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Transcript of Chapter 5, Problem 93 - NOTES · PDF file Chapter 5, Problem 3 Determine the output voltage...

• 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.

Figure 5.43 for Prob. 5.1

8x104vd

Chapter 5, Solution 1. (a) Rin = 1.5 MΩ (b) Rout = 60 Ω (c) 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 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 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)

v2 - v1 = V2 10x2 4

A v

6 0 μ−=

− =

v2 - v1 = -2 µV = –0.002 mV 1 mV - v1 = -0.002 mV

v1 = 1.002 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 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

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, Solution 5.

+ - Avd

- +vi

I

Rin

R0 -

vd

+ +

v0 -

-vi + Avd + (Ri + R0) I = 0 (1) But vd = RiI, -vi + (Ri + R0 + RiA) I = 0

I = i0

i R)A1(R

v ++

(2)

-Avd - R0I + v0 = 0

v0 = Avd + R0I = (R0 + RiA)I = i0

ii0

R)A1(R v)ARR(

++ +

45 54

i0

i0

i

0 10 )101(100

10x10100 R)A1(R

ARR v v

⋅ ++

+ =

++ +

=

≅ ( ) =⋅+ 4

5

9

10 101

10 =

001,100 000,100 0.9999990

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 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 closed- loop gain vo/vs . Determine current i when vs = 2 V.

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, Solution 6.

+ - Avd

-+ vi

I

Rin

R0

+ vo -

- vd

+

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

I = i0

i

R)A1(R v ++

− (1)

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

v0 = ⎟⎟ ⎠

⎞ ⎜⎜ ⎝

⎛ ++

+ −

i0

i0

R)A1(R ARR

vi

= ( )( ) 65 356

10x2x10x2150 1010x2x10x250

++ ⋅+

− −

≅ mV 10x2x001,200 10x2x000,200

6

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

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, Solution 7.

– + AVd Rin

Rout = 100 Ω

+ Vout –

+ Vd –

– +VS

100 kΩ

10 kΩ 21

At node 1, (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

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 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 = k2 v0 0− v0 = -2V

(b)

- +

- +1V

- + 2V

10 kΩ

2 kΩ

ia

vb

va

+ vo

(a)

+ voi

(b)

10 kΩ

+ va

+ -

2V

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

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 9 Determine