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890 Resistor CI
(Obsolete)
1
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of 6
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Electronic
Devices and
laboratorio 8
circuitos
 Search document

Part 1
Electronic Principles
Electronic
Eighth Edition
Chapter 1
Introduction
SELF-TEST
12.
1. a
7. b
13. c
19. b
2. c
8. c
14. d
20. c
3. a
9. b
15. b
21. b
4. b
10. a
16. b
22. b
5. d
11. a
17. a
23. c
6. d
12. a
18. b
us more insight into how changes in load resis
load voltage.
It is usually easy to measure open-circuit volta
load current. By using a load resistor and mea
under load, it is easy to calculate the Theve
resistance.
PROBLEMS
1-1.
V = 12 V
RS = 0.1 Ω
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.
Solution:
RL = 100RS
RL = 100(0.1 Ω)
RL = 10 Ω
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.
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7. The expression cold-cranking amperes refers to the amount
of current a car battery can deliver in freezing weather
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Given:
Answer: The voltage source will appear sti
load resistance of ≥10 Ω.
1-2.
Given:
RLmin = 270 Ω
RLmax = 100 kΩ
Solution:
(Eq. 1-1)
RS < 0.01 RL
<
Ω
0.01(270
)
RS
<Free
ΩForon
2.7
RSup
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1-3.

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largest
internal
resistance
Answer:
Useful The
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haveCancel
is 2.7anytime.
Given: RS = 50 Ω
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1-5.
890 Resistor CI
(Obsolete)
of 6
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Given:
Solution:
RS = 0.05 Ω
I=2A
(Eq. 1-4)
RL = 0.01RS
RL = 0.01(250 kΩ)
RL = 2.5 kΩ
Answer: The load current is 4.80 mA, and, no, th
source is not stiff since the load resistance is not l
or equal to 2.5 kΩ.
Given:
1-12.
V=9V
RS = 0.4 Ω
Solution:
VTH = VR2
(Voltage divider
VR2 = VS[(R2)/(R1 + R2)]
VR2 = 36 V[(3 kΩ)/(6 kΩ + 3 kΩ)]
VR2 = 12 V
Solution:
(Ohm’s law)
I = V /R
I = (9 V)/(0.4 Ω)
I = 22.5 A
(Parallel resistance
RTH = [R1R2/R1 + R2]
RTH = [(6 kΩ)(3 kΩ)/(6 kΩ + 3 kΩ)]
RTH = 2 kΩ
Answer: The load current is 22.5 A.
Given:
Answer: The Thevenin voltage is 12 V, and the T
resistance is 2 kΩ.
IS = 10 mA
RS = 10 MΩ
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.

(Current divider form
IL = IT [(RS)/(RS + RL)]
IL = 5 mA [(250 kΩ)/(250 kΩ + 10 kΩ)]
IL = 4.80 mA
Answer: The voltage drop across the internal resistance
is 0.1 V.
1-7.
laboratorio 8
circuitos
 Search document
Solution:
(Ohm’s law)
V = IR
V = (2 A)(0.05 Ω)
V = 0.1 V
1-6.
Electronic
Devices and
6 kV
R1
3 kV
R2
6 kV
R1
3 kV
R2
36 V
Given:
VTH
RLmin = 270 Ω
RLmax = 100 kΩ
Solution:
(Eq. 1-3)
RS > 100 RL
RS > 100(100 kΩ)
RS > 10 MΩ
36 V
Answer: The internal resistance of the source is greater
than 10 MΩ.
1-9.
Given: RS = 100 kΩ
Solution:
= 0.01R
(Eq. 1-4)
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= 0.01(100
kΩ)
R = 1 kΩ
& The New
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The maximum
load resistance for the current
Answer: York
L
S
L
L
source
to appear
is 1 k Ω.
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1-10.
Given:
RTH
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(a) Circuit for
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fo
VTH in Prob. 1-12. ( b) Circuit

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Prob. 1-12.
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1-13.
Given:
V
12 V
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Solution:
RTH
(Eq. 1-10)
RN = RTH
RTH = 10 kΩ
VTH
RL
(Eq. 1-12)
IN = VTH/RTH
VTH = INRN
VTH = (10 mA)(10 kΩ)
VTH = 100 V
Thevenin equivalent circuit for Prob. 1-13.
Answer: RTH = 10 kΩ, and VTH = 100 V
1-14.
Given:
VS = 18 V
R1 = 6 kΩ
R2 = 3 kΩ
RTH
VTH
10 kV
Solution:
VTH = VR2
(Voltage divider formula)
VR2 = VS[(R2)/(R1 + R2)]
VR2 = 18 V[(3 kΩ)/(6 kΩ + 3 kΩ)]
VR2 = 6 V
(Parallel resistance formula)
RTH = [[((R1 × R2)/(R1 + R2)]
RTH = [(6 kΩ × 3 kΩ)/(6 kΩ + 3 kΩ)]
RTH = 2 kΩ
100 V
Thevenin circuit for Prob. 1-17.
1-18.
VTH = 12 V
RTH = 2 kΩ
Answer: The Thevenin voltage decreases to 6 V, and the
Thevenin resistance is unchanged.
1-15.
Solution:
(Eq. 1-10)
Given:
RN = RTH
RN = 2 kΩ
VS = 36 V
R1 = 12 kΩ
R2 = 6 kΩ
IN = VTH/RTH
IN = 12 V/2 kΩ
IN = 6 mA
Solution:
Answer: RN = 2 kΩ, and IN = 6 mA
VTH = VR2
(Voltage divider formula)
VR2 = VS[(R2)/(R1 + R2)]
VR2 = 36 V[(6 kΩ)/(12 kΩ + 6 kΩ)]
VR2 = 12 V
(Parallel resistance formula)
RTH = [(R1R2)/(R1 + R2)]
RTH = [(12 k Ω)(6 kΩ)/(12 kΩ + 6 kΩ)]
RTH = 4 kΩ
Answer: The Thevenin voltage is unchanged, and the
Thevenin resistance doubles.
1-16.
Given (from Prob. 1-12):
IN
RN
6 mA
2 kV
Norton circuit for Prob. 1-18.
1-19.
Given:
VTH = 12 V
RTH = 3 kΩ
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Solution:
=R
& The NewRR York
= 3 kΩ Times
N
TH
N
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IN = VOnly
TH/RTH
IN = 12 V/3 kΩ
(Eq. 1-12)
Shorted, which would cause load resistor t
across the voltage source seeing all of the
1-20. a. R1 is open, preventing any of the voltage
the load resistor. b. R2 is shorted, making i
zero. Since the load resistor is in parallel w
age drop would also be zero. 
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1-21. The battery or interconnecting wiring.

1-22.
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anytime.
= 2 kΩ
RTH Cancel
Solution:
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890 Resistor CI
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1-31.
(Kirchhoff’s law)
VS = VRS + VL
VRS = VS – VL
VRS = 10 V – 9 V
VRS = 1 V
RS = VRS/IRS
RS = 8.33 Ω
Given:
VS = 30 V
VL = 10 V
RL > 1 MΩ
RS < 0.01RL
Solution:
IRS = IL = VL/RL
IRS = 9 V/75 Ω
IRS = 120 mA

Answer: The value for R1 and R2 is 1 kΩ. Anot
sible solution is R1 = R2 = 4 kΩ. Note: The crit
be satisfied for any resistance value up to 4 kΩ
both resistors are the same value.
Given:
VS = 10 V
VL = 9 V
RL = 75 Ω
laboratorio 8
circuitos
 Search document
Answer: If an ideal 12 V voltage source is shorted and
provides 150 A, the internal resistance is 80 mΩ.
1-24.
Electronic
Devices and
(since the voltage source must
(Eq. 1-1)
Solution:
RS < 0.01RL
RS < 0.01(1 MΩ)
RS < 10 kΩ
(Ohm’s law)
Since the Thevenin equivalent resistance woul
series resistance, RTH < 10 kΩ.
(Ohm’s law)
Assume a value for one of the resistors. Si
Thevenin resistance is limited to 1 kΩ, pick a v
than 10 kΩ. Assume R2 = 5 kΩ.
(Eq. 1-1)
RS < 0.01 RL
8.33 Ω < 0.01(75 Ω)
8.33 Ω ≮ 0.75 Ω
1-25.
Answer: Disconnect the resistor and measure the voltage.
(Voltage divider form
VL = VS[R2/(R1 + R2)]
R1 = [(VS)(R2)/VL] – R2
R1 = [(30 V)(5 kΩ)/(10 V)] – 5 kΩ
R1 = 10 kΩ
1-26.
Answer: Disconnect the load resistor, turn the internal
voltage and current sources to zero, and measure the
resistance.
RTH = R1R2/(R1 + R2)
RTH = [(10 k Ω)(5 kΩ)]/(10 kΩ + 5 kΩ)
RTH = 3.33 kΩ
1-27.
Answer: Thevenin’s theorem makes it much easier to
solve problems where there could be many values of a
resistor.
Since RTH is one-third of 10 kΩ, we can use
values that are three times larger.
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.
R1 = 30 kΩ
R2 = 15 kΩ
Answer: a. The internal resistance (RS) is 8.33 Ω. b. The
source is not stiff since RS ≮ 0.01 RL.
1-28.
1-29.
Answer:
Note: The criteria will be satisfied as long as
R2 and R2 is not greater than 15 kΩ.
1-32.
Answer: First, measure the voltage across the te
This is the Thevenin voltage. Next, connect the
ter to the battery terminals—measure the curren
use the values above to find the total resistance.
subtract the internal resistance of the ammeter f
result. This is the Thevenin resistance.
1-33.
Answer: First, measure the voltage across the te
This is the Thevenin voltage. Next, connect a

across
terminals.
Next,
measure the voltag
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the resistor. Then, calculate the current 
through
Not
useful
 Useful

resistor.
Then,
subtract
the
load
voltage
from the
Cancel anytime.
nin voltage. Then, divide the difference voltag
current. The result is the Thevenin resistance.
Given:
RL = 1 kΩ
I = 1 mA
Solution:
RS > 100RL
RS > 100(1 kΩ)
RL > 100 kΩ
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V = IR
kΩ)Times
V = (1 mA)(100
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V
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Answer: A 100 V battery in series with a 100 kΩ resistor.
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890 Resistor CI
(Obsolete)
1
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Electronic
Devices and
2-4.
2-5.
2-6.
2-7.
Trouble:
1-38.
No supply voltage
1-39.
R4 open
1-40.
R2 shorted
a.
p-type
n-type
c. p-type
d. n-type
e. p-type
Given:
Barrier potential at 25°C is 0.7 V
Tmin = 25°C
Tmin = 75°C
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
R2 open
5 mA
5 mA
5 mA
b.
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.
1-37.
a.
c.
IL = 0.148 V/(6 kΩ + 6 kΩ)
IL = 12.3 μA
R1 shorted
500,000 free electrons
b.
IL = 0.148 V/(6 kΩ + 5 kΩ)
IL = 13.5 μA
1-36.

 Search document
IL = 0.148 V/(6 kΩ + 4 kΩ)
IL = 14.8 μA
1-35.
laboratorio 8
circuitos
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
ΔV = (–2 mV/°C) ΔT
(Eq. 2-4)
ΔV = (–2 mV/°C)(75°C – 25°C)
ΔV = –100 mV
ΔV
Vnew = Vold + ΔV
Vnew = 0.7 V – 0.1 V
Vnew = 0.6 V
Chapter 2 Semiconductor
Semiconductors
s
Answer: The barrier potential is 0.75
0.6 V at 75°C.
SELF-TEST
1. d
15. a
29. d
42. b
2. a
16. b
30. c
43. b
3. b
17. d
31. a
44. c
4. b
18. d
32. a
45. a
5. d
19. a
33. b
46. c
6. c
20. a
34. a
47. d
7. b
21. d
35. b
48. a
8. b
22. a
36. c
49. a
9. c
23. a
37. c
50. d
10. a
24. a
38. a
51. c
11. c
25. d
39. b
52. b
12. c
26. b
40. a
53. d
13. b
27. b
41. b
54. b
14. b
28. a
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9. Holes do not flow in a conductor. Conductors allow current
flow by virtue of their single outer-shell electron, which is
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)
[(0°C – 25°C)/10]
IS(new) = 2
IS(new) = 1.77 nA
10 nA
IS(new) = 2(∆T/10) IS(old)
[(75°C – 25°C) /10)]
IS(new) = 2
IS(new) = 320 nA
(Eq. 2-5)
10 nA
 is 1.77 n
Answer: The saturation current
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320 nA at 75°C.

2-9.
Useful
Given:
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ISL = 10 nA with a reverse voltage of 10 V
New reverse voltage = 100 V
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2-13.
R1 open
2-14.
D1 shorted
2-15.
D1 open
2-16.
V1 = 0 V
890 Resistor CI
(Obsolete)
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laboratorio 8
circuitos
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3-3.
Given:
Solution: Since the diodes are in series, the
through each is the same.
Answer: 400 mA
3-4.
SELF-TEST
Given:
VS = 20 V
VD = 0 V
RL = 1 kΩ
7. c
13. a
18. b
2. b
8. c
14. d
19. a
3. c
9. a
15. a
20. b
4. d
10. a
16. c
21. a
Solution:
5. a
11. b
17. b
22. c
6. b
12. b
(Kirchhoff’s law)
VS = VD + VL
20 V = 0 V + VL
VL = 20 V
JOB INTERVIEW QUESTIONS
4.
7.
8.
10.
11.
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.
Measure the voltage across a resistor in series with the diode.
Then, divide the voltage by the resistance.
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.
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.
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.
(Ohm’s law)
IL = VL/RL
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.
Solution:
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Given:
(Ohm’s law)
IL = VL/RL
/
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20
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2
k
IL =Read
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IL = 10 mA
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I=
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I = 6 V/220 Ω
Given:
VS = 20 V
VD = 0 V
RL = 2 kΩ
PROBLEMS
3-1.

VD1 = 0.75 V
VD2 = 0.8 V
ID1 = 400 mA
Chapter 3 Diode Theory
1. b
Electronic
Devices and
3-6.
Given:
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