7: Since VC is 10 V, there is an open below it or a
short above it. A shorted RC would not affect VB; therefore
7-59. No VCC
7-60. Transistor B-E open
Chapter 8 Basic BJT Amplifiers
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
a
b
c
c
a
d
b
b
c
c
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
b
d
b
b
d
b
c
b
b
c
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
a
31. b
c
32. a
b
33. a
c
34. b
d
35. c
c
36. b
c
37. a
b
c
c
JOB INTERVIEW QUESTIONS
4. To permit the output voltage to swing over the largest possible voltage when the input signal is large enough to produce
a maximum output.
10. Very high input impedance to limit the current drawn from
the preceding stage and to prevent distortion. Also, high
current gain and low output impedance to provide a match
to a speaker.
PROBLEMS
8-1.
Given:
C = 47 μF
R = 10 kΩ
Solution:
XC = 1/(2πfC)
XC < 0.1R (Eq. 8-1)
1/(2πfC) = 0.1R
1/(2πC) = (0.1R)f
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f = 1/{[2π(47 μF)][0.1(10 kΩ)]}
f = 3.39 Hz
Solution:
REQ = (R1 × R2)/(R1 + R2) (Parallel resistance formula)
REQ = (10 kΩ × 10 kΩ)/(10 kΩ + 10 kΩ)
REQ = 5 kΩ
Answer: The lowest frequency where good coupling
exists is 3.39 Hz.
8-2.
XC = 1/(2πfC)
XC < 0.1REQ
(Eq. 8-1)
1/(2πfC) = 0.1REQ
f =1/{[2π(220 μF)][0.1(5 kΩ)]}
f = 1.45 Hz
Given:
C = 47 μF
R = 1 kΩ
Solution:
XC = 1/(2πfC)
XC < 0.1R (Eq. 8-1)
1/(2πfC) = 0.1R
1/(2πC) = (0.1R)f
f = 1/{[2π(47 μF)][0.1(1 kΩ)]}
f = 33.9 Hz
Answer: The lowest frequency where good coupling
exists is 1.45 Hz.
8-7.
Answer: The lowest frequency where good coupling
exists is 33.9 Hz.
8-3.
Given:
C = 100 μF
R = 10 kΩ
Solution:
REQ = (R1 × R2)/(R1 + R2) (Parallel resistance formula)
REQ = (2.2 kΩ × 10 kΩ)/(2.2 kΩ + 10 kΩ)
REQ = 1.8 kΩ
Solution:
XC = 1/(2πfC)
XC < 0.1R (Eq. 8-1)
1/(2πfC) = 0.1R
1/(2πC) = (0.1R)f
f = 1/{[2π(100 μF)][0.1(10 kΩ)]}
f = 1.59 Hz
Answer: The lowest frequency where good coupling
exists is 18.8 Hz.
Answer: The lowest frequency where good coupling
exists is 1.59 Hz.
8-4.
XC = 1/(2πfC)
XC < 0.1REQ
(Eq. 8-1)
1/(2 πfC) = 0.1REQ
f = 1/{[2π(47 μF)][0.1(1.8 kΩ)]}
f = 18.8 Hz
8-8.
Given:
f = 100 Hz
R = 10 kΩ
XC = 1/(2πfC)
XC < 0.1REQ
(Eq. 8-1)
1/(2πfC) = 0.1R
C = 1/[2π(1 kHz)(0.1)(1.8 kΩ)]
C = 0.88 μF
Answer: A capacitor value of 1.59 μF is required for good
coupling.
Given:
C = 220 μF
R1 = 2.2 kΩ
R2 = 10 kΩ
Solution:
REQ = (R1 × R2)/(R1 + R2) (Parallel resistance formula)
REQ = (2.2 kΩ × 10 kΩ)/(2.2 kΩ + 10 kΩ)
REQ = 1.8 kΩ
XC = 1/(2 πfC)
XC < 0.1REQ
(Eq. 8-1)
1/(2πfC) = 0.1REQ
f = 1/{[2π(220 μF)][0.1(1.8 kΩ)]}
f = 4 Hz
Answer: The lowest frequency where good coupling
exists is 4 Hz.
8-6.
Given:
C = 220 μF
R1 = 10 kΩ
R2 = 10 kΩ
Given:
f = 1 kHz
R1 = 2.2 kΩ
R2 = 10 kΩ
Solution:
REQ = (R1 × R2)/(R1 + R2) (Parallel resistance formula)
REQ = (2.2 kΩ × 10 kΩ)/(2.2 kΩ + 10 kΩ)
REQ = 1.8 kΩ
Solution:
XC = 1/(2πfC)
XC < 0.1R
(Eq. 8-1)
1/(2πfC) = 0.1R
C = 1/[2π(100 Hz)(0.1)(10 kΩ)]
C = 1.59 μF
8-5.
Given:
C = 47 μF
R1 = 2.2 kΩ
R2 = 10 k
Answer: A capacitor value of 0.88 μF is required for good
coupling.
8-9.
Given:
R1 = 1.5 kΩ
R2 = 330 Ω
RC = 1.2 kΩ
RE = 470 Ω
VCC = 15 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [330 Ω/(1.5 kΩ + 330 Ω)]15 V
VBB = 2.7 V
VE = VBB − VBE
VE = 2.7 V − 0.7 V
VE = 2.0 V
IE = VE/RE
IE = 2.0 V/470 Ω
IE = 4.26 mA
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ie(pp) < 0.1 IEQ
(Eq. 8-6)
ie(pp)max = 0.1 (4.26 mA)
ie(pp)max = 426 μA
RE = 470 Ω
VCC = 15 V
VBE = 0.7 V
Answer: The maximum ac emitter current for small
signal operation is 426 μA.
Solution:
VBB = [R2/(R1 + R2)]VCC (Voltage divider formula)
VBB = [330 Ω/(1.5 kΩ + 330 Ω)]15 V
VBB = 2.7 V
8-10. Given:
R1 = 1.5 kΩ
R2 = 330 Ω
RC = 1.2 kΩ
RE = 940 Ω
VCC = 15 V
VBE = 0.7 V
VE = VBB – VBE
VE = 2.7 V – 0.7 V
VE = 2.0 V
IE = VE/RE
IE = 2.0 V/470 Ω
IE = 4.26 mA
Solution:
VBB = [R2/(R1 + R2)]VCC (Voltage divider formula)
VBB = [330 Ω/(1.5 kΩ + 330 Ω)] 15 V
VBB = 2.7 V
VE = VBB – VBE
VE = 2.7 V – 0.7 V
VE = 2.0 V
IE = VE/RE (Eq. 7-6)
IE = 2.0 V/940 Ω
IE = 2.128 mA
r'e = 25 mV/IE (Eq. 8-10)
r'e =25 mV/4.26 mA
r'e = 5.88 Ω
Answer: The ac resistance of the emitter diode is 5.88 Ω.
8-15. Given:
IE = 2.13 mA (from Prob. 8-10)
Solution:
r'e = 25 mV/IE (Eq. 8-10)
r'e = 25 mV/2.13 mA
r'e = 11.7 Ω
ie(pp) < 0.1 IEQ (Eq. 8-6)
ie(pp)max = 0.1 (2.13 mA)
ie(pp)max = 213 μA
Answer: The maximum ac emitter current for small
signal operation is 213 μA.
8-11. Given:
ic = 15 mA
ib = 100 μA
Answer: The ac resistance of the emitter diode is 11.7 Ω.
8-16. Given:
r'e = 5.88 Ω (from Prob. 8-14)
β = 200
Solution:
zin(base) = βr'e (Eq. 8-11)
zin(base) = 200 (5.88 Ω)
zin(base) = 1.18 kΩ
Solutions:
β = ic/ib (Eq. 8-8)
β = 15 mA/100 μA
β = 150
Answer: The ac beta is 150.
8-12. Given:
β = 200
ib = 12.5 μA
Solutions:
β = ic/ib (Eq. 8-8)
ic = βib
ic = 200 (12.5 μA)
ic = 2.5 mA
Answer: The ac collector current is 2.5 mA.
8-13. Given:
β = 100
ic = 4 mA
Answer: The input impedance to the base is 1.18 kΩ.
8-17. Given:
r'e = 11.7 Ω (from Prob. 8-15)
β = 200
Solution:
zin(base) = βr'e (Eq. 8-11)
zin(base) = 200 (11.7 Ω)
zin(base) = 2.34 kΩ
Answer: The input impedance to the base is 2.34 kΩ.
8-18. Given: Since the collector resistor does not affect the dc
emitter current, the ac emitter resistance does not change.
Since the beta did not change either, the input resistance
remains the same as in problem 8-16.
Answer: The input impedance to the base is 1.18 kΩ.
Solutions:
β = ic/ib (Eq. 8-8)
ib = ib ⁄β
ib = 4 mA/100
ib = 40 μA
8-19. Answer:
Answer: The ac base current is 40 μA.
1
8-14. Given:
R1 = 1.5 kΩ
R2 = 330 Ω
RC = 1.2 kΩ
z in(base) 5 207 V
2
1.5 kV
z out 5 1.02 kV
330 V
b re9
b 5 150
lC
1.2 kV
6.8 kV
re9 5 5.86 V
1-38
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8-20. Answer:
3 kV
1
660 V
2
b re9
lC
2.4 kV
13.6 kV
8-21. Answer:
min hfe = 50
max hfe = 200
Current is 1 mA
Temperature 25°C
8-22. Given:
IE = IC = 5 mA
From Fig. 13 on the data sheet hie is 875 Ω at 5 mA; from
Fig. 11 on the data sheet hfe is 150 Ω at 5 mA.
Solution:
r'e = (25 mV)/IE (Eq. 8-10)
r'e = (25 mV)/5 mA
r'e = 5 Ω
r'e = hic/hfe
r'e = 875 Ω/150
r'e = 5.83
Answer: The value of r'e is 5.83 Ω. The calculated value is
larger than the ideal.
8-23. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
RL = 10 kΩ
VCC = 10 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V
VBB = 1.8 V
VE = VBB – VBE (Eq. 7-5)
VE = 1.8 V – 0.7 V
VE = 1.1 V
IE = VE/RE (Eq. 7-6)
IE = 1.1 V/1 kΩ
IE = 1.1 mA
r'e = (25 mV)/IE (Eq. 8-10)
r'e = (25 mV)/1.1 mA
r'e = 22.7 Ω
rc = RC || RL (Eq. 8-15)
rc = 3.6 kΩ || 10 kΩ
rc = 2.65 kΩ
Av = rc/r'e (Eq. 8-16)
Av = 2.65 kΩ/22.7 Ω
Av = 117
vout = Avin
vout = 117(2 mV)
vout = 234 mV
Answer: The output voltage is 234 mV.
8-24. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
RL = 5 kΩ
VCC = 10 V
VBB = 1.8 V VE = 1.1 V IE = 1.1 V r'e = 22.7 Ω (from Prob. 8-23)
(from Prob. 8-23)
(from Prob. 8-23)
(from Prob. 8-23)
Solution:
rc = RC || RL
rc = 3.6 kΩ || 5 kΩ
rc = 2093 kΩ
Av = rc/r'e
Av = 2093 kΩ/22.7 Ω
Av = 92.2
Answer: The voltage gain is 92.2.
8-25. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
RL = 10 V
VCC = 15 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V
VBB = 2.7 V
VE = VBB – VBE
VE = 2.7 V – 0.7 V
VE = 2.0 V
IE = VE/RE
IE = 2.0 V/1 kΩ
IE = 2 mA
r'e = (25 mV)/IE
r'e = (25 mV)/2 mA
r'e = 12.5 Ω
rc = RC || RL
rc = 3.6 kΩ || 10 kΩ
rc = 2.65 kΩ
Av = rc/r'e
Av = 2.65 kΩ/12.5 Ω
Av = 212
vout = Av(vin)
vout = 212(1 mV)
vout = 212 mV
Answer: The voltage gain is 212, the output voltage is
212 mV.
8-26. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
RL = 10 kΩ
RG = 600 Ω
VBE = 0.7 V
VCC = 15 V
Assume β = 100
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Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V
VBB = 2.7 V
VE = VBB – VBE
VE = 2.7 V – 0.7 V
VE = 2.0 V
IE = VE/RE
IE = 2.0 V/1 kΩ
IE = 2 mA
r'e = (25 mV)/IE
r'e = (25 mV)/2 mA
r'e = 12.5 Ω
rc = RC || RL
rc = 3.6 kΩ || 10 kΩ
rc = 2.65 kΩ
Av = rc/re'
Av = 2.65 kΩ/12.5 Ω
Av = 212
zin = R1 || R2 || βr'e
zin = 10 kΩ || 2.2 kΩ || 1.25 kΩ
zin = 738 Ω
vin = [zin/(RG + zin)]vg
vin = [738 Ω/(600 Ω + 738 Ω)]1 mV
vin = 551.57 μV
vout = Av(vin)
vout = 212(551.57 μV)
vout = 117 mV
Answer: The voltage gain is 212, the output voltage is
117 mV.
8-27. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 2 kΩ
RL = 10 kΩ
RG = 600 Ω
VCC = 10 V
VBE = 0.7 V
Assume β = 100
Solution:
VBB = [R1/(R1 + R2)]VCC (Eq. 7-4)
VBB = [10 kΩ/(10 kΩ + 2.2 kΩ)]10 V
VBB = 1.8 V
VE = VBB – VBE (Eq. 7-5)
VE = 1.8 V – 0.7 V
VE = 1.1 V
IE = VE/RE (Eq. 7-6)
IE = 1.1 V/2 kΩ
IE = 0.55 mA
r'e = (25 mV)/IE (Eq. 8-10)
r'e = (25 mV)/0.55 mA
r'e = 45.5 Ω
rc = RC || RL (Eq. 8-15)
rc = 3.6 kΩ || 10 kΩ
rc = 2.65 kΩ
Av = rc / r'e (Eq. 8-16)
Av = 2.65 kΩ/45.5 Ω
Av = 58
zin = R1 || R2 || βr'e
zin = 10 kΩ || 2.2 kΩ || 100(45.5 Ω)
zin = 1.29 kΩ
vin = [zin(RG + zin)]vg (Eq. 8-17)
vin = [1.29 kΩ/(600 Ω + 1.29 kΩ)]1 mV
vin = 0.683 mV
vout = Av(vin)
vout = 58(0.683 mV)
vout = 39.6 mV
Answer: The output voltage is 39.6 mV.
8-28. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
RL = 10 kΩ
RG = 300 Ω
VBE = 0.7 V
VCC = 10 V
Assume β = 100
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V
VBB = 1.8 V
VE = VBB – VBE
VE = 1.8 V – 0.7 V
VE = 1.1 V
IE = VE/RE
IE = 1.1 V/1 kΩ
IE = 1.1 mA
r'e = (25 mV)/IE
r'e = (25 mV)/1.1 mA
r'e = 22.7 Ω
rc = RC || RL
rc = 3.6 kΩ || 10 kΩ
rc = 2.65 kΩ
Av = rc/r'e
Av = 2.65 kΩ/22.7 Ω
Av = 117
zin = R1 || R2 || βr'e
zin = 10 kΩ || 2.2 kΩ || 2.27 kΩ
zin = 1 kΩ
vin = [zin/(RG + zin)]vg
vin = [1 kΩ/(300 Ω + 1 kΩ)]1 mV
vin = 769 μV
vout = Av(vin)
vout = 117(769 μV)
vout = 90 mV
Answer: The voltage gain is 117, the output voltage is
90 mV.
8-29. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 820 Ω
re = 180 Ω
RL = 10 kΩ
RG = 600 Ω
VBE = 0.7 V
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VCC = 10 V
Assume β = 100
Solution:
rc = RC || RL
rc = 3.6 kΩ || 10 kΩ
rc = 2.65 kΩ
Av = rc/re
Av = 2.65 kΩ/180 Ω
Av = 14.7
zin = R1 || R2 || βre
zin = 10 kΩ || 2.2 kΩ || 18 kΩ
zin = 1.64 kΩ
vin = [zin/(RG + zin)]vg
vin = [1.64 kΩ/(600 Ω + 1.64 kΩ)]25 mV
vin = 18.3 mV
vout = Av(vin)
vout = 14.7(18.3 mV)
vout = 269 mV
Answer: The voltage gain is 14.7, the output voltage
is 269 mV.
8-30. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 820 Ω
re = 180 Ω
RL = 10 kΩ
RG = 50 Ω
VBE = 0.7 V
VCC = 10 V
Assume β = 100
Solution:
rc = RC || RL
rc = 3.6 kΩ || 10 kΩ
rc = 2.65 kΩ
Solution:
rc = RC || RL
rc = 3.6 kΩ || 3.6 kΩ
rc = 1.8 kΩ
Av = rc/re
Av = 1.8 kΩ/180 Ω
Av = 10
Answer: The voltage gain is 10.
8-32. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 820 Ω
re = 180 Ω
RL = 10 kΩ
RG = 600 Ω
VCC = 30 V
VBE = 0.7 V
Solution:
rc = RC || RL (Eq. 8-15)
rc = 3.6 kΩ || 10 kΩ
rc = 2.65 kΩ
Av = rc/re (Eq. 8-19)
Av = 2.65 kΩ/180 Ω
Av = 14.7
Answer: The voltage gain is 14.7.
8-33. Answer: Since the capacitor is an open to direct current,
the dc voltages do not change and the ac voltage gain will
be drastically reduced.
8-34. Answer: Some of the possible causes are: open transistor, open emitter resistor, or open output coupling
capacitor.
CRITICAL THINKING
Av = rc/re
Av = 2.65 kΩ/180 Ω
Av = 14.7
8-35. Answer: The capacitor has a certain amount of leakage
current, and this current will flow through the resistor and
create a voltage drop across the resistor.
zin = R1 || R2 || βre
zin = 10 kΩ || 2.2 kΩ || 18 kΩ
zin = 1.64 kΩ
8-36. Answer: A wire has a very small inductance value. As
the frequency increases, the inductive reactance starts
to become significant. The wires connected to the
capacitor and the leads will start to have an inductive
reactance, causing the voltage to rise at the node.
vin = [zin/(RG + zin)]vg
vin = [1.64 kΩ /(50 Ω + 1.64 kΩ)]50 mV
vin = 48.52 mV
vout = Av(vin)
vout = 14.7(48.52 mV)
vout = 713 mV
Answer: The voltage gain is 14.7, the output voltage
is 713 mV.
8-31. Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 820 Ω
re = 180 Ω
RL = 3.6 kΩ
RG = 600 Ω
VBE = 0.7 V
VCC = 10 V
Assume β = 100
8-37. Given:
R = 30 Ω
f = 20 Hz to 20 kHz
Solution:
XC = 1/(2πfC)
XC < 0.1R (Eq. 8-5)
1/(2πfC) = 0.1R
1/(2πf) = (0.1R)(C)
1/(2πf)(0.1R) = C
C = 1/{[2π(20 Hz)][0.1(30 Ω)]}
C = 2653 μF
Answer: The capacitor would have to be at least 2653 μF,
or 2700 μF (standard value).
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8-38. Given:
R1 = 20 kΩ
R2 = 4.4 kΩ
RC = 7.2 kΩ
RE = 2 kΩ
RL = 20 kΩ
VCC = 10 V
VBE = 0.7 V
Trouble 2: Since the input voltage increased to 0.75 mV,
the problem is an open RE.
Trouble 3: Since there are no ac voltages and the base
voltage has changed, the problem is in the input circuit.
Since there is a 0.7 V drop across the BE diode, the transistor should be conducting and thus the collector voltage
should be less than 10 V. It appears that the BC diode is
open, except the base voltages are not consistent with that
problem. To make this problem correct for the BC diode
open, return VB, VE, and vb to the OK values.
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [4.4 kΩ/(20 kΩ + 4.4 kΩ)]10 V
VBB = 1.8 V
Trouble 4: Since the dc base voltage is 0 and there is an
ac base voltage, the problem is an R1 open.
Trouble 5: Since there is no output ac voltage, the
problem is C2 open.
VE = VBB – VBE (Eq. 7-5)
VE = 1.8 V – 0.7 V
VE = 1.1 V
IE = VE/RE (Eq. 7-6)
IE = 1.1 V/2 kΩ
IE = 0.55 mA
re' = (25 mV)/IE (Eq. 8-10)
re' = (25 mV)/0.55 mA
re' = 45.5 Ω
rc = RC || RL (Eq. 8-15)
rc = 7.2 kΩ || 20 kΩ
rc = 5.3 kΩ
Trouble 6: Since there are no ac voltages and the base
voltage has changed, the problem is in the input circuit.
The voltage points to an open R2.
8-41. Answers:
Trouble 7: All the dc voltages are OK; thus the transistor
and resistors are OK. Since the base and emitter ac
voltages are the same, the problem appears to be an open
bypass capacitor C3.
Trouble 8: Since there are no ac voltages and the base
voltage has changed, the problem is in the input circuit.
Since the collector voltage is so low, the collector resistor
is open.
Av = rc/r'e (Eq. 8-16)
Av = 5.3 kΩ/45.5 Ω
Av = 116
Trouble 9: Since there are no dc voltages, the problem
is no VCC.
Answer: The voltage gain is 116.
8-39. Given:
R1 = 20 kΩ
R2 = 4.4 kΩ
RC = 7.2 kΩ
RE = 2 kΩ
RL = 20 kΩ
RG = 1.2 kΩ
VCC = 10 V
VBE = 0.7 V
Assume β = 100
VBB = 1.8 V (from Prob. 8-38)
VE = 1.1 V (from Prob. 8-38)
IE = 0.55 mA (from Prob. 8-38)
re' = 45.5 Ω (from Prob. 8-38)
re' = 5.3 kΩ
Av = 116
Solution:
zin = R1 || R2 || βre'
zin = 20 kΩ || 4.4 kΩ || 100(45.5 Ω)
zin = 2.01 kΩ
vin = [zin/(RG + zin)]vg (Eq. 10-4)
vin = [2.01 kΩ/(1.2 kΩ + 2.01 kΩ)]1 mV
vin = 0.626 mV
vout = Av(vin)
vout = 116(0.626 mV)
vout = 72.6 mV
Answer: The output voltage is 72.6 mV.
8-40. Answers:
Trouble 1: Since all the ac voltages are 0, the problem
could be the generator, RG open, or C1 open.
Trouble 10: Since the emitter voltage is 0 and the base
voltage is near normal, the problem is an open BE diode.
Trouble 11: With all the dc voltages the same, the
problem is a shorted transistor in all three terminals.
Trouble 12: Since all the ac voltages are 0, the problem
could be the generator, RG open, or C1 open.
8-42. C3 is open
8-43. C1 is open
8-44. RE is shorted
8-45. Transistor is upside down.
8-46. No VCC
Chapter 9 Multistage, CC,
and CB Amplifiers
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
9.
a
b
c
b
c
b
c
d
c
10.
11.
12.
13.
14.
15.
16.
17.
18.
a
a
d
c
a
c
d
a
c
19.
20.
21.
22.
23.
24.
25.
26.
27.
c
a
c
c
a
a
d
a
d
28.
29.
30.
31.
32.
33.
34.
35.
36.
a
c
d
c
b
d
b
d
b
JOB INTERVIEW QUESTIONS
5. Voltage gain is always less than but usually near 1. The circuit is used as a current or power amplifier. Applications
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include stereo output stages, linear power-supply regulation,
and drivers for relays, LEDs.
7. They allow excellent impedance matching and maximum
power transfer to low-impedance loads.
11. None.
12. Power gain is the product of voltage gain and current gain.
Although the voltage gain is slightly less than 1, the current
gain is very large. Therefore, the power gain is very large.
PROBLEMS
9-1.
Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
RL = 10 kΩ
RG = 600 Ω
VCC = 10 V
VBE = 0.7 V
β = 100
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V
VBB = 1.8 V
VE = VBB – VBE (Eq. 7-5)
VE = 1.8 V – 0.7 V
VE = 1.1 V
IE = VE/RE (Eq. 7-6)
IE = 1.1 V/1 kΩ
IE = 1.1 mA
r'e = (25 mV)/IE (Eq. 8-10)
r'e = (25 mV) /1.46 mA
r'e = 22.7 Ω
zin = R1 || R2 || βre'
zin = 10 kΩ || 2.2 kΩ || 100(22.7 Ω)
zin = 1.0 kΩ
The input impedance for each stage is 1.0 kΩ.
vin(1) = [zin/(RG + zin)]vg (Eq. 8-17)
vin(1) = [1.0 kΩ/(600 Ω + 1.0 kΩ)]1 mV
vin(1) = 0.625 mV
The input impedance for the second stage is the load
resistance for the first stage.
rc = RC || RL (Eq. 8-15)
rc = 3.6 kΩ || 1.0kΩ
rc = 783 Ω
Av = rc/r'e (Eq. 8-16)
Av = 783 Ω/22.7 Ω
Av = 34.5
The output voltage of the first stage is the input voltage
for the second stage.
vout(1) = Av(vin)
vout(1) = 34.5(0.625 mV)
vout(1) = 21.6 mV
rc(2) = Rc || RL (Eq. 8-15)
rc(2) = 3.6 kΩ || 10 kΩ
rc(2) = 2.65 kΩ
Av(2) = rc/r'e (Eq. 8-16)
Av(2) = 2.65 kΩ/22.7Ω
Av(2) = 117
vout(2) = Av(vin)
vout(2) = 117(21.6 mV)
vout(2) = 2.53 V
Answer: The base voltage of the first stage is 0.625 mV,
the base voltage of the second stage is 21.6 mV, and the
voltage across the collector resistor is 2.53 V.
9-2.
Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
RL = 10 kΩ
RG = 600 Ω
VCC =12 V
VBE = 0.7 V
β = 100
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]12 V
VBB = 2.16 V
VE = VBB − VBE (Eq. 7-5)
VE = 2.16 V – 0.7 V
VE = 1.46 V
IE = VE/RE (Eq. 7-6)
IE = 1.46 V/1 kΩ
IE = 1.46 mA
r'e = (25 mV)/IE (Eq. 8-10)
r'e = (25 mV)/1.46 mA
r'e = 17.1 Ω
zin = R1 || R2 || βr'e
zin = 10 kΩ || 2.2 kΩ || 100(17.1 Ω)
zin = 878 Ω
The input impedance for each stage is 878 Ω.
vin(1) = [zin/(RG + zin)]vg (Eq. 8-17)
vin(1) = [878 Ω/ (600 Ω + 878 Ω)]1 mV
vin(1) = 0.594 mV
The input impedance for the second stage is the load
resistance for the first stage.
rc = RC || RL (Eq. 8-15)
rc = 3.6 kΩ || 878 kΩ
rc = 706 Ω
Av = rc/r'e (Eq. 8-19)
Av = 706 Ω/17.1 Ω
Av = 41.3
The output voltage of the first stage is the input voltage
for the second stage.
vout(1) = Av(vin)
vout(1) = 41.3(0.594 mV)
vout(1) = 24.5 mV
rc(2) = RC || RL (Eq. 8-15)
rc(2) = 3.6 kΩ || 10 kΩ
rc(2) = 2.65 kΩ
Av(2) = rc/r'e (Eq. 8-19)
Av(2) = 2.65 kΩ/17.1 Ω
Av(2) = 155
vout(2) = Av(vin)
vout(2) = 155(24.5 mV)
vout(2) = 3.80 V
Answer: The output voltage is 3.80 V.
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9-3.
Given:
R1 = 10 kΩ
R2 = 2.2 kΩ
RC = 3.6 kΩ
RE = 1 kΩ
RL = 10 kΩ
RG = 600 Ω
Vcc =10 V
VBE = 0.7 V
β = 300
Solution:
Av = rf /re (Eq. 9-2)
Av = 5 kΩ/50 Ω
Av = 100
Answer: The voltage gain is 100.
9-5.
Solution:
Av = rf /re (Eq. 9-2)
rf = 100(125 Ω)
rf = 12.5 kΩ
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V
VBB = 1.8 V
VE = VBB – VBE (Eq. 7-5)
VE = 1.8 V – 0.7 V
VE = 1.1 V
Answer: The feedback resistor would need to be 12.5 kΩ.
9-6.
IE = VE/RE (Eq. 7-6)
IE = 1.1 V/1 kΩ
IE = 1.1 mA
r'e = (25 mV)/IE (Eq. 8-10)
r'e = (25 mV) /1.1 mA
r'e = 22.7 Ω
zin = R1 || R2 || βr'e
zin = 10 kΩ || 2.2 kΩ || 300(22.7 Ω)
zin = 1.43 kΩ
VE = VBB – VBE (Eq. 7-5)
VE = 7.5 V – 0.7 V
VE = 6.8 V
IE = VE/RE (Eq. 7-6)
IE = 6.8 V/1 kΩ
IE = 6.8 mA
The input impedance for the second stage is the load
resistance for the first stage.
rc = RC || RL (Eq. 8-15)
rc = 3.6 kΩ || 1.43 kΩ
rc = 1.02 kΩ
r'e = 25 mV/IE (Eq. 8-10)
r'e = 25 mV/6.8 mA
r'e = 3.68 Ω
Av = rc/r'e (Eq. 8-16)
Av = 1.02 kΩ/22.7 Ω
Av = 45
re = RE || RL (Eq. 9-3)
re = 1 kΩ || 3.3 kΩ
re = 767 Ω
The output voltage of the first stage is the input voltage
for the second stage.
vout(1) = Av(vin)
vout(1) = 45(0.704 mV)
vout(1) = 31.7 mV
Av(2) = rc/r'e (Eq. 8-16)
Av(2) = 2.65 kΩ/22.7 Ω
Av(2) = 117
vout(2) = Av(vin)
vout(2) = 117(31.7 mV)
vout(2) = 3.71 V
Answer: The output voltage is 3.71 V.
9-4.
Given:
rf = 5 kΩ
re = 50 Ω
Given:
R1 = 2.2 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RL = 3.3 kΩ
RG = 50 Ω
β = 200
VCC = 15 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 kΩ/(2.2 kΩ + 2.2 kΩ)]15 V
VBB = 7.5 V
The input impedance for each stage is 1.43 kΩ.
vin(1) = [zin/(RG + zin)]vg (Eq. 8-17)
vin(1) = [1.43 kΩ/(600 Ω + 1.43 kΩ)]1 mV
vin(1) = 0.704 mV
rc(2) = RC || RL (Eq. 8-15)
rc(2) = 3.6 kΩ || 10 kΩ
rc(2) = 2.65 kΩ
Given:
re = 125 Ω
Av = 100
zin(base) = β(re + r'e) (Eq. 9-5)
zin(base) = 200(767 Ω + 3.48 Ω)
zin(base) = 154 kΩ
zin(stage) = 154 kΩ || 2.2 kΩ || 2.2 kΩ = 1.09 kΩ
Answer: The input impedance of the base is 154 kΩ, and
the input impedance of the stage is 1.09 kΩ.
9-7.
Given:
R1 = 2.2 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RL = 3.3 kΩ
RG = 50 Ω
β = 150
VCC = 15 V
VBE = 0.7 V
r'e = 3.68 Ω (from Prob. 9-6)
re = 767 Ω (from Prob. 9-6)
Solution:
zin = R1 || R2 || β(re + r'e) (Eq. 9-6)
zin = 2.2 kΩ || 2.2 kΩ || 150(767 Ω + 3.48 Ω)
zin = 1.09 kΩ
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vin = [zin/(zin + RG)]vg
vin = [1.09 kΩ/(1.09 kΩ + 50 Ω)]1 V
vin = 0.956 V
Answer: The input voltage is 0.956 V.
9-8.
Given:
R1 = 2.2 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RL = 3.3 kΩ
RG = 50 Ω
β = 175
VCC = 15 V
VBE = 0.7 V
r'e = 3.68 Ω (from Prob. 9-6)
re = 767 Ω (from Prob. 9-6)
Solution:
Av = re /(re + r'e) (Eq. 11-2)
Av = 767 Ω/(767 Ω + 3.48 Ω)
Av = 0.995
9-9.
RE = 2 kΩ
RL = 6.6 kΩ
RG = 100 Ω
β = 150
VCC = 15 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [4.4 kΩ/(4.4 kΩ + 4.4 kΩ)]15 V
VBB = 7.5 V
VE = VBB – VBE (Eq. 7-5)
VE = 7.5 V – 0.7 V
VE = 6.8 V
IE = VE/RE (Eq. 7-6)
IE = 6.8 V/2 kΩ
IE = 3.4 mA
r'e = 25 mV/IE (Eq. 8-10)
r'e = 25 mV/3.4 mA
r'e = 7.35 Ω
zin = R1 || R2 || β(re + r'e) (Eq. 9-6)
zin = 2.2 kΩ || 2.2 kΩ || 175(767 Ω + 3.48 Ω)
zin = 1.09 kΩ
re = RE || RL (Eq. 9-3)
re = 2 kΩ || 6.6 kΩ
re = 1.53 kΩ
vin = [zin/(zin + RG)]vg
vin = [1.09 kΩ/(1.09 kΩ + 50 Ω)]1 V
vin = 0.956 V
zin = R1 || R2 || β(re + r'e) (Eq. 9-6)
zin = 4.4 kΩ || 4.4 kΩ || 150(1.53 kΩ + 7.35 kΩ)
zin = 2.18 kΩ
vout = Av(vin) (Eq. 8-3)
vout = (0.995)(0.956 V)
vout = 0.951 V
vin = [zin/(zin + RG)]vg
vin = [2.18 kΩ/(2.18 kΩ + 100 Ω)]1 V
vin = 0.956 V
Answer: The gain is 0.995, and the output voltage is
0.951 V.
Answer: The input impedance doubles to 2.18 kΩ, and
the input voltage remains the same at 0.956 V.
Given:
R1 = 2.2 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RL = 3.3 kΩ
RG = 50 Ω
β = 50 to 300
VCC =15 V
VBE = 0.7 V
r'e = 3.68 Ω (from Prob. 9-6)
re = 767 Ω (from Prob. 9-6)
Solution:
zin(min) = R1 || R2 || β(re + r'e) (Eq. 9-6)
zin(min) = 2.2 kΩ || 2.2 kΩ || 50(767 Ω + 3.48 Ω)
zin(min) = 1.07 kΩ
zin(max) = R1 || R2 || β(re + r'e) (Eq. 9-6)
zin(max) = 2.2 kΩ || 2.2 kΩ || 300(767 Ω + 3.48 Ω)
zin(max) = 1.09 kΩ
vin(min) = [zin/(zin + RG)]VG
vin(min) = [1.07 kΩ/(1.07 kΩ + 50 Ω)]1 V
vin(min) = 0.955 V
vin(min) = [zin/(zin + RG)]vg
vin(min) = [1.09 kΩ/(1.09 kΩ + 50 Ω)]1 V
vin(min) = 0.956 V
Answer: The input voltage varies over the range of 0.955
to 0.956 V.
9-10. Given:
R1 = 4.4 kΩ
R2 = 4.4 kΩ
9-11. Given:
R1 = 100 Ω
R2 = 200 Ω
RE = 30 Ω
RL = 10 Ω
RG = 50 Ω
β = 200
VCC = 20 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [200 Ω/(100 Ω + 200 Ω)]20 V
VBB = 13.3 V
VE = VBB – VBE (Eq. 7-5)
VE = 13.3 V – 0.7 V
VE = 12.6 V
IE = VE/RE (Eq. 7-6)
IE = 12.6 V/30 Ω
IE = 420 mA
r'e = 25 mV/IE (Eq. 8-10)
r'e = 25 mV/420 mA
r'e = 0.06 Ω
re = RE || RL (Eq. 9-3)
re = 30 Ω || 10 Ω
re = 7.5 Ω
zin(base) = β(re + r'e) (Eq. 9-5)
zin(base) = 200(7.5 Ω + 0.06 Ω)
zin(base) = 1.51 kΩ
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zin (stage) = R1 || R2 || β(re + r'e) (Eq. 9-6)
zin (stage) = 100 Ω || 200 Ω || 1.51 kΩ
zin (stage) = 63.8 Ω
Answer: The input impedance of the base is 1.51 kΩ, and
the input impedance to the stage is 63.8 Ω.
9-12. Given:
R1 = 100 Ω
R2 = 200 Ω
RE = 30 Ω
RL = 50 Ω
RG = 50 Ω
β = 150
VCC = 20 V
VBE = 0.7 V
r'e = 0.06 Ω (from Prob. 9-11)
re = 7.5 Ω (from Prob. 9-11)
Solution:
zin = R1 || R2 || β(re + r'e) (Eq. 9-6)
zin = 100 Ω || 200 Ω || 150(7.5 Ω + 0.06 Ω)
zin = 63 Ω
vin = [zin/(zin + RG)]vg
vin = [63 Ω/(63 Ω + 50 Ω)]1 V
vin = 0.558 V
Answer: The input voltage is 0.558 V.
9-13. Given:
R1 = 100 Ω
R2 = 200 Ω
RE = 30 Ω
RL = 10 Ω
RG = 50 Ω
β = 175
VCC = 20 V
VBE = 0.7 V
r'e = 0.06 Ω (from Prob. 9-11)
re = 7.5 Ω (from Prob. 9-11)
Solution:
Av = re/(re + r'e) (Eq. 9-4)
Av = 7.5/(7.5 + 0.06)
Av = 0.992
zin = R1 || R2 || β(re + r'e) (Eq. 9-6)
zin = 100 Ω || 200 Ω || 175(7.5 Ω + 0.06 Ω)
zin = 63.5 Ω
vin = [zin/(zin + RG)]VG
vin = [63.5 Ω/(63.5 Ω + 50 Ω)]1 V
vin = 0.559 V
vout = Av(vin) (Eq. 8-3)
vout = (0.992)(0.559 V)
vout = 0.555 V
Answer: The gain is 0.992, and the output voltage is
0.555 V.
9-14. Given:
R1 = 2.2 kΩ
R2 = 2.2 kΩ
RE = 1 kΩ
RL = 3.3 kΩ
RG = 50 Ω
β = 200
VCC = 15 V
VBE = 0.7V
r'e = 3.68 Ω (from Prob. 9-6)
re = 767 Ω (from Prob. 9-6)
Solution:
zout = RE || [r'e + (RG || R1 || R2)/β] (Eq. 9-7)
zout = 1 kΩ || [3.68 Ω + (50 Ω || 2.2 kΩ || 2.2 kΩ)/200]
zout = 3.9 Ω
Answer: The output impedance is 3.9 Ω.
9-15. Given:
R1 = 100 Ω
R2 = 200 Ω
RE = 30 Ω
RL = 10 Ω
RG = 50 Ω
β = 100
VCC = 20 V
VBE = 0.7 V
r'e = 0.06 Ω (from Prob. 9-11)
re = 7.5 Ω (from Prob. 9-11)
Solution:
zout = RE || [r'e + (RG || R1 || R2)/β] (Eq. 9-7)
zout = 30 Ω || [0.06 Ω + (50 Ω || 100 Ω || 200 Ω)/100]
zout = 0.342 Ω
Answer: The output impedance is 0.342 Ω.
9-16. Given:
Q2 β = 200 (ac & dc)
R1 = 4.7 kΩ
R2= 1 kΩ
RC = 1.5 kΩ
RE = 330 Ω
RL = 150 Ω
VCC = 15 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [1 kΩ/(4.7 kΩ + 1 kΩ)]15 V
VBB = 2.63 V
VE = VBB − VBE
VE = 2.63 V − 0.7 V
VE = 1.93 V
IE = VE/RE
IE = 1.93 V/330 Ω
IE = 5.85 mA
r'e = (25 mV)/IE
r'e = (25 mV)/5.85 mA
r'e = 4.28 Ω
zinQ2 = βre2
zinQ2 = 200 (150 Ω)
zinQ2 = 30 kΩ
rc = RC || zinQ2
rc = 1.5 kΩ || 30 kΩ
rc = 1.429 kΩ
Av = rc/r'e
Av = 1.429 kΩ/4.28 Ω
Av = 335
Answer: The voltage gain is 335.
9-17. Given:
β = 150 (ac & dc)
R1 = 4.7 kΩ
R2 = 1 kΩ
RC = 1.5 kΩ
RE = 330 Ω
RL = 150 Ω
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VCC = 15 V
vg = 10 mV
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [1 kΩ/(4.7 kΩ + 1 kΩ)]15 V
VBB = 2.63 V
VE = VBB − VBE
VE = 2.63 V − 0.7 V
VE = 1.93 V
IE = VE/RE
IE = 1.93 V/330 Ω
IE = 5.85 mA
r'e = (25 mV)/IE
r'e = (25 mV)/5.85 mA
r'e = 4.28 Ω
zinQ2 = βRL
zinQ2 = 150 (150 Ω)
zinQ2 = 22.5 kΩ
rc = RC || zinQ2
rc = 1.5 kΩ || 22.5 kΩ
rc = 1.406 kΩ
Av1 = rc/r'e
Av1 = 1.4 kΩ/4.28 Ω
Av1 = 327
vout = Av1(vg) = 327(10 mV) = 3.27 V
Answer: The 1st stage voltage gain is 327, the 2nd stage
voltage gain is 1, vout = 3.27 V.
9-18. Given:
β = 200 (ac & dc)
R1 = 4.7 kΩ
R2 = 1 kΩ
RC = 1.5 kΩ
RE = 330 Ω
RL = 125 Ω
VCC = 15 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [1 kΩ/(4.7 kΩ + 1 kΩ)]15 V
VBB = 2.63 V
VE = VBB − VBE
VE = 2.63 V − 0.7 V
VE = 1.93 V
IE = VE/RE
IE = 1.93 V/330 kΩ
IE = 5.85 mA
r'e = (25 mV)/IE
r'e = (25 mV)/5.85 mA
r'e = 4.28 Ω
zinQ2 = βRL
zinQ2 = 200 (125 Ω)
zinQ2 = 25 kΩ
rc = RC || zinQ2
rc = 1.5 kΩ || 25 kΩ
rc = 1.415 kΩ
Av1 = rc/r'e
Av1 = 1.415 kΩ/4.28 Ω
Av1 = 331
9-19. Given:
β = 200 (ac & dc)
R1 = 4.7 kΩ
R2 = 1 kΩ
RC = 1.5 kΩ
RE = 330 Ω
RL = 150 Ω
VCC = 15 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [1 kΩ/(4.7 kΩ + 1 kΩ)]15 V
VBB = 2.63 V
VE = VBB − VBE
VE = 2.63 V − 0.7 V
VE = 1.93 V
IE = VE /RE
IE = 1.93 V/330 Ω
IE = 5.85 mA
r'e = (25 mV)/IE
r'e = (25 mV)/5.85 mA
r'e = 4.27 Ω
rc = RC || RL
rc = 1.5 kΩ || 150 Ω
rc = 136.5 Ω
Av1 = rc/r'e
Av1 = 136.5 Ω/4.27 Ω
Av1 = 31.9
Answer: The voltage gain drops to 31.9.
9-20. Given:
R1 = 150 kΩ
R2 = 150 kΩ
RE = 470 Ω
RL = 1 kΩ
RG = 5.1 kΩ
VCC = 15 V
β = 5000
Solution:
re = RE || RL (Eq. 9-3)
re = 470 Ω || 1 kΩ
re = 320 Ω
zin(base) = βre
zin(base) = (5000)(320)
zin(base) = 1.6 MΩ
Answer: The input impedance of the base is1.6 MΩ.
9-21. Given:
R1 = 150 kΩ
R2 = 150 kΩ
RE = 470 Ω
RL = 1 kΩ
RG = 5.1 kΩ
VCC = 15 V
β = 7000
r­e = 320 Ω (from Prob. 9-20)
Solution:
zin(base) = βre
zin(base) = (7000) (320)
zin(base) = 2.24 MΩ
Answer: The voltage gain remains at 331.
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zin = R1 || R2 || zin(base) (Eq. 9-6)
zin = 150 kΩ || 150 kΩ || 2.24 MΩ
zin = 72.6 kΩ
vin = [zin/(zin + RG)]vg
vin = [72.6 kΩ/(72.6 kΩ + 5.1 kΩ)]10 mV
vin = 9.34 mV
Answer: The input voltage is 9.34 mV.
9-22. Given:
R1 = 1 kΩ
R2 = 2 kΩ
RE = 10 Ω
RL = 8 Ω
RG = 600 Ω
VCC = 20 V
β1 = 150
β2 = 150
Solution:
re = RE || RL (Eq. 9-3)
re = 10 Ω || 8 Ω
re = 4.44 Ω
β = β1β2 (Eq. 9-9)
β = 150(150)
β = 22500
zin(base) = βre
zin(base) = (22500)(4.44 Ω)
zin(base) = 100 kΩ
Answer: The input impedance of the base is 100 kΩ.
9-23. Given:
R1 = 1 kΩ
R2 = 2 kΩ
RE = 10 Ω
RL = 8 Ω
RG = 600 Ω
VCC = 20 V
β = 2000
re = 4.44 Ω (from Prob. 9-22)
Solution:
zin(base) = βre
zin(base) = (2000)(4.44 Ω)
zin(base) = 8.88 kΩ
zin = R1 || R2 || zin(base) (Eq. 9-6)
zin = 1 kΩ || 2 kΩ || 8.88 kΩ
zin = 620 Ω
vin = [zin/(zin + RG)]vg
vin = [620 Ω/(620 Ω + 600 Ω)]1 V
vin = 0.508 V
Answer: The input voltage is 0.508 V.
9-24. Given:
VZ = 7.5 V
VBE = 0.7 V
RS = 1 kΩ
VCC = 15 V
Solution:
Vout = VZ – VBE (Eq. 9-11)
Vout = 7.5 V – 0.7 V
Vout = 6.8 V
IZ = (VCC – VZ)/RS
IZ = (15 V– 7.5 V)/1 kΩ
IZ = 7.5 mA
Answer: The output voltage is 6.8 V, and the zener
current is 7.5 mA.
9-25. Given:
VZ = 7.5 V
VBE = 0.7 V
RS = 1 kΩ
VCC = 25 V
Solution:
Vout = VZ – VBE (Eq. 9-11)
Vout = 7.5 V – 0.7 V
Vout = 6.8 V
Take the base current into account.
IZ = (VCC – VZ)/RS – Iout/β
IZ = (25 – 7.5)/1 kΩ – (6.8 V/33 Ω)/150
IZ = 17.5 mA – 1.37 mA
IZ = 16.1 mA
Answer: The output voltage is 6.8 V, and the zener
current is 16.1 mA.
9-26. Given: With the wiper in the middle, the voltage divider
is effectively two resistors: each has a value of 1.5 kΩ.
Vz = 7.5 V
VBE = 0.7 V
Solution:
Vout = [(R3 + R4)/R4](VZ + VBE) (Eq. 9-14)
Vout = [(1.5 kΩ + 1.5 kΩ)/1.5 kΩ](7.5 V + 0.7 V)
Vout = 16.4 V
Answer: The output voltage is 16.4 V.
9-27. Given: With the wiper all the way up, the voltage divider
is effectively two resistors: the top has a value of 1 kΩ,
and the bottom has a value of 2 kΩ.
VZ = 7.5 V
VBE = 0.7 V
With the wiper all the way down, the voltage divider is
effectively two resistors: the top has a value of 2 kΩ, and
the bottom has a value of 1 kΩ.
Solution:
Vout(top) = [(R3 + R4)/R4](VZ + VBE) (Eq. 9-14)
Vout(top) = [(1 kΩ + 2 kΩ)/1.5 kΩ](7.5 V + 0.7 V)
Vout(top) = 12.3 V
Vout(bottom) = [(R3 + R4)/R4](VZ + VBE) (Eq. 9-14)
Vout(bottom) = [(2 kΩ + 1 kΩ)/1 kΩ](7.5 V + 0.7 V)
Vout(bottom) = 24.6 V
Answer: The output voltage with the wiper all the way up
is 12.3 V, and all the way down is 24.6 V.
9-28. Given:
R1 = 10 kΩ
R2 = 2 kΩ
RC = 3.3 kΩ
RE = 2 kΩ
VCC =12 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2 kΩ/(10 kΩ + 2 kΩ)]12 V
VBB = 2 V
VE = VBB – VBE
VE = 2 V – 0.7 V
VE = 1.3 V
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IE = VE/RE
IE = 1.3 V/2 kΩ
IE = 650 μA
Answer: The emitter current is 650 μA.
9-29. Given:
R1 = 10 kΩ
R2 = 2 kΩ
RC = 3.3 kΩ
RE = 2 Ω
VCC = 12 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2 kΩ/(10 kΩ + 2 kΩ)]12 V
VBB = 2 V
VE = VBB – VBE
VE = 2 V – 0.7 V
VE = 1.3 V
IE = VE/RE
IE = 1.3 V/2 kΩ
IE = 650 μA
r'e = (25 mV)/I E
r'e = (25 mV)/650 μA
r'e = 38.46 Ω
rc = RC || RL
rc = 3.3 kΩ || 10 kΩ
rc = 2.48 kΩ
Av = rc/r'e
Av = 2.48 kΩ/38.46 Ω
Av = 64.4
Answer: The voltage gain at 64.4.
9-30. Given:
R1 = 10 kΩ
R2 = 2 kΩ
RC = 3.3 kΩ
RE = 2 Ω
VCC = 12 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2 kΩ/(10 kΩ + 2 kΩ)]12 V
VBB = 2 V
VE = VBB – VBE
VE = 2 V – 0.7 V
VE = 1.3 V
IE = VE/RE
IE = 1.3 V/2 kΩ
IE = 650 μA
9-31. Given:
R1 = 10 kΩ
R2 = 2 kΩ
RC = 3.3 kΩ
RE = 2 kΩ
RG = 50 Ω
VCC = 12 V
vg = 2 mV
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2 kΩ/(10 kΩ + 2 kΩ)]12 V
VBB = 2 V
VE = VBB – VBE
VE = 2 V – 0.7 V
VE = 1.3 V
IE = VE/RE
IE = 1.3 V/2 kΩ
IE = 650 μA
r'e = (25 mV)/I E
r'e = (25 mV)/650 μA
r'e = 38.46 Ω
re = RC || RL
re = 3.3 kΩ || 10 kΩ
re = 2.48 kΩ
Av = rc/r'e
Av = 2.48 kΩ/38.46 Ω
Av = 64.4
zin(stage) = RE || r'e
Since RE >> r'e
zin(stage) = r'e = 38.5 Ω
vin ≈ [zin/(RG + zin)]vg
vin = [38.5 Ω/(50 Ω + 38.5 Ω)] 2 mV
vin = 870 μV
vout = Av(vin)
vout = 64.4(870 μV)
vout = 56 mV
Answer: The output voltage is 56 mV.
9-32. Given:
R1 = 10 kΩ
R2 = 2 kΩ
RC = 3.3 kΩ
RE = 2 kΩ
RG = 50 Ω
VCC = 15 V
vg = 2 mV
r'e = (25 mV)/IE
r'e = (25 mV)/650 μA
r'e = 38.46 Ω = 38.5 Ω
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [2 kΩ/(10 kΩ + 2 kΩ)]15 V
VBB = 2.5 V
zin(emitter) = r'e
zin(emitter) = 38.5 Ω
zin(stage) = RE || r'e
VE = VBB – VBE
VE = 2.5 V – 0.7 V
VE = 1.8 V
Since RE >> r'e
zin(stage) ≅ r'e = 38.5 Ω
zout ≈ RC
zout = 3.3 kΩ
IE = VE/RE
IE = 1.8 V/2 kΩ
IE = 900 μA
Answer: The zin(emitter) = 38.5 Ω, the zin(stage) = r'e = 38.5 Ω.
r'e = (25 mV)/I E
r'e = (25 mV) /900 μA
r'e = 27.8 Ω
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rc = RC || RL
rc = 3.3 kΩ || 10 kΩ
rc = 2.48 kΩ
Av = rc/r'e
Av = 2.48 kΩ/27.8 Ω
Av = 89.3
zin(stage) = RE || r'e
Since RE >> r'e
zin(stage) = r'e = 27.8 Ω
vin ≈ [zin/(RG + zin)]vg
vin = [27.8 kΩ/(50 Ω + 27.8 Ω)]2 mV
vin = 715 μV
vout = Av(vin)
vout = 89.3(715 μV)
vout = 63.8 mV
Answer: The output voltage is 63.8 mV.
CRITICAL THINKING
9-33. Given:
VZ = 7.5 V
VCC = 15 V
Vout = 6.8 V (from Prob. 9-24)
RL = 33 Ω
Solution:
VCE = VCC – Vout
VCE = 15 V – 6.8 V
VCE = 8.2 V
IC = Iout = Vout/RL
IC = 6.8 V/33 Ω
IC = 206 mA
P = VCEIC
P = (8.2 V)(206 mA)
P = 1.69 W
Answer: 1.69 W
9-34. Given:
R1 = 4.7 kΩ
R2 = 2 kΩ
RC = 1 kΩ
RE = 1 kΩ
VCC = 15 V
β = 150
Solution:
VBB = [R2/(R1 + R2)] VCC (Eq. 7-4)
VBB = [2 kΩ/(4.7 kΩ + 2 kΩ)]15 V
VBB = 4.48 V
VE = VBB – VBE (Eq. 7-5)
VE = 4.48 V – 0.7 V
VE = 3.78 V
IE = VE/RE (Eq. 7-6)
IE = 3.78 V/1 kΩ
IE = 3.78 mA
IE = IC (Eq. 7-7)
VC = VCC – ICRC (Eq. 7-18)
VC = 15 V – 3.78 mA(1 kΩ)
VC = 11.22 V
IB = IC/β
IB = 3.78 mA/150
IB = 25.2 μA
Answer: The values are VB = 4.48 V, VE = 3.78 V, VC =
11.22 V, IE = 3.78 mA, IC = 3.78 mA, and IB = 25.2 μA.
9-35. Given:
R1 = 4.7 kΩ
R2 = 2 kΩ
RC = 1 kΩ
RE = 1 kΩ
VCC = 15 V
β = 150
vin = 5 mV
vout(2) is an emitter follower that has a gain of 1.
Solution:
rc = 1 kΩ
re = 1 kΩ
Av = rc/re (Eq. 8-19)
Av = 1 kΩ/1 kΩ
Av = 1
Answer: Both outputs are 5 mV; the top one is 180° out
of phase. The purpose of this circuit is to produce two
signals that are the same magnitude and 180° out of
phase.
9-36. Given:
R1 = 33 kΩ
R2 = 10 kΩ
RC = 4.7 kΩ
RE = 2.2 kΩ
VCC = 12 V
vin = 10 mV
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [10 kΩ/(33 kΩ + 10 kΩ)]12 V
VBB = 2.79 V
VE = VBB – VBE (Eq. 7-5)
VE = 2.79 V – 0.7 V
VE = 2.09 V
IE = VE/RE (Eq. 7-6)
IE = 2.09 V/2.2 kΩ
IE = 0.95 mA
rc = 4.7 kΩ
r'e = 25 mV/IE (Eq. 8-10)
r'e = 25 mV/0.95 mA
r'e = 26.3 Ω
Av = rc/re' (Eq. 8-16)
Av = 4.7 kΩ /26.3 Ω
Av = 179
vout = Av(vin)
vout = 179(10 mV)
vout = 1.79 V
When the control voltage is 5 V, the control transistor is
saturated and grounds the input; thus the output is zero.
Answer: With the control voltage at 0 V, the output is
1.79 V. With the control voltage at 5 V, the output is 0 V.
This circuit could be a mute circuit.
9-37. Given:
RL = 33 Ω
RS = 1 kΩ
VCC = 15 V
βdc = 200
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vin = [815 Ω/(270 Ω + 815 Ω)](100 mVp-p)
vin = 75 mVp-p
Solution:
IE = (VCC – VBE)/(RL + (RS/βdc))
IE = (15 V – 0.7 V)/(33 Ω + (1 kΩ/200))
IE = 376 mA
Vout = VE = IERE = (376 mA)(33 Ω)
Vout = 12.4 V
Answer: The output voltage is 12.4 V.
9-38. Given:
RL = 33 Ω
RS = 1 kΩ
VCC = 15 V
βdc = 100
Solution:
IB = (VCC – VBE)/RS
IB = (15 V – 0.7 V)/1 kΩ
IB = 14.3 mA
IE = βdcIB
IE = (100)(14.3 mA)
IE = 1.43 A
vout = (75 mVp-p)(4.49)
vout = 337 mVp-p
Answer: The output voltage would be 337 mVp-p.
9-42. Answer: The output would decrease to zero volts.
9-43. Given:
Trouble 1: Since there is voltage at H and none at I, the
trouble is an open C4.
Trouble 2: Since there is voltage at F and none at G, the
trouble is an open between F and G.
Trouble 3: Since there is voltage at A and none at B, the
trouble is an open C1.
9-44. Answers:
Trouble 4: Since E is not a ground, potential trouble is
an open C3.
Trouble 5: Since there is a voltage at B and none at C, the
trouble is an open between B and C.
PD = ICVCE
PD = (1.43 A)(15 V)
PD = 21.5 W
Answer: The transistor will dissipate 21.5 W and be
destroyed.
9-39. Given:
Wiper is at 50%
RL = 100 Ω
Solution:
Vout = [(1.5 kΩ + 1.5 kΩ)/1.5 kΩ](8.2 V)
Vout = 16.4 V
Iout = 16.4 V/100 Ω
Iout = 164 mA
PD = (164 mA)(25 V – 16.4 V)
PD = 1.41 W
Answer: The power dissipation of Q2 is 1.41 W when the
wiper is at 50%.
9-40. Given:
βdc = 100 for both transistors
Solution:
IE = (7.5 V – 0.7 V – 0.7 V)/470 Ω
IE = 13 mA
r'e = 25 mV/13 mA
r'e = 1.93 Ω
zout = r'e + [((5.1 kΩ || 150 kΩ || 150 kΩ)/10,000) || 470 Ω]
zout = (1.93 Ω + 0.478 Ω) || 470 Ω
zout = 2.4 Ω
Answer: The output impedance is 2.4 Ω.
9-41. Given:
vg = 100 mV
Solution:
Av1 = 1.5 kΩ/(4.27 Ω + 330 Ω)
Av1 = 4.49
zin = 4.7 kΩ || 1 kΩ || (200)(334)
zin = 815 Ω
Trouble 6: Since there is voltage at D and none at F, the
trouble is an open C2.
Trouble 7: Since there is voltage at G and none at H, the
trouble is an open Q2.
9-45. C3 is shorted
9-46. C2 is open
9-47. Q1 B-E short
9-48. Q2 is open
9-49. R2 is shorted
Chapter 10
Power Amplifiers
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
9.
b
b
c
a
c
d
d
b
b
10.
11.
12.
13.
14.
15.
16.
17.
18.
d
c
d
b
b
b
b
c
a
19.
20.
21.
22.
23.
24.
25.
26.
27.
a
c
b
d
a
a
b
c
c
28.
29.
30.
31.
32.
33.
34.
35.
a
d
d
b
c
d
c
a
JOB INTERVIEW QUESTIONS
6. Tuned RF amplifier. It would be impractical to use a Class-C
amplifier for an audio application because it would distort the
signal.
8. The lower the duty cycle is, the less the current drain.
11. Thermal conductive paste used to create a low thermal
resistance path between the case and the heat sink.
12. Class-A. No signal is lost in a Class-A amplifier: 360° in,
360° out. With Class-C, over half the signal is lost.
13. Narrowband.
PROBLEMS
10-1.
Given:
R1 = 2 kΩ
R2 = 470 Ω
RC = 680 Ω
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RE = 220 Ω
RL = 2.7 kΩ
VCC = 15 V
or
VCEQ = VC – VE
VCEQ = 8.36 V – 2.15 V
VCEQ = 6.21 V
Solution:
RC = 680 Ω
IC(sat) = VCC/(RC + RE) (Eq. 10-1)
IC(sat) = 15 V/(680 Ω + 220 Ω)
IC(sat) = 16.67 mA
Answer: The dc collector resistance 680 Ω, and the dc
saturation current is 16.67 mA.
10-2.
Answer: The maximum peak-to-peak voltage is 10.62 V.
10-4.
Given:
R1 = 2 kΩ
R2 = 470 Ω
RC = 680 Ω
RE = 220 Ω
RL = 2.7 kΩ
VCC = 15 V
VBE = 0.7 V
RG = 50 Ω
Solution:
rc = RC || RL (Eq. 8-15)
rc = 680 Ω || 2.7 kΩ
rc = 543 Ω
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [470 Ω/(2 kΩ + 470 Ω)]15 V
VBB = 2.85 V
VE = VBB – VBE (Eq. 7-5)
VE = 2.85 V – 0.7 V
VE = 2.15 V
IE = ICQ = VE/RE (Eq. 7-6)
ICQ = 2.15 V/220 Ω
ICQ = 9.77 mA
Since ICQ is the center of the load line, the load line is
linear, the other end is zero, and the ac saturation current
is double the Q point current. The ac saturation current
is 19.5 mA.
Answer: The ac collector resistance is 543 Ω, and the ac
saturation current is 19.5 mA.
10-3.
MPP = 2MP
MPP = 2(5.31 V)
MPP = 10.62 V
Given:
R1 = 2 kΩ
R2 = 470 Ω
RC = 680 Ω
RE = 220 Ω
RL = 2.7 kΩ
VCC = 15 V
VBE = 0.7 V
RG = 50 Ω
rc = 543 Ω (from Prob. 10-2)
ICQ = 9.77 mA (from Prob. 10-2)
VE = 2.15 V (from Prob. 10-2)
Solution:
VC = VCC – RCICQ
VC = 15 V – (680 Ω)(9.77 mA)
VC = 8.36 V
MP = ICQrc or VCEQ (Eq. 10-8)
MP = (9.77 mA)(543 Ω)
MP = 5.31 V
Given:
R1 = 4 kΩ
R2 = 940 Ω
RC = 1.36 kΩ
RE = 440 Ω
RL = 5.4 kΩ
VCC = 15 V
VBE = 0.7 V
RG = 100 Ω
Solution:
rc = RC || RL (Eq. 8-15)
rc = 1.36 kΩ || 5.4 kΩ
rc = 1086 Ω
Answer: The ac collector resistance is 1086 Ω.
10-5.
Given:
R1 = 6 kΩ
R2 =1.41 kΩ
RC = 2.04 kΩ
RE = 660 Ω
RL = 8.1 kΩ
VCC = 15 V
VBE = 0.7 V
RG = 150 Ω
Solution:
rc = RC || RL (Eq. 8-15)
rc = 2.04 kΩ || 8.1 kΩ
rc = 1.63 kΩ
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [1.41 kΩ/(6 kΩ + 1.41 kΩ)]15 V
VBB = 2.85 V
VE = VBB – VBE (Eq. 7-5)
VE = 2.84 V – 0.7 V
VE = 2.15 V
IE = ICQ = VE/RE (Eq. 7-6)
ICQ = 2.15 V/660 Ω
ICQ = 3.26 mA
VC = VCC – RCICQ
VC = 15 V – (2.04 kΩ)(3.26 mA)
VC = 8.35 V
MP = ICQrc or VCEQ (Eq. 10-8)
MP = (3.26 mA)(1.63 kΩ)
MP = 5.31 V
or
VCEQ = VC – VE
VCEQ = 8.35 V – 2.15 V
VCEQ = 6.2 V
MPP = 2MP
MPP = 2(5.31 V)
MPP = 10.62 V
Answer: The maximum peak-to-peak voltage is 10.62 V.
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10-6.
Given:
R1 = 200 Ω
R2 = 100 Ω
RC = 100 Ω
RE = 68 Ω
RL = 100 Ω
VCC = 30 V
Solution:
VC = VCC – RCICQ
VC = 30 V – (100 Ω)(137 mA)
VC = 16.3 V
Solution:
RC = 100 Ω
MP = VCEQ = 7 V
MPP = 2MP = 14 V
VCEQ = VC – VE
VCEQ = 16.3 V – 9.3 V
VCEQ = 7 V
IC(sat) = VCC/RC + RE
IC(sat) = 30 V/100 Ω + 68 Ω
IC(sat) = 179 mA
or
MP = ICQrc
MP = (137 mA)(50 Ω)
MP = 6.85 V
MPP = 2MP = 13.7 V
Answer: The dc collector resistance is 100 Ω, and the
saturation current is 179 mA.
10-7.
Given:
R1 = 200 Ω
R2 = 100 Ω
RC = 100 Ω
RE = 68 Ω
RL = 100 Ω
VCC = 30 V
VBE = 0.7 V
Solution:
rc = RC || RL
rc = 100 Ω || 100 Ω
rc = 50 Ω
VBB = [R2/(R1 + R2)]VCC
VBB = [100 Ω/(200 Ω + 100 Ω)]30 V
VBB = 10 V
VE = VBB – VBE
VE = 10 V – 0.7 V
VE = 9.3 V
IE = ICQ = VE/RE
ICQ = 9.3 V/68 Ω
ICQ = 137 mA
VC = VCC – ICQRC
VC = 30 V – (137 mA)(100 Ω)
VC = 16.3 V
VCEQ = VC – VE
VCEQ = 16.3 V – 9.3 V
VCEQ = 7 V
ic(sat) = ICQ + VCEQ/rc
ic(sat) = 137 mA + 7 V/50 Ω
ic(sat) = 277 mA
Answer: The ac collector resistance is 50 Ω, and the ac
saturation current is 277 mA.
10-8.
Given:
R1 = 200 Ω
R2 = 100 Ω
RC = 100 Ω
RE = 68 Ω
RL = 100 Ω
VCC = 30 V
VBE = 0.7 V
rc = 50 Ω (from Prob. 10-7)
ICQ = 137 mA (from Prob. 10-7)
ic(sat) = 277 mA (from Prob. 10-7)
VE = 9.3 V (from Prob. 10-7)
Answer: The maximum peak-to-peak voltage is 13.7 V.
10-9.
Given:
R1 = 400 Ω
R2 = 200 Ω
RC = 200 Ω
RE = 136 Ω
RL = 200 Ω
VCC = 30 V
VBE = 0.7 V
Solution:
rc = RC || RL
rc = 200 Ω || 200 Ω
rc = 100 Ω
Answer: The ac collector resistance is 100 Ω.
10-10. Given:
R1 = 600 Ω
R2 = 300 Ω
RC = 300 Ω
RE = 204 Ω
RL = 300 Ω
VCC = 30 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC
VBB = [300 Ω/(600 Ω + 300 Ω)]30 V
VBB = 10 V
VE = VBB – VBE
VE = 10 V – 0.7 V
VE = 9.3 V
IE = ICQ = VE/RE
ICQ = 9.3 V/204 Ω
ICQ = 45.59 mA
VC = VCC – RCICQ
VC = 30 V – (300 Ω)(45.59 mA)
VC = 16.3 V
VCEQ = VC – VE
VCEQ = 16.3 V – 9.3 V
VCEQ = 7 V
MP = VCEQ = 7 V
MPP = 2MP = 14 V
or
MP = ICQrc
MP = (45.59 mA) (150 Ω)
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MP = 6.85 V
MPP = 2MP = 13.7 V
Answer: The maximum peak-to-peak voltage is 13.7 V.
10-11. Given:
Pout = 2 W
Pin = 4 mW
Solution:
AP = Pout/Pin (Eq. 10-12)
AP = 2 W/4 mW
AP = 500
Answer: The power gain is 500.
10-12. Given:
Vout = 15 Vp-p
RL = 1 kΩ
Pin = 400 μW
Solution:
Pout = V2/8RL
Pout = (15 V)2/8 kΩ
Pout = 28.1 mW
AP = Pout/Pin (Eq. 10-12)
AP = 28.1 mW/400 μW
AP = 70.3
Answer: The power gain is 70.3.
10-13. Given:
R1 = 2 kΩ
R2 = 470 Ω
RC = 680 Ω
RE = 220 Ω
RL = 2.7 kΩ
VCC = 15 V
VBE = 0.7 V
RG = 50 Ω
VBB = 2.85 V (from Prob. 10-2)
Solution:
Ibias = VCC/(R1 + R2)
Ibias = 15 V/(2 kΩ + 470 Ω)
Ibias = 6.07 mA
VE = VBB – VBE (Eq. 7-5)
VE = 2.85 V – 0.7 V
VE = 2.15 V
IE = VE/RE (Eq. 7-6)
IE = 2.15 V/220 Ω
IE = 9.77 mA
Idc = Ibias + IE
Idc = 6.07 mA + 9.77 mA
Idc = 15.84 mA
Answer: The current drain is 15.84 mA.
10-14. Given:
Idc = 15.84 mA (from Prob. 10-13)
VCC = 15 V
Solution:
Pdc = IdcVCC (Eq. 10-17)
Pdc = (15.84 mA)(15 V)
Pdc = 237.6 mW
Answer: The dc input power is 237.6 mW.
10-15. Given:
MPP = 10.62 V (from Prob. 10-3)
RL = 2.7 kΩ
Pdc = 237.6 mW (from Prob. 10-14)
Solution:
Pout(max) = MPP2/8RL (Eq. 10-15)
Pout = (10.62 V)2/8(2.7 kΩ)
Pout = 5.22 mW
η = [Pout/Pin]100%
η = [5.22 mW/237.6 mW]100%
η = 2.2%
Answer: The efficiency is 2.2%.
10-16. Given:
ICQ = 9.77 mA (from Prob. 10-2)
VCEQ = 6.21 V (from Prob. 10-3)
Solution:
PDQ = VCEQ ICQ (Eq. 10-16)
PDQ = (6.21 V)(9.77 mA)
PDQ = 60.7 mW
Answer: The quiescent power dissipation is 60.7 mW.
10-17. Given:
R1 = 200 Ω
R2 = 100 Ω
RC = 100 Ω
RE = 68 Ω
RL = 100 Ω
VCC = 30 V
VBE = 0.7 V
VBB = 10 V (from Prob. 10-7)
Solution:
Ibias = VCC/(R1 + R2)
Ibias = 30 V/(200 Ω + 100 Ω)
Ibias = 100 mA
VE = VBB – VBE
VE = 10 V – 0.7 V
VE = 9.3 V
IE = VE/RE
IE = 9.3 V/68 Ω
IE = 136.8 mA ≈ 137 mA
Idc = Ibias + IE
Idc = 100 mA + 137 mA
Idc = 237 mA
Answer: The current drain is 237 mA.
10-18. Given:
Idc = 237 mA (from Prob. 10-17)
VCC = 30 V
Solution:
Pdc = IdcVCC
Pdc = (237 mA)(30 V)
Pdc = 7.11 W
Answer: The dc input power is 7.11 W.
10-19. Given:
MPP = 2MP = 13.7 V (from Prob. 10-10)
Pdc = 7.11 W (from Prob. 10-18)
RL = 100 Ω
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Solution:
Pout(max) = MPP2/8RL
Pout = (13.7 V)2/8(100 Ω)
Pout = 235 mW
η = [Pout/Pin]100%
η = [235 mW/7.11 W]100%
η = 3.3%
Answer: The efficiency is 3.3%.
10-20. Given:
ICQ = 137 mA (from Prob. 10-7)
VCEQ = 7 V (from Prob. 10-8)
Solution:
PDQ = VCEQICQ
PDQ = (7 V)(137 mA)
PDQ = 960 mW
Answer: The quiescent power dissipation is 960 mW.
10-21. Given:
R1 = 10 Ω
R2 = 2.2 Ω
RE = 1 Ω
VCC = 10 V
VBE = 0.7 V
Solution:
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 Ω/(10 Ω + 2.2 Ω)] 10 V
VBB = 1.80 V
VE = VBB – VBE (Eq. 7-5)
VE = 1.80 V – 0.7 V
VE = 1.10 V
IE = VE/RE (Eq. 7-6)
IE = 1.1 V/1 Ω
IE = 1.1 A
Answer: The dc emitter current is 1.1 A.
10-22. Given:
R1 = 10 Ω
R2 = 2.2 Ω
RE = 1 Ω
VCC = 10 V
VBE = 0.7 V
RC = 3.2 Ω
Vout = 5 Vp-p
Solution:
Pout = vout2/8RL (Eq. 10-14)
Pout = (5 V)2/8(3.2 Ω)
Pout = 0.977 W
Ibias = VCC/(R1 + R2)
Ibias = 10 V/(10 Ω + 2.2 Ω)
Ibias = 0.82 A
VBB = [R2/(R1 + R2)]VCC (Eq. 7-4)
VBB = [2.2 Ω/(10 Ω + 3.2 Ω)]10 V
VBB = 1.80 V
VE = VBB – VBE (Eq. 7-5)
VE = 1.80 V – 0.7 V
VE = 1.10 V
IE = VE/RE (Eq. 7-6)
IE = 1.10 V/1 Ω
IE = 1.1 A
Idc = Ibias + IE
Idc = 0.82 A + 1.1 A
Idc = 1.92 A
Pdc = IdcVCC (Eq. 10-17)
Pdc = (1.92 A)(10 V)
Pdc = 19.2 W
η = [Pout/Pin]100%
η = [0.977 W/19.2 W]100%
η = 5.1%
Answer: The output power is 0.977 W, and the efficiency
is 5.1%.
10-23. Given:
VCE(cutoff) = 12 V
Solution:
MPP = 12 VCE(cutoff)
MPP = 2(12 V)
MPP = 24 V
Answer: The maximum peak-to-peak voltage is 24 V.
10-24. Given:
VCC = MPP = 30 V
RL = 16 Ω
Solution:
PD(max) = MPP2/40RL
PD(max) = (30 V)2/40(16 Ω)
PD(max) = 1.41 W
Answer: The maximum power dissipation of each
transistor is 1.41 W.
10-25. Given:
VCC = MPP = 30 V
RL = 16 Ω
Solution:
Pout(max) = MPP2/8RL
Pout(max) = (30 V)2/8(16 Ω)
Pout(max) = 7.03 W
Answer: The maximum output power is 7.03 W.
10-26. Given:
R1 = 100 Ω
R2 = 100 Ω
RL = 50 Ω
VCC = 30 V
VDiode = 0.7 V
Solution:
Ibias = (VCC – 2VDiode)/(R1 + R2)
Ibias = 28.6 V/(100 Ω + 100 Ω)
Ibias = 143 mA
ICEQ ≈ Ibias = 143 mA
Answer: The quiescent collector current is 143 mA.
10-27. Given:
VCC = MPP = 30 V
RL = 50 Ω
Solution:
Ibias = (VCC – 2VDiode)/(R1 + R2)
Ibias = 28.6 V/(100 Ω + 100 Ω)
Ibias = 143 mA
Idc = 238 mA
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Pdc = IdcVCC
Pdc = 238 mA(30 V)
Pdc = 7.14 W
Pout(max) = MPP2/8RL
Pout(max) = (30 V)2/8(50 Ω)
Pout(max) = 2.25 W
η = [Pout/Pin]100%
η = [2.25 W/7.14 W]100%
η = 31.5%
Answer: The efficiency is 31.5%.
10-28. Given:
VCC = MPP = 30 V
RL = 50 Ω
Solution:
Ibias = (VCC – 2VDiode)/(R1 + R2)
Ibias = 28.6 V/(1 kΩ + 1 kΩ)
Ibias = ICQ = 14.3 mA
Idc = 110 mA
Pdc = IdcVCC
Pdc = 110 mA(30 V)
Pdc = 3.3 W
Pout(max) = MPP2/8RL
Pout(max) = (30 V)2/8(50 Ω)
Pout(max) = 2.25 W
η = [Pout/Pin]100%
η = [2.25 W/3.3 W]100%
η = 68.3%
Answer: The efficiency is 68.3% and the quiescent
collector current is 14.3 mA.
10-29. Given:
MPP = 30 V
RL = 100 Ω
Solution:
Pout(max) = MPP2/8RL
Pout(max) = (30 V)2/8(100 Ω)
Pout(max) = 1.13 W
Answer: The maximum power output is 1.13 W.
10-30. Given for 1st stage:
R1 = 10 kΩ
R2 = 5.6 kΩ
R3 = 1 kΩ
R4 = 1 kΩ
VBB = 10.7 V
VE = 10 V
Second Stage:
R5 = 12 kΩ
R6 = 1 kΩ (variable)
R7 = 1 kΩ
R8 = 100 Ω
β = 200
VCC = 30 V
Solution:
r'e = 25 mV/IE
r'e = 25 mV/(10 V/1 kΩ)
r'e = 2.5 Ω
re = R8 (second stage)
re = 100 Ω
rc = R3 || zin(stage 2)
zin(stage 2) = 12 kΩ || 910 Ω || β r'e
rc = 1 kΩ || 12 kΩ || 910 Ω || 200(100 Ω)
rc = 496 Ω
Av(stage 1) = rc / r'e
Av(stage 1) = 496 Ω/2.5 Ω
Av(stage 1) = 188
Answer: The voltage gain of the first stage is 188.
10-31. Given for 2nd Stage:
R5 = 12 kΩ
R6 = 1 kΩ (variable)
R8 = 100 Ω
R7 = 1 kΩ
VE = 1.43 V
3rd Stage:
β = 200
VCC = 30 V
RL = 100 Ω
Solution:
IE = VE/RE (Eq. 7-6)
IE = 1.43 V/100 Ω
IE = 14.3 mA
r'e = 25 mV/IE (Eq. 8-10)
r'e = 25 mV/(14.3 mA)
r'e = 1.75 Ω
re = R8 (second stage)
re = 100 Ω
zin(base) = β re (Eq. 8-21)
zin(base) = 200(100 Ω)
zin(base) = 20 kΩ
rc = R7 || zin(base)
rc = 1 kΩ || 20 kΩ
rc = 952 Ω
Av = rc/(re + r'e)
Av = 952 Ω/(100 Ω + 1.75 Ω)
Av = 9.36
Answer: The gain of the second stage is 9.36.
10-32. Given:
IE = 14.3 mA
Solution:
ICQ = Ibias = 14.3 mA
Answer: The quiescent collector current is 14.3 mA.
10-33. Given:
Av1 = 188 (from Prob. 10-30)
Av2 = 9.36 (from Prob. 10-31)
Solution:
Av3 = 1 (Eq. 10-25)
Av = Av1Av2Av3
Av = (188)(9.36)(1)
Av = 1679
Answer: The total voltage gain is 1679.
10-34. Given: νin = 5 Vrms.
Solution:
Vp-p = 2.828 Vrms.
Vp-p = 2.828(5 V)
Vp-p = 14.14 Vp-p
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Since the input is clamped at 0.7 V, the negative peak
is –13.44 V. The average value is –6.37 V, so the DMM
will read –6.37 V.
Answer: The input voltage is 14.14 Vp-p, and the base
voltage is –6.37 V.
10-35. Given:
L = 1 μH
C = 220 pF
Solution:
___
fr = 1/(2π​√_____________
LC ​) (Eq. 10-29)
fr = 1/[2π​√(1
μH)(220 pF) ​]
fr = 10.73 MHz
Answer: The resonant frequency is 10.73 MHz.
10-36. Given:
L = 2 μH
C = 220 pF
Solution:
____
fr = 1/(2π​√_____________
LC) ​ (Eq. 10-29)
fr = 1/[2π​√
(2 μH)(220 pF) ​]
fr = 7.59 MHz
Answer: The resonant frequency is 7.59 MHz.
10-37. Given:
L = 1 μH
C = 100 pF
Solution:
___
fr = 1/(2π​√_____________
LC ​) (Eq. 10-29)
fr = 1/[2π​√(1
μH)(100 pF) ​]
fr = 15.92 MHz
Answer: The resonant frequency is 15.92 MHz.
10-38. Given:
Pout = 11 mW
Pin = 50 μW
Solution:
Ap = Pout/Pin (Eq. 10-12)
Ap = 11 mW/50 μW
Ap = 220
Answer: The power gain is 220.
10-39. Given:
νout = 50 Vp-p
RL = 10 kΩ
Solution:
Pout = ν2out/8RL (Eq. 10-14)
Pout = (50 Vp-p)2/8(10 kΩ)
Pout = 31.25 mW
Answer: The output power is 31.25 mW.
10-40. Given: VCC = 30 V.
Solution:
MPP = 2 VCC (Eq. 10-38)
MPP = 2(30 V)
MPP = 60 V
Pout = MPP2/8RL (Eq. 10-15)
Pout = (60 Vp-p)2/8(10 kΩ)
Pout = 45 mW
Answer: The maximum output power is 45 mW.
10-41. Given:
Idc = 0.5 mA
VCC = 30 V
Solution:
Pdc = VCCIdc (Eq. 10-17)
Pdc = (30 V)(0.5 mA)
Pdc = 15 mW
Answer: The dc input power is 15 mW.
10-42. Given:
Idc = 0.4 mA
VCC = 30 V
vout = 30 Vp-p
RL = 10 kΩ
Solution:
Pdc = VCCIdc (Eq. 10-17)
Pdc = (30 V)(0.4 mA)
Pdc = 12 mW
Pout = vout2/8RL (Eq. 10-14)
Pout = (30 Vp-p)2 /8(10 kΩ)
Pout = 11.25 mW
η = (Pout/Pin)100% (Eq. 10-18)
η = (11.25 mW/12 mW)100%
η = 93.75%
Answer: The efficiency is 93.75%.
10-43. Given:
Q = 125
fr = 10.73 MHz (from Prob. 10-35)
Solution:
B = fr/Q
B = 10.73 MHz/125
B = 85.84 kHz
Answer: The bandwidth is 85.84 kHz.
10-44. Given:
Q = 125
fr = 10.73 MHz (from Prob. 10-35)
RL = 10 kΩ
MPP = 60 V (from Prob. 10-40)
L = 1 μH
Solution:
XL = 2πfL
XL = 2(3.14)(10.73 MHz)(1 μH)
XL = 67.38 Ω
RP = QXL (Eq. 10-33)
RP = (125)(67.38 Ω)
RP = 8.42 kΩ
rC = RP || RL (Eq. 10-34)
rC = 8.42 kΩ || 10 kΩ
rC = 4.57 kΩ
PD = MPP2/40rC (Eq. 10-39)
PD = (60 V)2/40(4.57 kΩ)
PD = 19.7 mW
Answer: The worst-case power dissipation is 19.7 mW.
10-45. Given:
PD = 625 mW
D = 5 mW/°C
TA = 100°C
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Solution:
ΔP = D(TA – 25°C) (Eq. 10-40)
ΔP = (5 mW/°C)(100°C – 25°C)
ΔP = 375 mW
PD(max) = PD – ΔP
PD(max) = 625 mW – 375 mW
PD(max) = 250 mW
Answer: The worst-case power rating is 250 mW.
10-46. Given: Derating curve on Fig. 10-34.
Answer: The maximum dissipation at 100°C is 2 W.
10-47. Given:
PD = 115 W
D = 0.657 W/°C
TC = 90°C
Solution:
ΔP = D(TC – 25°C) (Eq. 10-40)
ΔP = (0.657 W/°C)(90°C – 25°C)
ΔP = 42.7 W
PD(max) = PD – ΔP
PD(max) = 115 W – 42.7 W
PD(max) = 72.3 W
Answer: The power rating is 72.3 W with a case temperature of 90°C.
CRITICAL THINKING
10-48. Answer: The input is larger than the maximum allowed
input for an undistorted output. The input is driving the
output into saturation, clipping the wave off, and turning
it into a square wave.
10-49. Answer: Electrically, it would be safe to touch, but it may
be hot and cause a burn.
10-50. Answer: No, the maximum efficiency of anything is
100 percent. It is impossible to get more power out of a
device than is put into the device.
10-51. Answer: No, the ac load line is more vertical because the
ac collector resistance is usually less than the dc collector resistance. If the collector had an inductor instead of
a resistor, the ac resistance would be greater than the dc
resistance and make the ac load line less vertical.
10-52. Given:
IC(sat) = 16.67 mA (from Prob. 10-1)
VCC = 15 V
ICQ = 9.77 mA (from Prob. 10-2)
MP = ICQ rc = 5.31 V (from Prob. 10-3)
VCEQ = 6.21 V (from Prob. 10-3)
Solution:
The left side of the dc load line is IC(sat), and the right side
is VCC. The Q point is ICQ, VCEQ. The ac load line passes
through the Q point. The right side of the ac load line is
ICQ rc above the Q point, or 11.52 V. This gives the line
a slope of ICQ/ICQ rc = 9.77 mA/5.31 V = 1.84 mA/V. To
find the ac saturation current, take the ac voltage maximum multiplied by the slope = (11.52 V) (1.84 mA/V) =
21.2 mA.
Answer: See the graph.
IC
20 mA
Alternating current
16 mA
12 mA
8 mA Direct
4 mA current
0 mA
Q
VCE
0V
11.5 V 15 V
10-53. C2 is shorted
10-54. D1 is open
10-55. VCC is now 20 V
10-56. Q1 B-E shorted
10-57. R6 is shorted
10-58. Class-B/AB push-pull power amplifier
10-59. Approximately 24 Vp-p
10-60. Compensation diodes used for temperature stability
10-61. 511 microamps
10-62. Approximately zero volts dc
Chapter 11
JFETs
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
a
d
c
d
b
b
d
8.
9.
10.
11.
12.
13.
c
d
c
c
a
c
14.
15.
16.
17.
18.
19.
d
a
b
c
c
a
20.
21.
22.
23.
24.
25.
c
c
b
b
d
d
JOB INTERVIEW QUESTIONS
7. The gate can be triggered using the static electricity of the
human hand to put the device into saturation briefly, enough
to trigger another circuit, such as a one-shot multivibrator.
10. It has low input capacitance that allows it to amplify higher frequencies (VHF and UHF) than are possible with a CS amplifier.
11. Although they do not have as much voltage gain as bipolar
transistors, they have a high input impedance and very low
noise. This is preferred in applications in which the incoming
signal may be a few microvolts to be followed by an amplification of a million or more.
PROBLEMS
11-1.
Given:
IG = 1 nA
Reverse voltage = –15 V
Solution:
Rin = Reverse voltage/IG
Rin = 15 V/1 nA
Rin = 15 GΩ
Answer: The input resistance is 15 GΩ.
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11-2.
Given:
IG = 1 μA
Reverse voltage = –20 V
Ambient Temperature = 100°C
Solution:
VGS/VGS(off) = 1/2
VGS = 1/2(VGS(off))
VGS = 1/2(–6 V)
VGS = –3 V
Solution:
Rin = Reverse voltage/IG
Rin = 20 V/1 nA
Rin = 20 MΩ
ID /IDSS = 1/4
ID = 1/4 16 mA
ID = 4 mA
Answer: The gate voltage at the 1/2 cutoff point is –3 V,
and the drain current is 4 mA.
Answer: The input resistance of the gate is 20 MΩ at
100°C.
11-3.
Given:
IDSS = 20 mA
VP = 4 V
11-7.
Solution:
VGS/VGS(off) = 1/2
VGS = 1/2(VGS(off))
VGS = 1/2(–4 V)
VGS = –2 V
Solution:
IDSS = Maximum drain current = 20 mA
VGS(off) = –VP (Eq. 11-2)
VGS(off) = –4 V
RDS = VP /IDSS (Eq. 11-1)
RDS = 4 V/20 mA
RDS = 200 Ω
Answer: The maximum drain current is 20 mA, the
gate-source cutoff voltage is –4 V, and the value of RDS
is 200 Ω.
11-4.
Given:
IDSS = 16 mA
VGS(off) = –2 V
ID/IDSS = 1/4
ID = 1/4 10 mA
ID = 2.5 mA
Answer: The gate voltage at the 1/2 cutoff point is –2 V,
and the drain current is 2.5 mA.
11-8.
Solution:
VGS(off) = –VP (Eq. 11-2)
VP = 2 V
ID = IDSS[1 – (VGS(2)/VGS(off))]2 (Eq. 11-3)
ID = 14 mA[1 – (–3 V/–4V)]2
ID = 0.88 mA
Answer: The pinch-off voltage is 2 V, and the value of
RDS is 125 Ω.
Given:
IDSS(min) = 1 mA
IDSS(max) = 5 mA
VGS(off)min = –0.5 V
VGS(off)max = –6 V
Solution:
VGS(off) = –VP (Eq. 11-2)
VP(min) = 0.5 V
VP(max) = 6 V
RDS(min) = VP(min)/IDSS(min) (Eq. 11-1)
RDS(min) = 0.5 V/1 mA
RDS(min) = 500 Ω
RDS(max) = VP(max)/IDSS(max) (Eq. 11-1)
RDS(max) = 6 V/5 mA
RDS(max) = 1.1 kΩ
11-6.
Given:
IDSS = 14 mA
VGS(off) = –4 V
VGS(1) = –1 V
VGS(2) = –3 V
Solution:
ID = IDSS[1 – (VGS(1)/VGS(off))]2 (Eq. 11-3)
ID = 14 mA[1 – (–1 V/–4V)]2
ID = 7.88 mA
RDS = VP /IDSS (Eq. 11-1)
RDS = 2 V/16 mA
RDS = 125 Ω
11-5.
Given:
IDSS = 10 mA
VGS(off) = –4 V
Answer: The drain current is 7.88 mA when the gate
voltage is –1 V, and 0.88 mA when the gate voltage
is –3 V.
11-9.
Given:
VDD = 15 V
RD = 10 kΩ
VGS(off) = –3 V
IDSS = 5 mA
Solution:
ID(sat) = VDD /RD
ID(sat) = 15 V/10 kΩ
ID(sat) = 1.5 mA
VGS(off) = –VP (Eq. 11-2)
VP = 3 V
Answer: The minimum value of RDS is 500 Ω, and the
maximum value is 1.1 kΩ.
RDS = VP /IDSS (Eq. 11-1)
RDS = 3 V/5 mA
RDS = 600 Ω
Given:
IDSS = 16 mA
VGS(off) = –6 V
VD = [RDS/(RDS + RD)]VDD
VD = [600 Ω/(600 Ω + 10 kΩ)]15 V
VD = 0.849 V
Answer: The drain saturation current is 1.5 mA, and the
drain voltage is 0.849 V.
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11-10. Given:
VDD = 15 V
RD = 20 kΩ
IDSS = 5 mA
VGS(off) = –3 V
Solution:
VG = [R2/(R1 + R2)]VDD
VG = [1 MΩ/(1.5 MΩ + 1 MΩ)] 25 V
VG = 10 V
ID = VG/RS (Eq. 11-10)
ID = 10 V/22 kΩ
ID = 0.455 mA
Solution:
VGS(off) = –VP (Eq. 11-2)
VP = 3 V
VD = VDD – IDRD (Eq. 11-4)
VD = 25 V – (0.455 mA)(10 kΩ)
VD = 20.45 V
RDS = VP /IDSS (Eq. 11-1)
RDS = 3 V/5 mA
RDS = 600 Ω
VD = [RDS/(RDS + RD)]VDD
VD = [600 Ω/(600 Ω + 20 kΩ)]15 V
VD = 0.437 V
Answer: The drain voltage is 0.437 V.
11-11. Given:
VDD = 20 V
RD = 20 kΩ
VGS(off) = –6 V
IDSS = 30 mA
Answer: The drain voltage is 20.45 V.
11-14. Given:
R1 = 1.5 MΩ
R2 = 1 MΩ
RS = 22 kΩ
RD = 10 kΩ
VDD = 25 V
VG = 10 V (from Prob. 11-13)
ID = 0.455 mA (from Prob. 11-13)
VD = 20.45 V (from Prob. 11-13)
Solution:
Solution:
VGS(off) = –VP (Eq. 11-2)
VP = 6 V
ID(sat) = VDD/(RD + RS)
ID(sat) = 25 V/(10 kΩ + 22 kΩ)
ID(sat) = 0.781 mA
RDS = VP/IDSS (Eq. 11-1)
RDS = 6 V/30 mA
RDS = 200 Ω
VS ≈ VG
VDSQ = VD – VS
VDSQ = 20.45 V – 10 V
VDSQ = 10.45 V
VD = [RDS/(RDS + RD)]VDD
VD = [200 Ω/(200 Ω + 20 kΩ)]20 V
VD = 0.198 V
Answer: The drain voltage is 0.198 V.
11-12. Given:
VDD = 20 V
RD = 10 kΩ
VGS(off) = –6 V
IDSS = 30 mA
Solution:
ID(sat) = VDD/RD
ID(sat) = 20 V/10 kΩ
ID(sat) = 2 mA
VGS(off) = VP (Eq. 11-2)
VP = 6 V
RDS = VP/IDSS (Eq. 11-1)
RDS = 6 V/30 mA
RDS = 200 Ω
VD = [RDS/(RDS + RD)]VDD
VD = [200 Ω/(200 Ω + 10 kΩ)]20 V
VD = 0.392 V
Answer: The drain saturation current is 2 mA, and the
drain voltage is 0.392 V.
11-13. Given:
R1 = 1.5 MΩ
R2 = 1 MΩ
RS = 22 kΩ
RD = 10 kΩ
VDD = 25 V
ID (mA)
0.781
0.455
Q
10.45 VDS (V)
25
DC load line and Q point for Prob. 11–14.
11-15. Given:
VDD = 25 V
VSS = –25 V
RD = 7.5 kΩ
RS = 18 kΩ
Solution:
ID = VSS/RS (Eq. 11-12)
ID = –25 V/18 kΩ
ID = 1.39 mA
VD = VDD – IDRD (Eq. 11-4)
VD = 25 V – (1.39 mA)(7.5 kΩ)
VD = 14.58 V
Answer: The drain voltage is 14.58 V.
11-16. Given:
VDD = 25 V
VSS = –25 V
RD = 7.5 kΩ
RS = 30 kΩ
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Solution:
ID = VSS/RS (Eq. 11-12)
ID = –25 V/30 kΩ
ID = 0.833 mA
Solution:
ID = VS/RS
ID = 1.5 V/1 kΩ
ID = 1.5 mA
VD = VDD – IDRD (Eq.11-4)
VD = 25 V – (0.833 mA)(7.5 kΩ)
VD = 18.75 V
VD = VDD – IDRD
VD = 25 V – (1.5 mA)(8.2 kΩ)
VD = 12.7 V
Answer: The drain voltage is 18.75 V.
Answer: The drain voltage is 12.7 V.
11-17. Given:
VDD = 15 V
VEE = –9 V
RD = 7.5 kΩ
RE = 8.2 kΩ
VBE = 0.7 V
Solution:
ID = (VEE – VBE)/RE (Eq.11-13)
ID = (9 V – 0.7 V)/8.2 kΩ
ID = 1.01 mA
VD = VDD – IDRD (Eq.11-4)
VD = 15 V – (1.01 mA)(7.5 kΩ)
VD = 7.43 V
Answer: The drain voltage is 7.43 V, and the drain current is 1.01 mA.
11-18. Given:
VDD = 15 V
VEE = –9 V
RD = 4.7 kΩ
RE = 8.2 kΩ
VBE = 0.7 V
Solution:
ID = (VEE – VBE)/RE (Eq.11-13)
ID = (9 V – 0.7 V)/8.2 kΩ
ID = 1.01 mA
VD = VDD – IDRD (Eq.11-4)
VD = 15 V – (1.01 mA)(4.7 kΩ)
VD = 10.25 V
Answer: The drain voltage is 10.25 V, and the drain current is 1.01 mA.
11-19. Given:
VDD = 25 V
RD = 8.2 kΩ
RS = 1 kΩ
ID = 1.5 mA
Solution:
VGS = –IDRS (Eq.11-7)
VGS = –(1.5 mA)(1 kΩ)
VGS = –1.5 V
VD = VDD – IDRD – IDRS
VD = 25 V – (1.5 mA)(8.2 kΩ) – (1.5 mA)(1 kΩ)
VD = 11.2 V
Answer: The gate-source voltage is –1.5 V, and the drainsource voltage is 11.2 V.
11-20. Given:
VDD = 25 V
RD = 8.2 kΩ
RS = 1 kΩ
VS = 1.5 V
11-21. Given:
VDD = 25 V
RD = 10 kΩ
RS = 22 kΩ
R1 = 1.5 MΩ
R2 = 1 MΩ
Answer: The gate-source voltage is –2.5 V, and the drain
current is 0.55 mA from the transconductance curve.
11-22. Given:
VDD = 15 V
RG = 2.2 MΩ
RE = 8.2 kΩ
VEE = –9 V
Answer: The gate-source voltage is –2.0 V, and the drain
voltage is 7.5 V from the transconductance curve.
11-23. Given:
VDD = 25 V
RG = 1.5 MΩ
RS = 1 kΩ
Answer: The gate-source voltage is –1.5 V, and the drain
current is 1.5 mA.
11-24. Given:
VDD = 25 V
RG = 1.5 MΩ
RS = 2 kΩ
Answer: The gate-source voltage is –2.0 V, and the drain
current is 1 mA and the drain-source voltage is 14.8 V.
11-25. Given:
gm0 = 4000 μs
IDSS = 10 mA
Solution:
VGS(off) = –2IDSS/gm0 (Eq.11-15)
VGS(off) = –2(10 mA)/4000 μs
VGS(off) = –5 V
gm = gm0 [1 – (VGS/VGS(off))] (Eq.11-16)
gm = 4000 μs[1 – (–1 V/–5V)]
gm = 3200 μs
Answer: The gate-source cutoff voltage is –5 V, and the
gm0 for VGS = –1 V is 3200 μs.
11-26. Given:
gm0 = 1500 μs
IDSS = 2.5 mA
VGS = –1 V
Solution:
VGS(off) = –2IDSS/gm0 (Eq.11-15)
VGS(off) = –2(2.5 mA)/1500 μs
VGS(off) = –3.33 V
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gm = gm0 [1 – (VGS/VGS(off))] (Eq.11-16)
gm = 1500 μs[1 – (–1 V/–3.3 V)]
gm = 1045 μs
Answer: The gm for VGS = –1 V is 1045 μs.
11-27. Given:
gm0 = 6000 μs
IDSS = 12 mA
VGS = –2 V
Solution:
VGS(off) = –2IDSS/gm0 (Eq.11-15)
VGS(off) = –2(12 mA)/6000 μs
VGS(off) = –4 V
Since the ratio of VGS to VGS(off) is one-half, the following
equation can be used:
ID/IDSS = 1/4
ID = 1/4(IDSS)
ID = 1/4(12 mA)
ID = 3 mA
gm = gm0 [1 – (VGS /VGS(off))] (Eq.11-16)
gm = 6000 μs[1 – (–2 V/–4 V)]
gm = 3000 μs
Answer: The drain current is 3 mA, and the transconductance is 3000 μs.
11-28. Given:
VDD = 30 V
R1 = 20 MΩ
R2 = 10 MΩ
RD = 1 kΩ
RS = 2 kΩ
RL = 10 kΩ
vg = 2 mV
gm = 3000 μs
Solution:
rd = RD || RL
rd = 1 kΩ || 10 kΩ
rd = 909 Ω
zin = R1 || R2
zin = 20 MΩ || 10 MΩ
zin = 6.67 MΩ
Av = gmrd (Eq.11-17)
Av = (3000 μs)(909 Ω)
Av = 2.73
vout = Av(vin)
vout = (2.73)(2 mV)
vout = 5.46 mV
Answer: zin = 6.67 MΩ and the output voltage is 5.46 mV.
11-29. Given:
VDD = 30 V
R1 = 20 MΩ
R2 = 10 MΩ
RD = 1 kΩ
RS = 2 kΩ
RL = 10 kΩ
vg = 2 mV
IDSS = 12 mA (from the graph)
VGS(off) = –4 V (from the graph)
Solution:
rd = RD || RL
rd = 1 kΩ || 10 kΩ
rd = 909 Ω
VGS(off) = –2IDSS/gm0 (Eq.11-15)
gm0 = –2IDSS/VGS(off)
gm0 = –2(12 mA)/–4 V
gm0 = 6000 μs
VG = [R2/(R1 + R2)]VDD
VG = [10 MΩ/(20 MΩ + 10 MΩ)] 30 V
VG = 10 V
ID = VG/RS (Eq.11-10)
ID = 10 V/2 kΩ
ID = 5 mA
From the graph, VGS is approximately –1.4 V when ID is
5 mA.
With Eq.11-16, gm0 = 3900 μs. Then:
Av = gmrd (Eq.11-17)
Av = (3900 μs)(909 Ω)
Av = 3.54
vout = Av(vin)
vout = 3.54(2 mV)
vout = 7.09 mV
Answer: The output voltage is 7.09 mV.
11-30. Given:
VDD = 30 V
R1 = 20 MΩ
R2 = 10 MΩ
RS = 3.3 kΩ
RL = 1 kΩ
vin = 5 mV
gm = 2000 μs
Solution:
rS = RS || RL
rS = 3.3 kΩ || 1 kΩ
rS = 767 Ω
Av = (gmrs)/(1 + gmrs) (Eq.11-21)
Av = (2000 μs)(767 Ω)/[1 + (2000 μs)(767 Ω)]
Av = 0.605
vout = Av(vin)
vout = (0.605)(5 mV)
vout = 3.03 mV
zout = RS || 1/gm
zout = 3.3 kΩ || 1/2000 μs
zout = 434 Ω
Answer: The output voltage is 3.03 mV and zout is 434 Ω.
11-31. Given:
VDD = 30 V
R1 = 20 MΩ
R2 = 10 MΩ
RS = 3.3 kΩ
RL = 1 kΩ
vin = 5 mV
IDSS = 6 mA (from the graph)
VGS(off) = –4 V (from the graph)
rS = 767 Ω (from Prob.11-30)
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Solution:
VGS(off) = –2IDSS/gm0 (Eq.11-15)
gm0 = –2IDSS/VGS(off)
gm0 = –2(6 mA)/ –4 V
gm0 = 3000 μs
VG = [R2/(R1 + R2)]VDD
VG = [10 MΩ/(20 MΩ + 10 MΩ)] 30 V
VG = 10 V
ID = VG/RS (Eq.11-7)
ID = 10 V/3.3 kΩ
ID ≅ 3 mA
From the graph, VGS is roughly –1.25 V when ID = 3 mA.
With Eq. (11-16), gm = 2060 μs.
With Eq. (11-21):
gmrS = (2060 μs)(767 Ω) = 1.58
Av = 1.58/(1 + 1.58) = 0.612
vout = Av(vin)
vout = (0.612)(5 mV)
vout = 3.06 mV
Answer: The output voltage is 3.06 mV.
11-32. Given:
RD = 22 kΩ
vin = 50 mVp-p
IDSS = 10 mA
VP = 2 V
Solution:
RDS = VP/IDSS (Eq.11-1)
RDS = 2 V/10 mA
RDS = 200 Ω
With VGS at –10 V the JFET is cut off and appears as an
open; thus vout = vin = 50 mVp-p.
With VGS at 0 V, the JFET is conducting and a voltage
divider is created with RD.
vout = [RDS/(RDS + Rin)]vin
vout = [200 Ω/(200 Ω + 22 kΩ)]50 mVp-p
vout = 0.45 mVp-p
On-off ratio = vout(max)/vout(min) (Eq.11-23)
On-off ratio = 50 mVp-p/0.45 mVp-p
On-off ratio = 111
Answer: The output voltage at a VGS of –10 V is 50 mVp-p,
the output voltage at a VGS of 0 V is 0.45 mVp-p, and the
on-off ratio is 111.
11-33. Given:
RD = 33 kΩ
vin = 25 mVp-p
IDSS = 5 mA
VP = 3 V
Solution:
RDS = VP/IDSS (Eq.11-1)
RDS = 3 V/5 mA
RDS = 600 Ω
With VGS at –10 V the JFET is cut off and appears as an
open; thus vin = 0 mVp-p.
With VGS at 0 V, the JFET is conducting and a voltage
divider is created with the output resistance.
vout = [RD/(RDS + Rout)]vin
vout = [33 kΩ/(600 Ω + 33 kΩ)]25 mVp-p
vout = 24.55 mVp-p
On-off ratio = vout(max)/vout(min) (Eq.11-23)
On-off ratio = 24.55 mVp-p/0 mVp-p
On-off ratio = ∞
Answer: The output voltage at a VGS of –10 V is 0 mVp-p,
the output voltage at a VGS of 0 V is 24.55 mVp-p, and the
on-off ratio is ∞.
CRITICAL THINKING
11-34. Answer:
IDSS = 20 mA
VDS(max) = 5 V for the ohmic region
VDS = 5 to 30 V in the active range
11-35. Given:
VGS(off) = –8 V (from the graph)
IDSS = 32 mA (from the graph)
VGS(1) = –4 V
VGS(2) = –2 V
Solution:
VGS(off) = –2IDSS/gm0 (Eq.11-15)
gm0 = –2IDSS/VGS(off)
gm0 = –2(32 mA)/–8 V
gm0 = 8000 μs
gm = gm0 [1 – (VGS /VGS(off))] (Eq.11-16)
gm = 8000 μs[1 – (VGS/–8 V)]
ID = IDSS[1 – (VGS(1)/VGS(off))]2 (Eq.11-3)
ID = 32 mA[1 – (–4 V/–8 V)]2
ID = 8 mA
ID = IDSS[1 – (VGS(2)/VGS(off))]2 (Eq.11-3)
ID = 32 mA[1 – (–2 V/–8 V)]2
ID = 18 mA
Answer: The transconductance equation is gm = 8000 μs
[1 – (VGS/–8 V)], the drain current at –4 V is 8 mA, and
the drain current at –2 V is 18 mA.
11-36. Given:
VGS(off) = –5 V (from the graph)
IDSS = 12 mA (from the graph)
VGS = –1 V
Solution:
ID = IDSS[1 – (VGS/VGS(off))]2 (Eq.11-3)
ID = 12 mA[1 – (–1 V/–5 V)]2
ID = 7.68 mA
Answer: The drain current is 7.68 mA.
11-37. Given:
VDD = 15 V
VEE = –10 V
RD = 3.3 kΩ
RE = 4.7 kΩ
VBE = 0.7 V
gm = 2000 μs
vg = vin = 3 mV
Solution:
ID = (VEE – VBE)/RE (Eq.11-13)
ID = (10 V – 0.7 V)/4.7 kΩ
ID = 2 mA
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VD = VDD – IDRD (Eq.11-4)
VD = 15 V – (2 mA)(3.3 kΩ)
VD = 8.4 V
rd = RD || RL
rd = 3.3 kΩ || 15 kΩ
rd = 2.7 kΩ
Av = gmrd (Eq.11-17)
Av = (2000 μs)(2.7 kΩ)
Av = 5.4
vout = Av(vin)
vout = (5.4)(3 mV)
vout = 16.2 mV
Answer: The drain voltage is 8.4 V, and the output
voltage is 16.2 mV.
11-38. Answer:
a. Multiply 4 mA and 510 Ω to get 2.04 V.
b. It must equal 2.04 V.
c. Because of the linearity of the circuit, the meter reads
half of maximum, or 0.5 mA.
11-39. Given:
IDSS = 16 mA
RDS = 200 Ω
RL = 10 kΩ
VDD = 30 V
vd = 200 mV
vout = 200 mV
Answer: Open R1.
11-42. Given:
Normal Operation
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
vd = 357 mV
vout = 357 mV
With Trouble
VGS = –0.6 V
ID = 7.58 mA
VDS = 1.25 V
vg = 100 mV
vS = 0 V
vd = 29 mV
vout = 29 mV
Answer: Open R2.
11-43. Given:
Normal Operation
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
vd = 357 mV
vout = 357 mV
With Trouble
VGS = –0.56 V
ID = 0 mA
VDS = 0 V
vg = 100 mV
vS = 0 V
vd = 0 mV
vout = 0 mV
Answer: Open RD.
Solution: Since VGS is 0 V, it is operating in the active
region. The JFET appears to be a current source, but
since the load is so large, the power supply cannot supply
enough voltage to produce that current and it drops into
the ohmic region and the JFET acts like resistor.
I = VDD/(RDS + RL)
I = 30 V/(200 Ω + 10 kΩ)
I = 2.94 mA
VDS = IRDS
VDS = (2.94 mA)(200 Ω)
VDS = 0.59 V
If the load is shorted, RL = 0 Ω and the JFET operates in
the active region.
I = IDSS
I = 16 mA
VDS = VDD
VDS = 30 V
Answer: During normal operation, the current is 2.94 mA
and the voltage across the JFET is 0.59 V. With the load
shorted, the current is 16 mA and the voltage is 30 V.
11-40. Answer:
a. The gm0 is 6000 μs. Multiply this by 1 kΩ to get a
voltage gain of 6.
b. At –1 V, the gm is 4500 μs and the voltage gain is 4.5.
c. 3
d. 1.5
e. 0.75
11-41. Given:
Normal Operation
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
vd = 357 mV
vout = 357 mV
11-44. Given:
Normal Operation
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
vd = 357 mV
vout = 357 mV
With Trouble
VGS = –8 V
ID = 0 mA
VDS = 8 V
vg = 100 mV
vS = 0 V
vd = 0 mV
vout = 0 mV
Answer: Open RS.
11-45. Given:
Normal Operation
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
vd = 357 mV
vout = 357 mV
With Trouble
VGS = +8 V
ID = 0 mA
VDS = 24 V
vg = 100 mV
vS = 0 V
vd = 0 mV
vout = 0 mV
Answer: Open G-S.
11-46. Given:
Normal Operation
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
vd = 357 mV
vout = 357 mV
With Trouble
VGS = –1.61 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 87 V
vd = 40 mV
vout = 40 mV
Answer: Open Bypass Capacitor.
With Trouble
VGS = –2.75 V
ID = 1.38 mA
VDS = 19.9 V
vg = 100 mV
vS = 0 V
11-47. Given:
Normal Operation
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
With Trouble
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
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vd = 357 mV
vout = 357 mV
vd = 397 mV
vout = 0 mV
Solution:
ID = IDSS(1 – (VGS/VGS(off)))2
ID = 4 mA(1 – (–0.5 V/–2.0 V))2
ID = 2.25 mA
Answer: Open Drain Coupling Capacitor.
11-48. Given:
Normal Operation
VGS = –1.6 V
ID = 4.8 mA
VDS = 9.6 V
vg = 100 mV
vS = 0 V
vd = 357 mV
vout = 357 mV
ID = IDSS(1 – (VGS/VGS(off)))2
ID = 4 mA(1 – (–1 V/–2.0 V))2
ID = 1 mA
With Trouble
VGS = 0 V
ID = 7.58 mA
VDS = 1.5 V
vg = 1 mV
vS = 0 V
vd = 0 V
vout = 0 V
ID = IDSS(1 – (VGS/VGS(off)))2
ID = 4 mA(1 – (–1.5 V/–2.0 V))2
ID = 0.25 mA
Answer:
VGS = –0.5 V, ID = 2.25 mA
VGS = –1 V, ID = 1 mA
VGS = –1.5 V, ID = 0.25 mA
Answer: Open Drain Coupling Capacitor.
11-49. R2 is shorted
12-2.
11-50. C2 is open
11-51. C3 is shorted
11-52. vg is 1 mVp-p
11-53. Q1 is shorted D-S
Chapter 12
MOSFETs
Solution:
ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 12-1)
ID = 4 mA(1 – (+0.5 V/–2.0 V))2
ID = 6.25 mA
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
c
d
d
c
c
d
d
8.
9.
10.
11.
12.
13.
14.
c
b
d
a
b
d
c
15.
16.
17.
18.
19.
20.
21.
a
b
d
d
c
d
a
22.
23.
24.
25.
26.
b
d
d
c
d
ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 12-1)
ID = 4 mA(1 – (+1 V/–2.0 V))2
ID = 9 mA
ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 12-1)
ID = 4 mA(1 – (+1.5 V/–2.0 V))2
ID = 12.25 mA
JOB INTERVIEW QUESTIONS
5. MOS technology, especially CMOS, consumes low power
and requires a small space. The result is complex circuits that
are lightweight, will last a long time on batteries, and are
suitable for solar power.
9. Because the thin insulating layer within the device is easily
destroyed by static electricity.
10. Ship or store the devices in antistatic foam material or wire
wrap around leads. Also, technicians should be grounded by
using an antistatic wrist strap, touching the chassis, or standing on antistatic (grounding) mats. Finally, use grounded soldering irons and test equipment.
11. MOSFETs have faster switching times, resulting in less time
spent in the active region, which produces higher efficiency
and reduced heat-sink requirements. MOSFETs are also
immune to thermal runaway and are easily connected in parallel for greater power dissipation.
PROBLEMS
12-1.
Given:
VGS = –0.5 V
VGS = –1.0 V
VGS = –1.5 V
VGS = + 0.5 V
VGS = +1.0 V
VGS = +1.5 V
VGS(off) = –2 V
IDSS = 4 mA
Given:
VGS = –0.5 V
VGS = –1.0 V
VGS = –1.5 V
VGS = +0.5 V
VGS = +1.0 V
VGS = +1.5 V
VGS(off) = –2 V
IDSS = 4 mA
Answer:
VGS = 0.5 V, ID = 6.25 mA
VGS = 1 V, ID = 9 mA
VGS = 1.5 V, ID = 12.25 mA
12-3.
Given:
VGS = –1.0 V
VGS = –2.0 V
VGS = 0 V
VGS = +1.5 V
VGS = +2.5 V
VGS(off) = +3 V
IDSS = 12 mA
Solution:
ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 12-1)
ID = 4 mA(1 – (+1.5 V/+3.0V))2
ID = 3 mA
ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 12-1)
ID = 4 mA(1 – (2.5 V/+3.0 V))2
ID = 0.333 mA
Answer:
VGS = 1.5 V, ID = 3 mA
VGS = 2.5 V, ID = 333 μA
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12-4.
Given:
VGS(off) = –3 V
IDSS = 12 mA
Solution:
VDS = VDD – (IDSSRD) (Eq. 12-2)
VDS = 12 V – ((12 mA)(470 Ω))
VDS = 6.36 V
ID = IDSS = 12 mA
Answer: The drain current is 12 mA and the drainsource
voltage is 6.36 V.
12-5.
Given:
gm0 = 4000 μs
RD = 470 Ω
RL = 2 kΩ
vin = 100 mV
Solution:
rd = RD || RL
rd = 470 Ω || 2 kΩ
rd = 381 Ω
Av = gmrd
Av = (4000 μs)(381 Ω)
Av = 1.52
vout = vinAv
vout = (100 mV)(1.52)
vout = 152 mV
Answer: The voltage gain is 1.52, the voltage out is
152 mV, and rd is 381 Ω.
12-6.
Given:
gm0 = 4000 μs
RD = 680 Ω
RL = 10 kΩ
vin = 100 mV
Solution:
rd = RD || RL
rd = 680 Ω || 10 kΩ
rd = 637 Ω
Av = gmrd
Av = (4000 μs)(637 Ω)
Av = 2.55
vout = vinAv
vout = (100 mV)(2.55)
vout = 255 mV
Answer: The voltage gain is 2.55, the voltage out is
255 mV, and rd is 637 Ω.
12-7.
Given:
gm0 = 4000 μs
RD = 680 Ω
RL = 2 kΩ
Vin = 100 mV
RG = 1 MΩ
Solution:
zin ≈ RG ≈ 1MΩ
Answer: The input impedance is approximately 1 MΩ.
12-8a. Given:
VDS(on) = 0.1 V
ID(on) = 10 mA
Solution:
RDS(on) = VDS(on)/ID(on) (Eq. 12-1)
RDS(on) = 0.1 V/10 mA
RDS(on) = 10 Ω
Answer: The drain-source resistance is 10 Ω.
12-8b. Given:
VDS(on) = 0.25 V
ID(on) = 45 mA
Solution:
RDS(on) = VDS(on)/ID(on) (Eq. 12-1)
RDS(on) = 0.25 V/45 mA
RDS(on) = 5.56 Ω
Answer: The drain-source resistance is 5.56 Ω.
12-8c. Given:
VDS(on) = 0.75 V
ID(on) = 100 mA
Solution:
RDS(on) = VDS(on)/ID(on) (Eq. 12-1)
RDS(on) = 0.75 V/100 mA
RDS(on) = 7.5 Ω
Answer: The drain-source resistance is 7.5 Ω.
12-8d. Given:
VDS(on) = 0.15 V
ID(on) = 200 mA
Solution:
RDS(on) = VDS(on)/ID(on) (Eq. 12-1)
RDS(on) = 0.15 V/200 mA
RDS(on) = 0.75 Ω
Answer: The drain-source resistance is 0.75 Ω.
12-9a. Given:
VGS(on) = 3 V
ID(on) = 500 mA
RDS(on) = 2 Ω
ID(sat) = 25 mA
Solution:
VDS = ID(sat)RDS(on)
VDS = 25 mA (2 Ω)
VDS = 0.05 V
Answer: The voltage across the E-MOSFET is 0.05 V.
12-9b. Given:
VGS(on) = 3 V
ID(on) = 500 mA
RDS(on) = 2 Ω
ID(sat) = 50 mA
Solution:
VDS = ID(sat)RDS(on)
VDS = 50 mA (2 Ω)
VDS = 0.1 V
Answer: The voltage across the E-MOSFET is 0.1 V.
12-9c. Given:
VGS(on) = 3 V
ID(on) = 500 mA
RDS(on) = 2 Ω
ID(sat) = 100 mA
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Solution:
VDS = ID(sat)RDS(on)
VDS = 100 mA (2 Ω)
VDS = 0.2 V
Answer: The voltage across the E-MOSFET is 0.2 V.
12-9d. Given:
VGS(on) = 3 V
ID(on) = 500 mA
RDS(on) = 2 Ω
ID(sat) = 200 mA
Solution:
VDS = ID(sat)RDS(on)
VDS = 200 mA (2 Ω)
VDS = 0.4 V
Answer: The voltage across the E-MOSFET is 0.4 V.
12-10. Given:
VGS(on) = 2.5 V (from Table 12-1)
ID(on) = 100 mA (from Table 12-1)
RDS(on) = 10 Ω (from Table 12-1)
VDD = 20 V
RD = 390 Ω
Solution:
VD = [RDS(on)/(RDS(on) + RD)]VDD
VD = [10 Ω/(10 Ω + 390 Ω)]20 V
VD = 0.5 V
Answer: The voltage across the E-MOSFET is 0.5 V.
12-11. Given:
VGS(on) = 2.6 V (from Table 12-1)
ID(on) = 20 mA (from Table 12-1)
RDS(on) = 28 Ω (from Table 12-1)
VDD = 15 V
RD = 1.8 kΩ
Solution:
VD = [RDS(on)/(RDS(on) + RD)]VDD
VD = [28 Ω/(28 Ω + 1.8 kΩ)]15 V
VD = 0.23 V
Answer: The drain voltage is 0.23 V.
12-12. Given:
VGS(on) = 5 V (from Table 12-1)
ID(on) = 200 mA (from Table 12-1)
RDS(on) = 7.5 Ω (from Table 12-1)
VDD = 25 V
RD = 150 Ω
Solution:
VD = [RD/(RDS(on) + RD)]VDD
VD = [150 Ω/(7.5 Ω + 150 Ω)] 25 V
VD = 23.8 V
Answer: The drain voltage is 23.8 V.
12-13. Given:
VGS(on) = 10 V (from Table 12-1)
ID(on) = 1 A (from Table 12-1)
RDS(on) = 0.9 Ω (from Table 12-1)
VDD = 12 V
RD = 18 Ω
Solution:
VD = [RDS(on)/(RDS(on) + RD)]VDD
VD = [0.9 Ω/(0.9 Ω + 18 Ω)]12 V
VD = 0.57 V
Answer: The drain voltage is 0.57 V.
12-14. Given:
VGS(on) = 5 V (from Table 12-1)
ID(on) = 200 mA (from Table 12-1)
RDS(on) = 7.5 Ω (from Table 12-1)
VDD = 30 V
RD = 1 kΩ
VLED = 2 V
Solution:
ID = (VDD – VLED)/(RDS(on) + RD)
ID = (30 V – 2 V)/(7.5 Ω + 1 kΩ)
ID = 27.8 mA
Answer: The LED current is 27.8 mA.
12-15. Given:
VGS(on) = 2.6 V (from Table 12-1)
ID(on) = 20 mA (from Table 12-1)
RDS(on) = 28 Ω (from Table 12-1)
VDD = 20 V
RD = 1 kΩ
Solution:
ID = (VDD)/(RDS(on) + RD)
ID = (20 V)/(28 Ω + 1 kΩ)
ID = 19.5 mA
IL = VDD/RL
IL = 20 V/2 Ω
IL = 10 A
Answer: The MOSFET current is 19.5 mA. The load
current is 10 A.
12-16. Given:
ID(active) = 1 mA
VDS(active) = 10 V
Solution:
RD = VDS(active) /ID(active) (Eq. 12-6)
RD = 10 V/1 mA
RD = 10 kΩ
Answer: The drain resistance is 10 kΩ.
12-17. Given:
RDS(on) = 300 Ω
VDD = 12 V
RD = 8 kΩ
Solution: When the input is low, the lower MOSFET is
open and the output voltage is pulled up to the supply
voltage. When the input is high, the lower MOSFET has
a resistance of 300 Ω.
vout = [RDS(on)/(RDS(on) + RD)]VDD
vout = [300 Ω/(300 Ω + 8 kΩ)]12 V
vout = 0.43 V
Answer: When the input voltage is low, the output voltage is 12 V; when the input voltage is high, the output
voltage is 0.43 V.
12-18. Given:
RDS(on) = 150 Ω
VDD = 18 V
RD = 2 kΩ
Solution: When the input is low, the lower MOSFET is
open and the output voltage is pulled up to the supply
voltage. When the input is high, the lower MOSFET has
a resistance of 150 Ω.
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vout = [RDS(on)/(RDS(on) + RD)]VDD
vout = [150 Ω/(150 Ω + 2 kΩ)]18 V
vout = 1.26 V
Answer: When the input voltage is low, the output voltage is 18 V; when the input voltage is high, the output
voltage is 1.26 V.
12-19. Answer: The output waveform is a square wave with an
upper peak of +12 V and a lower peak of 0.43 V.
12-20. Answer: Inverted square wave from +12 V to 0 V.
12-21. Answer: The on MOSFET has an RDS(on) of 10 V divided
by 1 mA, which equals 10 kΩ. The off MOSFET has an
RDS(off) of 10 V divided by 1 μA, which equals 10 MΩ.
When the input voltage is high, the lower MOSFET is on,
and the output voltage is given by:
vout = _________
​  10 kΩ  ​ 12 V ≅ 0.012V
10.01 MΩ
When the input voltage is low, the lower MOSFET is off,
and the output voltage is given by:
vout = _________
​  10 MΩ  ​ 12 V ≅ 12V
10.01 MΩ
12-22. Given:
12-V peak square-wave input
f = 1 kHz
Assume the same values from the previous problem.
Answer: The signal will be 180° out of phase and have
a maximum value of 12 V and a minimum value of 0 V.
12-23. Given:
VDD = 12 V
RDS(on) = 5 kΩ
Solution:
ID = VDD/2(RDS(on))
ID = 12 V/2(5 kΩ)
ID = 1.2 mA
Answer: The current is 1.2 mA.
12-24. Given:
VGS(on) = 10 V (from Table 12-2)
ID(on) = 2 A (from Table 12-2)
RDS(on) = 1.95 Ω (from Table 12-2)
VDD = 12 V
RD = 10 Ω
Solution: When the input is low, the MOSFET is open
and no current flows. When the input is high, the MOSFET has a resistance of RDS(on) = 1.95 Ω.
ID = (VDD)/(RDS(on) + RD)
ID = (12 V)/(1.95 Ω + 10 Ω)
ID = 1 A
Answer: The current is 0 A when the input is low, and
1 A when the input is high.
12-25. Given:
VGS(on) = 10 V (from Table 12-2)
ID(on) = 2 A (from Table 12-2)
RDS(on) = 1.95 Ω (from Table 12-2)
VDD = 12 V
RD = 6 Ω
Solution: When the input is high, the MOSFET has a
resistance of RDS(on) = 1.95 Ω.
ID = (VDD)/(RDS(on) + RD)
ID = (12 V)/(1.95 Ω + 6 Ω)
ID = 1.51 A
Answer: The current is 1.51 A when the input is high.
12-26. Given:
VGS(on) = 10 V (from Table 12-2)
ID(on) = 5 A (from Table 12-2)
RDS(on) = 1.07 Ω (from Table 12-2)
VDD = 15 V
RD = 3 Ω
Solution: When the input is low, the MOSFET is open
and no current flows. When the input is high, the MOSFET has a resistance of RDS(on) = 1.07 Ω.
ID = (VDD)/(RDS(on) + RD)
ID = (15 V)/(1.07 Ω + 3 Ω)
ID = 3.69 A
Answer: The current is 0 A when the input is low, and
3.69 A when the input is high.
12-27. Given:
VGS(on) = 10 V (from Table 12-2)
ID(on) = 5 A (from Table 12-2)
RDS(on) = 1.07 Ω (from Table 12-2)
VDD = 15 V
RD = 5 Ω
Solution: When it is dark, the photodiode acts like an
open and the gate voltage is 10 V. When the input is 10 V,
the MOSFET has a resistance of RDS(on) = 1.07 Ω.
ID = (VDD)/(RDS(on) + RD)
ID = (15 V)/(1.07 Ω + 5 Ω)
ID = 2.47 A
P = (2.47 A)2 (5 Ω) = 30.5 W
Answer: The power is 30.5 W when it is dark.
12.28. Given:
VGS(on) = 10 V (from Table 12-2)
ID(on) = 2 A (from Table 12-2)
RDS(on) = 1.95 Ω (from Table 12-2)
VDD = 24 V
RD = 12 Ω
Solution: When the input is low, the MOSFET is open
and no current flows. When the input is high, the MOSFET has a resistance of RDS(on) = 1.95 Ω.
ID = (VDD)/(RDS(on) + RD)
ID = (24 V)/(1.95 Ω +12 Ω)
ID = 1.72 A
Answer: The current is 0 A when the input is low, and
1.72 A when the input is high.
12-29. Given:
VGS(on) = 10 V (from Table 12-2)
ID(on) = 2 A (from Table 12-2)
RDS(on) = 1.95 Ω (from Table 12-2)
VDD = 12 V
RD = 18 Ω
Solution: When the probes are underwater, their resistance is low and the gate voltage is also low. When the
input is low, the MOSFET is open and no current flows.
When the probes are out of the water, their resistance is
high and the gate voltage is also high. When the input is
high, the MOSFET has a resistance of RDS(on) = 1.95 Ω.
ID = (VDD)/(RDS(on) + RD)
ID = (12 V)/(1.95 Ω + 18 Ω)
ID = 0.6 A
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Answer: The current is 0 A when the probes are underwater, and 0.6 A when the probes are above the water.
12-30. Given:
VGS(on) = 10 V (from Table 12-2)
ID(on) = 5 A (from Table 12-2)
RDS(on) = 1.07 Ω (from Table 12-2)
VDD = 20 V
RD = 4 Ω
R1 = R2 = 1 MΩ
C = 20 μF
Solution:
τ = RC
τ = (1 MΩ || 1 MΩ)(20 μF)
τ = 10 s
At full brightness, the FET appears to have a resistance
of RDS(on) = 1.07 Ω.
ID = (VDD)/(RDS(on) + RD)
ID = (20 V)/(1.07 Ω + 4 Ω)
ID = 3.94 A
P = I2R
P = (3.94 A)2(4 Ω)
P = 62.1 W
Answer: The time constant is 10 s, and the lamp power
dissipation at full brightness is 62.1 W.
12-31. Given:
VGS(on) = 10 V (from Table 12-2)
ID(on) = 5 A (from Table 12-2)
RDS(on) = 1.07 Ω (from Table 12-2)
VDD = 20 V
RD = 6 Ω
R1 = R2 = 2 MΩ
C = 20 μF
Solution:
τ = RC
τ = (2 MΩ || 2 MΩ)(20 μF)
τ = 20 s
At full brightness, the FET appears to have a resistance
of RDS(on) = 1.07 Ω.
ID = (VDD)/(RDS(on) + RD)
ID = (20 V)/(1.07 Ω + 6 Ω)
ID = 2.83 A
Answer: The time constant is 20 s, and the lamp current
at full brightness is 2.83 A.
12-32. Given:
Vin = 15 V
Q1 RDS(on) = 1 Ω
R1 = 10 kΩ
EN signal = 0 V and +5 V
Solution: When the input EN is low (0 V), Q1 is off and
ID = 0 A. When the EN signal is high (+5 V), Q1 is on.
ID = Vin/(R1 + RDS(on))
ID = 15 V/(10 kΩ + 1 Ω)
ID = 1.5 mA
Answer: The current of Q1 is 0 A when the EN signal is
low and 1.5 mA when the enable signal is high.
12-33. Given:
Vin = 15 V
Q2 RDS(on) = 0.1 Ω
RL = 5 Ω
EN signal = +5 V
Solution: When the enable signal is high, the output voltage across the load is divided between RDS(on) and RL.
VL = [RL/(RL + RDS(on))]Vin
VL = [5 Ω/(5 Ω + 0.1 Ω)]15 V
VL = 14.7 V
Answer: When the enable signal is high, the output voltage is 14.7 V.
12-34. Given:
Vin = 15 V
Q2 RDS(on) = 0.1 Ω
RL = 5 Ω
EN signal = +5 V
Solution:
IL = Vin/(RDS(on) + RL)
IL = 15 V/(0.1 Ω + 5 Ω)
IL = 2.94 A
Q2 PLOSS = (IL)2(RDS(on))
Q2 PLOSS = (2.94 A)2(0.1 Ω)
Q2 PLOSS = 865 mW
PL = (IL)2(RL)
PL = (2.94 A)2(5 Ω)
PL = 43.2 W
Answer: When the enable signal is +5.0 V, Q2's power
loss is 865 mW and the output load power is 43.2 W.
12-35. Given:
ID(on) = 75 mA
VGS(on) = 4.5 V
VGS(th) = 0.8
Solution:
k = IDS(on)/(VGS(on) – VGS(th))2
k = 75 mA/(4.5 V – 0.8 V)2
k = 5.48 × 10–3 A/V2
ID = k[VGS – VGS(th)]2
ID = 5.48 × 10–3 A/V2[3 – 0.8]2
ID = 26 mA
Answer: The value for k is 5.48 × 10–3 A/V2 and the drain
current is 26 mA.
12-36. Given:
RD = 150 Ω
RL = 1 kΩ
vin = 50 mV
Solution:
rd = RD || RL
rd = 150 Ω || 1 kΩ
rd = 130 Ω
k = 5.48 × 10–3 A/V2 (from Problem 12-35)
gm = 2k[VGS – VGS(th)]
gm = 24 mS
Av = gmrd
Av = (24 mS)(130 Ω)
Av = 3.14
vout = vgAv
vout = (50 mV)(3.14)
vout = 157 mV
Answer: The voltage gain is 3.14, the voltage out is
157 mV, and gm is 24 mS.
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12-37. Given:
RD = 50 Ω
ID(on) = 600 mA
VGS(on) = 4.5 V
VGS(th) = 2.1 V
Solution:
k = IDS(on)/(VGS(on) – VGS(th))2
k = 600 mA/(4.5 V – 2.1 V)2
k = 104 × 10–3 A/V2
ID = k[VGS – VGS(th)]2
ID = 104 × 10-3 A/V2[3 – 2.1]2
ID = 84.4 mA
Answer: The value for k is 104 × 10–3 A/V2 and the drain
current is 84.4 mA.
12-38. Given:
RD = 15 Ω
RL = 1 kΩ
vin = 50 mV
VDD = 12 V
Solution:
rd = RD || RL
rd = 15 Ω || 1 kΩ
rd = 14.8 Ω
Av = gmrd
Av = (395.8 mS)(14.8 Ω)
Av = 5.85
vout = VinAv
vout = (50 mV)(5.85)
vout = 292 mV
Answer: The voltage gain is 5.85, the voltage out is
292 mV, and gm is 395.8 mS.
CRITICAL THINKING
12-39. Given:
VGS(on) = 5 V (from Table 12-1)
ID(on) = 200 mA (from Table 12-1)
RDS(on) = 7.5 Ω (from Table 12-1)
VDD = 25 V
RD = 150 Ω
f = 1 kHz
Solution:
ID = (VDD)/(RDS(on) + RD)
ID = (25 V)/(7.5 Ω + 150 Ω)
ID = 158.7 mA
Since the signal is a square wave, the off time equals the
on time, which gives it a duty cycle of 0.5.
Duty cycle = on time/total time
P = ID2 RD
P = (158.7 mA)2(150 Ω)
P = 3.78 W
Pave = P (duty cycle)
Pave = 3.78 W(0.5)
Pave = 1.89 W
Answer: The average power dissipation in the load resistor is 1.89 W.
12-40. Given:
VGS(on) = 10 V (from Table 12-1)
ID(on) = 1 A (from Table 12-1)
RDS(on) = 0.9 Ω (from Table 12-1)
VDD = 12 V
RD = 18 Ω
Solution:
ID = (VDD)/(RDS(on) + RD)
ID = (12 V)/(0.9 Ω + 18 Ω)
ID = 0.635 A
P = ID2 RD
P = (0.635 A)2(18 Ω)
P = 7.26 W
Pave = P (duty cycle)
Pave = 7.26 W(0.25)
Pave = 1.81 W
Answer: The average power dissipation in the load resistor is 1.81 W.
12-41. Given:
VDD = 12 V
RDS(on) = 100 Ω
RDS(off) = 10 MΩ
Iave = 50 μA
Solution: Since the FETs are complementary, one is off
and the other is on. Thus the current drawn from the
power supply is going to be controlled by the off device.
ID = VDD /RDS(off)
ID = 12 V/10 MΩ
ID = 1.2 μA
P = VDDID
P = (12 V)(1.2 μA)
P = 14.4 μW
P = VDDIave
P = (12 V)(50 μA)
P = 600 μW
Answer: The quiescent power drain is 14.4 μW, and the
average power drain is 600 μW.
12-42. Given:
VG = 3 V
VDD = 15 V
R1 = 1 MΩ
Solution:
VR = VDD – VG
VR = 15 V – 3 V
VR = 12 V
I = VR/R1
I = 12 V/1 MΩ
I = 12 μA
The current through R2 is 3 V divided by 2 MΩ, which
equals 1.5 μA. Therefore, the photodiode current is 12 μA
minus 1.5 μA, or 10.5 μA.
Answer: The diode current is 10.5 μA.
12-43. Given: RDS(on) = 0.17 Ω at 25°C.
Solution: As the temperature rises 100°C, the normalized
resistance increases by a factor of 2.25. Thus 2.25/100°C =
0.0225/°C. The temperature increases 75°C. Thus the resistance increases by a factor of 75°C(0.0225/°C) = 1.69.
0.17(1.69) = 0.29 Ω
Answer: The resistance at 100°C is 0.29 Ω.
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12-44. Given:
Vin = 12 V
Turns ratio = 4:1
13-2.
Solution: The primary voltage will be 12 V.
N1/N2 = 4
N1/N2 = V1/V2
V2 = V1/(N1/N2)
V2 = 12 V/4
V2 = 3 V
Solution: Just before breakover, the capacitor voltage is
VB.
I = (V – VB)/R1
I = (19 V – 12 V)/5 kΩ
I = 1.4 mA
Answer: The output voltage is 3 V.
While the diode is conducting, the voltage across it is
0.7 V.
12-45. C1 is open
12-46. VCC has failed and is now 0 V
I = (V – VD)/R1
I = (19 V – 0.7 V)/5 kΩ
I = 3.66 mA
12-47. Q1 has failed
12-48. VG is set to 5 Vp, not 50 mVp
Answer: The current through the resistor just before
breakover is 1.4 mA, and during conduction is 3.66 mA.
12-49. R1 is shorted
Chapter 13
13-3.
Thyristors
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
c
b
d
c
b
b
a
b
9.
10.
11.
12.
13.
14.
15.
16.
b
c
a
b
d
d
d
d
17.
18.
19.
20.
21.
22.
23.
24.
d
a
a
b
c
b
c
b
25.
26.
27.
28.
29.
30.
31.
d
d
b
a
c
b
a
5. The SCR remains latched once the initial stimulus is
removed; the transistor does not. This prevents silencing the
alarm by a clever burglar or destruction of the sending unit by
fire or flood, etc.
6. In every section of the field.
7. Power-handling capability: The SCR can handle the most
current, and the power FET the least current. Efficiency:
The SCR is the most efficient since the control signal can be
removed once SCR is conducting, and the power FET is the
next-most efficient since its control current is low. Control
input: The power FET and BJT are easier to control because
they can be shut off using the control input. Maximum frequency: The power FET switches the fastest.
f = 1/T
f = 1/0.1 ms
f = 10 kHz
Answer: The RC time constant is 0.1 ms, and the frequency is 10 kHz.
13-4.
Solution:
V = IH RS + 0.7 V (Eq. 13-2)
V = (4 mA)(1 kΩ) + 0.7 V
V = 4.7 V
Answer: The power supply voltage will be 4.7 V at
dropout.
Given:
VD = 0.7 V
VB = 20 V
IH = 3 mA
RS = 1 kΩ
Solution: Since the diode is open before breakover, no
current flows before the device breaks over. Thus when
the power supply reaches breakover voltage, the device
will break over.
V = IHRS + 0.7 V (Eq. 13-2)
V = (3 mA)(1 kΩ) + 0.7 V
V = 3.7 V
PROBLEMS
Given:
VD = 0.7 V
IH = 4 mA
RS = 1 kΩ
Given:
VD = 0.7 V
VB = 12 V
V = 19 V
R1 = 5 kΩ
C1 = 0.02 μF
Solution:
RC = (5 kΩ)(0.02 μF)
RC = 0.1 ms
T = 0.1 ms since the period equals the RC time constant
JOB INTERVIEW QUESTIONS
13-1.
Given:
VD = 0.7 V
VB = 12 V
V = 19 V
R1 = 5 kΩ
Answer: The power supply voltage will be 20 V at breakover and 3.7 V at dropout.
13-5.
Given:
VD = 0.7 V
VB = 12 V
V = 19 V
R1 = 10 kΩ
C1 = 0.06 μF
Solution: The maximum voltage across the capacitor
will be breakover voltage, because as soon as the device
breaks over, the voltage drops to about 0.7 V.
RC = (10 kΩ)(0.06 μF)
RC = 0.6 ms
Answer: The maximum voltage across the capacitor is
12 V, and the time constant is 0.6 ms.
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13-6.
Given:
VGT = 1.0 V
IGT = 2 mA
IH = 12 mA
VCC = 12 V
RG = 2.2 kΩ
RL = 47 Ω
Solution: When the SCR is off, no current flows. The output voltage when the SCR is off is the same as the power
supply voltage.
Vin = VGT + IGTRG (Eq. 13-1)
Vin = 1 V + (2 mA)(2.2 kΩ)
Vin = 5.4 V
V = IHRL + 0.7 V (Eq. 13-2)
VCC = (12 mA)(47 Ω) + 0.7 V
VCC = 1.26 V
Answer: The output voltage when the SCR is off is
12 V. The input voltage required to turn on the SCR is
5.4 V, and the supply voltage required to turn the SCR
off is 1.26 V.
13-7.
Given:
VGT = 0.7 V
IGT = 1.5 mA
IH = 2 mA
VCC = 12 V
RG = 4.4 kΩ
RL = 94 Ω
Solution:
Vin = VGT + IGTRG
(Eq. 13-1)
Vin = 0.7 V + (1.5 mA)(4.4 kΩ)
Vin = 7.3 V
Answer: The input voltage required to turn on the SCR
is 7.3 V.
13-8.
Answer: The highest output occurs when 0.8 V is across
the 500-Ω resistor. The current through this resistor is
0.8 V divided by 500 Ω, which equals 1.6 mA. This
1.6 mA must flow through the 3.3-kΩ resistor. The
200 μA of gate current must also flow through the
3.3-kΩ resistor. If we ignore the 200 μA on the grounds
that it is much smaller than 1.6 mA, we get an approximate
answer of:
V = 0.8 V + (1.6 mA)(3.3 kΩ) = 6.08 V
If we include the 200 μA, we get a slightly larger output
voltage:
V = 0.8 V + (1.6 mA + 200 μA)(3.3 kΩ) = 6.74 V
13-9.
Given:
VGT = 1.5 V
IGT = 15 mA
IH = 10 mA
VCC = 12 V
RG = 2.2 kΩ
RL = 47 Ω
Solution:
Vin = VGT + IGTRG (Eq. 13-1)
Vin = 1.5 V + (15 mA)(2.2 kΩ)
Vin = 34.5 V
VCC = IHRL + 0.7 V (Eq. 13-2)
VCC = (10 mA)(47 Ω) + 0.7 V
VCC = 1.17 V
Answer: The input voltage required to turn on the SCR is
34.5 V, and the supply voltage required to turn the SCR
off is 1.17 V.
13-10. Given:
VGT = 2 V
IGT = 8 mA
IH = 2 mA
VCC = 12 V
RG = 6.6 kΩ
RL = 141 Ω
Solution:
Vin = VGT + IGTRG (Eq. 13-1)
Vin = 2 V + (8 mA)(6.6 kΩ)
Vin = 54.8 V
Answer: The input voltage required to turn on the SCR
is 54.8 V.
13-11. Given:
R1 = 3.3 kΩ
R2 = 6.8 kΩ
R3 = 750 Ω
C1 = 4.7 μF
Solution:
RC = RTH(cap)C1
RC = (2.54 kΩ)(4.7 μF)
RC = 11.9 msec
Rth = R || R1
Rth = 750 Ω || 3.3 kΩ
Rth = 611 Ω
Answer: The charging time constant is 11.9 ms. and the
Thevenin resistance is 611 Ω.
13-12. Given:
R1 = 1 kΩ
R2 = 4.6 kΩ
C = 0.47 μF
Solution:
XC = 1/(2πfC)
XC = 1/2π(60 Hz)(0.47 μF)
XC = 5644 Ω
________
Z = ​√_________________
R2 + XC2 ​
Z = ​√5.6
kΩ2 + 5.644 kΩ2 ​
Z = 7.95 kΩ
X
θZ = ∠ – arctan​  ​ ___C ​  ​
R
5.644 kΩ
θZ = ∠ – arctan ​  ​ ________
​  ​
5.6 kΩ
θZ = ∠ – 45º
Vin
IC ∠ θ = _______________
​
​
X
ZT ∠ – arctan​  ​ ___C ​  ​
R
120
V
∠
0º
______________
IC ∠ θ = ​
​
7.95 kΩ ∠ – 45º
IC ∠ θ = 15 mA ∠ 45º
( )
(
)
( )
VC = (IC ∠ θ) (XC ∠ – 90º)
VC = (15 mA ∠ 45º)(5644 Ω ∠ – 90º)
VC = 85 V ∠ – 45º
θºcond =180º– θºfiring
θºcond =180º – 45º
θºcond =135
Answer: The firing angle is 45º, the conduction angle is
135º, and the voltage across the capacitor is 85 Vac.
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13-13. Given:
R1 = 1 kΩ
R2 = 50 kΩ pot
C = 0.47 μF
Solution:
VCC = VZ + VGT (Eq. 13-3)
VCC = 10 V + 0.8 V
VCC = 10.8 V
Solution: Perform the following calculations with an R
value of 1 kΩ and 51 kΩ.
XC = 1/(2πfC)
XC = 1/(2π(60 Hz))(0.47 μF)
XC = 5644 Ω
_______
Z=√
​ _______________
R2 + XC ​2
√
Z = ​ 1
kΩ2+ 5.644 kΩ2 ​
Z = 5.732 kΩ
(X )
θZ = ∠ – arctan​ __
​  RC ​  ​
(
)
kΩ
θZ = ∠ – arctan​ _______
​  5.644
​  ​
1 kΩ
θZ = ∠ – 80º
Vin
IC ∠ θ = ​ ______________
​
X
ZT ∠ –arctan​ __
​  RC ​  ​
120 V ∠ 0º
IC ∠ θ =
​ _______________
​
5.732 kΩ ∠ – 80º
IC ∠ θ = 20.9 mA ∠ 80º
( )
VC = (IC ∠ θ) (XC ∠ – 90º)
VC = (20.9 mA ∠ 80º)(5644 Ω ∠ – 90º)
VC = 118 V ∠ – 10º
Answer: The minimum firing angle is 10º, and the
maximum firing angle is 83.7º.
13-14. Given:
R1 = 1 kΩ
R2 = 50 kΩ pot
C = 0.47 μF
13-16. Given:
VGT = 1.5 V
IGT = 200 μA
VZ = 10 V ± 10%
Solution:
VZ(max) = VZ + 0.1(VZ)
VZ(max) = 10 V + 0.1(10 V)
VZ(max) = 11 V
VCC = VZ + VGT (Eq. 13-3)
VCC = 11 V + 1.5 V
VCC = 12.5 V
Answer: The voltage needed to trigger the crowbar is
12.5 V.
13-17. Given:
VGT = 0.8 V
IGT = 200 μA
VZ = 12 V
Solution:
Vtrig = VZ + VGT (Eq. 13-3)
Vtrig = 12 V + 0.8 V
Vtrig = 12.8 V
Answer: The SCR will trigger at 12.8 V.
Solution: Perform the following calculations with an R
value of 1 kΩ and 51 kΩ.
XC = 1/(2πfC)
XC = 1/(2π(60 Hz))(0.47 μF)
XC = 5644 Ω
________
Z=√
​ _______________
R2 + XC2 ​
√
Z = ​ 1
kΩ2+ 5.644 kΩ2 ​
Z = 5.732 kΩ
(X )
θZ = ∠ – arctan​ __
​  RC ​  ​
θZ = ∠ – arctan​ ________
​  5.644 kΩ
​  ​
1 kΩ
θZ = ∠ – 80º
Vin
IC ∠ θ = ​ _______________
​
X
ZT ∠ – arctan​ __
​  RC ​  ​
120 V ∠ 0º
IC ∠ θ = ​ _______________
​
5.732 kΩ ∠ – 80º
IC ∠ θ = 20.9 mA ∠ 80º
(
)
( )
VC = (IC ∠ θ) (XC ∠ – 90º)
VC = (20.9 mA ∠ 80º)(5644 Ω ∠ – 90º)
VC = 118 V ∠ – 10º
θºcond = 180º – θºfiring
θºcond = 180º – 10º
θºcond = 170º
Answer: The minimum conduction angle is 96.3º, and the
maximum conduction angle is 170º.
13-15. Given:
VGT = 0.8 V
IGT = 200 μA
VZ = 10 V
Answer: The voltage needed to trigger the crowbar is
10.8 V.
13-18. Given:
VGT = 0.8 V
IGT = 200 μA
VZ = 11 V
Solution:
VCC = VZ + VGT (Eq. 13-3)
VCC = 11 V + 0.8 V
VCC = 11.8 V
Answer: The voltage needed to trigger the crowbar is
11.8 V.
13-19. Given:
VB = 20 V
VGT = 2.5 V
Solution: Ignore the gate current in the triac. Then
VC = VB + VGT
VC = 20 V + 2.5 V
VC = 22.5 V
Answer: The capacitor voltage required to turn on the
triac is 22.5 V.
13-20. Given:
Vin = 100 V
RL = 15 Ω
Solution: Ideally, when the triac is conducting, the voltage drop across it is 0 V.
I = Vin/RL
I = 100 V/15 Ω
I = 6.67 A
Answer: The load current is 6.67 A.
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13-21. Given:
VB = 28 V
VGT = 2.5 V
Solution: Ignore the current through the diac and triac.
Then
VC = VB + VGT
VC = 28 V + 2.5 V
VC = 30.5 V
Answer: The capacitor voltage required to turn on the
triac is 30.5 V.
13-22. Given:
VCC = 15 V
R2 = 1 kΩ
R3 = 2 kΩ
Solution:
Vgate = [R3/(R2 + R3)]VCC (Voltage divider formula)
Vgate = [2 kΩ/(1 kΩ + 2 kΩ)]15 V
Vgate = 10 V
Vanode = VTrig + 0.7 V
Vanode = 10 V + 0.7
Vanode = 10.7 V
Answer: The gate trigger voltage is 10 V and the anode
is 10.7 V.
13-23. Given:
VCC = 15 V
Vgate = 10 V
Vanode = 10.7 V
Solution:
VR4 = Vanode – 0.7 V
VR4 = 10.7 V – 0.7
VR4 = 10 V
Answer: The peak voltage across R4 = 10 V.
RCmin = RmaxC1
RCmin = (1 kΩ)(0.1 μF)
RCmin = 0.1 msec
Tmax = 0.2(RC1 max)
Tmax = 0.2(5.1 ms)
Tmax = 1.02 ms
Tmin = 0.2(RC1 min)
Tmin = 0.2(0.1 ms)
Tmin = 0.02 ms
fmax = 1/Tmin
fmax = 1/0.02 ms
fmax = 50 kHz
fmin = 1/Tmax
fmin = 1/1.02 ms
fmin = 980 Hz
Answer: The maximum frequency is 50 kHz, and the
minimum is 980 Hz.
13-28. Given:
RL = 100 Ω
VCC = 15 V
Solution: In a dark room the SCR is off and the output
voltage is 15 V. Once the SCR fires, its voltage drops to
0.7 V.
I = (VCC – 0.7 V)/RL
I = (15 V – 0.7 V)/100 Ω
I = 143 mA
Answer: The output voltage when it is dark is 15 V and
when it is light is 0.7 V, and the current through the resistor is 143 mA when it is light.
13-29. Answer:
Trouble 1: Since there is voltage at D and not at E, the
wire connecting the two is open.
13-24. Answer: The output waveform will be a sawtooth waveform from 0 V to 10.7 V.
Trouble 2: No supply voltage.
CRITICAL THINKING
Trouble 4: Since there is voltage at A and not at B, the
fuse is open.
13-25. Answer: The breakover voltage of the diode, which is
10 V.
Trouble 3: Since there is voltage at B and not at C, the
transformer is the problem.
13-30. Answer:
13-26. Answer: The breakover voltage of the diode, which is
10 V.
Trouble 5: Since there is an overvoltage and the crowbar
is off, the problem is the crowbar.
13-27. Given:
R1 = 0 to 50 kΩ
R2 = 1 kΩ
C1 = 0.1 μF
T = 20%(RC)
Trouble 6: Since there is voltage at C and not at D and the
load resistor is not shorted, the rectifier is the problem.
Solution:
Rmax = R2 + R1(max)
Rmax = 1 kΩ + 50 kΩ
Rmax = 51 kΩ
Trouble 7: Since there is voltage at E and not at F, the
wire connecting the two is open.
Trouble 8: Since there is voltage at A and not at B, the
fuse is open.
13-31. The fuse is open
13-32. The SCR D3 has been triggered on
Rmin = R2 + R1(min)
Rmin = 1 kΩ + 0 kΩ
Rmin = 1 kΩ
13-33. Open transformer secondary
RCmax = RmaxC1
RCmax = (51 kΩ)(0.1 μF)
RCmax = 5.1 ms
13-35. VS is 0 V
13-34. C1 is open
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Chapter 14
Frequency Effects
14-2.
SELF-TEST
1.
2.
3.
4.
5.
a
b
c
c
b
6.
7.
8.
9.
10.
c
b
c
c
d
11.
12.
13.
14.
15.
c
c
d
a
c
16.
17.
18.
19.
20.
a
d
b
c
a
Answer: See the figure below.
Av
500,000
353,000
JOB INTERVIEW QUESTIONS
1. Too much stray-wiring capacitance. Shorten the leads as
much as possible.
2. With a sine wave, find the frequency at which the voltage
gain is down 3 dB. With a square wave, use the step-response
method.
5. Some oscilloscopes (with plug-in vertical preamps) specify
the risetime of the main frame. A risetime of 7.0 ns converts
to a bandwidth of 50 MHz.
6. Use a step voltage and measure the risetime of the output
signal.
8. Maximum power transfer.
9. dBm is referenced to 1 mW, whereas dB is not referenced to
any standard.
10. Because it amplifies down to 0 Hz, which is the frequency of
a dc signal.
11. Semilogarithmic
12. It is a computer program that provides electronic circuit
simulation. It is used to build, test, and analyze simulated
circuits.
PROBLEMS
14-1.
15 Hz
Frequency response for Prob. 14-2.
14-3.
Given:
Av(mid) = 200
f2 = 10 kHz
f = 100 kHz, 200 kHz, 500 kHz, 1 MHz
Solution: Substitute in the appropriate value for f.
Av = Av(mid)
______________
​  ____________ ​
√
( )
f  ​  2​  ​
​ 1 + ​ ​ ________
f2
Av = 200
______________
​
____________ ​
√
(
)
​ 1
+ ​ ________
​  100 kHz ​  2​  ​
10 kHz
Av = 19.9
Answer: Av = 19.9 at 100 kHz, Av = 9.98 at 200 kHz,
Av = 4 at 500 kHz, Av = 2 at 1 MHz.
14-4.
Given:
Av(mid) = 1000
f1 = 100 Hz
f2 = 100 kHz
f
Solution:
_________
Av(20K) = Av(mid)/[​√_________________
1 + (f1/f)2] ​ (Eq. 14-3)
Av(20K) = 1000/[​√1
+ (100 Hz/20 Hz)2 ​]
Av(20K) = 196
_________
Av(300K) = Av(mid)/[​√_____________________
1 + (f /f2)2 ​
] (Eq. 14-3)
Given: AP = 5, 10, 20, 40
Solution:
AP(dB) = 10 logAP
AP(dB) = 10 log(5)
AP(dB) = 7 dB
AP(dB) = 10 logAP
AP(dB) = 10 log(10)
AP(dB) = 10 dB
AP(dB) = 10 logAP
AP(dB) = 10 log(20)
AP(dB) = 13 dB
Av(300K) = 1000/[​√1
+ (300 kHz/100 kHz)2 ​]
Av(300K) = 316
AP(dB) = 10 logAP
AP(dB) = 10 log(40)
AP(dB) = 16 dB
Answer: The frequency response looks like the figure
below; the gain at 20 Hz is 196, and at 300 kHz is 316.
Answer: The decibel power gain is 7 dB at a power gain
of 5, 10 dB at 10, 13 dB at 20, and 16 dB at 40.
Av
14-5.
1000
707
100 Hz
Frequency response for Prob. 14-1.
300 kHz
Given: Ap = 0.4, 0.2, 0.1, 0.05
Solution:
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(0.4)
AP(dB) = –3.98
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(0. 2)
AP(dB) = –6.99
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(0.1)
AP(dB) = –10
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(0.05)
AP(dB) = –13
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Answer: The decibel power gain is –3.98 at a power gain
of 0.4, –6.99 at 0.2, –10 at 0.1, and –13 at 0.05.
14-6.
Given: Ap = 2, 20, 200, 2000
Solution:
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(2)
AP(dB) = 3 dB
Av = antilog(Av(dB)/20) (Eq. 14-15)
Av = antilog(82 dB/20)
Av = 12,589
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(200)
AP(dB) = 23 dB
Answer: The decibel voltage gain is 82, and the voltage
gain is 12,589.
Answer: The decibel power gain is 3 dB at a power gain
of 2, 13 dB at 20, 23 dB at 200, and 33 dB at 2000.
Given: AP = 0.4, 0.04, 0.004
Solution:
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(0.4)
AP(dB) = –3.98
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(0.04)
AP(dB) = –13.98
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(0.004)
AP(dB) = –23.98
Answer: The decibel power gain is –3.98 dB at a power
gain of 0.4, –13.98 dB at 0.04, and –23.98 dB at 0.004.
14-8.
Given:
Av1 = 200
Av2 = 100
Solution:
Av = Av1Av2 (Eq. 14-10)
Av = (200)(100)
Av = 20,000
Av(dB) = 20 logAv (Eq. 14-8)
Av(dB) = 20 log(20,000)
Av(dB) = 86 dB
Answer: The voltage gain is 20,000, and the decibel voltage gain is 86 dB.
14-9.
Solution:
Av(dB) = Av1(dB) + Av2(dB) (Eq. 14-11)
Av(dB) = 30 dB + 52 dB
Av(dB) = 82 dB
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(20)
AP(dB) = 13 dB
AP(dB) = 10 logAP (Eq. 14-8)
AP(dB) = 10 log(2000)
AP(dB) = 33 dB
14-7.
14-10. Given:
Av1(dB) = 30 dB
Av2(dB) = 52 dB
Given:
Av1 = 200
Av2 = 100
Solution:
Av1(dB) = 20 logAv1 (Eq. 14-8)
Av1(dB) = 20 log(200)
Av1(dB) = 46 dB
Av2(dB) = 20 logAv2 (Eq. 14-8)
Av2(dB) = 20 log(100)
Av2(dB) = 40 dB
Answer: The decibel voltage gain for stage 1 is 46 dB,
and stage 2 is 40 dB.
14-11. Given:
Av1(dB) = 30 dB
Av2(dB) = 52 dB
Solution:
Av1 = antilog(Av1(dB)/20) (Eq. 14-15)
Av1 = antilog(30 dB/20)
Av1 = 31.6
Av2 = antilog(Av2(dB)/20) (Eq. 14-15)
Av2 = antilog(52 dB/20)
Av2 = 398
Answer: The voltage gain of the first stage is 31.6, and
the second stage is 398.
14-12. Given: Av = 100,000
Solution:
Av(dB) = 20 logAv (Eq. 14-9)
Av(dB) = 20 log(100,000)
Av(dB) = 100
Answer: The decibel voltage gain is 100 dB.
14-13. Given: AdB = 34 dB
Solution:
Av = antilog(Av(dB)/20) (Eq. 14-15)
Av = antilog(34 dB/20)
Av = 50.1
Answer: The voltage gain is 50.1.
14-14. Given:
Av1 = 25.8
Av2 = 117
Solution:
Av1(dB) = 20 logAv1 (Eq. 14-9)
Av1(dB) = 20 log(25.8)
Av1(dB) = 28.2 dB
Av2(dB) = 20 logAv2 (Eq. 14-9)
Av2(dB) = 20 log(117)
Av2(dB) = 41.4 dB
Av(dB) = Av1(dB) + Av2(dB) (Eq. 14-11)
Av(dB) = 28.2 dB + 41.4 dB
Av(dB) = 69.6 dB
Answer: The decibel voltage gain for the first stage
is 28.2 dB, the second stage is 41.4 dB, and overall is
69.6 dB.
14-15. Given:
AP1(dB) = 23 dB
AP2(dB) = 18 dB
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Solution:
AP1(dB) = Av1(dB) = 23 dB
AP2(dB) = Av2(dB) = 18 dB
Av(dB) = Av1(dB) + Av2(dB) (Eq. 14-11)
Av(dB) = 23 dB + 18 dB
Av(dB) = 41 dB
Answer: The total decibel voltage gain is 41 dB, the firststage decibel voltage gain is 23 dB, and the second stage
voltage gain is 18 dB.
PdBm = 10 log(P/1 mW) (Eq. 14-16)
PdBm = 10 log(4.87 W/1 mW)
PdBm = 36.9 dBm
Answer: 25 mW is 14 dBm, 93.5 mW is 19.7 dBm, and
4.87 W is 36.9 dBm.
14-20. Given:
V = 1 μV, 34.8 mV, 12.9 V, and 345 V
Solution:
VdBV = 20 logV (Eq. 14-18)
VdBV = 20 log(1 μV)
VdBV = –120 dBV
14-16. Given:
Vin = 10 μV
RG = 300 Ω
AP1(dB) = 23 dB
AP2(dB) = 18 dB
AP(dB) = 41 dB (from Prob. 14-15)
VdBV = 20 logV (Eq. 14-18)
VdBV = 20 log(34.8 mV)
VdBV = –29.2 dBV
VdBV = 20 logV (Eq. 14-18)
VdBV = 20 log(12.9 V)
VdBV = 22.2 dBV
Solution:
AP = antilog(APdB/10) (Eq. 14-15)
AP = antilog(41 dB/10)
AP = 12,589
Pin = V2/R
Pin = (5 μV)2/300 Ω (because half of the source voltage appears at the input)
Pin = 0.0833 pW
Pout = APPin
Pout = (12,589)(0.0833 pW)
Pout = 1.05 nW
_____
vout = √
​ _______________
Pout R ​
vout = √
​ (1.05
nW)(300 Ω) ​
vout = 0.56 mV
VdBV = 20 logV (Eq. 14-18)
VdBV = 20 log(345 V)
VdBV = 50.8 dBV
Answer: 1 μV is –120 dBV, 34.8 mV is –29.2 dBV, 12.9 V
is 22.2 dBV, and 345 V is 50.8 dBV.
14-21. Given:
Av(mid) = 200,000
f2 = 10 Hz
Roll-off = 20 dB/decade
Answer:
Av(dB) = 20 log Av (Eq. 14-9)
Av(dB) = 20 log(200,000)
Av(dB) = 106 dB
Answer: The load voltage is 0.56 mV, and the load power
is 1.05 nW.
From the graph below, the gain at 1 MHz is 6 dB.
14-17. Given:
PdBm = 20 dBm
Solution:
P = antilog(PdBm/10) (Eq. 14-17)
P = antilog(20 dBm/10)
P = 100 mW
Answer: The output power is 100 mW.
14-18. Given:
VdBV = –45
Solution:
V = antilog(VdBV/20) (Eq. 14-19)
V = antilog(–45 dBV/20)
V = 5.6 mV
Answer: The output voltage is 5.6 mV.
14-19. Given:
P = 25 mW, 93.5 mW, and 4.87 W
Solution:
PdBm = 10 log(P/1 mW) (Eq. 14-16)
PdBm = 10 log(25 mW/1 mW)
PdBm = 14 dBm
PdBm = 10 log(P/1 mW) (Eq. 14-16)
PdBm = 10 log(93.5 mW/1 mW)
PdBm = 19.7 dBm
Av = antilog(Av(dB)/20) (Eq. 14-15)
Av = antilog(6 dB/20)
Av = 2
Answer: The voltage gain at 1 MHz is 2.
Av(dB)
106 dB
86 dB
66 dB
46 dB
26 dB
6 dB
10 Hz 100 Hz 1 kHz 10 kHz 100 kHz 1 MHz
Ideal Bode plot for Prob. 14-21.
14-22. Given:
Av(mid) = 316,000
f2 = 40 Hz
Roll-off = 20 dB/decade
Answer:
Av(dB) = 20 log Av (Eq. 14-9)
Av(dB) = 20 log(316,000)
Av(dB) = 110 dB
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14-25. Given:
R = 15 kΩ
C = 100 pF
Av(mid) = 400
Answer: See figure below.
Av(dB)
110 dB
Solution:
f2 = 1/(2πRC)
f2 = 1/[2π(15 kΩ)(100 pF)]
f2 = 106 kHz
90 dB
70 dB
50 dB
Av(dB) = 20 log Av (Eq. 14-9)
Av(dB) = 20 log(400)
Av(dB) = 52 dB
30 dB
10 dB
40 Hz
4 kHz
Answer: See figure at top of next column.
400 kHz
Av(dB)
Ideal Bode plot for Prob. 14-22.
14-23. Given:
C = 1000 pF
R = 10 kΩ
Solution:
f2 = 1/(2πRC)
f2 = 1/[2π (10 kΩ)(1000 pF)]
f2 = 15.9 kHz
52 dB
Answer: See figure below.
12 dB
32 dB
106 kHz
Av(dB)
1.06 MHz
10.6 MHz
Ideal Bode plot for Prob. 14-25.
15.9 kHz
0 dB
f
20 dB/decade
14-26. Given:
C = 5 pF
Av = 200,000
Solution:
Cin = C(Av + 1) (Eq. 14-26)
Cin = 5 pF(200,000 + 1)
Cin = 1 μF
Answer: The Miller input capacitance is 1 μF.
Ideal Bode plot for Prob. 14-23.
14-27. Given:
Cin(M) = 15 pF
Av = 250,000
RL = 10 kΩ
RG = 1 kΩ
14-24. Given:
R = 1 kΩ
C = 50 pF
Solution:
Solution:
Cin(M) = C(Av + 1) (Eq. 14-26)
Cin(M) = 15 pF(250,000 + 1)
Cin(M) = 3.75 μF
f2 = 1/(2πRC)
f2 = 1/[2π(1 kΩ)(50 pF)]
f2 = 3.18 MHz
f2 = 1/(2πRC)
f2 = 1/[2π (1 kΩ)(3.75 μF)]
f2 = 42 Hz
Answer: See figure below.
Av(dB)
3.18 MHz
0 dB
f
20 dB/decade
Av(dB) = 20 log Av (Eq. 14-9)
Av(dB) = 20 log(250,000)
Av(dB) = 108 dB
Answer: See figure below.
Ideal Bode plot for Prob. 14-24.
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Av(dB)
108 dB
14-31. Given:
TR = 0.25 μs
Solution:
f2 = 0.35/TR (Eq. 14-29)
f2 = 0.35/0.25 μs
f2 = 1.4 MHz
88 dB
68 dB
48 dB
Answer: The bandwidth is 1.4 MHz.
28 dB
14-32. Given:
f2 = 100 kHz
8 dB
42 Hz 420 Hz 4.2 kHz 42 kHz 420 kHz 4.2 MHz
Ideal Bode plot for Prob. 14-27.
14-28. Given:
C = 50 pF
Av = 200,000
Solution:
Cin(M) = C(Av + 1) (Eq. 14-26)
Cin(M) = 50 pF(200,000 + 1)
Cin(M) = 10 μF
Answer: The Miller input capacitance is 10 μF.
14-29. Given:
C = 100 pF
Av = 150,000
RL = 10 kΩ
RG = 1 kΩ
Solution:
Cin(M) = C(Av + 1) (Eq. 14-26)
Cin(M) = 100 pF(150,000 + 1)
Cin(M) = 15 μF
f2 = 1/(2πRC)
f2 = 1/[2π(1 kΩ)(15 μF)]
f2 = 11 Hz
Av(dB) = 20 log Av (Eq. 14-9)
Av(dB) = 20 log(150,000)
Av(dB) = 104 dB
Answer: See figure below.
Av(dB)
104 dB
Solution:
f2 = 0.35/ TR (Eq. 14-29)
TR = 0.35/f2
TR = 0.35/100 kHz
TR = 3.5 μs
Answer: The risetime is 3.5 μs.
14-33. Given:
Cin = 1 μF
RG = 50 Ω
Solution:
R = RG + zin(stage)
R = RG + R1 || R2 || βr'e
R = 1.34 kΩ
fC1 = 1/(2πRCin)
fC1 = 1/[2π(1.34 kΩ)(1 μF)]
fC1 = 119 Hz
Answer: The lower cutoff frequency for the base coupling circuit is 119 Hz.
14-34. Given:
Cout = 4.7 μF
RC = 3.6 kΩ
RL = 10 kΩ
Solution:
R = RC + RL
R = 36 kΩ + 10 kΩ
R = 13.6 kΩ
fC1 = 1/(2πRCout)
fC1 = 1/[2π(13.6 kΩ)(4.7 μF)]
fC1 = 2.49 Hz
Answer: The lower cutoff frequency for the collector
coupling circuit is 2.49 Hz.
84 dB
14-35. Given:
CE = 25 μF
RG = 50 Ω
RE = 1 kΩ
64 dB
44 dB
24 dB
4 dB
11 Hz 110 Hz 1.1 kHz 11 kHz 110 kHz 1.1 MHz
Ideal Bode plot for Prob. 14-29.
14-30. Given:
TR = 10 μs
Solution:
f2 = 0.35/TR (Eq. 14-29)
f2 = 0.35/10 μs
f2 = 35 kHz
Answer: The upper cutoff frequency is 35 kHz.
Solution:
fC1 = 1/(2πzoutCE)
fC1 = 1/[2π(22.4 Ω)(25 μF)]
fC1 = 284 Hz
Answer: The lower cutoff frequency for the emitter
bypass circuit is 284 Hz.
14-36. Given:
C'c = 2 pF
C'e = 10 pF
C'Stray = 5 pF
R1 = 10 kΩ
R2 = 2.2 kΩ
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RC = 3.6 kΩ
RL = 10 kΩ
RG = 50 Ω
β = 200
Solution:
rg = RG || R1 || R2
rg = 50 Ω || 10 kΩ || 2.2 kΩ
rg = 48 Ω
Cin(M) = 236 pF
C = Cin(M) + C'e = 246 pF
Base
f2 = 1/(2πrgC)
f2 = 1/[2π(48 Ω)(246 pF)]
f2 = 13.5 MHz
Collector
C = Cout(M) + Cstray
C = 2 pF + 5 pF = 7 pF
R = RC || RL
R = 3.6 kΩ || 10 kΩ
R = 2.65 kΩ
f2 = 1/(2πRC)
f2 = 1/[2π(2.65 kΩ)(7 pF)]
f2 = 8.59 MHz
Answer: The high cutoff frequency for the base is
13.5 MHz and the collector is 8.59 MHz.
Solution:
Av = gmrd
Av = (16.5 ms)(1 kΩ || 10 kΩ)
Av = 15
Cin(M) = Cgd(Av + 1) (Eq. 14-40)
Cin(M) = 5 pF(15 + 1)
Cin(M) = 80 pF
C = Cgs + Cin(M)
C = 25 pF + 80 pF
C = 105 pF
R = RG || R1 || R2
R = 50 Ω || 2 MΩ || 1 MΩ
R = 50 Ω
Gate
f2 = 1/(2πRC)
f2 = 1/[2π(50 Ω)(105 pF)]
f2 = 30.3 MHz
Collector
Cout(M) = Cgd[(Av + 1)/Av] (Eq. 14-41)
Cout(M) = 5 pF[(15 + 1)/15]
Cout(M) = 5.3 pF
14-37. Given:
gm = 16.5 ms
Ciss = 30 pF
Coss = 20 pF
Crss = 5 pF
Solution:
Cgd = Crss = 5 pF
Cgs = Ciss – Crss
Cgs = 30 pF – 5 pF
Cgs = 25 pF
Cds = Coss – Crss
Cds = 20 pF – 5 pF
Cds = 15 pF
Answer: Cgs = 25 pF, Cgd = Crss = 5 pF, Cds = 15 pF.
14-38. Given:
R1 = 2 MΩ
R2 = 1 MΩ
RD = 1 kΩ
RL = 10 kΩ
RG = 50 Ω
C1 = 0.01 μF
C2 = 1 μF
Solution:
Rin = zin(stage)
Rin = R1 || R2
Rin = 667 kΩ
14-39. Given:
R1 = 2 MΩ
R2 = 1 MΩ
RD = 1 kΩ
RL = 10 kΩ
RG = 50 Ω
Cgd = 5 pF (from Prob. 14-37)
Cgs = 25 pF (from Prob. 14-37)
Cds = 15 pF (from Prob. 14-37)
C = Cds + Cout(M)
C = 15 pF + 5.3 pF
C = 20.3 pF
Drain
R = RD || RL
f2 = 1/(2πRC)
f2 = 1/[2π(909 Ω)(20.3 pF)]
f2 = 8.61 MHz
Answer: The high frequency cutoff for the gate is
30.3 MHz and the drain is 8.61 MHz.
CRITICAL THINKING
14-40. Given:
f2 = 100 Hz
Av(dB) = 80 dB
(RG is considered insignificant)
fC1 = 1/(2πRinC1)
fC1 = 1/[2π(667 kΩ)(0.01 μF)]
fC1 = 23.9 Hz or 14.5 Hz (from the output coupling
capacitor)
Answer: The dominant low cutoff frequency is 23.9 Hz.
Solution:
Av(mid) = antilog(Av(dB)/20) (Eq. 14-15)
Av(mid) = antilog(80 dB/20)
Av(mid) = 10,000
________
Av(20K) = Av(mid)/[​√1__________________
+ (f/f2)2 ​
] (Eq. 14-3)
Av(20K) = 10,000/[​√1
+ (20 kHz/100 Hz)2 ​]
Av(20K) = 50
Av(dB) = 20 log A (Eq. 14-9)
Av(dB) = 20 log(50)
Av(dB) = 34 dB
________
Av(44.4K) = Av(mid)/[​√1____________________
+ (f/f2)2 ​
] (Eq. 14-3)
Av(44.4K) = 10,000/[​√ 1
+ (44.4 kHz/100 Hz)2 ​]
Av(44.4K) = 22.5
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Av(dB) = 20 log Av (Eq. 14-9)
Av(dB) = 20 log(22.5)
Av(dB) = 27 dB
Answer: The decibel voltage gain at 20 kHz is 34 dB, and
at 44.4 kHz is 27 dB.
14-41. Given:
f2 = 100 Hz
Second breakpoint is 10 kHz
Av(mid) = 120 dB
Solution: Since the roll-off is 20 dB/decade at a frequency
of 1 kHz (one decade above the cutoff frequency), the
gain is 100 dB (20 dB less than the midband), and at
10 kHz the gain is 80 dB. From this point the roll-off
increases to 40 dB/decade; thus at 100 kHz, the gain will
be 40 dB.
Answer: The voltage gain at 100 kHz is 40 dB.
14-42. Given:
vin = 20 mV
Av(mid) = 100
Solution:
vout(max) = Av(mid)vin
vout(max) = (100)(20 mV)
vout(max) = 2 V
14-48. CE is open
14-49. VCC is at 15 V, not 10 V
Chapter 15
SELF-TEST
1.
2.
3.
4.
5.
6.
b
c
a
c
b
a
13.
14.
15.
16.
17.
18.
c
a
a
b
d
c
19.
20.
21.
22.
23.
b
c
a
c
c
PROBLEMS
15-1.
Given:
VCC = 15 V
VEE = –15 V
RE = 270 kΩ
RC = 180 kΩ
Solution:
IT = VEE/RE (Eq. 15-5)
IT = 15 V/270 kΩ
IT = 55.6 μA
Answer: The voltage at the 10% point is 0.2 V, and at the
90% point is 1.8 V.
IE = 1/2 IT
IE = 1/2 (55.6 μA)
IE = 27.8 μA
14-43. Given:
R = 4 kΩ
C = 50 pF
VC = VCC – (27.8 μA)(180 kΩ)
VC = 10 V
Solution:
f2 = 1/(2πRC)
f2 = 1/[2π (4 kΩ)(50 pF)]
f2 = 796 kHz
Answer: The tail current is 55.6 μA, the emitter is
27.8 μA, and the quiescent voltage is 10 V.
15-2.
Answer: The risetime is 0.44 μs.
Given:
VCC = 15 V
VEE = –15 V
RE = 270 kΩ
RC = 180 kΩ
Solution:
IT = (VEE – VBE)/RE (Eq. 15-8)
IT = (15 V – 0.7 V)/270 kΩ
IT = 53 μA
14-44. Given:
f2 = 1 MHz
TR = 1 μs
Solution:
f2 = 0.35/TR (Eq. 14-29)
f2 = 0.35/1 μs
f2 = 350 kHz
IE = 1/2 IT
IE = 1/2(53 μA)
IE = 26.5 μA
VC = VCC – (26.5 μA)(180 kΩ)
VC = 10.2 V
Answer: The amplifier with the cutoff frequency of
1 MHz has the larger bandwidth.
Answer: The tail current is 53 μA, the emitter current is
26.5 μA, and the quiescent voltage is 10.2 V.
14-45. RG is 500 Ω instead of 50 Ω
14-47. Cin has changed to 0.1 μF
b
a
d
a
c
b
6. Use a transistor as a current source instead of a tail resistor.
It could be a regulator configuration or a current source.
9. A transistor acting as a current source.
11. Current sources and active loads.
12. Increased voltage gain and higher CMRR.
13. Trick question. You can’t test a 741 with an ohmmeter.
At the 90% point = 0.9 vout(max)
At the 90% point = 0.9(2 V)
At the 90% point = 1.8 V
14-46. RE is 2 kΩ instead of 1 kΩ
7.
8.
9.
10.
11.
12.
JOB INTERVIEW QUESTIONS
At the 10% point = 0.1 vout(max)
At the 10% point = 0.1(2 V)
At the 10% point = 0.2 V
f2 = 0.35/TR (Eq. 14-29)
TR = 0.35/f2
TR = 0.35/796 kHz
TR = 0.44 μs
Differential Amplifiers
15-3.
Given:
VCC = 12 V
VEE = –12 V
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RE = 200 kΩ
RC = 200 kΩ
vout = Av(v1 – v2) (Eq. 15-2)
vout = 207.3(2.5 mV – 0)
vout = 518 mV
Solution:
IT = (VEE)/RE
IT = (12 V)/200 kΩ
IT = 60 μA
IE = 1/2 IT
IE = 1/2(60 μA)
IE = 30 μA
zin = 2β r'e (Eq. 15-11)
zin = 2(275)(226.7 Ω)
zin = 125 kΩ
Answer: The output voltage is 518 mV, and the input
impedance is 125 kΩ.
15-6.
Right Side
VC = VCC – (30 μA)(200 kΩ)
VC = 6 V
Left Side
VC = 12 V
Answer: The tail current is 60 μA, the emitter current is
30 μA, and the quiescent voltage is 6 V on the right side
and 12 V on the left side.
15-4.
Solution:
IT = (VEE – VBE)/RE (Eq. 15-5)
IT = (15 V – 0.7 V)/68 kΩ
IT = 210.3 μA
Given:
VCC = 12 V
VEE = –12 V
RE = 200 kΩ
RC = 200 kΩ
IE = 1/2 IT (Eq. 15-6)
IE = 1/2(210.3 μA)
IE = 105.2 μA
r'e = 25 mV/ IE (Eq. 8-10)
r'e = 25 mV/105.2 μA
r'e = 237.6 Ω
Solution:
IT = (VEE – VBE)/RE
IT = (12 V – 0.7 V)/200 kΩ
IT = 56.5 μA
Av = RC/r'e (Eq. 15-10)
Av = 47 kΩ/237.6 Ω
Av = 197.8
IE = ½ IT
IE = 1/2(56.5 μA)
IE = 28.3 μA
vout = Av(v1 – v2) (Eq. 15-2)
vout = 197.8(2.5 mV – 0)
vout = 494 mV
Right Side
VC = VCC – (28.3 μA)(200 kΩ)
VC = 6.35 V
zin = 2β r'e (Eq. 15-11)
zin = 2(275)(237.6 Ω)
zin = 131 kΩ
Left Side
VC = 12 V
Answer: The tail current is 56.5 μA, the emitter current is
28.3 μA, and the quiescent voltage is 6.35 V on the right
side and 12 V on the left side.
15-5.
Given:
VCC = 15 V
VEE = –15 V
RE = 68 kΩ
RC = 47 kΩ
β = 275
v1 = 2.5 mV
Solution:
IT = (VEE/RE) (Eq. 15-5)
IT = (15 V)/68 kΩ
IT = 220.6 μA
IE = 1/2 IT (Eq. 15-6)
IE = 1/2 (220.6 μA)
IE = 110.3 μA
r'e = 25 mV/IE
r'e = 25 mV/110.3 μA
r'e = 226.7 Ω
Av = RC/r'e (Eq. 15-10)
Av = 47 kΩ/226.7 Ω
Av = 207.3
Given:
VCC = 15 V
VEE = –15 V
RE = 68 kΩ
RC = 47 kΩ
β = 275
v1 = 2.5 mV
Answer: The output voltage is 494 mV, and the input
impedance is 131 kΩ.
15-7.
Given:
VCC = 15 V
VEE = –15 V
RE = 68 kΩ
RC = 47 kΩ
β = 275
v1 = 0 mV
v1 = 1 mV
Solution:
IT = (VEE)/RE (Eq. 15-5)
IT = (15 V)/68 kΩ
IT = 220.6 μA
IE = 1/2 IT (Eq. 15-6)
IE = 1/2(220.6 μA)
IE = 110.3 μA
r'e = 25 mV/IE (Eq. 8-10)
r'e = 25 mV/110.3 μA
r'e = 226.7 Ω
Av = RC/r'e (Eq. 15-10)
Av = 47 kΩ /226.7 Ω
Av = 207.3
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vout = Av(v1 – v2) (Eq. 15-2)
vout = 207.3(0 V – 1 mV)
vout = –207 mV
zin = 2β r′e (Eq. 15-11)
zin = 2(275)(226.7 Ω)
zin = 125 kΩ
Answer: The output voltage is –207 mV, and the input
impedance is 125 kΩ.
15-8.
Given:
Av = 360
Iin(bias) = 600 nA
Iin(off) = 100 nA
Vin(off) = 1 mV
RB1 = 10 kΩ
Solution:
V1 err = (RBl – RB2)/Iin(bias) (Eq. 15-16)
V1 err = (10 kΩ – 0)600 nA
V1 err = 6 mV
V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 15-17)
V2 err = (10 kΩ + 0)(100 nA/2)
V2 err = 0.5 mV
V3 err = Vin(off) (Eq. 15-18)
V3 err = 1 mV
Verror = Av(V1 err + V2 err + V3 err) (Eq. 15-19)
Verror = 360(6 mV + 0.5 mV + 1 mV)
Verror = 2.7 V
With base resistors equal.
V1 err = 0.
V2 err = RBIin(off)
V2 err = (10 kΩ)(100 nA)
V2 err = 1 mV
V3 err = Vin(off) (Eq. 15-18)
V3 err = 1 mV
Verror = Av(V1 err + V2 err + V3 err) (Eq. 15-19)
Verror = 360(0 mV + 1 mV + 1 mV)
Verror = 0.72 μV
Answer: The output error voltage is 2.7 V. If the base
resistors are equal, the output error voltage is 0.72 V.
15-9.
Given
Av = 250
Iin(bias) = 1 μA
Iin(off) = 200 nA
Vin(off) = 5 mV
RB1 = 10 kΩ
Solution:
V1 err = (RB1 − RB2)Iin(bias) (Eq. 15-16)
V1 err = (10 kΩ − 0)1 μA
V1 err = 10 mV
V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 15-17)
V2 err = (10 kΩ + 0)(200 nA/2)
V2 err = 1 mV
V3 err = Vin(off) (Eq. 15-18)
V3 err = 5 mV
Verror = Av(V1 err + V2 err + V3 err) (Eq. 15-19)
Verror = 250(10 mV + 1 mV + 5 mV)
Verror = 4 V
With base resistors equal,
V1 err = 0
V2 err = RBIin(off)
V2 err = (10 kΩ)(200 nA)
V2 err = 2 mV
V3 err = Vin(off) (Eq. 15-18)
V3 err = 5 mV
Verror = Av(V1 err + V2 err + V3 err) (Eq. 15-19)
Verror = 250(0 mV + 2 mV + 5 mV)
Verror = 1.75 V
Answer: The output error voltage is 4 V. If the base resistors are equal, the output error voltage is 1.75 V.
15-10. Given:
RC = 500 kΩ
RE = 500 kΩ
vin(CM) = 20 μV
Solution:
Av(CM) = RC /2RE (Eq. 15-20)
Av(CM) = 500 kΩ/2(500 kΩ)
Av(CM) = 0.5
vout(CM) = Av(CM)(vin(CM))
vout(CM) = 0.5(20 μV)
vout(CM) = 10 μV
Answer: The common-mode voltage gain is 0.5, and the
common-mode output voltage is 10 μV.
15-11. Given:
RC = 500 kΩ
RE = 500 kΩ
VCC = 15 V
VEE = −15 V
vin(CM) = 5 mV
vin = 2 mV
Av(CM) = 0.5 (from Prob. 15-10)
Solution:
vout(CM) = Av(CM)(vin(CM))
vout(CM) = 0.5(5 mV)
vout(CM) = 2.5 mV
IT = (VEE − VBE)RE (Eq. 15-8)
IT = (15 V − 0.7)/500 kΩ
IT = 28.6 μA
IE = IC = IT /2 (Eq. 15-6)
IE = IC = 28.6 μA/2
IE = IC = 14.3 μA
r'e = 25 mV/IE
r'e = 25 mV/14.3 μA
r'e = 1.75 kΩ
Since it is a single-ended output:
Av = Rc/2r'e (Eq. 15-9)
Av = 500 kΩ/2(1.75 kΩ)
Av = 143
vout1 = Av(v1 − v2) (Eq. 15-2)
vout1 = 143(2 mV − 0)
vout1 = 286 mV
Answer: The output voltage is 286 mV (desired) and
2.5 mV (common mode).
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15-12. Given:
Av = 100,000
CMRRdB = 70 dB
vin(CM) = 5 μV
Solution:
CMRR = antilog(CMRRdB/20)
CMRR = antilog(70 dB/20)
CMRR = 3162
Av(CM) = Av/CMRR
Av(CM) = 100,000/3,162
Av(CM) = 31.6
vout(CM) = Av(CM)(vin(CM))
vout(CM) = 31.6(5 μV)
vout(CM) = 158 μV
vout(desired) = Av(vin(desired))
vout = (100,000)(5 μV)
vout = 500 mV
Answer: The common-mode voltage gain is 31.6, and the
output voltage is 158 μV. The desired output is 500 mV.
15-13. Given:
RC = 500 kΩ
RE = 500 kΩ
VCC = 10 V
VEE = −10 V
Av(CM) = 0.5 (from Prob. 15-10)
Solution:
IT = (VEE − VBE)RE (Eq. 15-8)
IT = (10 V − 0.7)/500 kΩ
IT = 18.6 μA
IE = IC = IT /2 (Eq. 15-6)
IE = IC = 18.6 μA/2
IE = IC = 9.3 μA
r'e = 25 mV/IE
r'e = 25 mV/9.3 μA
r'e = 2.69 kΩ
Since it is a single-ended output:
Av = RC /2r'e (Eq. 15-9)
Av = 500 kΩ/2(2.69 kΩ)
Av = 93
CMRR = Av/Av(CM)
CMRR = 93/0.5
CMRR = 186
CMRRdB = 20 logCMRR
CMRRdB = 20 log186
CMRRdB = 45.4 dB
Answer: The common-mode rejection ratio is 45.4 dB.
15-14. Given:
Av = 150,000
CMRRdB = 85 dB
15-15. Given:
VCC = 12 V
VEE = −12 V
RC = 51 kΩ
RE = 51 kΩ
RL = 27 kΩ
v1 = 5 mV
Solution:
IT = (VEE − VBE)/RE (Eq. 15-8)
IT = (12 V − 0.7 V)/51 kΩ
IT = 221.6 μA
IE = IC = IT /2 (Eq. 15-6)
IE = IC = 221.6 μA/2
IE = IC = 110.8 μA
r'e = 25 mV/IE
r'e = 25 mV/110.8 μA
r'e = 225.6 Ω
Since it is differential output:
Av = RC/r'e (Eq. 15-10)
Av = 51 kΩ/225.6 Ω
Av = 226
vout = Av(v1 − v2) (Eq. 15-2)
vout = 226(5 mV − 0 V)
vout = 1.13 V
RTH = 2RC
RTH = 2(51 kΩ)
RTH = 102 kΩ
The output voltage is divided between the Thevenin resistance and the load resistance.
vL = [RL/(RTH + RL)]vout
vL = [27 kΩ/(102 kΩ + 27 kΩ)]vout
vL = 237 mV
Answer: The load voltage is 237 mV.
15-16. Given:
VCC = 12 V
VEE = −12 V
RC = 51 kΩ
RE = 51 kΩ
RL = 27 kΩ
v1 = 5 mV
Solution:
IT = (VEE − VBE)/RE (Eq. 15-8)
IT = (12 V − 0.7 V)/51 kΩ
IT = 221.6 μA
IE = IC = IT /2 (Eq. 15-6)
IE = IC = 221.6 μA/2
IE = IC = 110.8 μA
r'e = 25 mV/IE
r'e = 25 mV/110.8 μA
r'e = 225.6 Ω
Solution:
CMRR = antilog(CMRRdB/20)
CMRR = antilog(85 dB/20)
CMRR = 17,783
Since it is differential output:
Av = RC/r'e (Eq. 15-10)
Av = 51 kΩ/225.6 Ω
Av = 226
Av(CM) = Av/CMRR
Av(CM) = 150,000/17,783
Av(CM) = 8.4
vout = Av(v1 − v2) (Eq. 15-2)
vout = 226(5 mV − 0 V)
vout = 1.13 V
Answer: The common-mode voltage gain is 8.4.
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RTH = 2RC
RTH = 2(51 kΩ)
RTH = 102 kΩ
IL = vout/RTH
IL = 1.13 V/102 kΩ
IL = 11.1 μA
Answer: The load current is 11.1 μA.
15-17. Answer: With the open base, the right transistor will go
into cutoff and its collector voltage will go high. Since
there is still a path for current through the left transistor,
it will conduct hard and its collector voltage will go low.
Thus the output will be high and the input will have little
effect. A diff amp or an op amp needs a current path to
ground for both bases.
15-18. Given:
VCC = 12 V
VEE = −12 V
RC = 20 kΩ
RE = 200 kΩ
Solution:
IT = (VEE − VBE)/RE
IT = (12 V − 0.7 V)/200 kΩ
IT = 56.5 μA
IE = 1/2 IT
IE = 1/2 (56.5 μA)
IE = 28.3 μA
Vout = VCC − ICRC
Vout = 12 V − (28.3 μA)(20 kΩ)
Vout = 11.4 V
Answer: The output voltage is 11.4 V.
15-19. Answer: C. When the left base is open, all the tail current
must flow through the right transistor. This will pull the
output voltage down to almost zero.
CRITICAL THINKING
15-20. Given:
VCC = 12 V
VEE = −12 V
RC = 200 kΩ
RE = 200 kΩ
Solution:
IT = (VEE − VBE)/RE
IT = (12 V − 0.7 V)/200 kΩ
IT = 56.5 μA
IE = 1/2 IT
IE = 1/2 (56.5 μA)
IE = 28.3 μA
Vout = VCC − ICRC
Vout = 12 V − (28.3 μA)(20 kΩ)
Vout = 11.4 V
Answer: The output voltage is 11.4 V.
15-21. Given: Both bases are connected to ground.
Answer: Both bases are at 0 V.
15-22. Answer: This is a current mirror. With Eq. (15-25),
you can calculate an ideal tail current of 2 mA and a
second-approximation current of 1.95 mA. The current
through the active load is half the tail current, ideally,
1 mA.
15-23. Answer: The resistor has to be changed to an ideal value
of 30 V/15 μA, which equals 2 MΩ.
15-24. Given:
VCC = 12 V
VEE = −12 V
RC = 200 kΩ
RE = 200 kΩ
VBE = 0.7 V
Room temperature = 25°C
Left transistor: −2 mV/°C
Right transistor: −2.1 mV/°C
Solution:
ΔT = 75°C − 25°C
ΔT = 50°C
VBE(L) = 0.7 V + (−2 mV/°C) ΔT
VBE(L) = 0.7 V + (−2 mV/°C)50°C
VBE(L) = 0.6 V
VBE(R) = 0.7 V + (−2.1 mV/°C) ΔT
VBE(R) = 0.7 V + (−2.1 mV/°C)50°C
VBE(R) = 0.595 V
Since each transistor has half the tail current through
it, the tail resistor would appear twice as large to each
transistor.
IE = (VEE − VBE(R))RE (Eq. 15-8)
IE = (12 V − 0.595 V)/400 kΩ
IE = 28.5 μA
Vout = VCC − ICRC
Vout = 12 V − (28.5 μA)(200 kΩ)
Vout = 6.3 V
Answer: The output voltage is 6.0 V ideally, and 6.3 V
using the 2nd approximation.
15-25. Given:
R1 = 5 kΩ
R2 = 10 kΩ
RC1 = RC2 = RE = 2 kΩ
VCC = 15 V
VEE = −15 V
Solution:
VB(3) = [R1/(R1 + R2)]VEE
VB(3) = [5 kΩ/(5 kΩ + 10 kΩ)] −15 V
VB(3) = −5 V
VE(3) = VB + 0.7
VE(3) = −5 V − 0.7 V
VE(3) = −5.7 V
IE(3) = IT = (VEE − VE)/RE
IE(3) = IT = (15 V − 5.7 V)/2 kΩ
IE(3) = IT = 4.65 mA
IE(1) = IC(1) = IT /2 (Eq. 15-6)
IE(1) = IC(1) = 4.65 mA/2
IE(1) = IC(1) = 2.33 mA
r'e = 25 mV/IE
r'e = 25 mV/2.33 mA
r'e = 10.7 Ω
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Av = RC/r'e
Av = 2 kΩ/10.7 Ω
Av = 187
PROBLEMS
16-1.
Answer: The r'e is 10.7 Ω, and the gain is 187.
Negative saturation voltage is −17 V
15-26. Given:
RC1 = RC2 = RE = 5 kΩ
R2 = R4 = 10 kΩ
R1 = R3 = 20 kΩ
VCC = 30 V
Av = 100,000 (from Table 16-1)
Solution:
v2 = 17 V/100,000
v2 = 170 μV
Solution:
VB = [R2/(R2 + R1)]VCC
VB = [10 kΩ/(10 kΩ + 20 kΩ)]30 V
VB = 10 V
Answer: The input voltage required to drive the 741C op
amp into negative saturation is 170 μV.
16-2.
VE = VB − 0.7
VE = 10 V − 0.7 V
VE = 9.3 V
IE = IT = VE/RE
IE = IT = 9.3 V/5 kΩ
IE = IT = 1.86 mA
16-3.
vout = VCC − ICRC
vout = 30 V − (930 μA)(5 kΩ)
vout = 25.35 V
15-27. Q1 is open from collector to emitter
15-28. VEE is at 0 V, not −15 V
15-29. VCC is at 25 V, not 15 V
15-30. RE has changed to 33.8 kΩ
15-31. Q2 is open from collector to emitter
Operational Amplifiers
SELF-TEST
d
b
a
b
d
a
b
a
9.
10.
11.
12.
13.
14.
15.
16.
b
c
c
d
d
d
d
c
17.
18.
19.
20.
21.
22.
23.
24.
c
c
b
a
c
b
c
d
25.
26.
27.
28.
29.
30.
31.
b
b
d
c
c
a
b
Answer: The voltage gain at 1 kHz is 19,900, at 10 kHz is
2000, and at 100 kHz is 200.
16-4.
JOB INTERVIEW QUESTIONS
9. The LM318 is preferable when slew-rate distortion is too
high with a 741C. Applications include high-frequency
(wideband) video signals, in which a low slew rate produces
a smeared picture, and digital interface circuits, in which timing slew would be objectionable. A disadvantage is that the
LM318 is more likely to break into oscillation unless good
design and construction practices are used.
11. Audio amps, video amps, IF amps, RF amps, and voltage
regulators.
12. Open feedback loop.
Given:
Av(mid) = 200,000 (from Table 16-1)
funity = 20 MHz
Av(unity) = 1
Solution:
________
Av(unity) = Av(mid)______________
/[​√1 + (f/f2)2 ​
] (Eq. 14-3)
1 =______________
200,000/[​√1
+ (20 MHz/f2)2 ​]
[​√
1 + (20 MHz/f2)2 ​] = 200,000
1 + (20 MHz/f2)2 = 40,000,000,000
(20 MHz/f2)2 = 40,000,000,001
20 MHz/f2 = 200,000
f2 = 20 MHz/200,000
f2 = 100 Hz
________
Av(1k) = Av(mid)/[​√1 +
(f/f2)2 ​
] (Eq. 14-3)
_________________
Av(1k) = 200,000/[​√1
+ (1 kHz/100 Hz)2 ​]
Av(1k) = 19,900
________
Av(10k) = Av(mid)/[​√1 +
(f/f2)2 ​
] (Eq. 14-3)
__________________
Av(10k) = 200,000/[​√1
+ (10 kHz/100 Hz)2 ​]
Av(10k) = 2000
________
Av(100k) = Av(mid)/[​√1 +___________________
(f/f2)2 ​
] (Eq. 14-3)
Av(100k) = 200,000/ [​√
1 + (100 kHz/100 Hz)2 ​]
Av(100k) = 200
Answer: The tail current is 1.86 mA, and the output
voltages are 25.35 V.
1.
2.
3.
4.
5.
6.
7.
8.
Given: CMRR = 100 dB (from Table 16-1)
Solution:
Av = antilog(AdB/20) (Eq. 14-15)
Av = antilog(100 dB/20)
Av = 100,000
Answer: The common-mode rejection ratio is 100 dB or
100,000.
IE = IC = IT /2 (Eq. 15-6)
IE = IC = 1.86 mA/2
IE = IC = 0.93 mA
Chapter 16
Given:
VCC = ±18 V
Given:
Δvout = 2 V
Δt = 0.4 μs
Solution:
SR = Δvout/Δt
SR = 2 V/0.4 μs
SR = 5 V/μs
Answer: The slew rate is 5 V/μs.
16-5.
Given:
SR = 70 V/μs
Vp = 7 V
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Av(dB) = 20 logAv(CL)
Av(dB) = 20 log(10)
Av(dB) = 20
Solution:
fmax = SR/2π (VP) (Eq. 16-2)
fmax = 70 V/μs/2π (7 V)
fmax = 1.59 MHz
Answer: The closed-loop voltage gain is 10, the bandwidth is 2 MHz, the output voltage at 1 kHz is 250 mVp-p,
and the output voltage at 10 MHz is 49 mVp-p.
Answer: The power bandwidth is 1.59 MHz.
16-6a. Given:
SR = 0.5 V/μs
VP = 1 V
Solution:
fmax = SR/2π(VP) (Eq. 16-2)
fmax = 0.5 V/μs/2π (1 V)
fmax = 79.6 kHz
Av(dB)
20 dB
20 dB/decade
Answer: The power bandwidth is 79.6 kHz.
16-6b. Given:
SR = 3 V/μs
VP = 5 V
Solution:
fmax = SR/2π (VP) (Eq. 16-2)
fmax = 3 V/μs/2π(5 V)
fmax = 95.5 kHz
2 MHz
Ideal Bode plot for Prob. 16-7.
16-8.
Answer: The power bandwidth is 95.5 kHz.
16-6c. Given:
SR = 15 V/μs
VP = 10 V
RB2 = Rf || R1 (Eq. 16-11)
RB2 = 300 kΩ || 15 kΩ
RB2 = 14.29 kΩ
Answer: The power bandwidth is 239 kHz.
Given:
R1 = 180 Ω
Rf = 1.8 kΩ
vin = 25 mVp-p
funity = 20 MHz
V1 err = (RB1 − RB2)Iin(bias) (Eq. 16-8)
V1 err = (0 − 14.29 kΩ)(30 pA)
V1 err = −429 nV
V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9)
V2 err = (0 + 14.29 kΩ)(3 pA/2)
V2 err = 21.4 nV
Solution:
Av(CL) = −Rf/R1 (Eq. 16-3)
Av(CL) = −1.8 kΩ/180 Ω
Av(CL) = −10
f2(CL) = funity/AvCL (Eq. 16-5)
f2(CL) = 20 MHz/10
f2(CL) = 2 MHz
At 1 kHz the voltage gain is the closed loop gain.
vout = Av(CL)(vin)
vout = 10(25 mVp-p)
vout = 250 mVp-p
At 10 MHz the voltage gain is reduced.
Av(CL)
Av(10 MHz) = __________
​  ________ ​
√
​ 1 + (f/f2)2 ​
10
____________________
Av(10 MHz) = ​
__________________  ​
​√ 1
+ (10 MHz/2 MHz)2 ​
Given:
Iin(bias) = 30 pA
Iin(off ) = 3 pA
Vin(off ) = 1 mV
Solution:
Av(CL) = −Rf /R1 (Eq. 16-3)
Av(CL) = −300 kΩ/15 kΩ
Av(CL) = −20
Solution:
fmax = SR/2π (VP) (Eq. 16-2)
fmax = 15 V/μs/2π(10 V)
fmax = 239 kHz
16-7.
f
20 MHz
V3 err = Vin(off) = 1 mV
Verror = ± Av(CL)(±V1 err ± V2 err ± V3 err)
Verror = 20(429 nV + 21.4 nV + 1 mV)
Verror = 20 mV
Answer: The output voltage is 20 mV.
16-9.
Given:
Iin(bias) = 50 pA
Iin(off) = 10 pA
Vin(off) = 2 mV
Solution:
Av(CL) = −Rf /R1 (Eq. 16-3)
Av(CL) = −300 kΩ/15 kΩ
Av(CL) = −20
Av(10 MHz) = 1.96
RB2 = Rf || R1 (Eq. 16-11)
RB2 = 300 kΩ || 15 kΩ
RB2 = 14.29 kΩ
vout = Av(10 MHz)(vin)
vout = 1.96(25 mVp-p)
vout = 49 mVp-p
V1 err = (RB1 − RB2)Iin(bias) (Eq. 16-8)
V1 err = (0 − 14.29 kΩ)(50 pA)
V1 err = −714.3 nV
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V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9)
V2 err = (0 + 14.29 kΩ)(10 pA/2)
V2 err = 71.4 nV
V3 err = Vin(off) = 2 mV
Verror = ± Av(CL)(± V1 err ± V2 err ± V3 err)
Verror = 20(714.3 nV + 71.4 nV + 2 mV)
Verror = 40 mV
Answer: The output voltage is 40 mV.
16-10. Given:
R1 = 150 Ω
Rf = 3 kΩ
vin = 25 mVp-p
funity = 20 MHz
Solution:
Av(CL) = (Rf /R1) + 1 (Eq. 16-12)
Av(CL) = (3 kΩ/150 Ω) + 1
Av(CL) = 21
f2(CL) = funity/Av(CL) (Eq. 16-5)
f2(CL) = 20 MHz/21
f2(CL) = 952 kHz
At 100 kHz the voltage gain is the closed loop gain.
vout = Av(CL)(vin)
vout = 21(25 mVp-p)
vout = 525 mVp-p
Answer: The closed-loop gain is 21, the bandwidth
is 952 kHz, and the output voltage at 100 kHz is
525 mVp-p.
16-11. Given:
Iin(bias) = 50 pA
Iin(off) = 10 pA
Vin(off) = 2 mV
Av(CL) = 21 (from Prob. 16-10)
Solution:
RB2 = R1 || Rf (Eq. 16-11)
RB2 = 150 Ω || 3 kΩ
RB2 = 142.9 Ω
V1 err = (RB1 − RB2)Iin(bias) (Eq. 16-8)
V1 err = (0 − 142.9 Ω)(50 pA)
V1 err = −7.15 nV
V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9)
V2 err = (0 + 142.9 Ω)(10 pA/2)
V2 err = 715 nV
V3 err = Vin(off) = 2 mV
Verror = ± Av(CL)(±V1 err ± V2 err ± V3 err)
Verror = 21(7.15 nV + 715 pV + 2 mV)
Verror = 42 mV
Answer: The output voltage is 42 mV.
16-12. Given:
R1 = 10 kΩ
R2 = 20 kΩ
R3 = 40 kΩ
Rf = 40 kΩ
v1 = 50 mVp-p
v2 = 90 mVp-p
v3 = 160 mVp-p
Solution:
Av1(CL) = −Rf /R1 (Eq. 16-3)
Av1(CL) = −40 kΩ/10 kΩ
Av1(CL) = −4
Av2(CL) = −Rf /R2 (Eq. 16-3)
Av2(CL) = −40 kΩ/20 kΩ
Av2(CL) = −2
Av3(CL) = −Rf /R3 (Eq. 16-3)
Av3(CL) = −40 kΩ/40 kΩ
Av3(CL) = −1
vout = Av1(CL)(vin1) + Av2(CL)(vin2) + Av3(CL)(vin3)
vout = −4(50 mVp-p) + −2(90 mVp-p) + −1(160 mVp-p)
vout = −540 mVp-p
RB2 = R1 || R2 || R3 || Rf (Eq. 16-14)
RB2 = 10 kΩ || 20 kΩ || 40 kΩ || 40 kΩ
RB2 = 5 kΩ
Answer: The output voltage is 540 mVp-p, and the compensating resistor should be 5 kΩ.
16-13. Given:
It is a voltage follower.
funity = 1 MHz (from Table 16-1)
vin = 50 mVp-p
Solution:
Av(CL) = 1 (Eq. 16-15)
f2(CL) = funity = 1 MHz
vout = Av(CL)(vin)
vout = 1(50 mVp-p)
vout = 50 mVp-p
Answer: The output voltage is 50 mVp-p, and the bandwidth is 1 MHz.
CRITICAL THINKING
16-14. Given:
R1 = 1 kΩ
Rf(max) = 101 kΩ
Rf(min) = 1 kΩ
funity = 20 MHz
Solution:
Av(CL)max = –Rf(max)/R1 (Eq. 16-3)
Av(CL)max = –101 kΩ/1 kΩ
Av(CL)max = 101 (the minus sign for phase inversion is
ignored here)
Av(CL)min = –Rf(min)/R1 (Eq. 16-3)
Av(CL)min = –1 kΩ/1 kΩ
Av(CL)min = 1
f2(CL)max = funity/(Av(CL)min + 1)
f2(CL)max = 20 MHz/2
f2(CL)max = 10 MHz
f2(CL)min = funity/Av(CL)max (Eq. 16-5)
f2(CL)min = 20 MHz/101
f2(CL)min = 198 kHz
Answer: The voltage gain has a range of 1 to 101 and a
bandwidth of 198 kHz to 10 MHz.
16-15. Given:
R1 = 2 kΩ
Rf(max) = 100 kΩ
Rf(min) = 0 kΩ
funity = 20 MHz
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Solution:
Av(CL)max = (Rf(max)/R1) + 1 (Eq. 16-12)
Av(CL)max = (100 kΩ/2 kΩ) + 1
Av(CL)max = 51
Av(CL)min = (Rf(min)/R1) + 1 (Eq. 16-12)
Av(CL)min = (0 kΩ/2 kΩ) + 1
Av(CL)min = 1
f2(CL)max = funity/Av(CL)min (Eq. 16-5)
f2(CL)max = 20 MHz/1
f2(CL)max = 20 MHz
f2(CL)min = funity/Av(CL)max (Eq. 16-5)
f2(CL)min = 20 MHz/51
f2(CL)min = 392 kHz
Answer: The voltage gain has a range of 1 to 51 and a
bandwidth of 392 kHz to 20 MHz.
16-16. Answer: The voltage across the closed-loop output
impedance is the difference between the ideal 50 mV and
the actual 49.98 mV. In other words, 0.02 mV is dropped
across the closed-loop output impedance. The load current is 49.98 mV divided by 2 Ω, which is approximately
25 mA. Divide 0.02 mV by 25 mA to get 0.0008 Ω for the
closed-loop output impedance.
16-17. Given:
f = 15 kHz
VP = 2 V
Solution:
SS = 2πfVP
SS = 2π(15 kHz)(2 V)
SS = 188 mV/μs
SS = 2πfVP
SS = 2π(30 kHz)(2 V)
SS = 376 mV/μs
Answer: The initial slope is 188 mV/μs, with a peak of
2 V, and 376 mV/μs, with a frequency of 30 kHz.
16-18. Answer:
a. OP-07A
b. TL082 and TL084
c. LM3876
d. LM7171
e. OP-07A
16-19. Answer:
CMRR = 38 dB (from Fig. 16-7a)
MPP = 21 V (from Fig. 16-7b)
Av = 1000 (from Fig. 16-7c)
16-20. Given:
R1 = 10 kΩ
R2 = 20 kΩ
R3 = 40 kΩ
Rf(max) = 100 kΩ
Rf(min) = 0 Ω
v1 = 50 mVp-p
v2 = 90 mVp-p
v3 = 160 mVp-p
Solution: When the resistance is zero, the voltage gains
are zero and the output voltage is zero.
–Av1(CL)max = – Rf /R1 (Eq. 16-3)
–Av1(CL)max = –100 kΩ/10 kΩ
–Av1(CL)max = –10
–Av2(CL)max = –Rf/R2 (Eq. 16-3)
–Av2(CL)max = –100 kΩ/20 kΩ
–Av2(CL)max = –5
–Av3(CL)max = –Rf /R3 (Eq. 16-3)
–Av3(CL)max = –100 kΩ/40 kΩ
–Av3(CL)max = –2.5
vout = Av1(CL)max(vin1) + Av2(CL)max(vin2) + Av3(CL)max(vin3)
vout = 10(50 mVp-p) + 5(90 mVp-p) + 2.5(160 mVp-p)
vout = 1.35 Vp-p
Answer: The maximum output voltage is 1.35 Vp-p, and
the minimum output voltage is zero.
16-21. Given:
R1 = 220 Ω
Rf1 = 47 kΩ
Rf2 = 18 kΩ
Rf3 = 39 kΩ
Solution:
–Av1(CL) = –Rf1/R1 (Eq. 16-3)
–Av1(CL) = –47 kΩ/220 Ω
–Av1(CL) = –214
–Av2(CL) = –Rf2/R1 (Eq. 16-3)
–Av2(CL) = –18 kΩ/220 Ω
–Av2(CL) = –82
–Av3(CL) = –Rf3/R1 (Eq. 16-3)
–Av3(CL) = –39 kΩ/220 Ω
–Av3(CL) = –177
Solution: The gain at position 1 is 214, at position 2 is 82,
and at position 3 is 177.
16-22. Given:
R1 = 6 kΩ at position 2
R1 = 6 kΩ || 3 kΩ at position 1 = 2 kΩ
Rf = 120 kΩ
funity = 1 MHz
Solution:
Av1(CL) = (Rf/R1) + 1 (Eq. 16-12)
Av1(CL) = (120 kΩ/2 kΩ) + 1
Av1(CL) = 61
Av2(CL) = (Rf/R1) + 1 (Eq. 16-12)
Av2(CL) = (120 kΩ/6 kΩ) + 1
Av2(CL) = 21
f2(CL)1 = funity/Av(CL1) (Eq. 16-5)
f2(CL)1 = 1 MHz/61
f2(CL)1 = 16.4 kHz
f2(CL)2 = funity/Av(CL1)(max) (Eq. 16-5)
f2(CL)2 = 1 MHz/21
f2(CL)2 = 47.6 kHz
Answer: The voltage gain at position 1 is 61, with a bandwidth of 16.4 kHz, and at position 2 is 21, with a bandwidth of 47.6 kHz.
16-23. Given:
R1 = ∞ at position 2
R1 = 3 kΩ at position 1
Rf = 120 kΩ
funity = 1 MHz
AVOL = 100,000
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Solution:
Av1(CL) = (Rf/R1) + 1 (Eq. 16-12)
Av1(CL) = (120 kΩ/3 kΩ) + 1
Av1(CL) = 41
V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9)
V2 err = (0 + 100 kΩ)(200 nA/2)
V2 err = 10 mV
At position 2, it becomes a voltage follower: Av2(CL) = 1.
Av(CL) = (Rf/R'1) + 1 (Eq. 16-12)
Av(CL) = (100 kΩ/∞) + 1
Av(CL) = 1
Answer: The voltage gain at position 1 is 41, and at
position 2 is 1.
16-24. Answer: The output will go to positive or negative
saturation.
16-25. Answer:
Position 1: The input voltage is applied directly to the
noninverting input. Because of the virtual short between
the noninverting and inverting input terminals, there is
no ac voltage across the left 10-kΩ resistor. Since there is
no ac voltage across the resistor, it can be removed from
the circuit without changing the operation. With the resistor removed, the circuit reduces to a voltage follower and
Av(CL) = 1 and a closed-loop bandwidth of
funity
f2(CL) = _____
​
​ = _____
​  1 MHz
​ = 1 MHz
Av(CL)
1
Position 2: The circuit is an inverting amplifier. The
magnitude of the voltage gain is Av(CL) = 1. Note that the
closed-loop bandwidth is only half as much because
funity
1 MHz ​ = 500 kHz
f2(CL) = ​  ________ ​ = ​ ______
Av(CL) + 1
1+1
This was covered briefly in the chapter. See the equation
at the top of p. 682 and the brief explanation that follows.
Chapter 17 discusses the closed-loop bandwidths in more
detail.
16-26. Answer:
Position 1: With the left resistor open, the circuit reduces
to a voltage follower and Av(CL) = 1.
Position 2: With the left resistor open, the voltage gain
is zero.
16-27. Answer: Go to positive or negative saturation.
16-28. Given:
Iin(bias) = 500 nA
Iin(off) = 200 nA
Vin(off) = 6 mV
R1 = 2 kΩ
Rf = 100 kΩ
C = 1 µF
Solution:
XC = 1/2πfC
XC = 1/[2π(0)(1 µF)]
XC = ∞
R'1 = XC + R1
R'1= ∞ + 2 kΩ
R'1= ∞
RB2 = R1 || Rf (Eq. 16-11)
RB2 = ∞ || 100 kΩ
RB2 = 100 kΩ
V1 err = (RB1 – RB2)Iin(bias) (Eq. 16-8)
V1 err = (0 – 100 kΩ)(500 nA)
V1 err = 50 mV
V3 err = Vin(off) = 6 mV
Verror = ± Av(CL)(±V1 err ±V2 err ±V3 err)
Verror = 1(50 mV + 10 mV + 6 mV)
Verror = 66 mV
Answer: The output error voltage is 66 mV.
16-29. Given:
R1 = 2 kΩ
Rf = 100 kΩ
C = 1 μF
vin = 50 mVp-p
f = 1 kHz
Solution:
XC = 1/2πfC
XC = 1/[2π(1 kHz)(1 μF)]
XC = 159 Ω
Since XC is less than one-tenth of 2 kΩ, the bottom of the
2 kΩ is approximately an ac ground.
Av(CL) = (Rf/R'1) + 1 (Eq. 16-12)
Av(CL) = (100 kΩ/2 kΩ) + 1
Av(CL) = 51
vout = Av(CL)vin
vout = 51(50 mVp-p)
vout = 2.55 Vp-p
Answer: The output voltage is 2.55 Vp-p.
16-30. Given:
Iin(bias) = 500 nA
Iin(off) = 200 nA
Vin(off) = 6 mV
R1 = 2 kΩ
Rf = 100 kΩ
Solution:
R'1 = XC = R1
R'1 = 0 + 2 kΩ
R'1 = 2 kΩ
RB2 = R1 || Rf (Eq. 16-11)
RB2 = 2 kΩ || 100 kΩ
RB2 = 1.96 kΩ
V1 err = (RB1 – RB2)Iin(bias) (Eq. 16-8)
V1 err = (0 – 1.96 kΩ)(500 nA)
V1 err = 980 μV
V2 err = (RB1 + RB2)(Iin(off)/2)
(Eq. 16-9)
V2 err = (0 + 1.96 kΩ)(200 nA/2)
V2 err = 196 μV
V3 err = Vin(off) = 6 mV
Av(CL) = (Rf /R'1) + 1 (Eq. 16-12)
Av(CL) = (100 kΩ/2 kΩ) + 1
Av(CL) = 51
Verror = ±Av(CL)(±V1 err ± V2 err ± V3 err)
Verror = 51(980 μV + 196 μV + 6 m∆V)
Verror = 366 mV
Answer: The output voltage is 366 mV.
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16-31. Rf changed to 9 kΩ
Solution:
B = R1/(R1 + Rf) (Eq. 17-6)
B = 2.7 kΩ/(2.7 kΩ + 39 kΩ)
B = 0.065
16-32. Rf changed to 94 kΩ
16-33. R1 4.7 kΩ, not 470 Ω
16-35. Op amp has failed
Av = 1/B (Eq. 17-4)
Av = 1/0.065
Av = 15.44
16-36. Output of the XR2206 (U6)
Answer: The feedback fraction is 0.065, and the closedloop voltage gain is 15.44.
16-34. Rf is 3.9 kΩ instead of 39 kΩ
16-37. Push-pull Class-B/AB power amp
17-3.
16-38. 24 Vp-p
16-39. 10
Solution:
B = R1/(R1 + Rf) (Eq. 17-6)
B = 4.7 kΩ/(4.7 kΩ + 68 kΩ)
B = 0.065
16-40. Output would latch at plus and minus Vmax
Chapter 17
Negative Feedback
Av = 1/B (Eq. 17-4)
Av = 1/0.065
Av = 15.47
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
b
d
a
a
a
c
b
8.
9.
10.
11.
12.
13.
14.
b
b
b
d
b
b
b
15.
16.
17.
18.
19.
20.
21.
b
d
c
b
c
b
c
22.
23.
24.
25.
26.
27.
28.
d
d
b
a
b
d
a
Answer: The feedback fraction is 0.065, and the closedloop voltage gain is 15.47.
17-4.
JOB INTERVIEW QUESTIONS
PROBLEMS
Av = 1/B (Eq. 17-4)
Av = 1/0.038
Av = 26.32
Given:
R1 = 2.7 kΩ
Rf = 68 kΩ
AVOL(dB) = 88 dB
Av = antilog(AVOL(dB)/20) (Eq. 14-15)
Av = antilog(108 dB/20)
Av = 251,189
Solution:
B = R1/(R1 + Rf) (Eq. 17-6)
B = 2.7 kΩ/(2.7 kΩ + 68 kΩ)
B = 0.038
% = 100%/(1 + AVOLB) (Eq 17-5)
%error = 100%/[1 + 251,189(0.038)]
%error = 0.01%
Av = 1/B (Eq. 17-4)
Av = 1/0.038
Av = 26.32
Av = AVOL/(1 + AVOLB) (Eq. 17-3)
Av = 251,189/[l + 251,189(0.038)]
Av = 26.31
Av = antilog(Av(dB)/20) (Eq. 14-15)
Av = antilog(88 dB/20)
Av = 25,119
%error = 100%(1 + AVOLB) (Eq 17-5)
%error = 100%[1 + 25,119(0.038)]
%error = 0.10%
Av = AVOL/(1 + AVOLB) (Eq. 17-3)
Av = 25,119/[l + 25,119(0.038)]
Av = 26.29
Answer: The feedback fraction is 0.038, the ideal closedloop voltage gain is 26.32, the percent error is
0.10%, and the exact voltage gain is 26.29.
17-2.
Given:
R1 = 2.7 kΩ
Rf = 39 kΩ
AVOL(dB) = 88 dB
Given:
R1 = 2.7 kΩ
Rf = 68 kΩ
AVOL(dB) = 108 dB
Solution:
B = R1/(R1 + Rf) (Eq. 17-6)
B = 2.7 kΩ/(2.7 kΩ + 68 kΩ)
B = 0.038
8. Increased voltage gain and possible oscillation.
12. Current amplifier and transconductance amplifier.
17-1.
Given:
R1 = 4.7 kΩ
Rf = 68 kΩ
AVOL(dB) = 88 dB
Answer: The feedback fraction is 0.038, the ideal closedloop voltage gain is 26.32, the percent error is 0.01%, and
the exact voltage gain is 26.31.
17-5.
Given:
R1 = 100 Ω
Rf = 7.5 kΩ
Rin = 3 MΩ
RCM = 500 MΩ
AVOL = 200,000
Solution:
B = R1/(R1 + Rf ) (Eq. 17-6)
B = 100 Ω/(100 Ω + 7.5 kΩ)
B = 0.013
zin(CL) = (1 + AVOLB)Rin || RCM (Eq. 17-8)
zin(CL) = [1 + 200,000(0.013)]3 MΩ || 500 MΩ
zin(CL) = 470 MΩ
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Answer: The closed-loop input impedance is 470 MΩ.
17-6.
Solution:
iout = vin/R1 (Eq. 17-19)
iout = 0.5 V/2.7 Ω
iout = 185 mArms
Given:
AVOL = 75,000
Rout = 50 Ω
B = 0.013 (from Prob. 17-5)
PL = (iout)2RL
PL = (185 mA)2(1 Ω)
PL = 34.2 mW
Solution:
zout(CL) = Rout/(1 + AVOLB) (Eq. 17-10)
zout(CL) = 50 Ω/[1 + (75,000)(0.013)]
zout(CL) = 0.051 Ω
Answer: The
0.051 Ω.
17-7.
closed-loop
output
impedance
Answer: The output current is 185 mArms, and the load
power is 34.2 mW.
is
Given:
AVOL = 200,000
B = 0.013 (from Prob. 17-5)
THDOL = 10%
Solution:
THDCL = THDVOL/(1 + AVOLB) (Eq. 17-12)
THDCL = 10%/[1 + 200,000(0.013)]
THDCL = 0.0038%
Answer: The closed-loop total harmonic distortion is
0.0038%.
17-8.
Given:
iin = 20 mApeak
Rf = 51 kΩ
f = 1 kHz
Solution:
vout = –(iinR2) (Eq. 17-14)
vout = –(20 μApeak)(51 kΩ)
vout = –1.02 Vpeak at 1 kHz
Answer: The output voltage is –1.02 Vpeak at 1 kHz.
17-9.
Given:
iin = 20 μApeak
Rf = 33 kΩ
f = 1 kHz
Solution:
vout = –(iinRf)
vout = (–20 μApeak)(33 kΩ)
vout = –0.660 Vpeak
Answer: The output voltage is –0.660 Vpeak.
17-10. Given:
iin = –10 μArms
Rf = 51 kΩ
f = 1 kHz
Solution:
vout = –(iinRf)
vout = (–10 μA)(51 kΩ)
vout = –510 Vrms
Vp-p = –1.44 Vp-p
Answer: The peak-to-peak output voltage is –1.44 V at
1 kHz.
17-11. Given:
R1 = 2.7 Ω
RL = 1 Ω
vin = 0.5 Vrms
17-12. Given:
R1 = 2.7 Ω
RL = 3 Ω
vin = 0.5 Vrms
Solution:
iout = vin/R1 (Eq. 17-19)
iout = 0.5 Vrms/2.7 Ω
iout = 185 mArms
PL = (iout)2RL
PL = (185 mArms)2(3 Ω)
PL = 277 mW
Answer: The output current is 185 mArms, and the load
power is 277 mW.
17-13. Given:
R1 = 4.7 Ω
RL = 1 Ω
vin = 0.5 Vrms
Solution:
iout = vin/R1 (Eq. 17-19)
iout = 0.5 V/4.7 Ω
iout = 106 mArms
PL = (iout)2RL
PL = (106 mA)2(1 Ω)
PL = 11.2 mW
Answer: The output current is 106 mArms, and the load
power is 11.2 mW.
17-14. Given:
R1 = 1.8 Ω
R2 = 1.5 kΩ
RL = 1 Ω
iin = 1 mAp-p
Solution:
Ai = R2/R1 + 1 (Eq. 17-23)
Ai = 1.5 kΩ/1.8 Ω + 1
Ai = 834
iout = Aiiin
iout = 834(1 mAp-p)
iout = 834 mAp-p
irms = ip-p /2.828
irms = 834 mAp-p/2.828
irms = 295 mArms
PL = (iout)2RL
PL = (295 mA)2(1 Ω)
PL = 87 mW
Answer: The current gain is 834, and the load power is
87 mW.
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17-15. Given:
R1 = 1.8 Ω
R2 = 1.5 kΩ
RL = 2 Ω
iin = 1 mAp-p
Ai = 834 (from Prob. 17-14)
Solution:
iout = Aiiin
iout = 834(1 mAp-p)
iout = 834 mAp-p
irms = ip-p /2.828
irms = 834 mAp-p/2.828
irms = 295 mArms
PL = (iout)2RL
PL = (295 mA)2(2 Ω)
PL = 174 mW
Answer: The output current is 834 mAp-p, and the load
power is 174 mW.
17-16. Given:
R1 = 7.5 Ω
R2 = 1.5 kΩ
RL = 1 Ω
iin = 1 mAp-p
Solution:
Ai = R2/R1 + 1 (Eq. 17-23)
Ai = (1.5 kΩ/7.5 Ω) + 1
Ai = 201
iout = AiIin
iout = (201)(1 mAp-p)
iout = 201 mAp-p
17-19. Given:
AVOL = 20,000
f2(OL) = 750 Hz
Solution:
f2(CL) = (1 + AVOL)f2(OL) (from Table 17-2)
f2(CL) = (1 + 20,000)(750 Hz)
f2(CL) = 15 MHz
Answer: The closed-loop bandwidth is 15 MHz.
17-20. Given:
(1 + AVOLB) = 5000
f2(OL) = 120 Hz
Solution:
f2(CL) = (1 + AVOLB) f2(OL) (from Table 17-2)
f2(CL) = 5000(120 Hz)
f2(CL) = 600 kHz
Answer : The closed-loop bandwidth is 600 kHz.
17-21. Given:
funity = 1 MHz
SR = 0.5 V/µs
Av(CL) = 10
Solution:
f2(CL) = funity/Av(CL) (Eq. 17-27)
f2(CL) = 1 MHz/10
f2(CL) = 100 kHz
VP(max) = SR/(2πf2(CL)) (Eq. 17-31)
VP(max) = 0.5 V/μs/(2π100 kHz)
VP(max) = 796 mVp
Answer: The closed-loop bandwidth is 100 kHz, and the
maximum peak voltage is 796 mV.
irms = ip-p/(2.828)
irms = 201 mAp-p/(2.828)
irms = 71.1 mArms
CRITICAL THINKING
PL = (iout)2RL
PL = (71.1 mArms)2(1 Ω)
PL = 5 mW
17-22. Given:
Rf = 150 kΩ
iin = 4 μA
Answer: The current gain is 201, and the load power is
5 mW.
17-17. Given:
(1 + AVOLB) = 1,000
f2(OL) = 2 Hz
Solution:
f2(CL) = (1 + AVOLB)f2(OL)
f2(CL) = (1000)(2 Hz)
f2(CL) = 2 kHz
Answer: The closed-loop bandwidth is 2 kHz.
17-18. Given:
AVOL = 316,000
f2(OL) = 4.5 Hz
Av(CL) = 75
Solution:
funity = AVOLf2(OL) = 316,000(4.5 Hz) = 1.42 MHz
f2(CL) = funity/Av(CL) (Eq. 17-27)
f2(CL) = 1.42 MHz/75
f2(CL) = 18.9 kHz
Answer: The closed-loop bandwidth is 18.9 kHz.
Solution:
vout = iinRf
vout = 4 μA(150 kΩ)
vout = 600 mV
Answer: The output voltmeter reads 600 mV.
17-23. Given:
Rf1 = 10 kΩ
iin = 1 μA
R1 = 1 kΩ
Rf2 = 99 kΩ
Solution:
vout(1) = iinRf (from Table 17-2)
vout(1) = 1 μA(10 kΩ)
vout(1) = 10 mV
Av(2) = Rf2/R1 + 1
Av(2) = 99 kΩ/1 kΩ + 1
Av(2) = 100
vout(2) = Av(vin(2))
vout(2) = 100(10 mV)
vout(2) = 1 V
Answer: The output voltage is 1 V.
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17-24. Given:
Rf = 50 kΩ
R1 = 1 kΩ
R2 = 25 kΩ
R3 = 100 kΩ
Solution:
Av1(CL) = Rf/R1 + 1
Av1(CL) = 50 kΩ/1 kΩ + 1
Av1(CL) = 51
B(2) = R2/(R2 + Rf) (Eq. 17-6)
B(2) = 25 kΩ/(25 kΩ + 50 kΩ)
B(2) = 0.333
zin 2(CL) = (1 + AVOLB(2))Rin (Eq. 17-8)
zin 2(CL) = (1 + (100,000)(0.333))2 MΩ
zin 2(CL) = 66,669 MΩ
zout 2(CL) = Rout/(1 + AVOLB(2)) (Eq. 17-10)
zout 2(CL) = 75 Ω/(1 + (100,000)(0.333))
zout 2(CL) = 2.5 mΩ
Av2(CL) = Rf/R2 + 1
Av2(CL) = 50 kΩ/25 kΩ + 1
Av2(CL) = 3
B(3) = R3/(R3 + Rf) (Eq. 17-6)
B(3) = 100 kΩ/(100 kΩ + 50 kΩ)
B(3) = 0.667
Av3(CL) = Rf/R3 + 1
Av3(CL) = 50 kΩ/100 kΩ + 1
Av3(CL) = 1.5
zin 3(CL) = (1 + AVOLB(3))Rin (Eq. 17-8)
zin 3(CL) = (1 + (100,000)(0.667))2 MΩ
zin 3(CL) = 133,335 MΩ
Answer: The voltage gains are 51 at the 1-kΩ position,
3 at the 25-kΩ position, and 1.5 at the 100-kΩ position.
zout3(CL) = Rout/(1 + AVOLB(3)) (Eq. 17-10)
zout3(CL) = 75 Ω/(1 + (100,000)(0.667))
zout3(CL) = 1.25 mΩ
17-25. Given:
Rf = 50 kΩ
R1 = 1 kΩ
R2 = 25 kΩ
R3 = 100 kΩ
Av1(CL) = 51 (from Prob. 17-24)
Av2(CL) = 3 (from Prob. 17-24)
Av3(CL) = 1.5 (from Prob. 17-24)
vin = 10 mV
Solution:
vout(1) = Av1(CL)(vin)
vout(1) = 51(10 mV)
vout(1) = 510 mV
vout(2) = Av2(CL)(vin)
vout(2) = 3(10 mV)
vout(2) = 30 mV
vout(3) = Av3(CL)(vin)
vout(3) = 1.5(10 mV)
vout(3) = 15 mV
Answer: The output voltages are 510 mV at the 1-kΩ
position, 30 mV at the 25-kΩ position, and 15 mV at the
100-kΩ position.
17-26. Given:
Rf = 50 kΩ
R1 = 1 kΩ
R2 = 25 kΩ
R3 = 100 kΩ
Av1(CL) = 51 (from Prob. 17-24)
Av2(CL) = 3 (from Prob. 17-24)
Av3(CL) = 1.5 (from Prob. 17-24)
AVOL = 100,000
Rin = 2 MΩ
Rout = 75 Ω
Solution:
B(1) = R1/(R1 + Rf) (Eq. 17-6)
B(1) = 1 kΩ/(1 kΩ + 50 kΩ)
B(1) = 0.0196
zin 1(CL) = (1 + AVOLB(1))Rin (Eq. 17-8)
z in 1(CL) = (1 + (100,000)(0.0196))2 MΩ
z in 1(CL) = 3924 MΩ
zout 1(CL) = Rout/(1 + AVOLB(1)) (Eq. 17-10)
zout 1(CL) = 75 Ω/(1 + (100,000)(0.0196))
zout 1(CL) = 38 mΩ
Answer: At the 1-kΩ position the input impedance is
3,924 MΩ and the output impedance is 38 mΩ. At the
25-kΩ position the input impedance is 66,669 MΩ
and the output impedance is 2.5 mΩ. At the 100-kΩ
position the input impedance is 133,335 MΩ and the
output impedance is 1.25 mΩ. Note: The RCM of the
op amp is not included in the calculations for input
impedance. See Example 17-2.
17-27. Given:
Iin(bias) = 80 nA
Iin(off) = 20 nA
Vin(off) = 1 mV
AVOL = 100,000
Rf = 100 kΩ
R1= 1 kΩ
R2 = 25 kΩ
R3 = 100 kΩ
Av1(CL) = 101
Av2(CL) = 5
Av3(CL) = 2
Solution:
RB2(1) = R1 || Rf (Eq. 16-11)
RB2(1) = 1 kΩ || 100 kΩ
RB2(1) = 990 Ω
V1 err(1) = (RB1 – RB2(1))Iin(bias) (Eq. 16-8)
V1 err(1) = (0 – 990 Ω)(80 nA)
V1 err(1) = – 79.2 µV
V2 err(2) = (RB1 + RB2(1))(Iin(off)/2) (Eq. 16-9)
V2 err(2) = (0 + 990 Ω)(20 nA/2)
V2 err(2) = 9.9 µV
V3 err(1) = Vin(off) = 1 mV
Verror(1) = ±Av(CL)(±V1 err(1) ± V2 err(1) ± V3 err(1))
Verror(1) = 101(79.2 µV + 9.9 µV + 1 mV)
Verror(1) = 110 mV
RB2(2) = R2 || Rf (Eq. 16-11)
RB2(2) = 25 kΩ || 100 kΩ
RB2(2) = 20 kΩ
V1 err(2) = (RB1 – RB2(2))Iin(bias) (Eq. 16-8)
V1 err(2) = (0 – 20 kΩ)(80 nA)
V1 err(2) = –1.6 µV
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V2 err(2) = (RB1 + RB2(2))(Iin(off)/2) (Eq. 16-9)
V2 err(2) = (0 + 20 kΩ)(20 nA/2)
V2 err(2) = 200 µV
V3 err(2) = Vin(off) = 1 mV
Verror(2) = ±Av(CL) (±V1 err(2) ± V2 err(2) ± V3 err(2))
Verror(2) = 5(1.6 mV + 200 µV + 1 mV)
Verror(2) = 14 mV
RB2(3) = R3 || Rf (Eq. 16-11)
RB2(3) = 100 kΩ || 100 kΩ
RB2(3) = 50 kΩ
V1 err(3) = (RB1 – RB2(3))Iin(bias) (Eq. 16-8)
V1 err(3) = (0 – 50 kΩ)(80 nA)
V1 err(3) = – 4 mV
V2 err(3) = (RB1 + RB2(3))(Iin(off)/2) (Eq. 16-9)
V2 err(3) = (0 + 50 kΩ)(20 nA/2)
V2 err(3) = 500 µV
V3 err(3) = Vin(off) = 1 mV
Verror(3) = ±Av(CL) (±V1 err(3) ± V2 err(3) ± V3 err(3))
Verror(3) = 2(4 mV + 500 µV + 1 mV)
Verror(3) = 11 mV
Answer: The output offset voltage is 110 mV at the 1-kΩ
position, 14 mV at the 25-kΩ position, and 11 mV at the
100-kΩ position.
17-28. Given:
Rf(1) = 100 Ω
Rf(2) = 1 kΩ
Rf(3) = 10 kΩ
iin = 1 mA
Solution:
vout(1) = iinRf (1) (from Table 17-2)
vout(1) = 1 mA(100 Ω)
vout(1) = 100 mV
vout(2) = iinRf (2) (from Table 17-2)
vout(2) = 1 mA(1 kΩ)
vout(2) = 1 V
vout(3) = iinRf (1) (from Table 17-2)
vout(3) = 1 mA(10 kΩ)
vout(3) = 10 V
Answer: The output offset voltage is 110 mV at
position A, 1 V at position B, and 10 V at position C.
17-29. Given:
Rf = 100 kΩ
iin = 2 µA
Solution:
vout = iinRf (from Table 17-2)
vout = 2 µA(100 kΩ)
vout = 200 mV
Answer: The output voltage is 200 mV.
17-30. Given:
Rf = 3.3 kΩ
iin = 1 mA
Solution:
vout = iinRf (from Table 17-2)
vout = 1 mA(3.3 kΩ)
vout = 3.3 V
17-31. Given:
vout = 2 V
iin = 1 mA
Solution:
vout = iinRf (from Table 17-2)
Rf = vout/iin
Rf = 2 V/1 mA
Rf = 2 kΩ
Answer: The unknown resistor is 2 kΩ.
17-32. Given:
R = 100 kΩ
V = 10 V
Rf (max) = 11 kΩ
Rf (min) = 9 kΩ
Solution:
iin = V/R
iin = 10 V/100 kΩ
iin = 0.1 mA
vout(max) = iinRf (max) (from Table 17-2)
vout(max) = 0.1 mA(11 kΩ)
vout(max) = 1.1 V
vout(min) = iin Rf (min) (from Table 17-2)
vout(min) = 0.1 mA(9 kΩ)
vout(min) = 0.9 V
Answer: The output voltage varies between 0.9 V and
1.1 V.
17-33. Given:
R = 100 kΩ
V = 10 V
Rf (max) = 10 kΩ
Rf (min) = 1 kΩ
Solution:
iin = V/R
iin = 10 V/100 kΩ
iin = 0.1 mA
vout(max) = iinRf(max) (from Table 17-2)
vout(max) = 0.1 mA(10 kΩ)
vout(max) = 1 V
vout(min) = iinRf(min) (from Table 17-2)
vout(min) = 0.1 mA(1 kΩ)
vout(min) = 0.1 V
Answer: The output voltage varies between 0.1 V and
1 V.
17-34. Given:
R1 = 10 Ω
R2 = 100 Ω
R3 = 1 kΩ
R4 = 10 kΩ
R5 = 100 kΩ
iout = 100 µA full scale
Solution:
iout = vinR1
vin = ioutR1
vin(1) = ioutR1
vin(1) = 100 µA(10 Ω)
vin(1) = 1 mV
Answer: The output voltage is 3.3 V.
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vin(2) = ioutR2
vin(2) = 100 µA(100 Ω)
vin(2) = 10 mV
vin(3) = ioutR3
vin(3) = 100 µA(1 kΩ)
vin(3) = 100 mV
vin(4) = ioutR4
vin(4) = 100 µA(10 kΩ)
vin(4) = 1 V
vin(5) = ioutR5
vin(5) = 100 µA(100 kΩ)
vin(5) = 10 V
Answer: The input voltages are 1 mV at the 10-Ω position, 10 mV at the 100-Ω position, 100 mV at the 1-kΩ
position, 1 V at the 10-kΩ position, and 10 V at the
100-kΩ position.
5. The problem is that the output is drawing too much current from the op amp. The circuit can be redesigned using a
higher-power op amp or a current booster.
8. The output is weighted, that is, proportional to the weighted
sum of the inputs.
9. Refer to Fig. 18-35. To modify the circuit for a dc response,
accept a dc offset and correct it later, or use two batteries as a
split supply.
10. Use an emitter follower on the output or employ a Class-B
stage for bidirectional amplification.
11. Because of the high open-loop voltage gain of the op amp,
the slightest input voltage immediately biases one of the
output transistors. In effect, the knee voltage is divided by the
open-loop gain.
PROBLEMS
18-1.
17-35. Answer:
Trouble 1: Since there is voltage at C and not at D, the
trouble is an open between C and D.
Trouble 2: Since all the voltages are zero, the trouble is
a shorted R2.
Solution:
First stage:
Trouble 3: The trouble is a shorted R4.
17-36. Answer:
Trouble 4: Since the output of the first stage is very high,
the trouble is an open R2.
Av = –R2/R1
Av = –20 MΩ/10 MΩ
Av = –2
Trouble 5: Since there is voltage at F and not at G, the
trouble is an open between F and G.
Second stage:
Av = –R4/R3
Av = –15 kΩ/15 kΩ
Av = –1
Trouble 6: Since there is voltage at F and not at E, the
trouble is an open R3.
17-37. Answer:
Trouble 7: Since there is voltage at A and not at B, the
trouble is an open between A and B.
Av = –R5/R3
Av = –75 kΩ/15 kΩ
Av = –5
Trouble 8: Since the second stage has no gain, the trouble
is a shorted R3.
Total:
Av = (–2)(–1)
Av = 2 for the switch position pointing to R4
Trouble 9: The trouble is R4 open.
Av = (–5)(–2)
Av = 10 for the switch position pointing to R5
17-38. R1 is shorted
17-39. R2 is actually 500 Ω, not 1 kΩ
Answer: The gain is 2 in the switch position pointing to
R4, and the gain is 10 for the switch position pointing
to R5.
17-40. R3 is actually 51 kΩ, not 100 kΩ
17-41. R2 is actually 10 kΩ, not 1 kΩ
18-2.
17-42. The op amp U2 has failed
Chapter 18 Linear Op-Amp Circuit
Applications
SELF-TEST
1.
2.
3.
4.
5.
6.
b
b
a
c
c
b
7.
8.
9.
10.
11.
12.
b
d
d
a
b
c
13.
14.
15.
16.
17.
18.
Given:
R1 = 10 MΩ
R2 = 20 MΩ
R3 = 15 kΩ
R4 = 15 kΩ
R5 = 75 kΩ
d
c
b
c
d
d
19.
20.
21.
22.
23.
24.
b
a
d
b
c
a
JOB INTERVIEW QUESTIONS
4. The first stage provides a high input impedance and voltage
gain, and the second stage produces a high CMRR.
Given:
R1 = 1.5 kΩ
Rf = 75 kΩ
RL = 15 kΩ
C1 = 1 µF
C2 = 4.7 µF
funity = 1 MHz
Solution:
Av = –Rf /R1
Av = –75 kΩ/1.5 kΩ
Av = –50
fC1 = 1/(2πR1C1)
fC1 = 1/[2π (1.5 kΩ)(1 µF)]
fC1 = 106 Hz
fC2 = 1/(2π RLC2)
fC2 = 1/[2π (15 kΩ)(4.7 µF)]
fC2 = 2.26 Hz
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Answer: The gain is 50 (inverted), and the cutoff frequencies are 2.26 Hz and 106 Hz.
18-3.
Solution:
Av = (Rf/R1) + 1
Av = (82 kΩ/2 kΩ) + 1
Av = 42
Given:
R1 = 10 kΩ
Rf = 180 kΩ
Rmin = 130 Ω
Rmax = 25.13 kΩ
funity = 1 MHz
f2 = funity/Av
f2 = 3 MHz/42
f2 = 71.4 kHz
fC1 = 1/(2πR3C1)
fC1 = 1/[2π(100 kΩ)(2.2 µF)]
fC1 = 0.72 Hz
Solution:
Bmin = (10 kΩ || 130 Ω)/(10 kΩ || 130 Ω + 180 kΩ)
Bmin = 0.000712
fC2 = 1/(2πRLC2)
fC2 = 1/[2π(25 kΩ)(4.7 µF)]
fC2 = 1.35 Hz
Bmax = (10 kΩ || 25.13 kΩ)/(10 kΩ || 25.13 kΩ + 180 kΩ)
Bmax = 0.0382
fC3 = 1/(2πR1C3)
fC3 = 1/[2π(2 kΩ)(1 µF)]
fC3 = 79.6 Hz
f2(min) = Bfunity
f2(min) = 0.000712(1 MHz)
f2(min) = 712 Hz
f2(max) = Bfunity
f2(max) = 0.0382(1 MHz)
f2(max) = 38.2 kHz
–Rf –180 kΩ
Av = ____
​   ​= ________
​
​= –18
R1
10 kΩ
Answer: The midband voltage gain is 42, the upper cutoff
frequency is 71.4 kHz, and the lower cutoff frequency is
79.6 Hz.
18-6.
Answer: The voltage gain is 18 with an inverted output.
The minimum bandwidth is 712 Hz and the maximum
bandwidth is 38.2 kHz.
18-4.
Given:
R1 = 1.5 kΩ
Rf = 100 kΩ
Rmin = 100 Ω
Rmax = 5.1 kΩ
funity = 1 MHz
Solution:
Av = (Rf/R1) + 1
Av = (150 kΩ/3.3 kΩ) + 1
Av = 46.5
Solution:
Bmin = (R1 || Rmin)/(R1 || Rmin + Rf)
Bmin = (1.5 kΩ || 100 Ω)/(1.5 kΩ || 100 Ω + 100 kΩ)
Bmin = 0.000937
f2 = funity/Av
f2 = 1 MHz/46.5
f2 = 21.5 kHz
Bmax = (R1 || Rmax)/(R1 || Rmax + Rf)
Bmax = (1.5 kΩ || 5.1 kΩ)/(1.5 kΩ || 5.1 kΩ + 100 kΩ)
Bmax = 0.01146
fC1 = 1/(2πR2C1)
fC1 = 1/[2π(100 kΩ)(1 µF)]
fC1 = 1.59 Hz
f2(min) = Bminfunity
f2(min) = 0.000937(1 MHz)
f2(min) = 937 Hz
fC2 = 1/(2πRLC2)
fC2 = 1/[2π(10 kΩ)(10 µF)]
fC2 = 1.59 Hz
f2(max) = Bmaxfunity
f2(max) = 0.01146(1 MHz)
f2(max) = 11.5 kHz
fC3 = 1/(2πR1C3)
fC3 = 1/[2π(3.3 kΩ)(4.7 µF)]
fC3 = 10.3 Hz
Av = –Rf/R1
Av = –100 kΩ/1.5 kΩ
Av = –66.7
vout = Avvin
vout = –66.7(4 mV)
vout = –266.8 mV
Answer: The minimum bandwidth is 937 Hz and the
maximum bandwidth is 11.5 kHz. The output voltage is
–266.8 mV.
18-5.
Given:
R1 = 2 kΩ
Rf = 82 kΩ
RL = 25 kΩ
C1 = 2.2 µF
C2 = 4.7 µF
funity = 3 MHz
Given:
R1 = 3.3 kΩ
Rf = 150 kΩ
R2 = 100 kΩ
RL = 10 kΩ
C1 = 1 µF
C2 = 10 µF
C3 = 4.7 µF
funity = 1 MHz
Answer: The midband voltage gain is 46.5, the upper cutoff frequency is 21.5 kHz, and the lower cutoff frequency
is 10.3 Hz.
18-7.
Given:
R1 = 2 kΩ
Rf = 100 kΩ
vin = 10 mV
Solution:
Av = (Rf/R1) + 1
Av = (100 kΩ/2 kΩ) + 1
Av = 51
vout = Avvin
vout = 51(10 mV)
vout = 510 mV
Answer: The output voltage at A, B, and C is 510 mV.
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18-8.
Given:
R1 = 91 kΩ
Rf = 12 kΩ
R2 = 1 kΩ
vin = 2 mV
Solution:
Low gate:
Av = (Rf/R1) + 1
Av = (12 kΩ/91 kΩ) + 1
Av = 1.13
vout = Avvin
vout = 1.13(2 mV)
vout = 2.26 mV
High gate:
Av = [Rf/(R1 || R2)] + 1
Av = [12 kΩ/(91 kΩ || 1 kΩ)] + 1
Av = 13.1
18-9.
18-11. Given:
R1 = 1 kΩ
R2 = 10 kΩ
Solution:
–R2/R1 < Av < 0
–10 kΩ/1 kΩ < Av < 0
–10 < Av < 0
Answer: The maximum inverting gain is –10, and the
maximum positive gain is 0.
18-12. Given: R1 = R2
Solution: At ground the circuit is an inverting amplifier.
Av = –Rf/R1
Av = –1
vout = Avvin
vout = 13.1(2 mV)
vout = 26.2 mV
When the wiper is 10% away from ground, so that the
noninverting gain will be 10% of its maximum of 2.
Av(non) = 10% (2) = 0.2
Av = Av(in) + Av(non)
Av = –1 + 0.2
Av = –0.8
Answer: When the gate is low, the output is 2.26 mV;
when the gate is high, the output is 26.2 mV.
Answer: The gain with the wiper at ground is –1, and
10% away is –0.8.
Given:
R1 = 20 kΩ
Rf = 68 kΩ
R2 = 1 kΩ
vin = 1 mV
Solution:
Low gate:
Av = (Rf/R1) + 1
Av = (68 kΩ/20 kΩ) + 1
Av = 4.4
vout = Avvin
vout = 4.41(1 mV)
vout = 4.4 mV
High gate:
Av = [Rf/(R1 || R2)] + 1
Av = [68 kΩ/(20 kΩ || 1 kΩ)] + 1
Av = 72.4
vout = Avvin
vout = 72.4(1 mV)
vout = 72.4 mV
Answer: When the gate is low, the output is 4.4 mV, and
when the gate is high, the output is 72.4 mV.
18-10. Given:
R1 = 10 kΩ
Rf = 10 kΩ
Vin = 2.5 V
Solution:
Av = (Rf/R1) + 1
Av = (10 kΩ/10 kΩ) + 1
Av = 2
Vout = Av(vin)
Vout = 2(2.5 V)
Vout = 5 V
Answer: The new output reference voltage is 5 V.
18-13. Given:
R = 5 kΩ
nR = 75 kΩ
nR/(n – 1)R = 5.36 kΩ
Solution:
Av = –nR/R
Av = –75 kΩ/5 kΩ
Av = –15
Answer: The maximum positive gain is 15, and the maximum negative gain is –15.
18-14. Given:
R' = 10 kΩ
R = 22 kΩ
C = 0.02 µF
fin = 100 Hz, 1 kHz, 10 kHz
Solution:
fC = 1/(2π RC)
fC = 1/[(2π22 kΩ)(0.02 µF)]
fC = 362 Hz
ϕ = –2 arctan (f/fC)
ϕ = –2 arctan (100 Hz/362 Hz)
ϕ = –30.9°
ϕ = –2 arctan (f/fC)
ϕ = –2 arctan (1 kHz/362 Hz)
ϕ = –140°
ϕ = –2 arctan (f/fC)
ϕ = –2 arctan (10 kHz/362 Hz)
ϕ = –176°
Answer: The phase shift is –30.9° at 100 Hz, –140° at
1 kHz, and –176° at 10 kHz.
18-15. Given:
R1 = 1.5 kΩ
R2 = 30 kΩ
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Solution:
Av(inv) = –R2/R1 (Eq. 18-6)
Av(inv) = –30 kΩ/1.5 kΩ
Av(inv) = –20
Av(non) = [(R2/R1) + 1][R'2/(R'1 + R'2)] (Eq. 18-7)
Av(non) = [(30 kΩ/1.5 kΩ) + 1][30 kΩ/(1.5 kΩ + 30 kΩ)]
Av(non) = 20
Av(CM) = ±4(0.1%) = ±4(0.001) = ±0.004
Answer: The differential voltage gain is –20, and the
common mode gain is ±0.004.
18-16. Given:
R1 = 1 kΩ
R2 = 20 kΩ
Solution:
Av(inv) = –R2/R1 (Eq. 18-6)
Av(inv) = –20 kΩ/1 kΩ
Aiv(inv) = –20
Av(CM) = ±4 ΔR/R (Eq. 18-5)
Av(CM) = ±4 (1%) = ±4(0.01)
Av(CM) = ±0.04
Answer: The differential voltage gain is –20, and the
common-mode gain is ±0.04.
18-17. Given:
R1 = 10 kΩ
R2 = 20 kΩ
R3 = 20 kΩ
R4 = 10 kΩ
Solution:
V2 = [R2/(R1 + R2)]VCC
V2 = [20 kΩ/(10 kΩ + 20 kΩ)]15 V
V2 = 10 V
V4 = [R4/(R3 + R4)]VCC
V4 = [10 kΩ/(20 kΩ + 10 kΩ)]15 V
V4 = 5
Answer: No, the bridge is not balanced.
18-18. Given:
R1 = 1 kΩ
ΔR = 15 Ω
Av = –100
Solution:
vin = (ΔR/4R)VCC
vin = (15 Ω/4 (1 kΩ))15 V
vin = 56.3 mV
vout = Av(vin)
vout = (–100)(56.3 mV)
vout = –5.63 V
Answer: The output voltage is –5.63 V.
18-19. Given:
R1 = 1 kΩ
R2 = 99 kΩ
R = 10 kΩ ± 0.5%
vin = 2 mV
Solution:
Av = (R2/R1) + 1
Av = (99 kΩ/1 kΩ) + 1
Av = 100
vout = Avvin
vout = 100(2 mV)
vout = 200 mV at preamp output and –200 mV at diff amp
output.
Av(CM) = ±2(ΔR/R)
Av(CM) = ±2(0.005)
Av(CM) = ±0.01
CMRR = |Av|/|Av(CM)|
CMRR = 100/0.01
CMRR = 10,000
Answer: The output voltage is –200 mV, and the CMRR
is 10,000.
18-20. Given: vin(CM) = 5 V
Solution: Since the first stage has a common-mode gain
of 1, both sides have the same voltage of 5 V. The guard
voltage is 5 V.
Answer: The guard voltage is 5 V.
18-21. Given:
RG = 1008 Ω
vin = 20 mV
Solution:
Av = (49.4 kΩ/RG) + 1 (Eq. 18-17)
Av = (49.4 kΩ/1008 Ω) + 1
Av = 50
vout = Av(vin)
vout = 50(20 mV)
vout = 1 V
Answer: The output voltage is 1 V.
18-22. Given:
R = 10 kΩ
v1 = –50 mV
v2 = –30 mV
Solution:
vout = v1 – v2
vout = (–50 mV) – (–30 mV)
vout = –20 mV
Answer: The output voltage is –20 mV.
18-23. Given:
R1 = 10 kΩ
R2 = 20 kΩ
R3 = 15 kΩ
R4 = 15 kΩ
R5 = 30 kΩ
Rf = 75 kΩ
v1 = 1 mV
v2 = 2 mV
v3 = 3 mV
v4 = 4 mV
Solution:
Av(1) = –Rf/R1
Av(1) = –75 kΩ/10 kΩ
Av(1) = –7.5
Av(2) = –Rf/R2
Av(2) = –75 kΩ/20 kΩ
Av(2) = –3.75
Av(3) = {[Rf/(R1 || R2)] + 1}{(R4 || R5)/[R3 + (R4 || R5)]}
Av(3) = {[75 kΩ/(10 kΩ || 20 kΩ)] + 1}{(15 kΩ || 30 kΩ)/
[15 kΩ + (15 kΩ || 30 kΩ)]}
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Av(3) = (12.25)(0.455)
Av(3) = 5.57
Av(4) = {[Rf/(R1 || R2)] + 1}{(R3 || R5)/[R4 + (R3 || R5)]}
Av(4) = {[75 kΩ/(10 kΩ || 20 kΩ)] + 1}{(15 kΩ || 30 kΩ)/
[15 kΩ + (15 kΩ || 30 kΩ)]}
Av(4) = (12.25)(0.4)
Av(4) = 4.9
vout = Av(1)v1 + Av(2)v2 + Av(3)v3 + Av(4)v4
vout = –7.5(1 mV) + –3.75(2 mV) + 4.9(3 mV) + 4.9
(4 mV)
vout = 19.3 mV
Answer: The output voltage is 19.3 mV.
18-24. Given:
R = 10 kΩ
v1 = 1.5 V
v2 = 2.5 V
v3 = 4 V
18-28. Given:
D3 – D0 = 0001
Solution:
BIN = (D0 × 20) + (D1 × 21) + (D2 × 22) + (D3 × 23)
BIN = (1 × 20) + (0 × 21) + (0 × 22) + (0 × 23) + (0 × 24)
BIN = 1
Vout = –​ ____
​  BIN
​× Vref ​​
2N
Vout = – ​ __
​  14 ​ × (2.5 V) ​ (Eq. 18-19)
2
Vout = –312.5 mV
(
(
)
)
Answer: The smallest output voltage is –312.5 mV.
18-29. Given:
R1 = 2 kΩ
R2 = 47 kΩ
β = 100
ISC = 25 mA
Solution:
vout = –(v1 + v2 + v3)/3
vout = –(1.5 V + 2.5 V + 4 V)/3
vout = –2.67 V
Solution:
Av = (R2/R1) + 1
Av = (47 kΩ/2 kΩ) + 1
Av = 24.5
Answer: The output voltage is –2.67 V.
Imax = βISC
Imax = (100) 25 mA
Imax = 2.5 A
18-25. Given:
v0 = 5 V
v1 = 0 V
v2 = 5 V
v3 = 0 V
Answer: The voltage gain is 24.5, and the maximum
current is 2.5 A (assumes 25 mA output maximum
of IC).
Solution:
vout = –(v3 + 0.5v2 + 0.25v1 + 0.125v0)
vout = –(0 + 0.5(5 V) + 0 + 0.125(5 V))
vout = –3.125 V
Answer: The output voltage is –3.125 V.
18-26. Given:
D7 – D0 = 10100101
Solution:
BIN = (D0 × 20) + (D1 × 21) + (D2 × 22) + (D3 × 23)
+ (D4 × 24) + (D5 × 25) + (D6 × 26) + (D7 × 27)
BIN = (1 × 20) + (0 × 21) + (1 × 22) + (0 × 23) + (0 × 24)
+ (1 × 25) + (0 × 26) + (1 × 27)
BIN = 165
Answer: The decimal equivalent value is 165.
18-27. Given:
D7 – D0 = 01100110
Vref = +5 V
Solution:
BIN = (D0 × 20) + (D1 × 21) + (D2 × 22) + (D3 × 23)
+ (D4 × 24) + (D5 × 25) + (D6 × 26) + (D7 × 27)
BIN = (0 × 20) + (1 × 21) + (1 × 22) + (0 × 23) + (0 × 24)
+ (1 × 25) + (1 × 26) + (0 × 27)
BIN = 102
Vout = –​ ____
​  BIN
​× Vref ​
2N
102 ​ × 2 (5 V) (Eq. 18-19)
Vout = –​  ​ ____
28
Vout = –3.98 V
(
(
)
)
18-30. Given:
R1 = 1 kΩ
R2 = 10 kΩ
β = 125
ISC = 25 mA
Solution:
Av = –(R2/R1)
Av = –(10 kΩ/1 kΩ)
Av = –10
Imax = βIsc
Imax = 125 (25 mA)
Imax = 3.125 A
Answer: The voltage gain is –10, and the maximum
current is 3.125 A.
18-31. Given:
R = 2 kΩ
RL = 75 Ω
vin = 1 V
VCC = 15 V
Solution:
iout = vin/R
iout = 1 V/2 kΩ
iout = 0.5 mA
RL(max) = R[(VCC/vin) – 1]
RL(max) = 2 kΩ[(15 V/1 V) – 1]
RL(max) = 28 kΩ
Answer: The output current is 0.5 mA, and the maximum
load resistance is 28 kΩ.
Answer: The output voltage is –3.98 V.
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18-32. Given:
R = 3.3 kΩ
RL = 150 Ω
vin = 5 V
VCC = 15 V
Solution:
iout = (VCC – vin)/R
iout = (15 V – 5 V)/3.3 kΩ
iout = 3.03 mA
RL(max) = R/[(VCC/vin) – 1]
RL(max) = 3.3 kΩ/[(15 V/5 V) – 1]
RL(max) = 1.65 kΩ
Answer: The output current is 3.03 mA, and the maximum load resistance is 1.65 kΩ.
18-33. Given:
R = 10 kΩ
vin = 3 V
VCC = 15 V
Solution:
iout = vin/R
iout = 3 V/10 kΩ
iout = 0.3 mA
RL(max) = R[(VCC/vin) – 1]
RL(max) = 10 kΩ[(15 V/3 V) – 1]
RL(max) = 40 kΩ
Answer: The output current is 0.3 mA, and the maximum
load resistance is 40 kΩ.
18-34. Given:
R = 2 kΩ
RL = 500 Ω
vin = 6 V
vin(max) = 7.5 V
Solution:
iout = –vin/R
iout = 6 V/2 kΩ
iout = 3 mA
RL(max) = (R/2)[(VCC/vin) – 1]
RL(max) = (2 kΩ/2)[(15 V/7.5 V) – 1]
RL(max) = 1 kΩ
Answer: The output current is 3 mA, and the maximum
load resistance is 1 kΩ.
18-35. Given:
R1 = 10 kΩ
R2 = 100 kΩ
R3 = 100 kΩ
R4 = 10 kΩ
rds(min) = 200 Ω
rds(max) = 1 MΩ
Solution:
Av(min) = [(R2/R1) + 1][rds(min)/(rds(min) + R3)]
Av(min) = [(100 kΩ/10 kΩ) + 1][200/(200 + 100 kΩ)]
Av(min) = 0.02
18-36. Given:
R1 = 5.1 kΩ
R2 = 51 kΩ
R5 = 68 kΩ
R6 = 1 kΩ
rds(min) = 120 Ω
rds(max) = 5 MΩ
Solution:
Av(min) = [–(R2/R1)][(R6 + rds(min))/(R5 + R6 + rds(min))]
Av(min) = [–(51 kΩ/5.1 kΩ)][(1 kΩ + 120 Ω)/
(68 kΩ + 1 kΩ + 120 Ω)]
Av(min) = –0.16
Av(max) = [–(R2/R1)][(R6 + rds(max))/(R5 + R6 + rds(max))]
Av(max) = [–(51 kΩ/5.1 kΩ)][(1 kΩ + 5 MΩ)/
(68 kΩ + 1 kΩ + 5 MΩ)]
Av(max) = –9.87
Answer: The maximum voltage gain is –9.87, and the
minimum voltage gain is –0.16.
18-37. Given:
R1 = 10 kΩ
R2 = 10 kΩ
R5 = 75 kΩ
R6 = 1.2 kΩ
R7(min) = 180 Ω
R7(max) = 10 MΩ
Solution:
Av(min) = [–(R2/R1)][(R6 + R7(min))/(R5 + R6 + R7 (min))]
Av(min) = [–(10 kΩ/10 kΩ)][(1.2 kΩ + 180 Ω)/(75 kΩ
+ 1.2 kΩ + 180 Ω)]
Av(min) = –0.018
Av(max) = [–( R2/R1 )][(R6 + R7(max))/(R5 + R6 + R7(max))]
Av(max) = [–(10 kΩ/10 kΩ)][(1.2 kΩ + 10 MΩ)/(75 kΩ
+ 1.2 kΩ + 10 MΩ)]
Av(max) = –0.99
Answer: The maximum voltage gain is –0.99, and the
minimum voltage gain is –0.018.
18-38. Given:
R1 = 3.3 kΩ
R2 = 82 kΩ
RL = 10 kΩ
R = 91 kΩ
C1 = 4.7 µF
C2 = 10 µF
C3 = 4.7 µF
Solution:
Av = –R2/R1
Av = –82 kΩ/3.3 kΩ
Av = –24.8
f1 = 1/(2π R1C1)
f1 = 1/[2π (3.3 kΩ)(4.7 µF)]
f1 = 10.26 Hz
f2 = 1/(2π RLC2)
f2 = 1/[2π (10 kΩ)(10 µF)]
f2 = 1.59 Hz
Av(max) = [(R2/R1) + 1][(rds(max)/rds(max) + R3)]
Av(max) = [(100 kΩ/10 kΩ) + 1][1 MΩ/(1 MΩ + 100 kΩ)]
Av(max) = 10
f3 = 1/[2π (R/2)C3]
f3 = 1/[2π (91 kΩ/2)(4.7 µF)]
f3 = 0.74 Hz
Answer: The maximum voltage gain is 10, and the minimum voltage gain is 0.02.
Answer: The gain is –24.8, and the cutoff frequencies are
f1 = 10.26 Hz, f2 = 1.59 Hz, and f3 = 0.74 Hz.
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18-39. Given:
R1 = 1.5 kΩ
R2 = 15 kΩ
RL = 15 kΩ
R = 68 kΩ
C1 = 1 µF
C2 = 2.2 µF
C3 = 3.3 µF
Solution:
Av = (R2/R1) + 1
Av = (15 kΩ/1.5 kΩ) + 1
Av = 11
f1 = 1/[2π(R/2)C1]
f1 = 1/[2π(68 kΩ/2)(1 µF)]
f1 = 4.68 Hz
f2 = 1/(2πRLC2)
f2 = 1/[2π(15 kΩ)(2.2 µF)]
f2 = 4.82 Hz
f3 = 1/(2πR1C3)
f3 = 1/[2π(1 kΩ)(3.3 µF)]
f3 = 32.2 Hz
Answer: The gain is 11, and the cutoff frequencies are
f1 = 4.68 Hz, f2 = 4.82 Hz, and f3 = 32.2 Hz.
CRITICAL THINKING
18-40. Answer: Since the terminal is floating, the output would
be saturated or VCC. To fix this problem, a large-value
resistor could be connected to the noninverting terminal.
This would keep it at ground potential during the transition and prevents a spike.
18-41. Given:
R1(min) = 990 Ω
R1(max) = 1010 Ω
Rf(min) = 99 kΩ
Rf(max) = 101 kΩ
Solution:
Av(min) = –Rf(min)/R1(max)
Av(min) = –99 kΩ/1010 Ω
Av(min) = –98
Av(max) = –Rf(max)/R1(min)
Av(max) = –101 kΩ/990 Ω
Av(max) = –102
Answer: The minimum gain is 98, and the maximum gain
is 102.
18-42. Given:
Transistor:
R1 = 22 kΩ
R2 = 10 kΩ
RS = 1 kΩ
RE = 5.6 kΩ
RC = 6.8 kΩ
VCC = 15 V
Op amp
R3 = 1 kΩ
Rf = 47 kΩ
Solution:
VBB = [R2/(R1 + R2 + RS)]VCC
VBB = [10 kΩ/(22 kΩ + 10 kΩ + 1 kΩ)]15 V
VBB = 4.54 V
VE = VBB – VBE
VE = 4.54 V – 0.7 V
VE = 3.84 V
IE = VE/RE
IE = 3.84 V/5.6 kΩ
IE = 0.685 mA
r'e = 25 mV/IE
r'e = 25 mV/0.685 mA
r'e = 36.5 Ω
rc = Rc
rc = 6.8 kΩ
Av = rc/r'e
Av = 6.8 kΩ/36.5 Ω
Av = 186
Op amp:
Av = (Rf/R3) + 1
Av = (47 kΩ/1 kΩ) + 1
Av = 48
Av = (186)(48)
Av = 9114
Answer: The voltage gain is 9114.
18-43. Given:
R1 = 1 kΩ
R2 = 10 kΩ
RL = 100 Ω
β = 50
vin = 0.5 V
Solution:
Av = –R2/R1
Av = –10 kΩ/1 kΩ
Av = –10
vout = Av(vin)
vout = –10(0.5 V)
vout = –5 V
Iout = vout/RL
Iout = –5 V/100 Ω
Iout = 50 mA
IB = Iout/β
IB = 50 mA/50
IB = 1 mA
Answer: The base current is 1 mA.
18-44. Answer:
Trouble 1: Since there is voltage at E and not at F, there
is an open between E and F.
Trouble 2: Since the output is only 200 mV, which is the
amplified output of A, R2 is open.
Trouble 3: Since the input is 2 mV and the output is maximum, R1 is shorted.
18-45. Answer:
Trouble 4: Since there is no voltage at B, there is an open
between K and B.
Trouble 5: Since the voltage at C is 3 mV and the voltage
at D is zero, there is an open between C and D.
Trouble 6: Since the voltage at A is zero, there is an open
between J and A.
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Q = f0/BW (Eq. 19-3)
Q = 1.86 kHz/7.36 kHz
Q = 0.25
18-46. Answer:
Trouble 7: Since the input voltage is 3 mV and the output
is maximum, R3 is open.
Since Q < 1, it is wideband.
Trouble 8: Since the output is only 250 mV, which is the
amplified output of B, R1 is open.
Trouble 9: Since there is input voltage and no voltage
reading a points C and D, the op amp is shorted at the
input.
Trouble 10: Since the input is 5 mV and the output is
maximum, R2 is shorted.
18-47. R1 is actually 10 kΩ not 1 kΩ
18-48. Rf is 10 kΩ not 100 kΩ
18-49. Rf is open
18-50. Rf is shorted
18-51. The closed loop feedback for the op amp U2 has opened
18-52. Noninverting voltage amplifier
18-53. Adjusts the op amp output to zero volts when the input
signal is zero.
18-54. Vout = (Av)(Vin) = (11)(0.5 Vp-p) = 5.5 Vp-p
18-55. Av = AVOL; The output would be clipped at (+) and (–)
12 V
18-56. 0 Hz to 1.36 MHz
Chapter 19
Active Filters
SELF TEST
1.
2.
3.
4.
5.
6.
7.
8.
c
b
d
c
c
b
c
d
9.
10.
11.
12.
13.
14.
15.
16.
d
d
d
b
c
d
a
b
17.
18.
19.
20.
21.
22.
23.
24.
a
b
d
d
a
d
b
b
25.
26.
27.
28.
29.
30.
31.
b
c
b
d
a
d
b
JOB INTERVIEW QUESTIONS
6. Low attenuation and the edge frequency. High attenuation
and the edge frequency.
7. A filter designed to control the phase of a signal rather than
its amplitude.
8. It compares the voltage gain to the frequency.
PROBLEMS
19-1.
Given:
f1 = 445 Hz
f2 = 7800 Hz
Solution:
BW = f2 – f1 (Eq. 19-1)
BW = 7800 Hz – 445 Hz
BW = 7355
____
f0 = √
​ f1 f2 ​
________________
f0 = √
​ (445
Hz)(7800 Hz) ​ (Eq. 19-2)
f0 = 1.86 kHz
Answer: The bandwidth is 7.36 kHz, the center frequency
is 1.86 kHz, the Q is 0.25, and it is wideband.
19-2.
Given:
f1 = 20 kHz
f2 = 22.5 kHz
Solution:
BW = f2 – f1 (Eq. 19-1)
BW = 22.5 kHz – 20 kHz
BW = 2.5 kHz
____
f0 = ​√f1 f2 ​
_________________
f0 = √
​ (20
kHz)(22.5 kHz)  (Eq. 19-2)
f0 = 21.2 kHz
Q = f0/BW (Eq. 19-3)
Q = 21.2 kHz/2.5 kHz
Q = 8.48
Since Q > 1, it is narrowband.
Answer: The bandwidth is 2.5 kHz, the center frequency
is 21.2 kHz, the Q is 8.48, and it is narrowband.
19-3a. Given:
f1 = 2.3 kHz
f2 = 4.5 kHz
Solution:
BW = f2 – f1 (Eq. 19-1)
BW = 4.5 kHz – 2.3 kHz
BW = 2.2 kHz
____
f0 = √f1 f2 ​
________________
f0 = ​√(2.3
kHz)(4.5 kHz) ​ (Eq. 19-2)
f0 = 3.2 kHz
Q = f0/BW (Eq. 19-3)
Q = 3.2 kHz/2.2 kHz
Q = 1.45
Since Q > 1, it is narrowband.
Answer: Narrowband.
19-3b. Given:
f1 = 47 kHz
f2 = 75 kHz
Solution:
BW = f2 – f1 (Eq. 19-1)
BW = 75 kHz – 47 kHz
BW = 28 kHz
____
f0 = √
​ _______________
f1 f2 ​
f0 = √(75
kHz)(47 kHz) ​ (Eq. 19-2)
f0 = 59.4 kHz
Q = f0/BW (Eq. 19-3)
Q = 59.4 kHz/28 kHz
Q = 2.12
Since Q > 1, it is narrowband.
Answer: Narrowband.
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19-3c. Given:
f1 = 2 Hz
f2 = 5 Hz
19-7.
Solution:
BW = f2 – f1 (Eq. 19-1)
BW = 5 Hz – 2 Hz
BW = 3 Hz
____
f0 = ​√f1 f2 ​
___________
f0 = ​√ (2 Hz)(5 Hz) ​ (Eq. 19-2)
f0 = 3.16 Hz
Solution:
___
f0 = 1/(2π​√LC ​) (Eq. 19-6)
_____________
f0 = 1/[2π​√
(20 mH)(5 μF) ​]
f0 = 503 Hz
XL = 2πf0L
XL = 2π(503 Hz)(20 mH)
XL = 63.2 Ω
Q = f0 /BW (Eq. 19-3)
Q = 3.16 Hz/3 Hz
Q = 1.05
Q = R/XL (Eq. 19-7)
Q = 600/63.2
Q = 9.5
Since Q > 1, it is narrowband.
Answer: Narrowband.
19-3d. Given:
f1 = 80 Hz
f2 = 160 Hz
Answer: The resonant frequency is 503 Hz, and the Q
is 9.5.
19-8.
Solution:
BW = f2 – f1 (Eq. 19-1)
BW = 160 Hz – 80 Hz
BW = 80 Hz
____
f0 = ​√f1 f2 ​
______________
f0 = ​√
(160 Hz)(80 Hz) ​ (Eq. 19-2)
f0 = 113 Hz
XL = 2πf0L
XL = 2π(712 Hz)(10 mH)
XL = 44.7 Ω
Q = R/XL (Eq. 19-7)
Q = 600/44.7
Q = 13.4
Since Q > 1, it is narrowband.
Answer: The resonant frequency is 712 Hz, and the Q is
13.4.
Answer: Narrowband.
Given: Seven capacitors
Answer: Seventh order
19-5.
Given: 10 capacitors
Solution:
n = number of capacitors (Eq. 19-4)
n = 10
Roll-off = 20n dB/decade (Eq. 19-4a)
Roll-off = 20(10) dB/decade
Roll-off = 200 dB/decade
Roll-off = 6n dB/octave (Eq. 19-4a)
Roll-off = 6(10) dB/octave
Roll-off = 60 dB/octave
Answer: The roll-off rate is roll-off = 200 dB/decade or
roll-off = 60 dB/octave.
19-6.
Given: 14 capacitors
Solution:
n = number of capacitors (Eq. 19-4)
n = 14
Number of ripples = n/2 (Eq. 19-5)
Number of ripples = 14/2
Number of ripples = 7
Answer: There are seven ripples.
Given:
L = 10 mH
C = 5 μF
R = 600 Ω
Solution:
___
f0 = 1/(2π​√LC ​) (Eq. 19-6)
____________
f0 = 1/[2π​√(10
mH)(5 µF) ​]
f0 = 712 Hz
Q = f0/BW (Eq. 19-3)
Q = 113 Hz/80 Hz
Q = 1.4
19-4.
Given:
L = 20 mH
C = 5 μF
R = 600 Ω
19-9.
Given:
R1 = 15 kΩ
C = 270 nF
Solution:
fC = 1/(2πR1C1) (Eq. 19-9)
fC = 1/[2π(15 kΩ)(270 nF)]
fC = 39.3 Hz
Answer: The cutoff frequency is 39.3 Hz.
19-10. Given:
R1 = 7.5 kΩ
R2 = 33 kΩ
R3 = 20 kΩ
C = 680 pF
Solution:
Av = (R2 /R1) + 1 (Eq. 19-10)
Av = (33 kΩ/7.5 kΩ) + 1
Av = 5.4
fC = 1/(2πR3C1) (Eq. 19-11)
fC = 1/[2π (20 kΩ)(680 pF)]
fC = 11.7 kHz
Answer: The voltage gain is 5.4, and the cutoff frequency
is 11.7 kHz.
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19-11. Given:
R1 = 2.2 kΩ
R2 = 47 kΩ
C = 330 pF
Solution:
Av = –R2/R1 (Eq. 19-12)
Av = –47 kΩ/2.2 kΩ
Av = –21.4
fC = 1/(2πR2C1) (Eq. 19-13)
fC = 1/[2π(47 kΩ)(330 pF)]
fC = 10.3 kHz
Answer: The voltage gain is –21.4, and the cutoff
frequency is 10.3 kHz.
19-12. Given:
R1 = 10 kΩ
C1 = 15 nF
Solution:
fC = 1/(2πR1C1) (Eq. 19-14)
fC = 1/[2π(10 kΩ)(15 nF)]
fC = 1.06 kHz
Answer: The cutoff frequency is 1.06 kHz.
19-13. Given:
R1 = 12 kΩ
R2 = 24 kΩ
R3 = 20 kΩ
C = 220 pF
Solution:
Av = (R2/R1) + 1 (Eq. 19-15)
Av = (24 kΩ/12 kΩ) + 1
Av = 3
fC = 1/(2πR3C1) (Eq. 19-16)
fC = 1/[2π(20 kΩ)(220 pF)]
fC = 36.2 kHz
Answer: The voltage gain is 3, and the cutoff frequency
is 36.2 kHz.
19-14. Given:
R1 = 8.2 kΩ
C1 = 560 pF
C2 = 680 pF
Solution:
Av = –C1/C2 (Eq. 19-17)
Av = –560 pF/680 pF
Av = –0.824
fC = 1/(2πR1C2) (Eq. 19-18)
fC = 1/[2π(8.2 kΩ)(680 pF)]
fC = 28.5 kHz
Answer: The voltage gain is –0.824, and the cutoff
frequency is 28.5 kHz.
19-15. Given:
R = 75 kΩ
C1 = 100 pF
C2 = 200 pF
Solution:
______
fp = 1/(2π​R√C1C2 ​
) (Eq. 19-20)
_______________
fp = 1/[2π(75 kΩ)​√
(100 pF)(200 pF) ​]
fp = 15 kHz
_______
Q = 0.5​√_______________
(C2/C1)  (Eq. 19-19)
Q = 0.5​√(200
pF)/(100 pF) ​
Q = 0.707
Since it is a Butterworth response, the cutoff and 3-dB
frequencies are the same as the pole frequency.
Answer: The frequencies are 15 kHz, and the Q is 0.707.
19-16. Given:
R = 51 kΩ
C1 = 100 pF
C2 = 680 pF
Solution:
_____
fp = 1/(2π​R​√C1C2 ​) (Eq. 19-20)
_______________
fp = 1/[2π(51 kΩ) √
​
(100 pF)(680 pF)] ​
fp = 12 kHz
_______
Q = 0.5​√(C2/C1)  (Eq. 19-19)
_______________
Q = 0.5​√(680
pF)/(100 pF) ​
Q = 1.3
fc = Kc fp (Eq. 19-23)
fc = 1.12(12 kHz)
fc = 13.44 kHz
f3dB = K3 fp (Eq. 19-24)
f3dB = 1.36 (12 kHz)
f3dB = 16.32 kHz
Answer: The pole frequency is 12 kHz, the cutoff frequency is 13.44 kHz, the 3-dB frequency is 16.32 kHz,
and the Q is 1.3
19-17. Given:
R1 = 51 kΩ
R2 = 30 kΩ
R = 33 kΩ
C = 220 pF
Solution:
Av = (R2/R1) + 1 (Eq. 19-29)
Av = (30 kΩ/51 kΩ) + 1
Av = 1.58
Q = 1/(3 – Av) (Eq. 19-30)
Q = 1/(3 – 1.58)
Q = 0.707
fp = 1/(2πRC)
fp = 1/[2π(33 kΩ) (220 pF)]
fp = 21.9 kHz
Since it is a Butterworth response, the cutoff and 3-dB
frequencies are the same as the pole frequency.
Answer: The frequencies are 21.9 kHz, and the Q is
0.707.
19-18. Given:
R1 = 33 kΩ
R2 = 33 kΩ
R = 75 kΩ
C = 100 pF
Solution:
Av = (R2/R1) + 1 (Eq. 19-29)
Av = (33 kΩ/33 kΩ) + 1
Av = 2
Q = 1/(3 – Av) (Eq. 19-30)
Q = 1/(3 – 2)
Q=1
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fp = 1/(2πRC)
fp = 1/[2π(75 kΩ)(100 pF)]
fp = 21.2 kHz
f3 = fp/K3
f3 = 9.89 kHz/1.30
f3 = 7.61 kHz
Kc = 1.000 (from Table 19-3)
K3 = 1.272 (from Table 19-3)
Answer: The pole frequency is 9.89 kHz, the cutoff frequency is 9.51 kHz, the 3-dB frequency is 7.61 kHz, and
the Q is 1.18.
fc = Kc fp (Eq. 19-23)
fc = 1.000(21.2 kHz)
fc = 21.2 kHz
f3 = K3 fp (Eq. 19-24)
f3 = 1.272(21.2 kHz)
f3 = 27 kHz
Answer: The pole frequency is 21.2 kHz, the cutoff frequency is 21.2 kHz, the 3-dB frequency is 27 kHz, and
the Q is 1.
19-19. Given:
R1 = 75 kΩ
R2 = 56 kΩ
R = 68 kΩ
C = 120 pF
Solution:
Av = (R2/R1) + 1 (Eq. 19-29)
Av = (56 kΩ/75 kΩ) + 1
Av = 1.75
19-21. Given:
R1 = 91 kΩ
R2 = 15 kΩ
C = 220 pF
Solution:
_____
fp = 1/(2πC​√
​ R1 R2 ​)
_____________
fp = 1/[2π(220 pF)​√
(15 kΩ)(91 kΩ) ​]
fp = 19.6 kHz
______
Q = 0.5√(R1/R2)
______________
Q = 0.5 ​√
(91 kΩ)/(15 kΩ) ​
Q = 1.23
Kc = 1.06 (from Fig. 19-26)
K3 = 1.32 (from Fig. 19-26)
fc = fp/Kc (Eq. 19-31)
fc = 19.6 kHz/1.06
fc = 18.5 kHz
Q = 1/(3 – Av) (Eq. 19-30)
Q = 1/(3 – 1.75)
Q = 0.8
f3 = fp/K3
f3 = 19.6 kHz/1.32
f3 = 14.8 kHz
fp = 1/(2πRC)
fp = 1/[2π (68 kΩ)(120 pF)]
fp = 19.5 kHz
Answer: The pole frequency is 19.6 kHz, the cutoff frequency is 18.5 kHz, the 3-dB frequency is 14.8 kHz, and
the Q is 1.23.
Kc = 0.661 (from Table 19-3)
K3 = 1.115 (from Table 19-3)
fc = Kc fp (Eq. 19-23)
fc = 0.661(19.5 kHz)
fc = 12.89 kHz
f3 = K3 fp (Eq. 19-24)
f3 = 1.115(19.5 kHz)
f3 = 21.74 kHz
Answer: The pole frequency is 19.5 kHz, the cutoff frequency is 12.89 kHz, the 3-dB frequency is 21.74 kHz,
and the Q is 0.8.
19-20. Given:
R1 = 56 kΩ
R2 = 10 kΩ
C = 680 pF
Solution:
_____
fp = 1/(2πC​√R1 R2 ​)
______________
fp = 1/[2π (680 pF)​√
(10 kΩ)(56 kΩ) ​]
fp = 9.89 kHz
______
Q = 0.5 √
​ (R1/R2) ​
______________
Q = 0.5 √
​
(56 kΩ)/(10 kΩ) ​
Q = 1.18
Kc = 1.04 (from Fig. 19-26)
K3 = 1.30 (from Fig. 19-26)
fc = fp/Kc (Eq. 19-31)
fc = 9.89 kHz/1.04
fc = 9.51 kHz
19-22. Given:
R1 = 2 kΩ
R2 = 56 kΩ
C = 270 pF
Solution:
Av = –R2/2R1 (Eq. 19-32)
Av = –56 kΩ/2(2 kΩ)
Av = –14
______
Q = 0.5 ​√_____________
(R2 /R1) ​ (Eq. 19-33)
Q = 0.5 ​√
(56 kΩ)/(2 kΩ) ​
Q = 2.65
_____
f0 = 1/[2πC √
​ (R1R2)] (Eq.
19-36)
_____________
f0 = 1/[2π(270 pF) √
​
(2 kΩ)(56 kΩ)​]
f0 = 55.7 kHz
Answer: The Q is 2.65, the voltage gain is –14, and the
center frequency is 55.7 kHz.
19-23. Given:
R1 = 3.6 kΩ
R2 = 7.5 kΩ
R3 = 27 Ω
C = 22 nF
Solution:
Av = –R2/2R1 (Eq. 19-32)
Av = –7.5 kΩ/2(3.6 kΩ)
Av = –1.04
___________
Q = 0.5 ​√____________________
[R
2/(R1 || R3)] ​ (Eq. 19-37)
Q = 0.5 ​√
​[7.5 kΩ/(2 kΩ || 27 Ω)] ​
Q = 8.39
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___________
f0 = 1/[2πC​√(R1 || R3) R2] ​ (Eq. 19-38)
___________________
f0 = 1/[2π(22 nF)√ (2
kΩ || 27 Ω) (7.5 kΩ) ​]
f0 = 16.2 kHz
Answer: The Q is 8.39, the voltage gain is –1.04, and the
center frequency is 16.2 kHz.
19-24. Given:
R1 = 28 kΩ
R3 = 1.8 kΩ
C = 1.8 nF
Av = –1
Solution:
_____________
Q = 0.707 ​√[(R
1 + R3) /R3] ​ (Eq. 19-40)
______________________
Q = 0.707 √
​
[(28 kΩ + 1.8 kΩ)/1.8 kΩ] ​
Q = 2.88
_____________
f0 =1/(2πC √
​
[2R1 (R1 || R3)]) ​ (Eq. 19-41)
_______________________
f0 =1/{2π(1.8 nF) √
​ [2(28
kΩ)(28 kΩ || 1.8 kΩ)]} ​
f0 = 9.09 kHz
Answer: The Q is 2.88, the voltage gain is –1, and the
center frequency is 9.09 kHz
19-25. Given:
R1 = 20 kΩ
R2 = 10 kΩ
R = 56 kΩ
C = 180 nF
Solution:
Av = (R2/R1) + 1 (Eq. 19-43)
Av = (10 kΩ/20 kΩ) + 1
Av = 1.5
f0 = 1/(2πRC) (Eq. 19-44)
f0 = 1/[2π(56 kΩ)(180 nF)]
f0 = 15.8 Hz
Q = 0.5/(2 – Av) (Eq. 19-45)
Q = 0.5/(2 – 1.5)
Q=1
BW = f0/Q
BW = 15.8 Hz/1
BW = 15.8 Hz
Answer: The voltage gain is 1.5, the Q is 1, the resonant
frequency is 15.8 Hz, and the bandwidth is 15.8 Hz.
19-26. Given:
R = 3.3 kΩ
C = 220 nF
Solution:
f0 = 1/(2πRC)
f0 = 1/[2π(3.3 kΩ)(220 nF)]
f0 = 219 Hz
f = 2(219 Hz)
f = 438 Hz
ϕ = –2 arctan f/f0
ϕ = –2 arctan (438 Hz/219 Hz)
ϕ = –127°
19-27. Given:
R = 47 kΩ
C = 6.8 nF
Solution:
f0 = 1/(2πRC)
f0 = 1/[2π(47 kΩ)(6.8 nF)]
f0 = 498 Hz
f = 0.5(498 Hz)
f = 249 Hz
ϕ = 2 arctan f/f0
ϕ = 2 arctan (498 Hz/249 Hz)
ϕ = 127°
19-28. Given:
R1 = 24 kΩ
R2 = 100 kΩ
R3 = 10 kΩ
R4 = 15 kΩ
C = 3.3 nF
Solution:
Av = –R2/R1
Av = –100 kΩ/24 kΩ
Av = –4.17
Q = R2/R3
Q = 100 kΩ/10 kΩ
Q = 10
f0 = 1/(2πR3C )
f0 = 1/[2π(10 kΩ)(3.3 nF)]
f0 = 4.82 kHz
BW = f0/Q (Eq. 19-34)
BW = 4.82 kHz/10
BW = 482 Hz
Answer: The voltage gain is – 4.17, the Q is 10, the center
frequency is 4.82 kHz, and the bandwidth is 482 Hz.
19-29. Given:
R1 = 24 kΩ
R2 = 100 kΩ
R3(min) = 2 kΩ
R3(max) = 10 kΩ
R4 = 15 kΩ
C = 3.3 nF
Solution:
Q(min) = R2/R3(max)
Q(min) = 100 kΩ/10 kΩ
Q(min) = 10
f0(min) = 1/(2πR3(max)C)
f0(min) = 1/[2π(10 kΩ)(3.3 nF)]
f0(min) = 4.82 kHz
Q(max) = R2/R3(min)
Q(max) = 100 kΩ/2 kΩ
Q(max) = 50
f0(max) = 1/(2πR3(min)C)
f0(max) = 1/[2π(2 kΩ)(3.3 nF)]
f0(max) = 24.1 kHz
1  ​ = _________________
1
BW = ​ ______
​
​= 482 Hz
2πR2C 2π (100 kΩ)(3.3 nF)
Answer: The maximum center frequency is 24.1 kHz, the
maximum Q is 50, the minimum bandwidth is 482 Hz,
and the maximum bandwidth 482 Hz.
19-30. Given:
R = 6.8 kΩ
R1 = 6.8 kΩ
R2 = 100 kΩ
C = 5.6 nF
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Solution:
Av = 1/3[(R2/R1) + 1]
Av = 1/3[(100 kΩ/6.8 kΩ) + 1]
Av = 5.23
19-34. Given:
n=2
R = 10 kΩ
fc = 5 kHz
Q = Av
Q = 5.23
Solution:
______
Q = 0.5 ​√C2 /C1 ​ (from Fig. 19-24)
_____
√ C2/C1 ​ (Butterworth response)
0.707 = 0.5​
_____
1.414 = √
​ C2/C1 ​
2 = C2/C1
C2 = 2C1
_______
fp = 1/2πR √
​ C1__(2C1) ​
fp = 1/2πRC1 √
​ __2 ​
C1 = 1/2πRfp√
​ 2
__
​C1 = 1/2π(10 kΩ)(5 kHz) √
​ 2 ​
C1 = 2.25 nF
C2 = 4.5 nF
f0 = 1/(2πRC)
f0 = 1/[2π(6.8 kΩ)(5.6 nF)]
f0 = 4.18 kHz
Answer: The voltage gain and Q are 5.23, and the center
frequency is 4.18 kHz.
CRITICAL THINKING
19-31. Given:
f0 = 50 kHz
Q = 20
Solution:
BW = f0/Q
BW = 50 kHz/20
BW = 2.5 kHz
19-35. Given:
n=2
R = 25 kΩ
fc = 7.5 kHz
Ap = 12 dB
f1 = f0 – ½ BW
f1 = 50 kHz – ½ (2.5 kHz)
f1 = 48.75 kHz
Solution:
Since Ap = 12 dB, Kc = 1.391 and Q = 4 (from Table 19-3)
fc = Kcfp (Eq. 19-23)
fp = fc/Kc
fp = 7.5 kHz/1.391
fp = 5.39 kHz
______
Q = 0.5 ​√______
C2/C1  (from Fig. 19-25)
√
4 = 0.5
​
C
_____2/C1 ​ (Chebyshev response)
8 = ​√C2/C1 ​
64 = C2/C1
C2 = 64C1 _________
fp = 1/2πR √
​ C1 (64C1)​
fp = 1/16πRC1​
C1 = 1/16πRfp
C1 = 1/16π (25 kΩ)(5.39 kHz)
C1 = 148 pF
C2 = 9.47 nF
f2 = f0 + ½ BW
f2 = 50 kHz + ½ (2.5 kHz)
f2 = 51.25 kHz
Answer: The cutoff frequencies are 48.75 kHz and
51.25 kHz.
19-32. Given:
f2 = 84.7 kHz
BW = 12.3 kHz
Solution:
f1 = f2 – BW
f1 = 84.7 kHz – 12.3 kHz
f1 = 72.4 kHz
Answer: The lower cutoff frequency is 72.4 kHz.
19-33. Given:
n = 10
fc = 2 kHz
Solution:
Roll-off = 6n dB/octave (Eq. 19-4b)
Roll-off = 6(10) dB/octave
Roll-off = 60 dB/octave
Roll-off = 20n dB/decade (Eq. 19-4a)
Roll-off = 20(10) dB/decade
Roll-off = 200 dB/decade
4 kHz is 1 octave above
Attenuation = 60 dB
8 kHz is 2 octaves above
Attenuation = 120 dB
20 kHz is 1 decade above
Attenuation = 200 dB
Answer: The attenuation is 60 dB at 4 kHz, 120 dB at
8 kHz, and 200 dB at 20 kHz.
19-36. R3 is open
19-37. U1 has failed
19-38. C4 is shorted
19-39. C3 is open
19-40. U2 has failed
Chapter 20 Nonlinear Op-Amp
Circuit Applications
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
d
a
c
b
c
a
a
b
9.
10.
11.
12.
13.
14.
15.
16.
c
b
c
b
b
b
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b
17.
18.
19.
20.
21.
22.
23.
24.
a
c
b
c
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30.
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a
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JOB INTERVIEW QUESTIONS
20-7.
5. It means using back-to-back zener diodes or other circuits to
limit the output voltage swing.
8. An IC comparator does not have an internal compensating
capacitor.
Solution:
vref = [R2/(R1 + R2)]VEE (Eq. 20-2)
vref = [7.5 kΩ/(15 kΩ + 7.5 kΩ)]−12 V
vref = −4 V
PROBLEMS
20-1.
Given:
Av(dB) = 106 dB
Vsat = ±20 V
Solution:
AVOL = antilog(Av(dB)/20)
AVOL = antilog(l06 dB/20)
AVOL = 200,000
vin(min) = ±Vsat/AVOL (Eq. 20-1)
vin(min) = ±20 V/200,000
vin(min) = ±100 µV
fc = 1/[2π (R1 || R2)C] (Eq. 20-3)
fc = 1/[2π (15 kΩ || 7.5 kΩ)1 μF]
fc = 31.8 Hz
Answer: The reference voltage is −4 V, and the cutoff
frequency is 31.8 Hz.
20-8.
Answer: An input voltage of 100 µV will produce positive saturation, assuming rail-to-rail output.
20-2.
20-3.
Given:
VS = ±12 V
R1 = 15 kΩ
R2 = 7.5 kΩ
C = 1 μF
Given:
VCC = 9 V
R1 = 22 kΩ
R2 = 4.7 kΩ
vin(peak) = 7.5 V
Given:
vin = 50 V
R = 10 kΩ
Solution:
vref = [R2/(R1 + R2)]VCC (Eq. 20-2)
vref = [4.7 kΩ/(22 kΩ + 4.7 kΩ)]9 V
vref = 1.58 V
Answer:
ID = (vin – 0.7 V)/R
ID = (50 V – 0.7 V)/10 kΩ
ID = 4.93 mA
vin = vin(peak)sinθ
θ = arcsin(vin/vin(peak))
θ = arcsin(1.58 V/7.5 V)
θ = 12° and 168°
Solution: The diode current is 4.93 mA.
D = conduction angle/360°
D = (168° − 12°)/360°
D = 43.3%
Given:
VZ = 6.8 V
VS = ±15 V
Solution:
vout = ±(VZ + VD)
vout = ±(6.8 V + 0.7 V)
vout = ±7.5 V
Answer: The output voltage will be limited to ±7.5 V.
Answer: The duty cycle is 43.3%.
20-9.
Given:
VCC = 15 V
R1 = 33 kΩ
R2 = 3.3 kΩ
vin(peak) = 5 V
20-4.
Given: VS = ±12 V
Answer: The output voltage would vary between 0.7 V
and –12 V.
Solution:
vref = [R2/(R1= R2)]VCC (Eq. 20-2)
vref = [3.3 kΩ/(33 kΩ + 3.3 kΩ)]15 V
vref = 1.36 V
20-5.
Given: VS = ±12 V
Answer: When the strobe is high, the output is zero.
When the strobe is low, the output will vary between
0.7 V and –9 V.
vin = vin(peak)sinθ
θ = arcsin(vin/vin(peak))
θ = arcsin(1.36 V/5 V)
θ = 16° and 164°
20-6.
Given:
VS = ±15 V
R1 = 47 kΩ
R2 = 12 kΩ
CBY = 0.5 µF
Solution:
vref = [R2/(R1 + R2)]VCC (Eq. 20-2)
vref = [12 kΩ /(47 kΩ + 12 kΩ)]15 V
vref = 3.05 V
fc = 1/[2π(R1 || R2)CBY] (Eq. 20-3)
fc = 1/[2π(47 kΩ || 12 kΩ) 0.5 µF]
fc = 33.3 Hz
Answer: The reference voltage is 3.05 V, and the cutoff
frequency is 33.3 Hz.
D = conduction angle/360°
D = (164° − 16°)/360°
D = 41%
Answer: The duty cycle is 41%.
20-10. Given:
Vsat = 14 V
R1 = 2.2 kΩ
R2 =18 kΩ
Solution:
B = R1/(R1 + R2) (Eq. 20-4)
B = 2.2 kΩ/(2.2 kΩ + 18 kΩ)
B = 0.1089
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UTP = BVsat (Eq. 20-6)
UTP = 0.1089(14 V)
UTP = 1.52 V
LTP = −BVsat (Eq. 20-7)
LTP = −0.1089(14 V)
LTP = −1.52V
H = 2BVsat (Eq. 20-9)
H = 2(0.1089)(14 V)
H = 3.05 V
Answer: The upper trip point is 1.52 V, the lower trip
point is −1.52 V. The hysteresis is 3.05 V.
20-11. Given:
R1 = 1 kΩ
R2 = 20 kΩ
Vsat = 15 V
Solution:
UTP = (R1/R2)Vsat (Eq. 20-10)
UTP = (1 kΩ/20 kΩ)15 V
UTP = 0.75 V
LTP = −(R1/R2)Vsat (Eq. 20-11)
LTP = −(1 kΩ/20 kΩ)15 V
LTP = −0.75 V
H = UTP − LTP (Eq. 20-8)
H = 0.75 V − (−0.75 V)
H = 1.5 V
Answer: The maximum peak-to-peak noise allowed is
1.5 V.
20-12. Given:
R1 = 1 kΩ
R2 = 18 kΩ
C1 = 3 pF
Solution:
C2 = (R1/R2)C1 (Eq. 20-12)
C2 = (1 kΩ/18 kΩ)3 pF
C2 = 0.167 pF
Answer: The speed-up capacitor should be at least
0.167 pF.
20-13. Given:
R1 = 1.5 kΩ
R2 = 68 kΩ
Vsat = 13.5 V
Solution:
B = R1/(R1 + R2) (Eq. 20-4)
B = 1.5 kΩ/(1.5 kΩ + 68 kΩ)
B = 0.0216
UTP = BVsat (Eq. 20-6)
UTP = 0.0216(13.5 V)
UTP = 0.292 V
LTP = −BVsat (Eq. 20-7)
LTP = −0.0216(13.5 V)
LTP = −0.292 V
H = 2BVsat (Eq. 20-9)
H = 2(0.0216)(13.5 V)
H = 0.584 V
Answer: The upper trip point is 0.292 V, the lower trip
point is −0.292, and the hysteresis is 0.584 V.
20-14. Given:
R1 = 2.2 kΩ
R2 = 68 kΩ
Vsat = 14 V
Solution:
UTP = (R1/R2)Vsat (Eq. 20-10)
UTP = (2.2 kΩ/68 kΩ)14 V
UTP = 0.453 V
LTP = −(R1/R2)Vsat (Eq. 20-11)
LTP = −(2.2 kΩ/68 kΩ)14 V
LTP = −0.453 V
H = UTP − LTP (Eq. 20-8)
H = 0.453 V − (−0.453 V)
H = 906 mV
Answer: The upper trip point is 0.453 V, the lower trip
point is −0.453 V, and the hysteresis is 906 mV.
20-15. Given:
LTP = 3.5 V
UTP = 4.75 V
vin(peak) = 10 V
Vsat = 12 V
Solution: The output voltage will be low when the input
voltage is between 3.5 and 4.75 V.
20-16. Given: VCC = 12 V
Solution:
UTP = [R/(4R + R)]12 V
UTP = 2.4 V
LTP = [R/(6R + R)]12 V
LTP = 1.71 V
Answer: The lower trip point is 1.71 V, and the upper trip
point is 2.4 V.
20-17. Given:
Vin = 5 V
R =1 kΩ
Solution:
Iin = Vin /R
Iin = 5 V/1 kΩ
Iin = 5 mA
Answer: The charging current is 5 mA.
20-18. Given:
Vin = 5 V
R = 1 kΩ
C = 10 μF
T = 1 ms
Solution:
V = [T/(RC)]Vin (Eq. 20-13)
V = [1 ms/(1 kΩ)(10 μF)]5 V
V = 0.5 V
Answer: The output voltage is 0.5 V.
20-19. Given:
Vin = 0.1 V
R = 1 kΩ
C(1) = 0.1 μF
C(2) = 1 μF
C(3) = 10 μF
C(4) = 100 μF
T = 1 ms
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Solution:
V(1) = [T/(RC(1))]Vin (Eq. 20-13)
V(1) = [1 ms/(1 kΩ)(0.1 μF)](0.1 V)
V(1) = 1 V
Solution:
vref = [R2/(R2 +R1)]VCC
vref = [10 kΩ/(10 kΩ + 5 kΩ)]15 V
vref = 10 V
V(2) = [T/(RC(2))]Vin (Eq. 20-13)
V(2) = [1 ms/(1 kΩ)(1 μF)](0.1 V)
V(2) = 0.1 V
Since the reference voltage is the same as the maximum
voltage, the output does not go high and the duty cycle
is 0%.
V(3) = [T/(RC(3))]Vin (Eq. 20-13)
V(3) = [1 ms/(1 kΩ)(10 μF)](0.1 V)
V(3) = 10 mV
With the wiper at the top, the reference voltage is zero
and the output is high half of the time. Therefore, the duty
cycle is 0.50.
V(4) = [T/(RC(4))]Vin (Eq. 20-13)
V(4) = [1 ms/(1 kΩ)(100 μF)](0.1 V)
V(4) = 1 mV
Answer: The duty cycle is 0.5 at the top and 0 at the
bottom.
Answer: The output voltages are 1 V with a 0.1-μF capacitor, 0.1 V with a 1-μF capacitor, 10 mV with a 10-μF
capacitor, and 1 mV with a 100-μF capacitor.
20-20. Given:
f = 10 kHz
R = 4.7 kHz
C = 6.8 μF
VP = 5 V
Solution:
Vout(p-p) = VP/[2fRC] (Eq. 20-17)
Vout(p-p) = 5 V/[2(10 kHz)(4.7 kΩ)(6.8 ΩF)]
Vout(p-p) = 7.8 mV
Answer: The output voltage is a triangular wave with a
peak-to-peak voltage of 7.8 mV.
20-21. Given:
f = 10 kHz
R = 4.7 kHz
C = 0.068 μF
VP = 5 V
Solution:
Vout(p-p) = VP/[2fRC] (Eq. 20-17)
Vout(p-p) = VP/[2(10 kHz)(4.7 kΩ)(0.068 μF)]
Vout(p-p) = 0.782 Vp-p
Answer: The output voltage is a triangular waveform
with a peak-to-peak voltage of 0.782 V.
20-22. Given:
f(1) = 5 kHz
f(2) = 20 kHz
R = 4.7 kHz
C = 6.8 μF
VP = 5 V
Solution:
Vout(p-p)1 = VP /[2fRC] (Eq. 20-17)
Vout(p-p)1 = 5 V/[2(5 kHz)(4.7 kΩ)(6.8 μF)]
Vout(p-p)1 = 15.6 mV
Vout(p-p)2 = VP /[2fRC] (Eq. 20-17)
Vout(p-p)2 = 5 V/[2(20 kHz)(4.7 kΩ)(6.8 μF)]
Vout(p-p)2 = 3.9 mV
Answer: The output voltage is 15.6 mV at 5 kHz, and
3.9 mV at 20 kHz.
20-23. Given:
R1 = 5 kΩ
R2 = 0 Ω (top) and 10 kΩ (bottom)
VCC = 15 V
VP = 10 V
20-24. Given:
R1 = 5 kΩ
R2 = 5 kΩ
VCC = 15 V
VP = 10 V
Solution:
vref = [R2/(R2 + R1)]VCC
vref = [5 kΩ/(5 kΩ + 5 kΩ)]15 V
vref = 7.5 V
D = (VP − vref)/(VP − (−VP))
D =(10 V − 7.5 V)/(10 V − (−10 V))
D = 0.125
Answer: The output is high for 1/8 the cycle, therefore the
duty cycle is 0.125.
20-25. Given:
R1 = 33 kΩ
R2 = 4.7 kΩ
R = 2 kΩ
C = 0.1 μF
Solution:
B = R1/(R1 + R2) (Eq. 20-4)
B = 33 kΩ/(33 kΩ + 4.7 kΩ)
B = 0.875
T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18)
T = 2(2 kΩ)(0.1 μF)ln[(1+ 0.875)/(1 − 0.875)]
T = 1.08 ms
f = 1/T
f = 1/1.08 ms
f = 923 Hz
Answer: The frequency is 923 Hz.
20-26. Given:
R1 = 66 kΩ
R2 = 9.4 kΩ
R = 4 kΩ
C = 0.1 μF
Solution:
B = R1/(R1 + R2) (Eq. 20-4)
B = 66 kΩ/(66 kΩ + 9.4 kΩ)
B = 0.875
T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18)
T = 2(4 kΩ)(0.1 μF)ln[(1 + 0.875)/(1 − 0.875)]
T = 2.16 ms
f = 1/T
f = 1/2.16 ms
f = 463 Hz
Answer: The frequency is reduced by half.
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20-27. Given:
R1 = 33 kΩ
R2 = 4.7 kΩ
R = 2 kΩ
C = 0.47 μF
Solution:
B = R1/(R1 + R2) (Eq. 20-4)
B = 33 kΩ/(4.7 kΩ + 33 kΩ)
B = 0.875
T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18)
T = 2(2 kΩ)(0.47 μF)ln[(1 + 0.875)/(1 − 0.875)]
T = 5.09 ms
f =1/T
f = 1/5.09 ms
f = 196 Hz
Answer: The frequency is 196 Hz.
20-28. Given:
R1 = 2.2 kΩ
R2 = 22 kΩ
Vsat = 12 V
Solution:
UTP = (R1/R2)Vsat (Eq. 20-10)
UTP = (2.2 kΩ/22 kΩ)12 V
UTP = 1.2 V
LTP = −(R1/R2)Vsat (Eq. 20-11)
LTP = −(2.2 kΩ/22 kΩ)12 V
LTP = −1.2 V
H = UTP – LTP (Eq. 20-8)
H = 1.2 V − (−1.2 V)
H = 2.4 V
Answer: The upper trip point is 1.2 V, and the lower trip
point is −1.2 V. The hysteresis is 2.4 V.
20-29. Given:
f = 5 kHz
R3 = 2.2 kΩ
R4 = 22 kΩ
C = 4.7 μF
Vp-p = 28 V
Vp = 14 V
Solution:
Vout(p-p) = VP/[2fR3C] (Eq. 20-17)
Vout(p-p) = 14 V/[2(5 kHz)(2.2 kΩ)(4.7 μF)]
Vout(p-p) = 135 mVp-p
Answer: The output voltage is 135 mVp-p.
20-30. Given: vin = 100 mVP
Answer: The output will be a half-wave signal with a
peak voltage of 100 mV.
20-31. Given:
vin = 75 mVrms
f = 20 kHz
Solution:
vin(p) = 1.414(vin(rms))
vin(p) = 1.414(75 mVrms)
vin(p) = 106 mV peak
20-32. Given:
RL = 33 kΩ
C = 6.8 μF
Solution:
RLC > 10T (Eq. 20-20)
RLC = 10T for the highest period or lowest frequency
[(33 kΩ)(6.8 μF)]/10 = T
T = 22.4 ms
f = 1/T
f = 122.4 ms
f = 45 Hz
Answer: The lowest frequency is 45 Hz.
20-33. Answer: With the diode reversed, it becomes a negative
peak detector and the output voltage is –106 mV.
20-34. Given: vin = 150 mV peak-to-peak
Answer: The output voltage is 75 mV.
20-35. Given: vin = 100 mV peak
Solution:
vout = vin + V peak (Eq. 20-21)
vout = 100 mV peak + 100 mV peak
vout = 200 mV peak
Answer: The output swings from 0 V to 200 mV peak.
20-36. Given:
RL = 10 kΩ
C = 4.7 μF
Solution:
RLC > 10T (Eq. 20-20)
RLC = 10T for the highest period or lowest frequency
[(10 kΩ)(4.7 μF)]/10 = T
T = 4.7 ms
f = 1/T
f = 1/4.7 ms
f = 213 Hz
Answer: The lowest frequency is 213 Hz.
20-37. Given: f = 10 kHz
Solution:
1 Hz = 1 cycle/second
10 kHz = 10,000 cycles/second or 10,000 cycles in
1 second
Each cycle has two transitions (low to high and high to
low); thus there are 2 pulses per cycle.
10,000 cycles in 1 second × 2 pulses/cycle = 20,000
pulses in 1 second.
Answer: There are 20,000 pulses in 1 second.
20-38. Given: f = 1 kHz
Solution:
Since there are 2 pulses per cycle, a pulse occurs every
T/2.
T = 1/f
T = 1/1 kHz
T = 1 ms
Answer: A pulse occurs every T/2 or 0.5 ms.
Answer: The output voltage is 106 mV.
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CRITICAL THINKING
20-39. Answer: Make the 3.3-kΩ resistor a variable so that it can
be adjusted to any desired value.
20-40. Given:
R = 1 kΩ
C = 50 pF
Solution: Risetime is from the 10% point to the 90% point
(discussed in Chap. 14). Using the universal time constant chart, it takes about 3 time constants to go from 10%
to 90%.
TR ≈ 2.2(RC) (Eq. 14-28)
TR ≈ 2.2(1 kΩ)(50 pF)
TR ≈ 110 ns
Answer: The risetime is 110 ns.
20-41. Given:
R1 = 33 kΩ
R2 = 3.3 kΩ
C = 47 μF
Vripple = 1 Vrms
Solution:
fC = 1/[2π(R1 || R2)C] (from Fig. 20-11)
fC = 1/[2π(33 kΩ || 3.3 kΩ)47 μF]
fC = 1.1 Hz
The power supply ripple is 120 Hz because rectification.
This is 2 decades above the cutoff frequency. Since there
is one capacitor, the roll-off is 20 dB/decade. The input is
attenuated by 40 dB, equivalent to 0.01.
3.3 kΩ
vtb = ______________
​
(1V) = 0.1V
33 kΩ + 3.3 kΩ
vout = (0.01)(0.1 V) = 0.001 V
Answer: The cutoff frequency is 1.1 Hz, and the ripple
voltage at the inverting input is 0.001 Vrms.
20-42. Given:
VCC = 15 V
R1 = 33 kΩ
R2 = 3.3 kΩ
vin(peak) = 5 V
vref = 1.36 V (from Prob. 20-9)
θ = 16° and 164° (from Prob. 20-9)
D = 41% (from Prob. 20-9)
Solution:
Ihigh = V/R
Ihigh = 5 V/1 kΩ
Ihigh = 5 mA
Since the output is high only 41% of the time, the average
current is:
Iave = DIhigh
Iave = (0.41)(5 mA)
Iave = 2.05 mA
Answer: The average current is 2.05 mA.
20-43. Given:
R1 = 1.5 kΩ ± 5%
R2 = 68 kΩ ± 5%
Vsat = 13.5 V
R2(max) = 68 kΩ + 5%(68 kΩ) = 71.4 kΩ
R2(min) = 68 kΩ – 5%(68 kΩ) = 64.6 kΩ
B(min) = R1(min)/(R1(min) + R2(max)) (Eq. 20-4)
B(min) = 1425 Ω/(1425 Ω + 71.4 kΩ)
B(min) = 0.0196
H(min) = 2B(min) Vsat (Eq. 20-9)
H(min) = 2(0.0196)(13.5 V)
H(min) = 0.529 V
Answer: The minimum hysteresis is 0.529 V.
20-44. Given:
LTP = 3.5 V
UTP = 4.75 V
vin(peak) = 10 V
Vsat = 12 V
Solution:
vin = vin(peak) sinθ
θ = arcsin(vin/vin(peak))
θ = arcsin(3.5 V/10 V)
θ = 21° and 159°
vin = vin(peak) sinθ
θ = arcsin(vin/vin(peak))
θ = arcsin(4.75 V/10 V)
θ = 28° and 152°
The output will be high from 21° to 28° and from 152° to
159°, or a total of 14°
D = conduction angle/360°
D = 14°/360°
D = 3.9%
Answer: The duty cycle is 0.039 or 3.9%.
20-45. Given: With Eq. (20-13), the output voltage at the end of
the pulse is
V = [(1 ms)/(1 kΩ)(10 μF)](5 V) = 0.5 V
C = (T/RV)Vin
C = [(1 ms)/(1 kΩ)(10 V)](5 V) = 0.5 μF
The foregoing is for T = 1 ms. Therefore, we need
0.05 μF for T = 0.1 ms and 5 μF for T = 10 ms.
Answer: Switch in different capacitors of 0.05, 0.5, and
5 μF. Also, cascade an inverter to get a positive-going
output.
20-46. Given:
f = 20 kHz
R1 = 33 kΩ
R2 = 4.7 kΩ
Solution:
B = R1/(R1 + R2) (Eq. 19-6)
B = 33 kΩ/(33 kΩ + 4.7 kΩ)
B = 0.875
T = 1/f
T = 1/20 kHz
T = 50 μs
T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18)
C = T/2Rln[(1 + B)/(1 − B)]
C = 50 μs/2(2 kΩ)ln[(1 + 0.875)/(1 − 0.875)]
C = 4.6 nF
Answer: Change the capacitor to 4.7 nF.
Solution:
R1(max) = 1.5 kΩ + 5%(1.5 kΩ) = 1575 Ω
R1(min) = 1.5 kΩ – 5%(1.5 kΩ) = 1425 Ω
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20-47. Given:
R1 = 2.2 kΩ
R2 = 82 kΩ
Vsat = 14 V
Solution:
UTP = (R1/R2) Vsat (Eq. 20-10)
UTP = (2.2 kΩ/82 kΩ)14 V
UTP = 0.376 V
H = 2 UTP = 0.751 V
R1 = 1/0.751 (2.2 kΩ) = 2.93 kΩ
Use nearest standard value of 3 kΩ.
To increase hysteresis to 1 V, increase R1 by a factor of
1 V/0.751 V, to get R1 ≅ 3 kΩ. For a safety factor, use
R1 = 3.3 kΩ.
Answer: Increase R1 to 3.3 kΩ.
20-48. Answer: Use a window comparator with an upper trip
point of 5 V and a lower trip point of −5 V.
20-49. Answer: Use a comparator with hysteresis and a lightdependent resistor in a voltage divider as the input.
20-50. Answer: Rectify the incoming signal and send it to a comparator. When the input drops below the reference output,
the alarm sounds.
20-51. Given:
Vin = 5 V
R = 1 MΩ
C = 10 μF
V = 1.23 V
Solution:
V = [T/(RC)]Vin (Eq. 20-13)
T = (V/Vin)RC
T = (1.23 V/5 V)(1 M Ω)(10 μF)
T = 2.46 s
20-59. U1 has failed
20-60. The trip point has changed from −10 V to −7.3 V
Chapter 21
SELF-TEST
1. a
9. a
17. b
2. b
10. b
18. d
3. a
11. b
19. b
4. c
12. d
20. d
5. b
13. c
21. b
6. a
14. b
22. a
7. c
15. a
23. d
8. d
16. b
24. B
PROBLEMS
21-1.
20-52. Answer:
Trouble 1: The positive clamper circuit is the trouble.
20-54. Answer:
Trouble 6: The comparator circuit is the trouble.
Trouble 7: The integrator circuit is the trouble.
20-55. Answer:
Trouble 8: The peak detector circuit is the trouble.
Trouble 9: The integrator circuit is the trouble.
Trouble 10: The comparator circuit is the trouble.
20-56. C4 is open
20-57. D1 is shorted
20-58. D2 is open
Given: Rf = 1 kΩ
Solution: The oscillator becomes stable with a lamp
resistance of 500 Ω and from the graph a lamp voltage
of 3 Vrms.
IL = 3 V/500 Ω
IL = 6 mA
Vout = IL(Rf + RL)
Vout = 6 mA(1 kΩ + 500 Ω)
Vout = 9 Vrms
Trouble 2: The integrator circuit is the trouble.
Trouble 5: The positive clamper circuit is the trouble.
a
d
c
b
d
d
a
d
c
5. The monostable timer has pins 6 and 7 connected together
and is externally triggered, whereas the astable timer has
a resistor between pins 6 and 7 and requires no external
trigger.
8. There must be an unwanted positive feedback path between
the output and the input of the three-stage amplifier. Lowfrequency oscillations may be caused by the high powersupply impedance. You can try using a large filter capacitor at
the supply point for each stage. If this docs not work, then
a power supply with better regulation is needed. For highfrequency oscillations, you can try shielding the stages, using
a single ground point, filter capacitors on each stage supply,
and ferrite beads on each base or gate lead.
Answer: The moon is 228,780 miles away.
Trouble 4: The peak detector circuit is the trouble.
25.
26.
27.
28.
29.
30.
31.
32.
33.
JOB INTERVIEW QUESTIONS
(2.46 s)186,000 miles/s = 457,560 miles for a round trip.
Therefore, the distance is half this amount: 228,780 miles.
20-53. Answer:
Trouble 3: The relaxation oscillator circuit is the trouble.
Oscillators
Answer: The output voltage is 9 Vrms.
21-2.
Given:
C = 200 pF
Rmin = 2 kΩ
Rmax = 24 kΩ
Solution:
fr(max) = 1/[2πRminC] (Eq. 21-4)
fr(max) = 1/[2π(2 kΩ)(200 pF)]
fr(max) = 398 kHz
fr(min) = 1/[2πRmaxC] (Eq. 21-4)
fr(min) = 1/[2π(24 kΩ)(200 pF)]
fr(min) = 33.2 kHz
Answer: The maximum frequency is 398 kHz, and the
minimum frequency is 33.2 kHz.
21-3a. Given:
C = 0.2 μF
Rmin = 2 kΩ
Rmax = 24 kΩ
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Solution:
fr(max) = 1/[2πRminC] (Eq. 21-4)
fr(max) = 1/[2π(2 kΩ)(0.2 μF)]
fr(max) = 398 Hz
fr(min) = 1/[2πRmaxC] (Eq. 21-4)
fr(min) = 1/[2π(24 kΩ)(0.2 μF)]
fr(min) = 33.2 Hz
21-5.
Given: Maximum frequency is 398 kHz, from Prob. 21-3.
Solution: 1 decade above 398 kHz is 3.98 MHz.
Answer: The cutoff frequency is 3.98 MHz.
21-6.
Given:
R = 10 kΩ
C = 0.01 μF
Solution:
fr = 1/[2πRC] (Eq. 21-4)
fr = 1/[2π(10 kΩ)(0.01 μF)]
fr = 1.59 kHz
Answer: The maximum frequency is 398 Hz, and the
minimum frequency is 33.2 Hz.
21-3b. Given:
C = 0.02 μF
Rmin = 2 kΩ
Rmax = 24 kΩ
Answer: The resonant frequency is 1.59 kHz.
21-7.
Solution:
fr(max) = 1/[2πRminC] (Eq. 21-4)
fr(max) = 1/[2π (2 kΩ)(0.02 μF)]
fr(max) = 3.98 kHz
Solution:
fr = 1/[2πRC] (Eq. 21-4)
fr = 1/[2π(20 kΩ)(0.02 μF)]
fr = 398 Hz
fr(min) = 1/[2πRmaxC] (Eq. 21-4)
fr(min) = 1/[2π (24 kΩ)(0.02 μF)]
fr(min) = 332 Hz
Answer: The maximum frequency is 3.98 kHz, and the
minimum frequency is 332 kHz.
Answer: The resonant frequency is 398 Hz.
21-8.
21-3c. Given:
C = 0.002 μF
Rmin = 2 kΩ
Rmax = 24 kΩ
Given:
R1 = 10 kΩ
R2 = 5 kΩ
RE = 1 kΩ
VBE = 0.7 V
VCC = 12 V
Solution:
fr(max) = 1/[2πRminC] (Eq. 21-4)
fr(max) = 1/[2π(2 kΩ)(0.002 μF)]
fr(max) = 39.8 kHz
Solution:
VB = [R2/(R1 + R2)]VCC
VB = [5 kΩ/(10 kΩ + 5 kΩ)]12 V
VB = 4 V
fr(min) = 1/[2πRmaxC] (Eq. 21-4)
fr(min) = 1/[2π(24 kΩ)(0.002 μF)]
fr(min) = 3.32 kHz
VE = VB − VBE
VE = 4 V – 0.7 V
VE = 3.3 V
Answer: The maximum frequency is 39.8 kHz, and the
minimum frequency is 3.32 kHz.
IE = VE/RE
IE = 3.3 V/1 kΩ
IE = 3.3 mA
21-3d. Given:
C = 200 pF
Rmin = 2 kΩ
Rmax = 24 kΩ
Solution:
fr(max) = 1/[2πRminC] (Eq. 21-4)
fr(max) = 1/[2π(2 kΩ)(200 pF)]
fr(max) = 398 kHz
fr(min) = 1/[2πRmaxC] (Eq. 21-4)
fr(min) = 1/[2π(24 kΩ)(200 pF)]
fr(min) = 33.2 kHz
Answer: The maximum frequency is 398 kHz, and the
minimum frequency is 33.2 kHz.
21-4.
Given:
R = 20 kΩ
C = 0.02 μF
Given:
Vout = 6 Vrms
Rf = 2Rlamp
Solution: Since the lamp resistance is one-third of the
total resistance, its voltage will be one-third of the total
voltage, or 2 Vrms. According to the graph, the lamp resistance would be 350 Ω, so the feedback resistor would
need to be twice that, or 700 Ω.
Answer: Change the feedback resistor to 700 Ω.
Since the RF choke is a short to direct current, the
collector voltage is 12 V.
VCE = VC – VE
VCE = 12 V – 3.3 V
VCE = 8.7 V
Answer: The emitter current is 3.3 mA, and the collectorto-emitter voltage is 8.7 V.
21-9.
Given:
C1 = 0.001 μF
C2 = 0.01 μF
L = 10 μH
Solution:
C = C1C2/(C1 + C2) (Eq. 21-6)
C = (0.001 μF)(0.01 μF)/(0.001 μF + 0.01 μF)
C = 909 pF
___
fr = 1/(2π​√______________
LC ​) (Eq.21-5)
fr = 1/(2π​√(10
μH)(909 pF) ​)
fr = 1.67 MHz
B = C1/C2 (Eq. 21-7)
B = 0.001 μF/0.01 μF
B = 0.10
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Av(min) = C2/C1 (Eq. 21-8)
Av(min) = 0.01 μF/0.001 μF
Av(min) = 10
Answer: The frequency is 1.67 MHz, the feedback
fraction is 0.10, and the minimum gain is 10.
21-10. Given:
C1 = 0.001 μF
C2 = 0.01 μF
Solution:
B = C1/(C1 + C2)
B = 0.001 μF/(0.001 μF + 0.01 μF)
B = 0.091
Answer: The feedback fraction is 0.091.
21-11. Given:
C1 = 0.001 μF
C2 = 0.01 μF
L = 20 μH
Solution:
C = C1C2/(C1 + C2) (Eq. 21-6)
C = (0.001 μF)(0.01 μF)/(0.001 μF + 0.01 μF)
C = 909 pF
___
fr = 1/(2π​√LC ​) (Eq.21-5)
______________
fr = 1/(2π​√(20
μH)(909 pF) ​)
fr = 1.18 MHz
Answer: The frequency is 1.18 MHz.
21-12. Answer: Reduce the inductance by a factor of 4 (since
there is a square root in the denominator).
21-13. Given:
C1 = 0.001 μF
C2 = 0.01 μF
C3 = 47 pF
L = 10 μH
Solution:
____
fr = 1/(2π​√_____________
LC3 ​
) (Eq.21-18)
fr = 1/(2π​√
(10 μH)(47 pF) ​)
fr = 7.34 MHz
Answer: The frequency is 7.34 MHz.
21-14. Given:
L1 = 1 μH
L2 = 0.2 μH
C = 1000 pF
Solution:
B = L2/L1 (Eq. 21-16)
B = 0.2 μH/1 μH
B = 0.2
L = L1 + L2
L = 1 μH + 0.2 μH
L = 1.2 μH
___
fr = 1/[2π​√_______________
LC ​] (Eq. 21-5)
fr = 1/[2π​√(1.2
μH)(1000 pF) ​]
fr = 4.59 MHz
Av(min) = L1/L2
Av(min) = 1 μH/0.2 μH
Av(min) = 5
21-15. Given:
M = 0.1 μH
L = 3.3 μH
Solution:
B = M/L (Eq. 21-14)
B = 0.1 μH/3.3 μH
B = 0.030
Av(min) = L/M
Av(min) = 3.3 μH/0.1 μH
Av(min) = 33
Answer: The feedback fraction is 0.03, and the minimum
gain is 33.
21-16. Given: f = 5 MHz
Answer: The first overtone is 10 MHz, the second overtone is 15 MHz, and the third overtone is 20 MHz.
21-17. Answer: Since the frequency is inversely proportional to
thickness, if thickness is reduced by 1% the frequency
will increase by 1%.
21-18. Given:
L=1H
Cs = 0.01 pF
R = 1 kΩ
Cm = 20 pF
Solution:
____
fs = 1/[2π​√____________
LCs ​] (Eq.21-20)
fs = 1/[2π​√
(1 H)(0.01 pF) ​]
fs = 1.5915 MHz
Cp = CmCs/(Cm + Cs) (Eq. 21-21)
Cp = (20 pF)(0.01 pF)/(20 pF + 0.01 pF)
Cp = 0.009995 pF
____
fp = 1/[2π​√LCp ​] (Eq.21-22)
________________
fp = 1/[2π​√(1
H)(0.009995 pF) ​]
fp = 1.5919 MHz
XL = 2πfL
XL = 2π (1.5915 MHz)(1 H)
XL = 10 MΩ
Q = XL/R
Q = 10 MΩ/1 kΩ
Q = 10,000
Answer: The series frequency is 1.5915 MHz, the parallel
frequency is 1.5919 MHz, and the Q is 10,000.
21-19. Given:
R = 10 kΩ
C = 0.047 μF
Solution:
W = 1.1RC (Eq. 21-25)
W = 1.1(10 kΩ)(0.047 μF)
W = 517 μs
Answer: The pulse width is 517 μs.
21-20. Given:
VCC = 10 V
R = 2.2 kΩ
C = 0.2 μF
Answer: The frequency is 4.59 MHz, the feedback fraction is 0.2, and the minimum gain is 5.
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Solution:
LTP = VCC/3 (Eq. 21-24)
LTP = 10 V/3
LTP = 3.33 V
UTP = 2VCC/3 (Eq. 21-23)
UTP = 2(10 V)/3
UTP = 6.67 V
W = 1.1RC (Eq. 21-25)
W = 1.1(2.2 kΩ)(0.2 μF)
W = 484 μs
Answer: The minimum trigger voltage is 3.33 V, the
maximum capacitor voltage is 6.67V, and the pulse width
is 484 μs.
21-21. Given:
R1 = 10 kΩ
R2 = 2 kΩ
C = 0.0022 μF
Solution:
1.44
f = __________
​
​ (Eq. 21-28)
(R1 + 2R2)C
1.44
_________________________
f =
​
​
[10 kΩ + 2(2 kΩ)](0.0022 μF)
f = 46.8 kHz
Answer: The frequency is 46.8 kHz.
21-22. Given:
R1 = 20 kΩ
R2 = 10 kΩ
C = 0.047 μF
Solution:
1.44
f = __________
​
​ (Eq.21-18)
(R1 + 2R2)C
1.44
_________________________
f = ​
​
[20 kΩ + 2(10 kΩ)](0.047 μF)
f = 766 Hz
D = (R1 + R2)/(R1 + 2R2) (Eq. 21-29)
D = (20 kΩ + 10 kΩ)/[20 kΩ + 2(10 kΩ)]
D = 0.75
Answer: The frequency is 766 Hz, and the duty cycle is
0.750.
21-23. Given:
VCC = 10 V
R = 5.1 kΩ
C = 1 nF
f = 10 kHz
vmod = 1.5 V
Solution:
T = 1/f
T = 1/10 kHz
T = 100 μs
W = 1.1RC (Eq. 21-25)
W = 1.1(5.1 kΩ)(1 nF)
W = 5.61 μs
UTPmax = 2VCC /3 + vmod (Eq. 21-35)
UTPmax = 2(10 V)/3 + 1.5 V
UTPmax = 8.17 V
UTPmin = 2VCC /3 − vmod (Eq. 21-35)
UTPmin = 2(10 V)/3 − 1.5 V
UTPmin = 5.17 V
Wmax = −RCln(1 − UTPmax/VCC) (Eq. 21-34)
Wmax = −(5.1 kΩ)(1 nF)ln[1 − (8.17 V/10 V)]
Wmax = 8.66 μs
Wmin = −RCln(1 − UTPmin/VCC) (Eq. 21-34)
Wmin = −(5.1 kΩ)(1 nF)ln[1 − (5.17 V/10 V)]
Wmin = 3.71 μs
Dmax = Wmax/T
Dmax = 8.66 μs/100 μs
Dmax = 0.0866
Dmin = Wmin/T
Dmin = 3.71 μs/100 μs
Dmin = 0.0371
Answer: The period is 100 μs, the quiescent pulse width
is 5.61 μs, the maximum pulse width is 8.66 μs, the minimum pulse width is 3.71 μs, the maximum duty cycle is
0.0866, and the minimum duty cycle is 0.0371.
21-24. Given:
VCC = 10 V
R1 = 1.2 kΩ
R2 = 1.5 kΩ
C = 4.7 nF
vmod = 1.5 V
Solution:
W = 0.693(R1 + R2)C (Eq. 21-26)
W = 0.693(1.2 kΩ + 1.5 kΩ)(4.7 nF)
W = 8.79 μs
T = 0.693(R1 + 2R2)C (Eq. 21-27)
T = 0.693[1.2 kΩ + 2(1.5 kΩ)](4.7 nF)
T = 13.68 μs
UTPmax = 2VCC/3 + vmod (Eq. 21-35)
UTPmax = 2(10 V)/3 + 1.5 V
UTPmax = 8.17 V
UTPmin = 2VCC/3 − vmod (Eq. 21-35)
UTPmin = 2(10 V)/3 – 1.5 V
UTPmin = 5.17 V
Wmax = – (R1 + R2)C1n[(VCC − UTPmax)/(VCC – 0.5 UTPmax)]
(Eq. 21-36)
Wmax = – {[(1.2 kΩ + 1.5 kΩ)(4.7 nF)]ln[(10 – 8.17 V)/
(10 V) – 0.5(8.17 V)]}
Wmax = 14.89 μs
Wmin = – (R1 + R2)C1n[(VCC – UTPmin)/(VCC – 0.5 UTPmin)]
(Eq. 21-36)
Wmin = –{[(1.2 kΩ + 1.5 kΩ)(4.7 nF)]ln[(10 – 5.17 V)/
(10 V) – 0.5(5.17 V)]}
Wmin = 5.44 μs
Space = 0.693R2C
Space = 0.693(1.5 kΩ)(4.7 nF)
Space = 4.89 μs
Answer: The quiescent pulse width is 8.79 μs, the quiescent period is 13.68 μs, the maximum pulse width is
14.89 μs, the minimum pulse width is 5.44 μs, and the
space between pulses is 4.89 μs.
21-25. Given:
IC = 0.5 mA
VCC = 10 V
C = 47 nF
Solution:
S = IC/C (Eq. 21-40)
S = 0.5 mA/47 nF
S = 10.6 V/ms
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V = 2VCC/3 (Eq. 21-41)
V = 2(10 V)/3
V = 6.67 V
T = 2VCC/3S (Eq. 21-42)
T = 2(10 V)/3(10.6 V/ms)
T = 0.629 ms
Answer: The slope is 10.6 V/ms, the peak value is 6.67 V,
and the duration is 0.629 ms.
21-26. Given:
S1 = Closed
R = 20 kΩ
R3 = 40 kΩ
C = 0.1 μF
Solution:
The waveform is a sine wave.
f0 = 1/RC (Eq. 21-31)
f0 = 1/(20 kΩ)(0.1 μF)
f0 = 500 Hz
Amplitude = 2.4 Vp (from Fig. 21-53c)
Amplitude = 4.8 Vp-p
Answer: The output is a sine wave at a frequency of
500 Hz and a peak voltage of 2.4 V.
21-27. Given:
S1 = Open
R = 10 kΩ
R3 = 40 kΩ
C = 0.01 μF
Solution:
The waveform is a triangle wave.
f0 = 1/RC (Eq. 21-31)
f0 = 1/(10 kΩ)(0.01 μF)
f0 = 10 kHz
Amplitude = 5 Vp (from Fig. 21-53c)
Amplitude = 10 Vp-p
Answer: The output is a triangle wave at a frequency of
10 kHz and a peak voltage of 5 V.
Solution:
f = __
​  2  ​​ _______
​  1  ​  ​
C R1 + R2
1
f = ______
​  2  ​​ ____________
​
​  ​
0.1 μF 2 kΩ + 10 kΩ
f = 1.67 kHz
(
21-29d. Same. Unless the supply falls low enough for clipping.
23-29e. Same. Only a very small change at the output.
21-30.
Answer:
1. Shorted inductor
2. Open inductor
3. Shorted capacitors
4. Open capacitors
5. Open in the feedback path
6. Loss of the power supply
21-31. Answer: The fuzz is probably oscillations. To correct this,
make sure that the leads are short and are not running
really close to each other. Also, a ferrite bead in the feedback path may dampen them out.
CRITICAL THINKING
21-32. Given:
f = 20 Hz to 20 kHz
Vout = 5 Vrms
Rmin = 2.2 kΩ
Solution: One of many possible designs is C = 0.22 μF,
0.022 μF, and 0.0022 μF. Change 2 kΩ in Fig. 21-53a to
3.3 kΩ and use a 50-kΩ potentiometer. Use a 1-kΩ potentiometer in the place of the 1 kΩ in series with the lamp.
Adjust 1 kΩ to get an output of 5 V.
21-33. Given:
f = 2.5 MHz
C1 = 0.001 μF
C2 = 0.01 μF
C = 909 pF (from Prob. 21-9)
Solution:
___
fr = 1/(2π​√LC ​) (Eq. 21-5)
L = [1/(2πfr)]2/C
L = {1/[2π (2.5 MHz)]2/(909 pF)}
L = 4.46 μH
Answer: The inductor would have to be 4.46 μH.
21-28. Given:
R1 = 2 kΩ
R2 = 10 kΩ
C = 0.1 μF
(
21-29c. Same. The upper potentiometer affects frequency, not
output voltage.
)
)
D = R1/(R1 + R2)
D = 2 kΩ/(2 kΩ + 10 kΩ)
D = 0.167
Answer: The frequency is 1.67 kHz, and the duty cycle
is 0.167.
21-29a. Decrease. With the lamp open, there is no path for feedback current. Thus the voltage at the inverting terminal
will equal the output voltage and it should be driven
to 0 V.
21-29b. Increase. With the inverting input grounded, there is no
feedback and the gain is open-loop gain and the output
will be saturation.
21-34. Given:
C = 0.001 μF
R1 = 10 kΩ
R2 = 10 kΩ
R3 = 5 kΩ
f2(CL) = 15.9 kHz
Solution:
XC = 1/(2πfC)
XC = 1/[2π (15.9 kHz)(0.001 μF)]
XC = 10 kΩ
ϕ = −arctan(R/XC)
ϕ = −arctan(10 kΩ/10 kΩ)
ϕ = −45°
Below its critical frequency, the op amp has 180° of
phase shift. At 15.9 kHz, the op amp has an additional
phase shift of 90° because it is well above its critical frequency. This means that the op amp has approximately
270° of phase shift at 15.9 kHz. Each lag circuit has a
phase shift of 45° at 15.9 kHz. Therefore, the total or
loop phase shift is 270° plus 45° plus 45°, or 360°.
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21-35. Given:
f = 1 kHz
D = 0.75
PROBLEMS
22-1.
Solution:
Pick a value for R1 = 10 kΩ.
D = (R1 + R2)/(R1 + 2R2) (Eq. 21-29)
D(R1 + 2R2) = (R1 + R2)
DR1 + 2DR2 = R1 + R2
2DR2 − R2 = R1 − DR1
R2(2D − 1) = (R1 − DR1)
R2 = (R1 − DR1)/(2D − 1)
R2 = [10 kΩ − 0.75(10 kΩ)]/[2(0.75) − 1]
R2 = 5 kΩ
Given:
VNL = 15 V
VFL = 14.5 V
Solution:
Load regulation = (VNL − VFL)/VFL × 100%
Load regulation = (15 V − 14.5 V)/14.5 V × 100%
Load regulation = 3.45%
Answer: The load regulation is 3.45%.
22-2.
f = 1.44/[(R1 + 2R2)C] (Eq. 21-28)
C = 1.44/[(R1 + 2R2)f]
C = 1.44/{[(10 kΩ + (2)(5 kΩ)(1 kHz)]}
C = 72 nF
Solution:
Line regulation = (VHL − VLL)/VLL × 100%
Line regulation = (20 V − 19 V)/19 V × 100%
Line regulation = 5.26%
Answer: R1 is 10 kΩ , R2 is 5 kΩ , and the capacitor is
72 nF.
Answer: The line regulation is 5.26%.
22-3.
21-36. C1 is shorted
Given:
VHL = 20 V
VLL = 19 V
21-37. VCC has failed
21-38. U1 has failed
Given:
VHL = 12.3 V
VLL = 12 V
Solution:
Line regulation = (VHL − VLL)/VLL × 100%
(Eq. 22-2)
Line regulation = (12.3 V − 12 V)/12 V × 100%
Line regulation = 2.5%
21-39. R2 is shorted
21-40. R1 is 1 kΩ instead of 75 kΩ
21-41. Sinewave
21-42. 10 Hz
Answer: The line regulation is 2.5%.
21-43. VR7 = 1.8 kΩ
22-4.
21-45. Output of pin 11 to Q3
Given:
RTH = 2 Ω
RL(min) = 50 Ω
Chapter 22
Supplies
Solution:
Load regulation = RTH/RL(min) × 100%
Load regulation = (2 Ω/50 Ω) × 100%
Load regulation = 4%
21-44. VR5 and VR8
Regulated Power
Answer: The load regulation is 4%.
SELF-TEST
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
a
b
b
b
c
a
c
c
b
a
22-5.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
c
21.
b
22.
b
23.
b
24.
a
25.
c
26.
b
27.
d
28.
c
29.
a
c
b
a
a
b
a
c
d
b
30.
31.
32.
33.
34.
35.
36.
37.
38.
d
c
a
a
c
a
c
b
d
JOB INTERVIEW QUESTIONS
6. The first is a member of the LM78XX family and is a positive voltage regulator with an output voltage of 6 V. The second is a member of the LM79XX family and is a negative
voltage regulator with an output voltage of −12 V.
10. Thermal shutdown means that the chip automatically shuts
itself off when it overheats.
Given:
Vin = 25 V
RS = 22 Ω
VZ = 18 V
VBE = 0.75 V
RL = 100 Ω
Solution:
Vout = VZ + VBE (Eq. 22-5)
Vout = 18 V + 0.75 V
Vout = 18.75 V
IS = (Vin − Vout)/RS (Fig. 22-4)
IS = (25 V − 18.75 V)/22 Ω
IS = 484 mA
IL = Vout/RL
IL = 18.75 V/100 Ω
IL = 187.5 mA
IC = IS − IL
IC = 284 mA − 187.5 mA
IC = 96.5 mA
Answer: The output voltage is 18.75 V, the input current
is 484 mA, the load current is 187.5 mA, and the collector
current is 96.5 mA.
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22-6.
Solution:
Vout = [(R1 + R2)/R1](VZ + VBE) (Eq. 22-6)
Vout = [(330 Ω + 680 Ω)/330 Ω](5.6 V + 0.77 V)
Vout = 19.5 V
IS = (Vin − Vout)/RS (Fig. 22-4)
IS = (25 V – 19.5 V)/15 Ω
IS = 367 mA
IL = Vout/RL
IL = 19.5 V/80 Ω
IL = 244 mA
IC = IS − IL
IC = 367 mA − 244 mA
IC = 123 mA
Answer: The output voltage is 19.5 V, the input current
is 367 mA, the load current is 244 mA, and the collector
current is 123 mA.
22-7.
Given:
Vin = 25 V
RS = 8.2 Ω
VZ = 5.6 V
RL = 50 Ω
R1 = 2.7 kΩ
R2 = 6.2 kΩ
Solution:
Vout = [(R1 + R2)/R1]VZ (Eq. 22-7)
Vout = [(2.7 kΩ + 6.2 kΩ)/2.7 kΩ]5.6 V
Vout = 18.46 V
IS = (Vin − Vout)/RS (Fig. 22-6)
IS = (25 V − 18.46 V)/8.2 Ω
IS = 798 mA
IL = Vout/RL (Fig. 22-6)
IL = 18.46 V/50 Ω
IL = 369 mA
IC = IS − IL (Fig. 22-6)
IC = 798 mA − 369 mA
IC = 429 mA
Answer: The output voltage is 18.46 V, the input current
is 798 mA, the load current is 369 mA, and the collector
current is 429 mA.
22-8.
IL = Vout/RL
IL = 16.9 V/50 Ω
IL = 339 mA
Given:
Vin = 25 V
RS = 15 Ω
VZ = 5.6 V
VBE = 0.77 V
RL = 80 Ω
R1 = 330 Ω
R2 = 680 Ω
Given:
Vin = 20 V
VZ = 4.7 V
VBE = 0.77 V
RL = 50 Ω
R1 = 2.2 kΩ
R2 = 4.7 kΩ
R3 = 1.5 kΩ
R4 = 2.7 kΩ
Solution:
Vout = [(R1 + R2)/R1](VZ + VBE) (Fig. 22-8)
Vout = [(2.2 kΩ + 4.7 kΩ)/2.2 kΩ](4.7 V + 0.7 V)
Vout = 16.9 V
PD = (Vin − Vout)IL (Fig. 22-8)
PD = (20 V − 16.9 V) 336 mA
PD = 1.05 W
Answer: The output voltage is 16.9 V, and the power dissipation is 1.05 W.
22-9.
Given:
Vin = 20 V
Vout = 16.9 V (from Prob. 22-8)
Solution:
Efficiency = Vout /Vin × 100% (Eq. 22-13)
Efficiency = 16.9 V/20 V × 100%
Efficiency = 84.5%
Answer: The efficiency is 84.5%.
22-10. Given:
VZ = 6.2 V
RL = 4 Ω
R1 = 2.7 kΩ
R2 = 2.2 kΩ
Solution:
Vout = [(R1 + R2)/R1](VZ) (Eq. 22-14)
Vout = [(2.7 kΩ + 2.2 kΩ)/2.7 kΩ](6.2 V)
Vout = 11.25 V≈ 11.3 V
Answer: The output voltage is 11.25 V≈ 11.3 V.
22-11. Given:
VZ = 4.7 V
Vin(max) = 30 V
R3 = 820 Ω
Solution:
IZ(max) = (Vin(max) − VZ)/R3
IZ(max) = (30 V − 4.7 V)820 Ω
IZ(max) = 30.9 mA
Answer: The maximum zener current is 30.9 mA.
22-12. Given:
VZ = 4.7 V
Radj = 1.5 kΩ
R1(min) = 750 Ω
R1(max) = 2.25 kΩ
R2(min) = 750 Ω
R2(max) = 2.25 kΩ
Solution:
Vout(max) = [(R1(min) + R2(max))/R1(min)]VZ (Eq. 22-14)
Vout(max) = [(750 Ω + 2.25 kΩ)/750 Ω](4.7 V)
Vout(max) =18.8 V
Vout(min) = [(R1(max) + R2(min))/R1(max)]VZ (Eq. 22-14)
Vout(min) = [(2.2 kΩ + 750 Ω)/2.25 k Ω](4.7 V)
Vout(min) = 6.27 V
Answer: The maximum output voltage is 18.8 V, and the
minimum is 6.27 V.
22-13. Given:
Vout(reg) = 10 V
R4 = 3 Ω
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Solution:
The current limiting starts at a VBE of 0.6 V.
IL = 0.6 V/R4
IL = 0.6 V/3 Ω
IL = 200 mA
Solution:
Maximum efficiency = (Vout/Vin(min)) × 100%
(Eq. 22-13)
Maximum efficiency = (15 V/18 V) × 100%
Maximum efficiency = 83.3%
RL = Vout(reg)/IL
RL = 10 V/200 mA
RL = 50 Ω
Minimum efficiency = (Vout/Vin(max)) × 100%
(Eq. 22-13)
Minimum efficiency = (15 V/25 V) × 100%
Minimum efficiency = 60%
ISL = 0.7 V/R4
ISL = 0.7 V/3 Ω
ISL = 233 mA
Answer: Current limiting starts at a load resistance of
50 Ω and the short-circuit current is 233 mA.
22-14. Given:
Vout = 15 V
Vin = 20 V
RL = 20 Ω
Solution:
IL = Vout/RL
IL = 15 V/20 Ω
IL = 750 mA
Headroom voltage = Vin − Vout (Eq. 22-11)
Headroom voltage = 20 V − 15 V
Headroom voltage = 5 V
PD = (Headroom voltage)(IL) (Eq. 22-12)
PD = (5 V)(750 mA)
PD = 3.75 W
Answer: The load current is 750 mA, the headroom voltage is 5 V, and the power dissipation is 3.75 W.
22-15. Given:
f = 120 Hz
C = 4700 μF
IL = 750 mA (from Prob. 22-14)
RRdB = 70 dB (from Table 22-1)
Solution:
VR(in) = IL/fC (Eq. 4-10)
VR(in) = 750 mA/[(120 Hz)(4700 μF)]
VR(in) = 1.33 V
RR = antilog(RRdB/20)
RR = antilog(70 dB/20)
RR = 3162
VR(out) = VR(in)/RR
VR(out) = 1.33 V/3162
VR(out) = 421 μV
Answer: The output ripple voltage is 421 μV.
22-16. Given:
R1 = 2.7 kΩ
R2 = 20 kΩ
Solution:
Vout = [(R1 + R2)/R1](1.25 V) (Eq. 22-19)
Vout = [(2.7 kΩ + 20 kΩ)/2.7 kΩ](1.25 V)
Vout = 10.5 V
Answer: The output voltage is 10.5 V.
22-17. Given:
Vout = 15 V
Vin(min) = 18 V
Vin(min) = 25 V
Answer: The maximum efficiency is 83.3%, and the minimum efficiency is 60%.
22-18. Given:
Vout = 12 V
Vin = 5 V
Iout = 0.25 A
Iin = 1 A
Solution:
Pout = VoutIout
Pout = (12 V)(0.25 A)
Pout = 3 W
Pin = VinIin
Pin = (5 V)(1 A)
Pin = 5 W
Efficiency = Pout/Pin × 100%
Efficiency = 3 W/5 W × 100%
Efficiency = 60%
Answer: The efficiency is 60%.
22-19. Given:
Vout = 5 V
Vin = 12 V
Iin = 2 A
Efficiency is 80%
Solution:
Pin = VinIin
Pin = 12 V(2 A)
Pin = 24 W
Efficiency = Pout/Pin × 100%
(Efficiency/100%)Pin = Pout
Pout = (80%/100%)24 W
Pout = 19.2 W
Pout = VoutIout
Iout = Pout/Vout
Iout = 19.2 W/5 V
Iout = 3.84 A
Answer: The output current is 3.84 A.
22-20. Given:
R1 = 1.5 kΩ
R2 = 10 kΩ
Vref = 2.5 V
Solution:
Vout = [(R1 +R2)/R1](Vref) (Eq. 22-22)
Vout = [(1.5 kΩ + 10 kΩ)/1.5 kΩ](2.5 V)
Vout = 19.2 V
Answer: The output voltage is 19.2 V.
22-21. Given:
D = 30%
Vin = 20 V
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Solution:
Vout = DVin (Eq. 22-21)
Vout = (0.3)(20 V)
Vout = 6 V
Answer: The output voltage is 6 V.
22-22. Given:
R1 = 1.2 kΩ
R2 = 15 kΩ
Vref = 1.25 V
Solution:
Vout = [(R1 + R2)/R1](Vref) (Eq. 22-22)
Vout = [(1.2 kΩ + 15 kΩ)/1.2 kΩ](1.25 V)
Vout = 16.9 V
Answer: The output voltage is 16.9 V.
22-23. Given:
R1 = 2.1 kΩ
R2 = 12 kΩ
Vref = 2.1 V
Solution:
Vout = [(R1 + R2)/R1](Vref) (Eq. 22-22)
Vout = [(2.1 kΩ + 12 kΩ)/2.1 kΩ](2.1 V)
Vout = 14.1 V
Answer: The output voltage is 14.1 V.
CRITICAL THINKING
22-24. Given:
R1 = 240 Ω
R2 = 0 to 5 kΩ
Vin = 30 V
Solution:
Vout(min) = [(R1 + R2)/R1]1.25 (Eq. 22-19)
Vout(min) = [(240 Ω + 5 kΩ)/240 Ω]1.25 V
Vout(min) = 27.3 V
As R2 approaches 0 Ω, the voltage approaches 1.25 V.
Since the LM317 has a dropout of 2 V, the highest regulated output is 28 V with an input of 30 V. With the transistor saturated, the output is 1.25 V.
Answer: The output voltage range with the shutdown
signal low is 1.25 to 27.3 V and is 1.25 V when the shutdown signal is high.
22-25. Given:
R1 = 240 Ω
R2 = 0 to 5 kΩ
Vin = 30 V
Vout = 18 V
Solution:
Using “Vout = [(R1 + R2)/R1]1.25” solve for R2
Vout = [(R1 + R2)/R1]1.25
Vout/1.25 V = [(R1 + R2)/R1]
R1(Vout/1.25 V) = R1 + R2
R1(Vout/1.25 V) − R1 = R2
R1(Vout/1.25 V) − 1] = R2
R2 = R1 [(Vout/1.25 V) − 1]
R2 = 240 Ω[(18 V/1.25 V) − 1]
R2 = 3.22 kΩ
Answer: The adjustable resistor needs to be 3.22 kΩ.
22-26. Answer: Since the output current and voltage are essentially constant, it is logical to assume (since the input
voltage is essentially constant) that the input current is
essentially constant. If the capacitor is supplying a constant current, it should discharge at a linear rate.
22-27. Given:
Load regulation is 5%
VNL = 12.5 V
Solution:
Load regulation = (VNL − VFL)/VFL × 100% (Eq. 22-1)
Load regulation/100% = (VNL − VFL)/VFL
VFL(load regulation/100%) = VNL − VFL
VFL(load regulation/100%) + VFL = VNL
VFL[(load regulation/100%) + 1] = VNL
VFL = VNL/[(load regulation/100%) + 1]
VFL = 12.5 V/[(5%/100%) + 1]
VFL = 11.9 V
Answer: The full load voltage is 11.9 V.
22-28. Given:
Line regulation is 3%
VLL = 16 V
Solution:
Line regulation = (VHL − VLL)/VLL × 100% (Eq. 22-2)
Line regulation/100% = (VHL − VLL)/VLL
VLL(line regulation/100%) = VHL − VLL
VLL(line regulation/100%) + VLL = VHL
VLL[(line regulation/100%) + 1] = VHL
VHL = 16 V/[(3%/100%) + 1]
VHL = 16.48 V
Answer: The high line voltage is 16.48 V.
22-29. Given:
Load regulation = 1%
RL(min) = 10 Ω
Solution:
Load regulation = RTH/RL(min) × 100%
Load regulation/100% = RTH/RL(min)
RTH = RL(min)[load regulation/100%]
RTH = 10 Ω[1%/100%]
RTH = 0.1 Ω
Answer: The power supply output resistance is 0.1 Ω.
22-30. Given:
Vin = 35 V
IC = 60 mA
IL = 140 mA
RS = 100 Ω
Solution:
IC = IS − IL
IS = IC + IL
IS = 60 mA + 140 mA
IS = 200 mA
Vout = Vin − ISRS
Vout = 35 V − (200 mA)(100 Ω)
Vout = 15 V
RL = Vout/IL
RL = 15 V/140 mA
RL =107 Ω
Answer: The load resistance is 107 Ω.
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22-31. Given:
Start current limiting at 250 mA
VBE where current limiting starts is 0.6 V
Solution:
R4 = VBE/I
R4 = 0.6 V/250 mA
R4 = 2.4 Ω
Answer: R4 needs to be 2.4 Ω.
22-32. Given:
Vout = 10 V
VBE = 0.7 V
R4 = 1 Ω
K = 0.70
Solution:
ISL = VBE/KR4
ISL = 0.7 V/0.70 (1 Ω)
ISL = 1 A
Imax = ISL + [(1 − K)Vout]/KR4
Imax = 1 A + [(1 − 0.70)10 V]/0.70(1 Ω)
Imax = 5.29 A
Answer: The shorted load current is 1 A, and the maximum current is 5.29 A.
22-33. Given:
R5 = 7.5 kΩ
R6 = 1 kΩ
R7 = 9 kΩ
C3 = 0.001 μF
Solution:
Vout(mid) = [(R1(mid) + R2(mid))/R1(mid)]VZ (Eq. 22-14)
Vout(mid) = [(1250 Ω + 1250 Ω)/1250 Ω]4.7 V
Vout(mid) = 9.2 V
Answer: The output voltage is 9.2 V.
22-35. Answer: Since the voltage at D is stuck low, the trouble is
the triangle-to-pulse converter.
22-36. Answer: Since the voltage at A is stuck low, the trouble is
the relaxation oscillator.
22-37. Answer: Since the voltage at D is high, the voltage at E
should be 12.8 V, but since it is 0 V, the trouble is Q1.
22-38. Answer: Since the voltage at C is stuck low, the trouble is
the integrator.
22-39. Answer: Since the voltage at A is stuck high, the trouble
is the relaxation oscillator.
22-40. Answer: Since the voltage at E is higher than the desired
output of 5 V, the transistor should be off more of the
time. Thus the output of the comparator should be high.
But because it is low, the problem is the comparator.
22-41. Answer: Since the voltage at D is stuck at 0 V, the trouble
is the triangle-to-pulse converter.
22-42. Answer: Since the voltage at E is low, the output of the
comparator should be low. But because it is high, the
trouble is the comparator.
22-43. Answer: The voltage at D is stuck high, since the output
of the comparator is high, as it should be, given the voltage at E. The trouble is the triangle-to-pulse converter.
Solution:
B = R7/(R6 + R7)
B = 9 kΩ/(1 kΩ + 9 kΩ)
B = 0.9
22-44. Q1 is open
T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18)
T = 2(7.5 kΩ)(0.001 μF)ln[(1 + 0.9)/(1 − 0.9)]
T = 44.16 μs
22-47. D1 is shorted
f = 1/T
f = 1/44.16 μs
f = 22.6 kHz
Answer: The switching frequency is 22.6 kHz.
22-34. Given:
VZ = 4.7 V
R1(mid) = 1250 Ω
R2(mid) = 1250 Ω
22-45. R4 is open
22-46. R2 is open
22-48. U1 has failed
22-49. U1 0 to 20 V; U2 +12 V; U3 +5 V; U4 −12 V; U5 0 to −20 V
22-50. 417 mVp-p
22-51. Vin = 14−15 V
22-52. 20 μVp-p
22-53. VR1 = 840 Ω
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