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3–48 For the circuit in Figure P3–48 find its Th!evenin andNorton equivalent circuits.

100 Ω

vT, RT, iN

1 A

100 Ω

FIGURE P3–48

3–49(a) Find the Th!evenin orNorton equivalent circuit seen byRL in Figure P3–49.(b) Use the equivalent circuit found in (a) to find iL if RL ¼22 kV.

iL

RL50 V 15 kΩ

5.6 kΩ10 kΩ

+−

FIGURE P3–49

3–50(a) Find the Th!evenin orNorton equivalent circuit seen byRL in Figure P3–50.(b) Use the equivalent circuit found in (a) to find iL in termsof iS, R1, R2, and RL.(c) Check your answer in (b) using current division.

iS R1

R2

RL

B

AiL

FIGURE P3–50

3–51 Find the Th!evenin equivalent circuit seen by RL in FigureP3–51. Find the voltage across the load whenRL ¼ 5V, 10V,and 20 V.

20 V20 Ω

20 Ω

20 Ω20 Ω

20 Ω

RL+−

FIGURE P3–51

3–52 Find the Norton equivalent seen by RL in Figure P3–52.Find the current through the load for when RL ¼ 4.7 kV,15 kV, and 68 kV.

iL

RL10 mA 15 kΩ

8.1 kΩ

15 kΩ

FIGURE P3–52

3–53 You need to determine the Th!evenin Equivalent circuit ofa more complex linear circuit. A technician tells you shemade two measurements using her DMM. The first was witha 10-kV load and the load current was 91 mA, the second waswith a 10-kV load and the load voltage was 5 V. Calculate theTh!evenin equivalent circuit as shown in Figure P3–53.

vT

RT

10 kΩ 1 kΩ+−

iL=91µA

+vL=124 mV−

FIGURE P3–53

3–54 Find the Th!evenin equivalent seen by RL in Figure P3–54.Find the power delivered to the load when RL ¼ 50 kV and200 kV.

RL

pL

+− 33 kΩ

47 kΩ 91 kΩ

24 V

FIGURE P3–54

3–55(a) Use OrCAD to find the Norton equivalent at termi-nals A and B in Figure P3–55. Hint: Find the open-circuitvoltage and short-circuit current at the requisite terminals.(b) Use the Norton equivalent circuit found in (a) to deter-mine the power dissipated in RL when it is equal to 37 kV.(c) UseOrCAD to simulate both the original and theNortonequivalent circuits with RL ¼ 37 kV. Verify that the powerdissipated by the load is the same in both situations.

A

B

RL+− 100 A

47 kΩ

56 kΩ

100 kΩ

33 kΩ

10 V

FIGURE P3–55

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3–56 The purpose of this problem is to use Th!evenin equivalentcircuits to find the voltage vX in Figure P3–56. Find theTh!evenin equivalent circuit seen looking to the left of ter-minals A and B. Find the Th!evenin equivalent circuit seenlooking to the right of terminals A and B. Connect theseequivalent circuits together and find the voltage vX.

+−

A

B

6 kΩ 1 kΩ 3 kΩ

250 Ω3 kΩ

vx 15 V

+

−

FIGURE P3–56

3–57 The circuit in Figure P3–57 was solved earlier usingsupermeshes (Prob.(3–19). In this problem solve for thevoltage across RL by first finding the Th!evenin equivalentthat the load resistor sees, the for RL ¼ 2.5 kV find vL.

+vL

−20 mA

10 mA

1 kΩ

2 kΩ

RL1.5 kΩ

FIGURE P3–57

3–58 Assume that Figure P3–58 represents a model of the auxil-iary output port of a car. The output current is i¼ 1Awhen v ¼0 V. The output voltage is v ¼ 12 V when a 20-V resistor isconnected between the terminals. Suppose you wanted tocharge a 12-V battery by connecting the battery at the port.How much current would the port deliver to the battery?

−

+

v

i

FIGURE P3–58

3–59 The i-v characteristic of the active circuit in Figure P3–58is 5v + 500i ¼ 100. Find the output voltage when a 500-Vresistive load is connected.

3–60 You successfully completed Circuits I and as an under-graduatework-study, yourpast professor askedyou tohelphergrade aCircuits I quiz.On thequiz, studentswereasked to findthe power supplied by the source to both the 10 kV load (RL)

and to the entire circuit as shown in Figure P3–60. Yourprofessor asks to help her by creating a grading sheet.(a) Solve he quiz and establish reasonable A, B . . . etc. cutsfor incorrect solutions.(b) A particular student correctly finds the Th!evenin equiv-alent circuit that the resistive load sees and calculates thepower to the load using VT

2 /RL. He then does a sourcetransformation correctly finding the Norton equivalent ofthe circuit. He calculates the source power by VT"IN. Whatgrade would you give him?(c) Another student finds PL¼ 5.625 mWand PS¼#22.5 mW,but provides no work to justify her answers. What grade wouldyou give her?(d) A third student first finds the Norton equivalent, findsthe current through the load using a current divider andcalculates the power in the load using IL

2RL. He figurescorrectly what the parallel voltage would be across theNorton circuit and the load vL, and then calculatesPs=IN"vL.What grade would you give him?

RL

+− 10 kΩ 10 kΩ

10 kΩ

10 kΩ

10 kΩ

15 V

FIGURE P3–60

3–61 The Th!evenin equivalent parameters of a practical volt-age source are vT ¼ 30 V RT ¼ 300 V. Find the smallest loadresistance for which the load voltage exceeds 10 V.

3–62 Use a sequence of source transformations to find theTh!evenin equivalent at terminals A and B in Figure P3–62.Then select a resistor to connect across A and B so that 5 Vappears across it.

1 A 3 A20 V

15 Ω

15 Ω

5 Ω−+

5 V

B

A

5 Ω

+−

FIGURE P3–62

3–63 The circuit in Figure P3–63 provides power to a number ofloads connected in parallel. The circuit is protected by a 3/4mA fuse with a nominal 100V resistance. Each load is 10 kV.

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20 mΑ

2 kΩ

3 kΩ RL

FIGURE P3–70

3–71 For the circuit of Figure P3–71 find the value of RL thatwill result in:(d) Maximum voltage. What is that voltage?(e) Maximum current. What is that current?(f) Maximum power. What is that power?

56 V 3.3 kΩ

2.2 kΩ 1 kΩ

RL+−

FIGURE P3–71

3–72 The resistance R in Figure P3–72 is adjusted until maxi-mum power is delivered to the load consisting of R and the15-kV resistor in parallel.(a) Find the required value of R.(b) How much power is delivered to the load?

3 kΩ

2 kΩ

1 kΩ

R 12 kΩ

Load

40 mA

FIGURE P3–72

3–73 When a 5-kV resistor is connected across a two-terminalsource a current of 15 mA is delivered to the load. When asecond 5-kV resistor is connected in parallel with the first, atotal current of 20mA is delivered. Find themaximum poweravailable from the source.

3–74 Find the value of R in the circuit of Figure P3–74 so thatmaximum power is delivered to the load.What is the value ofthe maximum power?

50 Ω R

5 kΩ 2 kΩ

Load

10 V +−

FIGURE P3–74

3–75 For the circuit of Figure P3–75 find the value of RL thatwill result in:

(g) Maximum voltage. What is that voltage?(h) Maximum current. What is that current?(i) Maximum power. What is that power?

RL3 kΩ10 mA

2.75 kΩ

1 kΩ

iL

+

vL

−

FIGURE P3–75

3–76 A 100 V-load needs 10 mA to operate correctly.Design a practical power source to provide the neededcurrent. The smallest source resistance you can practicallydesign for is 50V, but you can add any other series resistanceif you need to.

3–77 A practical source delivers 50 mA to a 300-V load. Thesource delivers 12 V to a 120-V load. Find the maximumpower available from the source.

O B J E C T I V E 3 – 5 I N T E R F A C E C I R C U I T D E S I G N A N D

E V A L U A T I O N ( S E C T . 3 – 6 )

(a) Given the signal transfer objective at a source-loadinterface, adjust the circuit parameters or design one ormore two-port interface circuits to achieve the specifiedobjectives within stated constraints.

(b) Given two or more circuits that perform the sameinterface function, rank-order the circuits using statedcriteria.

See Examples 3–23, 3–24, 3–25, 3–26, 3–27, 3–28, 3–29, 3–30, 3–31 and Exercises 3–34, 3–35, 3–36, 3–37, 3–38, 3–39, 3–40, 3–41,3–42, 3–43.

3–78 SelectRL and design an interface circuit for the circuitshown in Figure P3–78 so that the load voltage is 2 V.

10 kΩ

RL5 kΩ10 kΩ10 V +−

InterfaceCircuit

FIGURE P3–78

3–79 The source in Figure P3–79 has a 100 mA outputcurrent limit. Design an interface circuit so that the loadvoltage is v2 ¼ 20 V and the source current is i1 <50 mA.

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3–85 Design the interface circuit in Figure P3–85 so that thepower delivered to the load is 100 mW.

100 mW25 Ω50 Ω

50 Ω 50 Ω10 V +−

InterfaceCircuit

FIGURE P3–85

3–86 Design the interface circuit in Figure P3–85 so that thevoltage delivered to the load is 2.5 V.

3–87 Design the interface circuit in Figure P3–87 so thatRIN ¼ 100V and the current delivered to the 50-V load is i¼50 mA. Hint: Use an L-pad.

100 Ω

50 Ω15 V +−

+

RIN ROUT

−

vInterfacecircuit

i

FIGURE P3–87

3–88 Design the interface circuit in Figure P3–87 so thatROUT¼ 50V and the voltage delivered to the 50-V load is v¼2.5 V. Hint: Use an L-pad.

3–89 The circuit in Figure P3–89 has a source resistance of75 V and a load resistance of 300 V. Design the interfacecircuit so that the input resistance is RIN ¼ 75 V " 10% andthe output resistance is ROUT ¼ 300 V " 10%.

75 Ω

300 Ω+−

RIN ROUT

vSInterface

circuit

FIGURE P3–89

3–90 It is claimed that both interface circuits in FigureP3–90 will deliver v ¼ 4 V to the 75-V load. Verify this claim.Which interface circuit consumes the least power?Which hasan output resistance that best matches the 75-V load?

+−20 V

+

v

−

150 Ω

75 Ω

either

Circuit 1 Circuit 2

i

150 Ω 15 Ω

100 Ω

Interfacecircuit

ROUT

FIGURE P3–90

I N T E G R A T I N G P R O B L E M S

3–91 Audio Speaker Resistance-Matchign Network

A company is producing an interface network that theyclaim would result in an RIN of 600 V " 2 % and ROUT

of 16, 8, or 4 V " 2% – depending on whether the connectedspeakers are 16, 8, or 4 V – selectable via a built-in switch.Their design is shown in Figure P3–91. Prove or disprovetheir claim.

+−

600 Ω

vS

592 Ω

8 Ω

8 Ω

Audio matching network

4 Ω

4 Ω16, 8, or 4 Ω

4 Ω

16 Ω

FIGURE P3–91

3–92 Attenuation AnalysisIn Figure P3–92 a two-port attenuator connects a 600-V sourceto a 600-V load. Find the power delivered to the load in terms ofvS. Remove the attenuator and find the power delivered to theload when the source is directly connected to the load. By whatfraction does the attenuator reduce the power delivered to the600-V load? Express the fraction in dB.

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4–11 Select g in the circuit of Figure P4–11 so that the outputvoltage is 10 V.

+−

+

vO

−

+

vx

−

1 mVgvx

1 kΩ

2.2 kΩ

FIGURE P4–11

4–12 Designadependent source circuit thathas a closed-loopvoltage gain of!10 using a VCVS with a m of 100. The sourcecircuit is a voltage source vS in series with a 1-kV resistor,and the load is a 3.3-kV resistor. (Hint: See Figure P4–10.)

4–13 Find the Th!evenin equivalent circuit that the load RL seesin Figure P4–13. Repeat the problem with RF replaced by anopen circuit.

vS+−

RS RF RP

µvxRL

vT, RT

−+

+

vx

−

FIGURE P4–13

4–14 Find the Th!evenin equivalent circuit that the load RL seesin Figure P4–14.

vS+−

RS RP

RL

Thévenin circuit

iS

r.iS+−

FIGURE P4–14

4–15 Find RIN in Figure P4–15

vS+−

R

RIN

iS

r.iS+−

FIGURE P4–15

4–16 Find RIN in Figure P4–16.

iS βix

RIN

R

ix

FIGURE P4–16

4–17 Find the Norton Equivalent circuit seen by the load inFigure P4–17.

Loadβix ROvS Rx

RS i

+

−

v

ix+−

FIGURE P4–17

4–18 Find the Th!evenin Equivalent circuit seen by the load inFigure P4–18.

Loadgvx ROiS

i

+

−

v

+

−

vx

FIGURE P4–18

4–19 The circuit parameters in Figure P4–19 are RB¼ 100 kV,RC¼ 3.3 kV, b¼ 100,Vg¼ 0.7 V, andVCC¼ 15 V. Find iC andvCE for vS¼ 1 V. Repeat for vS¼ 5 V.

+−

RC

RB ++

−−vS

vCE VCC

iB

iC

FIGURE P4–19

4–20 The circuit parameters in Figure P4–19 are RC¼ 3 kV,b¼ 120, Vg¼ 0.7 V, and VCC¼ 5 V. Select a value of RB suchthat the transistor is in the saturation mode when vS # 2 V.

4–21 The parameters of the transistor in Figure P4–21 areb¼ 60 and Vg¼ 0.7 V. Find iC and vCE for vS¼ 0.8 V. Repeatfor vS¼ 2 V.

+−

++

−−vS

vCE 15 V

10 kΩ

20 kΩiB

10 kΩ

iC

FIGURE P4–21

228 C H A P T E R 4 ACTIVE CIRCUITS

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4–22 An emergency indicator light uses a 10 V, 2-W incan-descent lamp. It is to be ON when a digital output is high (5V). The digital circuit does not have sufficient power to turnon the lamp directly. However, as is common practice, atransistor driver is used as a digital switch. Select RB in thecircuit of Figure P4–22 so to drive the transistor into satura-tion causing it to act as a short circuit between the lamp andground when the digital output is high. The Th!evenin equiv-alent for the digital circuit is also shown in the figure.

+− Digital

circuit = 50Vγ = 0.7 V

Lamp = 10V @ 2W

+10 V

VD

RD RB

iLamp

VD = 5 V

RD = 500 Ω

FIGURE P4–22

O B J E C T I V E 4 – 2 O P A M P C I R C U I T A N A L Y S I S

( S E C T S . 4 – 3 , 4 – 4 )Given a linear resistance circuit containing OP AMPs, findselected output signals or input-output relationships.See Examples 4–12, 4–13, 4–14, 4–15, 4–16, 4–17, 4–18, 4–19, 4–20 and Exercises 4–14, 4–15, 4–16, 4–17, 4–18, 4–19, 4–20, 4–21,4–22, 4–23, 4–24, 4–25, 4–26, 4–27, 4–28.

4–23 Find the voltage gain of each OP AMP circuit shownin Figure P4–23.

+

−

330 kΩ

(a)

+

vO

+

vS

33 kΩ

+

−

330 kΩ

(b)

+

vO

33 kΩ

+

vS

FIGURE P4–23

4–24 Considering simplicity and standard 10% toleranceresistors as major constraints, design OP AMP circuits thatproduce the following voltage gains!10%:" 100,þ 200,þ 1,"0.5, þ0.5.

4–25 TwoOPAMP circuits are shown in Figure P4–25. Bothclaim to produce a gain of either ! 100.

(a) Show that the claim is true.(b) A practical source with a series resistor of 1 kV is

connected to the input of each circuit. Does theoriginal claim still hold? If it does not, explain why?

1 kΩ

1 kΩ

100 kΩ

99 kΩ

Source

VS

1 kΩ

Circuit 2

Circuit 1

vO

vO

vIN

vIN

−−

+

+ +

+

−

+

+

+

FIGURE P4–25

4–26 Suppose the output of the practical source shown inFigure P4–25 needs to be amplified by "104 and you can useonly the two circuits shown. How would you connect thecircuits to achieve this? Explain why.

4–27 (a) Find the voltage gain vO/vS in Figure P4–27.(b) Validate your answer by simulating the circuit in

OrCAD.

+

−

330 kΩ

+

vOvS 68 kΩ+−

22 kΩ 33 kΩ

VCC = ± 15 V

FIGURE P4–27

4–28 What is the range of the gain vO/vS in Figure P4–28?

vO

vS+

−+

+−

2 kΩ 100 kΩ 100 kΩ

VCC = ± 15 V

FIGURE P4–28

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4–29 Design a simple OP AMP circuit that has a variablegain from þ10 to þ100.

4–30 For the circuit in Figure P4–30:(a) Find vO in terms vS.(b) Find iO for vS¼ 1 V. Repeat for vS¼ 3 V.

+

−+−vS

iO

+vO−

10 kΩ

10 kΩ

150 kΩ 10 kΩ

VCC = ± 24 V

FIGURE P4–30

4–31 For the circuit in Figure P4–31:(a) Find vO in terms vS.(b) Find iO for vS¼ 0.5 V. Repeat for vS¼ 2 V.

vO

+

−+

+−

10 kΩ

10 kΩ

100 kΩ

100 kΩ

vS

iO

VCC = ± 18 V

FIGURE P4–31

4–32 A young designer needed to amplify a 2-V signal bythe factors of 1, 5, and 10. Find the problem with the designshown in Figure P4–32. Recommend a fix.

vO

vS

+

−+

+

10 kΩ

90 kΩ

40 kΩ1

2

3

VCC = ± 15 V

FIGURE P4–32

4–33 Design two circuits to produce the following output:vO¼ 2v1 # 3v2.

(a) In your first design use a standard subtractor.(b) In your second design both inputs must be into high

input resistance amplifiers to avoid loading.

4–34 For the circuit in Figure P4–34:(a) Find vO in terms of the inputs v1 and v2.(b) If v1¼ 1 V, what is the range of values v2 can have

without saturating the OP AMP?

vO

+

−

−

+

+−

50 kΩ50 kΩ

50 kΩ 100 kΩ

100 kΩ

+−

v2

v1

VCC = +15 V

FIGURE P4–34

4–35 The input-output relationship for a three-input invertingsummer is

vO ¼ # v1 þ 10v2 þ 100v3½ %

The resistance of the feedback resistor is 100 kV. Find thevalues of the input resistors R1, R2, and R3.

4–36 Find vO in terms of the inputs v1 and v2 in Figure P4–36.

vO+

−+

−

+−

10 kΩ

10 kΩ

33 kΩ

+−

v2

v1

33 kΩ

FIGURE P4–36

4–37 The switch in Figure P4–37 is open, find vO in terms of theinputs vS1 and vS2. Repeat with the switch closed.

vO+

−+

−

+−

30 kΩ

30 kΩ

Switch

+−

vS2

vS1

60 kΩ

60 kΩ

FIGURE P4–37

230 C H A P T E R 4 ACTIVE CIRCUITS

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The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual

Copyright © 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600.

Problem 3-39 Find the Thévenin equivalent seen by RL in Figure P3-39. Find the power delivered to the load when RL = 50 kΩ and 200 kΩ.

24 V RL30 kΩΩΩΩ

60 kΩΩΩΩ 40 kΩΩΩΩ iL

8 V RL

60 kΩΩΩΩ iL

Solution: The following MATLAB code presents the solution. clear all % Find the open-circuit voltage and the look-back resistance vT = 24*30e3/(30e3+60e3) RT = 40e3 + 1/(1/60e3 + 1/30e3) % Find the load power RL = [50e3 200e3]; iL = vT./(RT+RL); pL = iL.^2.*RL vT = 8.0000e+000 RT = 60.0000e+003 pL = 264.4628e-006 189.3491e-006 Answer: RT = 60 kΩ; vT = 8 V; pL = 264 µW for RL = 50 kΩ; pL = 189 µW for RL = 200 kΩ



Problem 3-42 Assume that Figure P3-42 represents a model of the auxiliary output port of a car. The output current is i = 1 A when v = 0 V. The output voltage is v = 12 V when a 20-Ω resistor is connected between the terminals. Suppose you wanted to charge a 12-V battery by connecting the battery at the port. How much current would the port deliver to the battery?

v

i

OutputPort

Solution: The short circuit current is 1 A, as stated in the problem. When a 20-Ω resistor is connected to the circuit, the equivalent resistance must be 12 Ω to create a 12-V drop. Use MATLAB to solve for the Norton resistance. Then find the Thévenin equivalent circuit and solve for the current into the battery, assuming the battery acts as an ideal source with no internal resistance clear all iN = 1; syms RN Eqn = '1/(1/RN+1/20) - 12'; RN = double(solve(Eqn,'RN')) vT = iN*RN RT = RN; iL = (vT-12)/RT RN = 30.0000e+000 vT = 30.0000e+000 iL = 600.0000e-003 Answer: i = 600 mA.



Problem 3-44 You successfully completed Circuits I and as an undergraduate work-study, your past professor asked you to help her grade a Circuits I quiz. On the quiz, students were asked to find the power supplied by the source to both the 10 kΩ load (RL) and to the entire circuit as shown in Figure P3-44. Your professor asks to help her by creating a grading sheet. (a) Solve the quiz and establish reasonable A, B…etc. cuts for incorrect solutions. (b) A particular student correctly finds the Thévenin equivalent circuit that the resistive load sees and calculates the power to the load using VT

2/RL. He then does a source transformation correctly finding the Norton equivalent of the circuit. He calculates the source power by VT×IN. What grade would you give him? (c) Another student finds PL = 5.625 mW and PS = –22.5 mW, but provides no work to justify her answers. What grade would you give her? (d) A third student first finds the Norton equivalent, finds the current through the load using a current divider and calculates the power in the load using IL2RL. He figures correctly what the parallel voltage would be across the Norton circuit and the load vL, and then calculates PS=IN×vL. What grade would you give him? Solution: (a) Use node-voltage analysis to solve for the required values as shown in the following MATLAB code. Let the center node be node A and the upper right node be node B. Ground the bottom node. clear all syms vA vB Eqn1 = '(vA-15)/10e3 + vA/10e3 + (vA-vB)/10e3'; Eqn2 = '(vB-vA)/10e3 + (vB-15)/10e3 + vB/10e3'; Soln = solve(Eqn1,Eqn2,'vA','vB'); vA = double(Soln.vA); vB = double(Soln.vB); vL = 15-vB; pL = vL^2/10e3 is = (15-vA)/10e3 + (15-vB)/10e3; ps = 15*is pL = 5.6250e-003 ps = 22.5000e-003 (b) The student deserves credit for finding the correct Thévenin and Norton equivalent circuits, but the power calculations are incorrect for both the load and the total power dissipated in the circuit. The grade for this solution should be low, such as a D.

10 kΩ

10 kΩRL

10 kΩ

10 kΩ10 kΩ15 V



(c) Give the student full credit and a grade of A. The solution is correct. If the solution had been wrong, then the grade would have been zero or F for not showing any work. (d) The load power is calculated correctly, but the total power calculation is not valid. Give the student a B grade. Answer: (a) pL = 5.625 mW and pTOTAL = 22.5 mW. The requirements for the grading sheet are instructor specific. Suggested answers for Parts (b), (c), and (d) are presented above.



Problem 3-54 The resistance R in Figure P3-54 is adjusted until maximum power is delivered to the load consisting of R and the 15-kΩ resistor in parallel. (a) Find the required value of R. (b) How much power is delivered to the load?

20 mA R

2 kΩΩΩΩ

3 kΩΩΩΩ1 kΩΩΩΩ

15 kΩΩΩΩ

Load Solution: The following MATLAB code presents the solution. clear all format short eng disp('Part (a)') % Find the Thevenin equivalent for the source circuit % Use the look-back technique to find the resistance RT = 2e3+3e3+1e3; % Open-circuit voltage vT = 20e-3*3e3; % Find an expression for R in terms of the total load resistance syms R RL Eqn1 = 'RL - 1/(1/R+1/15e3)'; R = solve(Eqn1,'R'); % The total load resistance should equal the Thevenin resistance % for maximum power transfer R_num = double(subs(R,RL,RT)) disp('Part (b)') % Use the Thevenin equivalent circuit to compute the power % dissipated by the load pL_max = vT^2/4/RT Part (a) R_num = 10.0000e+003 Part (b) pL_max = 150.0000e-003 Answer: (a) R = 10 kΩ (b) pL = 150 mW



Problem 3-56 Find the value of R in Figure P3-56 so that maximum power is delivered to the 2-kΩ load. Find the value of the maximum power.

10 V

R

10 kΩΩΩΩ 2 kΩΩΩΩ5 kΩΩΩΩ

Solution: The following MATLAB code presents the solution. clear all % Maximum power will be delivered if the Thevenin resistance % matches the load resistance of 2 kOhms % Find an expression for the Thevenin resistance in terms of R, solve for R, and % evaluate when RT = 2 kOhms syms RT R Eqn1 = 'RT - 1/(1/R+1/10e3+1/5e3)'; R = solve(Eqn1,'R'); R = double(subs(R,RT,2e3)) % Find the Thevenin voltage Req=1/(1/10e3+1/5e3) vT = 10*Req/(R+Req) % Calculate the maximum power transfer p_max = vT^2/4/2e3 R = 5.0000e+003 Req = 3.3333e+003 vT = 4.0000e+000 p_max = 2.0000e-003 Answer: R = 5 kΩ pMAX = 2 mW



Problem 3-61 The source in Figure P3-61 has a 100 mA output current limit. Design an interface circuit so that the load voltage is v2 = 2 V and the source current is i1 < 50 mA. Solution: The following MATLAB code presents the solution. clear all % Find the load current v2 = 2; R2 = 50; i2 = v2/R2 % Find the minimum total resistance seen by the source i1 = 50e-3; R_total_min = 10/i1 i2 = 40.0000e-003 R_total_min = 200.0000e+000 % The required current is less than the maximum current % Add a series resistor as the interface to create current of 40 mA Req = 10/i2 R_interface = Req - 50 - 50 Req = 250.0000e+000 R_interface = 150.0000e+000 The following OrCAD simulation verifies the solution.

v

i

10 V

50 ΩΩΩΩ

v1

i1 2

2 50 ΩΩΩΩInterface

circuit



Problem 3-70 It is claimed that both interface circuits in Figure 3-70 will deliver v = 4 V to the 75-Ω load. Verify this claim. Which interface circuit consumes the least power? Which has an output resistance that best matches the 75-Ω load? Solution: For each interface, calculate the load voltage, source power, and output resistance using MATLAB. clear all vs = 20; Rs = 150; RL = 75; disp('Circuit 1') R11 = 150; iL1 = vs/(Rs+R11+RL); vL1 = RL*iL1 ps1 = iL1*vs Rout1 = Rs+R11 disp('Circuit 2') R21 = 100; R22 = 15; Req2 = 1/(1/R21+1/(R22+RL)); veq2 = Req2*vs/(Rs+Req2); vL2 = RL*veq2/(R22+RL) is2 = vs/(Rs+Req2); ps2 = is2*vs Rout2 = 15+1/(1/Rs+1/R21) Circuit 1 vL1 = 4.0000e+000 ps1 = 1.0667e+000 Rout1 = 300.0000e+000 Circuit 2 vL2 = 4.0000e+000 ps2 = 2.0267e+000 Rout2 = 75.0000e+000

v

i

20 VInterface

circuit

150 ΩΩΩΩ

ROUT

75 ΩΩΩΩ

either

150 ΩΩΩΩ

100 ΩΩΩΩ

15 ΩΩΩΩ

Circuit 1 Circuit 2



Problem 4-11 Find the Thévenin equivalent circuit that the load RL sees in Figure P4-11.

vS

iS

r@iSvORL

RS

Thévenin circuit Solution: Since the circuit has a dependent source, we cannot reliably use the look-back technique to compute the Thévenin resistance. We need to find the open-circuit voltage and the short-circuit current. The solution is presented in the following MATLAB code. clear all syms vs Rs Rp r is vT RT isc Eqn1 = 'is - (vs-r*is)/Rs'; Eqn2 = 'vT - r*is'; Eqn3 = 'isc - r*is/Rp'; % Solve the equations Soln = solve(Eqn1,Eqn2,Eqn3,'is','vT','isc'); vT = simplify(Soln.vT) isc = simplify(Soln.isc); RT = simplify(vT/isc) vT = r*vs/(Rs+r) RT = Rp Answer:

RT = RP and rR

rVv+

=S

ST .



Problem 4-20 Suppose the output of the practical source shown in Figure P4-19 needs to be amplified by – 100 and you can use only the two circuits shown. How would you connect the circuits to achieve this? Explain why.

vOvS

9 kΩ

1 kΩ

Source

1 kΩ

Circuit 2

1 kΩ

10 kΩ

vIN

vINvO

Circuit 1 Solution: Connect the practical source to the input of Circuit 1 and then connect the output of Circuit 1 to the input of Circuit 2. The practical source does not change the gain of Circuit 1, so it still provides a gain of 10. The output of Circuit 1 will not change the gain of Circuit 2, since the output resistance of an OP AMP circuit is very small. The overall gain will be (10)(−10) = −100, as requested. Answer: Presented above in the Solution.



Problem 4-23 (a) Find vO in terms vS in Figure P4-23. (b) Find iO for vS = 1 V. Repeat for vS = 3 V. Solution: The solution is presented in the following MATLAB code. clear all disp('Part (a)') % No current flows into the OP AMP, so the voltage at the positive % input terminal is vs and the circuit is configured as a non-inverting % amplifier. The load resistor does not affect the output voltage. syms vs vo io R1 = 150e3; R2 = 10e3; RL = 10e3; % Compute the gain and the output voltage K = (R1+R2)/R2; vo = K*vs disp('Part (b)') % Compute the output voltage vs = [1 3]; vo = K*vs; % Adjust for saturation vcc = 24; for n = 1:length(vo) if vo(n) > vcc vo(n)=vcc; elseif vo(n)<-vcc vo(n)=-vcc; end end % Compute the output current io = vo/RL Part (a) vo = 16*vs Part (b) io = 1.6000e-003 2.4000e-003 Answer: (a) vO = 16vS (b) For vS = 1 V, iO = 1.6 mA. For vS = 3 V, iO = 2.4 mA because of saturation.

10 kΩ

10 kΩ 10 kΩ

150 kΩvSvO

iO

VCC=±24 V

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