, Part 2
Experiments Manual
Note: The tables in this part contain representative data for the has an internal resistance that is at least 100 times the load
experiments in the ninth edition. Variations from the data may be resistance.
expected, depending on the accuracy of the equipment used and 7. A shorted load has zero resistance. Ohm’s law says that the
the tolerance of the components. Measurements were made with load voltage equals the load current times the load resis-
both digital and analog test equipment, along with Multisim tance. The product of load current times zero gives zero load
simulation software. voltage.
8. With the load resistor open, the current through the series
resistor R of Fig. 1-1 is zero. Therefore, Ohm’s law tells us
Experiment 1 that no voltage is dropped across R. Kirchhoff’s voltage law
TABLE 1-1 VOLTAGE SOURCE tells us the sum of voltages around the loop is zero; therefore,
the load voltage equals the source voltage.
R Estimated VL Measured VL 9. 1 MΩ: For the current source to appear stiff, its internal resis-
tance must be at least 100 times the maximum load resis-
0Ω 10 V 10 V tance. In this case, 100(10 kΩ) is 1 MΩ.
10 Ω 10 V 9.99 V
100 Ω 9.9 V 9.91 V
470 Ω 9.5 V 9.53 V Experiment 2
TABLE 2-1 THEVENIN VALUES
TABLE 1-2 CURRENT SOURCE
VTH RTH IN
RL Estimated IL Measured IL
Calculated 5.43 V 2.39 kΩ 2.23 mA
0Ω 10 mA 9.36 mA Measured 5.67 V 2.36 kΩ 2.40 mA
10 Ω 9.9 mA 9.28 mA
47 Ω 9.5 mA 8.94 mA
TABLE 2-2 LOAD VOLTAGES
100 Ω 9 mA 8.56 mA
VL for 1 kΩ VL for 4.7 kΩ
TABLE 1-3 TROUBLESHOOTING
Calculated 1.6 V 3.6 V
Trouble Measured VL Measured 1.76 V 3.83 V
Shorted load 0V
Open load 10 V TABLE 2-3 TROUBLESHOOTING
Estimated Measured
TABLE 1-4 CRITICAL THINKING
Trouble VTH RTH VTH RTH
Type R Measured Quantity
Shorted 2.2 kΩ 1.5 V 1 kΩ 1.52 V 1.13 kΩ
Voltage source 100 Ω 9.91 V Open 2.2 kΩ 15 V 5.4 kΩ 15 V 5.13 kΩ
Current source 10 kΩ 0.98 mA
TABLE 2-4 CRITICAL THINKING
ANSWERS
1. d
2. b
3. b
4. b
5. a
6. A stiff voltage source has an internal resistance that is less
than 1⁄100 of the load resistance, whereas a stiff current source
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 2-1
, TABLE 2-5
RTH = 600 Ω
ANSWERS
1. b
2. c
3. d
4. d
5. a
6. The calculated Thevenin resistance is 2.39 kΩ. A load of 100
kΩ implies the load voltage will be down approximately 2
percent from the ideal open-load or Thevenin voltage. This
means the voltmeter reads slightly less than the ideal
Thevenin voltage.
7. In Fig. 2-1a, a shorted 2.2-kΩ resistor means the voltage
8. between point A and common is lower than it should be,
which implies a Thevenin voltage that is less than before.
Also, the shorted resistor means less Thevenin resistance.
9. In Fig. 2-1a, an open 2.2-kΩ resistor implies that all of the
source voltage will appear across the AB terminals when the
load is open. Furthermore, opening the resistor will increase
the Thevenin resistance.
10. If I wanted to stay in business, I had better produce batteries
with very low internal resistance because batteries are
supposed to act like stiff voltage sources for most load
resistances.
Experiment 3
TABLE 3-1 TROUBLES AND VOLTAGES
Trouble VA VB
Circuit OK 5.21 V 1.06 V
R1 open 0 0
R2 open 6.9 V 1.41 V
R3 open 6.81 V 0
R4 open 6.81 V 6.81 V
R1 shorted 10 V 2.04 V
R2 shorted 0 0
R3 shorted 2.72 V 2.72 V
R4 shorted 4.92 V 0
TABLE 3-2 TROUBLES AND VOLTAGES
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 2-2
, Experiment 5 Experiment 6
TABLE 5-1 FORWARD BIAS TABLE 6-1 TWO POINTS ON THE FORWARD CURVE
VS V I R I Estimated V Measured V
10 mA 0.7 V 0.69 V
0 0 0 (no entry)
50 mA 0.7 V 0.78 V
0.5 V 0.45 V 0.03 mA 15 kΩ
1V 0.55 V 0.4 mA 1.38 kΩ
2V 0.6 V 1.35 mA 444 Ω TABLE 6-2 DIODE VALUES
4V 0.64 V 3.4 mA 188 Ω
6V 0.66 V 5.5 mA 120 Ω Vknee = 0.69 V
8V 0.68 V 7.6 mA 89 Ω rB = 2.25 Ω
10 V 0.69 V 9.7 mA 71 Ω
15 V 0.71 V 14.2 mA 50 Ω
TABLE 6-3 DIODE CURRENT
TABLE 5-2 REVERSE BIAS Ideal I = 27.6 mA
Second I = 25.7 mA
VS V I R Third I = 25.6 mA
−1 V −1 V 0 Infinite Measured I = 24.1 mA
−5 V −5 V 0 Infinite
−10 V −10 V 0 Infinite TABLE 6-4 TROUBLESHOOTING
−15 V −15 V 0 Infinite
Estimated I Measured I
TABLE 5-3 DIODE CONDUCTION Shorted 470 Ω 0 0
Estimated I Measured I Open 470 Ω 30 mA 30.4 mA
Fig. 4-3a 45 mA 37.7 mA
Fig. 4-3b 0 0 TABLE 6-5 CRITICAL THINKING
Design 1 Design 2 Design 3
TABLE 5-4 TROUBLESHOOTING
VS 12.5 V 5.86 V 11.7 V
Estimated VL Measured VL
R1 470 Ω 220 Ω 220 Ω
Normal diode 0.7 V 0.71 V R2 220 Ω 470 Ω 220 Ω
Shorted diode 0 0 R3 220 Ω 220 Ω 470 Ω
Open diode 15 V 14.9 V I 7.87 mA 8.2 mA 8.59 mA
TABLE 5-5 CRITICAL THINKING
ANSWERS
VS = 10.7 V; RS = 1 kΩ; I = 10.2 mA
1. c
2. b
ANSWERS 3. c
1. b 4. c
2. a 5. c
3. c 6. The steeper the diode curve, the larger the changes in current
4. c for a given voltage change. This implies the bulk resistance
5. d decreases.
6. An ordinary resistor has the same resistance in either 7. With the diode shorted, there is no voltage to drive the diode.
direction. A diode has a low resistance when forward-biased 8. Two things happen. The Thevenin voltage increases from
and a high resistance when reverse-biased. 10.2 to 15 V, an increase of approximately 47 percent. The
7. Because the load is in parallel with a forward-biased diode Thevenin resistance increases from 370 to 440 Ω, an increase
whose typical voltage drop is 0.7 V. of approximately 19 percent. Therefore, the short-circuit cur-
8. Because the circuit becomes a voltage divider with 1 kΩ in rent is greater, which implies more diode current.
series with 100 kΩ. This implies that approximately 99 9. c
percent of the source voltage appears across the load resistor.
9. The higher the source voltage the better, because it swamps
out the effect of the diode offset voltage.
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 2-3
Experiments Manual
Note: The tables in this part contain representative data for the has an internal resistance that is at least 100 times the load
experiments in the ninth edition. Variations from the data may be resistance.
expected, depending on the accuracy of the equipment used and 7. A shorted load has zero resistance. Ohm’s law says that the
the tolerance of the components. Measurements were made with load voltage equals the load current times the load resis-
both digital and analog test equipment, along with Multisim tance. The product of load current times zero gives zero load
simulation software. voltage.
8. With the load resistor open, the current through the series
resistor R of Fig. 1-1 is zero. Therefore, Ohm’s law tells us
Experiment 1 that no voltage is dropped across R. Kirchhoff’s voltage law
TABLE 1-1 VOLTAGE SOURCE tells us the sum of voltages around the loop is zero; therefore,
the load voltage equals the source voltage.
R Estimated VL Measured VL 9. 1 MΩ: For the current source to appear stiff, its internal resis-
tance must be at least 100 times the maximum load resis-
0Ω 10 V 10 V tance. In this case, 100(10 kΩ) is 1 MΩ.
10 Ω 10 V 9.99 V
100 Ω 9.9 V 9.91 V
470 Ω 9.5 V 9.53 V Experiment 2
TABLE 2-1 THEVENIN VALUES
TABLE 1-2 CURRENT SOURCE
VTH RTH IN
RL Estimated IL Measured IL
Calculated 5.43 V 2.39 kΩ 2.23 mA
0Ω 10 mA 9.36 mA Measured 5.67 V 2.36 kΩ 2.40 mA
10 Ω 9.9 mA 9.28 mA
47 Ω 9.5 mA 8.94 mA
TABLE 2-2 LOAD VOLTAGES
100 Ω 9 mA 8.56 mA
VL for 1 kΩ VL for 4.7 kΩ
TABLE 1-3 TROUBLESHOOTING
Calculated 1.6 V 3.6 V
Trouble Measured VL Measured 1.76 V 3.83 V
Shorted load 0V
Open load 10 V TABLE 2-3 TROUBLESHOOTING
Estimated Measured
TABLE 1-4 CRITICAL THINKING
Trouble VTH RTH VTH RTH
Type R Measured Quantity
Shorted 2.2 kΩ 1.5 V 1 kΩ 1.52 V 1.13 kΩ
Voltage source 100 Ω 9.91 V Open 2.2 kΩ 15 V 5.4 kΩ 15 V 5.13 kΩ
Current source 10 kΩ 0.98 mA
TABLE 2-4 CRITICAL THINKING
ANSWERS
1. d
2. b
3. b
4. b
5. a
6. A stiff voltage source has an internal resistance that is less
than 1⁄100 of the load resistance, whereas a stiff current source
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 2-1
, TABLE 2-5
RTH = 600 Ω
ANSWERS
1. b
2. c
3. d
4. d
5. a
6. The calculated Thevenin resistance is 2.39 kΩ. A load of 100
kΩ implies the load voltage will be down approximately 2
percent from the ideal open-load or Thevenin voltage. This
means the voltmeter reads slightly less than the ideal
Thevenin voltage.
7. In Fig. 2-1a, a shorted 2.2-kΩ resistor means the voltage
8. between point A and common is lower than it should be,
which implies a Thevenin voltage that is less than before.
Also, the shorted resistor means less Thevenin resistance.
9. In Fig. 2-1a, an open 2.2-kΩ resistor implies that all of the
source voltage will appear across the AB terminals when the
load is open. Furthermore, opening the resistor will increase
the Thevenin resistance.
10. If I wanted to stay in business, I had better produce batteries
with very low internal resistance because batteries are
supposed to act like stiff voltage sources for most load
resistances.
Experiment 3
TABLE 3-1 TROUBLES AND VOLTAGES
Trouble VA VB
Circuit OK 5.21 V 1.06 V
R1 open 0 0
R2 open 6.9 V 1.41 V
R3 open 6.81 V 0
R4 open 6.81 V 6.81 V
R1 shorted 10 V 2.04 V
R2 shorted 0 0
R3 shorted 2.72 V 2.72 V
R4 shorted 4.92 V 0
TABLE 3-2 TROUBLES AND VOLTAGES
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 2-2
, Experiment 5 Experiment 6
TABLE 5-1 FORWARD BIAS TABLE 6-1 TWO POINTS ON THE FORWARD CURVE
VS V I R I Estimated V Measured V
10 mA 0.7 V 0.69 V
0 0 0 (no entry)
50 mA 0.7 V 0.78 V
0.5 V 0.45 V 0.03 mA 15 kΩ
1V 0.55 V 0.4 mA 1.38 kΩ
2V 0.6 V 1.35 mA 444 Ω TABLE 6-2 DIODE VALUES
4V 0.64 V 3.4 mA 188 Ω
6V 0.66 V 5.5 mA 120 Ω Vknee = 0.69 V
8V 0.68 V 7.6 mA 89 Ω rB = 2.25 Ω
10 V 0.69 V 9.7 mA 71 Ω
15 V 0.71 V 14.2 mA 50 Ω
TABLE 6-3 DIODE CURRENT
TABLE 5-2 REVERSE BIAS Ideal I = 27.6 mA
Second I = 25.7 mA
VS V I R Third I = 25.6 mA
−1 V −1 V 0 Infinite Measured I = 24.1 mA
−5 V −5 V 0 Infinite
−10 V −10 V 0 Infinite TABLE 6-4 TROUBLESHOOTING
−15 V −15 V 0 Infinite
Estimated I Measured I
TABLE 5-3 DIODE CONDUCTION Shorted 470 Ω 0 0
Estimated I Measured I Open 470 Ω 30 mA 30.4 mA
Fig. 4-3a 45 mA 37.7 mA
Fig. 4-3b 0 0 TABLE 6-5 CRITICAL THINKING
Design 1 Design 2 Design 3
TABLE 5-4 TROUBLESHOOTING
VS 12.5 V 5.86 V 11.7 V
Estimated VL Measured VL
R1 470 Ω 220 Ω 220 Ω
Normal diode 0.7 V 0.71 V R2 220 Ω 470 Ω 220 Ω
Shorted diode 0 0 R3 220 Ω 220 Ω 470 Ω
Open diode 15 V 14.9 V I 7.87 mA 8.2 mA 8.59 mA
TABLE 5-5 CRITICAL THINKING
ANSWERS
VS = 10.7 V; RS = 1 kΩ; I = 10.2 mA
1. c
2. b
ANSWERS 3. c
1. b 4. c
2. a 5. c
3. c 6. The steeper the diode curve, the larger the changes in current
4. c for a given voltage change. This implies the bulk resistance
5. d decreases.
6. An ordinary resistor has the same resistance in either 7. With the diode shorted, there is no voltage to drive the diode.
direction. A diode has a low resistance when forward-biased 8. Two things happen. The Thevenin voltage increases from
and a high resistance when reverse-biased. 10.2 to 15 V, an increase of approximately 47 percent. The
7. Because the load is in parallel with a forward-biased diode Thevenin resistance increases from 370 to 440 Ω, an increase
whose typical voltage drop is 0.7 V. of approximately 19 percent. Therefore, the short-circuit cur-
8. Because the circuit becomes a voltage divider with 1 kΩ in rent is greater, which implies more diode current.
series with 100 kΩ. This implies that approximately 99 9. c
percent of the source voltage appears across the load resistor.
9. The higher the source voltage the better, because it swamps
out the effect of the diode offset voltage.
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 2-3
