Solution Manual - Fundamentals of Electric Circuits 3rd Edition Chapter 05 PDF

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Chapter 5, Problem 1. The equivalent model of a certain op amp is shown in Fig. 5.43. Determine: (a) the input resistance. (b) the output resistance. (c) the voltage gain in dB.

8x104vd

Chapter 5, Solution 1. (a) (b) (c)

Rin = 1.5 MΩ Ω Rout = 60 Ω A = 8x104 Therefore AdB = 20 log 8x104 = 98.0 dB

Chapter 5, Problem 2 The open-loop gain of an op amp is 100,000. Calculate the output voltage when there are inputs of +10 µV on the inverting terminal and + 20 µV on the noninverting terminal.

Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 1V

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Chapter 5, Problem 3 Determine the output voltage when .20 µV is applied to the inverting terminal of an op amp and +30 µV to its noninverting terminal. Assume that the op amp has an open-loop gain of 200,000.

Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10-6 = 10V

Chapter 5, Problem 4 The output voltage of an op amp is .4 V when the noninverting input is 1 mV. If the open-loop gain of the op amp is 2 × 106, what is the inverting input?

Chapter 5, Solution 4. v0 = Avd = A(v2 - v1) v −4 = −2μV v2 - v 1 = 0 = A 2x106 v2 - v1 = -2 µV = –0.002 mV 1 mV - v1 = -0.002 mV v1 = 1.002 mV

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Chapter 5, Problem 5. For the op amp circuit of Fig. 5.44, the op amp has an open-loop gain of 100,000, an input resistance of 10 kΩ, and an output resistance of 100 Ω. Find the voltage gain vo/vi using the nonideal model of the op amp.

Figure 5.44 for Prob. 5.5

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Chapter 5, Solution 5.

I R0 Rin

vd + vi

+ -

Avd

+ v0

-

+

-

-vi + Avd + (Ri + R0) I = 0 But

(1)

vd = RiI, -vi + (Ri + R0 + RiA) I = 0 I=

vi R 0 + (1 + A) R i

(2)

-Avd - R0I + v0 = 0 v0 = Avd + R0I = (R0 + RiA)I =

( R 0 + R i A) v i R 0 + (1 + A) R i

v0 R 0 + R iA 100 + 10 4 x10 5 = = ⋅ 104 v i R 0 + (1 + A) R i 100 + (1 + 10 5 ) ≅

10 9 100,000 = 0.9999990 ⋅ 10 4 = 5 1 + 10 100,001

(

)

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Chapter 5, Problem 6 Using the same parameters for the 741 op amp in Example 5.1, find vo in the op amp circuit of Fig. 5.45.

Figure 5.45 for Prob. 5.6

Example 5.1 A 741 op amp has an open-loop voltage gain of 2×105, input resistance of 2 MΩ, and output resistance of 50Ω. The op amp is used in the circuit of Fig. 5.6(a). Find the closedloop gain vo/vs . Determine current i when vs = 2 V.

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Chapter 5, Solution 6. vi + -

R0 I

Rin

vd

+ -

+

Avd

+ vo

-

(R0 + Ri)R + vi + Avd = 0 But

vd = RiI, vi + (R0 + Ri + RiA)I = 0 I=

− vi R 0 + (1 + A) R i

(1)

-Avd - R0I + vo = 0 vo = Avd + R0I = (R0 + RiA)I Substituting for I in (1),

⎛ R 0 + R iA ⎞ ⎟⎟ vi v0 = − ⎜⎜ ⎝ R 0 + (1 + A) R i ⎠ 50 + 2 x10 6 x 2 x10 5 ⋅10 −3 = − 50 + 1+ 2 x105 x 2 x106

(

(

)

)

− 200,000 x 2 x10 mV 200,001x 2 x106 6

≅

v0 = -0.999995 mV PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 5, Problem 7 The op amp in Fig. 5.46 has Ri = 100 kΩ, Ro = 100 Ω, A = 100,000. Find the differential voltage vd and the output voltage vo.

+ –

Figure 5.46 for Prob. 5.7

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Chapter 5, Solution 7. Ω 100 kΩ

10 kΩ Ω

VS

+

–

Rout = 100 Ω

1

2

+ Vd

Rin

– +

AVd

–

At node 1,

+ Vout

–

(VS – V1)/10 k = [V1/100 k] + [(V1 – V0)/100 k] 10 VS – 10 V1 = V1 + V1 – V0 which leads to V1 = (10VS + V0)/12

At node 2,

(V1 – V0)/100 k = (V0 – (–AVd))/100

But Vd = V1 and A = 100,000, V1 – V0 = 1000 (V0 + 100,000V1) 0= 1001V0 + 99,999,999[(10VS + V0)/12] 0 = 83,333,332.5 VS + 8,334,334.25 V0 which gives us (V0/ VS) = –10 (for all practical purposes) If VS = 1 mV, then V0 = –10 mV Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = –100 nV

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Chapter 5, Problem 8 Obtain vo for each of the op amp circuits in Fig. 5.47.

Figure 5.47 for Prob. 5.8 Chapter 5, Solution 8. (a)

If va and vb are the voltages at the inverting and noninverting terminals of the op amp. va = vb = 0 1mA =

0− v 0 2k

v 0 = -2V

(b)

2V

10 kΩ Ω

+

ia

va

2V

+

vb 1V

+ vo

+

Ω 2 kΩ

+ va

10 kΩ Ω

+-

+ i

vo

(b)

(a)

Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ resistor. From Fig. (b), -va + 2 + v0 = 0

v0 = va - 2 = 1 - 2 = -1V

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Chapter 5, Problem 9 Determine vo for each of the op amp circuits in Fig. 5.48.

+ –

Figure 5.48 for Prob. 5.9 Chapter 5, Solution 9. (a) Let va and vb be respectively the voltages at the inverting and noninverting terminals of the op amp va = vb = 4V At the inverting terminal, 4 − v0 1mA = 2k

v0 = 2V

(b)

1V +-

+

+

vb

vo

-

-

Since va = vb = 3V, -vb + 1 + vo = 0

vo = vb - 1 = 2V

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Chapter 5, Problem 10 Find the gain vo/vs of the circuit in Fig. 5.49.

Figure 5.49 for Prob. 5.10

Chapter 5, Solution 10. Since no current enters the op amp, the voltage at the input of the op amp is vs. Hence ⎛ 10 ⎞ vo vs = vo ⎜ ⎟= ⎝ 10 + 10 ⎠ 2

vo =2 vs

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Chapter 5, Problem 11 Find vo and io in the circuit in Fig. 5.50.

Figure 5.50 for Prob. 5.11 Chapter 5, Solution 11.

− +

+

+ −

vo vb =

10 (3) = 2V 10 + 5

At node a, 3 − va v a − v o = 2 8

12 = 5va – vo

But va = vb = 2V, 12 = 10 – vo –io =

vo = –2V

va − v o 0 − v o 2 + 2 2 + = 1mA + = 8 4 8 4 i o = –1mA

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Chapter 5, Problem 12. Calculate the voltage ratio vo/vs for the op amp circuit of Fig. 5.51. Assume that the op amp is ideal. 25 kΩ

5 kΩ – + vs

+ vo

+ _ 10 kΩ

–

Figure 5.51

For Prob. 5.12.

Chapter 5, Solution 12. This is an inverting amplifier. 25 vo = − vs ⎯⎯ → 5

vo = −5 vs

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Chapter 5, Problem 13 Find vo and io in the circuit of Fig. 5.52.

Figure 5.52 for Prob. 5.13 Chapter 5, Solution 13.

+ −

+ + −

vo

By voltage division, va =

90 (1) = 0.9V 100

vb =

v 50 vo = o 3 150

But va = vb

i o = i1 + i2 =

v0 = 0. 9 3

vo = 2.7V

vo v + o = 0.27mA + 0.018mA = 288 μA 10 k 150k

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Chapter 5, Problem 14 Determine the output voltage vo in the circuit of Fig. 5.53.

Figure 5.53 for Prob. 5.14 Chapter 5, Solution 14. Transform the current source as shown below. At node 1,

10 − v 1 v1 − v 2 v1 − vo = + 5 20 10

− +

+ −

+ vo

But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo At node 2,

v1 − v 2 v 2 − v o = , 20 10

40 = 7v1 - 2vo

v 2 = 0 or v1 = -2vo

From (1) and (2), 40 = -14vo - 2vo

(1) (2)

vo = -2.5V

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Chapter 5, Problem 15 (a). Determine the ratio vo/is in the op amp circuit of Fig. 5.54. (b). Evaluate the ratio for R1 = 20 kΩ, R2 = 25 kΩ, R3 = 40 2kOmega$.

Figure 5.54 Chapter 5, Solution 15 (a) Let v1 be the voltage at the node where the three resistors meet. Applying KCL at this node gives

⎛ 1 v1 v1 − v o 1 ⎞ vo ⎟⎟ − = v 1 ⎜⎜ + + R2 R3 ⎝ R2 R3 ⎠ R3 At the inverting terminal, is =

0 − v1 ⎯ ⎯→ v1 = −i s R1 R1 Combining (1) and (2) leads to R ⎞ v ⎛ R i s ⎜⎜1 + 1 + 1 ⎟⎟ = − o ⎯ ⎯→ R3 ⎝ R2 R3 ⎠ is =

(1)

(2) RR ⎞ ⎛ = −⎜⎜ R1 + R3 + 1 3 ⎟⎟ R2 ⎠ is ⎝

vo

(b) For this case,

vo 20 x40 ⎞ ⎛ = − ⎜ 20 + 40 + ⎟ k Ω = - 92 k Ω is 25 ⎠ ⎝ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 5, Problem 16 Obtain ix and iy in the op amp circuit in Fig. 5.55.

Figure 5.55

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Chapter 5, Solution 16 10k Ω

5k Ω

ix va vb

+ 0.5V -

iy +

vo

2k Ω 8k Ω

Let currents be in mA and resistances be in k Ω . At node a, 0. 5 − v a v a − v o ⎯ ⎯→ 1 = 3v a − v o = 5 10

(1)

But 8 10 (2) vo ⎯ ⎯→ vo = v a 8+2 8 Substituting (2) into (1) gives 10 8 1 = 3v a − v a ⎯ ⎯→ va = 8 14 Thus, 0. 5 − v a = −1 / 70 mA = −14.28 μA ix = 5 v o −v b v o − v a 10 0. 6 8 + = 0.6(v o − v a ) = 0.6( v a − v a ) = x mA = 85.71 μA iy = 2 10 8 4 14 va = vb =

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Chapter 5, Problem 17 Calculate the gain vo/vi when the switch in Fig. 5.56 is in: (a) position 1 (b) position 2 (c) position 3

Figure 5.56

Chapter 5, Solution 17. (a) (b) (c)

G=

vo vi vo vi

vo R 12 = − 2 = − = -2.4 5 vi R1 80 =− = -16 5 2000 =− = -400 5

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* Chapter 5, Problem 18. For the circuit in Fig. 5.57, find the Thevenin equivalent to the left of terminals a-b. Then calculate the power absorbed by the 20-kΩ resistor. Assume that the op amp is ideal. 10 kΩ

2 kΩ 12 kΩ a

– 2 mV

+ _

8 kΩ 20 kΩ b

Figure 5.57

For Prob. 5.18.

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Chapter 5, Solution 18. We temporarily remove the 20-kΩ resistor. To find VTh, we consider the circuit below.

10 kΩ

2 kΩ 12 kΩ

– + 2 mV

+

+ _

8Ω

VTh –

This is an inverting amplifier. 10k VTh = − (2mV ) = − 10mV 2k

To find RTh, we note that the 8-kΩ resistor is across the output of the op amp which is acting like a voltage source so the only resistance seen looking in is the 12-kΩ resistor. The Thevenin equivalent with the 20-kΩ resistor is shown below.

12 kΩ

a I

–10 mV

+ _

20 k b

I = –10m/(12k + 20k) = 0.3125x10–6 A p = I2R = (0.3125x10–6)2x20x103 = 1.9531 nW

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Chapter 5, Problem 19 Determine io in ...