chapter6 selected answers

8/12/2019 Chapter6 Selected Answers

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CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. (a) -30 V; (b) -2.5 V; (c) 1.4 V

3. (a) ; (b)tvv inout 5sin2010 == tvv inout 5sin51010 ==

5. One possible design is to use a simple inverting op amp circuit with

Rf= 9.1 kand Rin= 5.1 k.

7. To get a positive output that is smaller than the input, the easiest way is to use

inverting amplifier with an inverted voltage supply to give a negative voltage,where Rf = 1.5 kand Rin= 5.1 k

9. (a) 1.7 V; (b) 3 V; (c) -2.4 V

11. (a) ; (b)tvv inout 10sin82 == tvv inout 10sin5.022 +==

13. -2.2 V

15. One possible solution of many: a non-inverting op amp circuit with themicrophone connected to the non-inverting input terminal, the switch connected

between the op amp output pin and ground, a feedback resistor Rf= 133 , and aresistor R1= 1 .

17. V1= 21 V

19. ; -5.6 V(out -4 1 sin 3 Vv t= + )

21. Rf= 236 kand R1= 1 k.

23. (a) Bmust be the non-inverting input; (b) Choose R2 = RB= 1 ; (c)A is theinverting input.

25. vout(0.25 s) = 0.93 V

27. 4.2 V

29. =

N

1

fR

R-i i

iv

31. Pick R1= 10 k. Then vS= -0.21 V.

Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved.


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33. One possible solution of many:

35. Set R = 10 k:

Then connect several into:

after setting Rf2 = Rf1 = Rin = R =10 k.

37. 1 kV

39. -179 kV

41. 1.7 V

43. Rf= 0, Rin= 100 k, R2= 51 .



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45. Rf = 120 kand Rin= 200 k, R = 560 .

47. R = 400 , R1= 82 .

I Is

49. R = 91 , R1= 560 , 467 > RL> 67 .

51. (a) 3.7 mV; (b) 28 mV; (c) 3.7 V.

53.A101

100A-

in

out

+=

v

v; A = 9999.

55. vout= -16 mV



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57. (a)

(b)

vout= 105(-0.00004v2 - 9.9998010

-6v1)+5v2= 1.00008v2 - 0.99998v1 = 0.0005

1.99996 sin t

(c) =0.99998v)2/(1010 255

adout vvvv == 2-0.99998v1= 1.99996 sin t

59. (a) V3= 27 V;

61. Positive voltage supply, negative voltage supply, inverting input, ground, output

pin.

63. This is a non-inverting op amp circuit, so we expect a gain of 214.

65. For vx= -10 mV, PSpice predicts vd= 6 V, where the hand calculations based onthe detailed model predict 50 V, which is about one order of magnitude larger.For the same input voltage, PSpice predicts an input current of -1 A, whereas the

hand calculations predict 99.5vxmA = -995 nA (which is reasonably close).

67. (a) Negative saturation begins at Vin= 4.72 V, and positive saturation begins at

Vin= +4.67 V. (b) 40.6 mA.

69.



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71. (a)

-15

-10

-5

0

5

10

15

-2 -1 0 1 2

Vactive(V)

Vout(V)

12 V

-12 V

73.

75. (a)

+

+==

Gauge

refoutRR

R

RR

RVVVV

3

3

21

221 ; (b) Vout= 0; (c) R = 4.3 kand R

= 4.7 k, gain of 5.39 for R = 4.7 k, so R = 11.5 k.


chapter6 selected answers

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