Opamp

Operational Amplifiers 1. Av =

3.5 4. In circuit shown in fig. P3.5.4, the input voltage vi is

vo =? vi

0.2 V. The output voltage vo is

400 kW

50 kW

40 kW

vi

150 kW

10 kW vi

vo

25 kW vo

R

Fig. P3.5.4

Fig. P3.5.1

(A) -10

(B) 10

(A) 6 V

(B) -6 V

(C) -11

(D) 11

(C) 8 V

(D) -8 V

2. Av =

5. For the circuit shown in fig. P3.5.5 gain is

vo =? vi

Av = vo vi = -10. The value of R is

400 kW

R

100 kW

40 kW

vi

100 kW

vo 60 kW

100 kW

vi

vo

Fig. P3.5.2

(A) -10

(B) 10

(C) 13.46

(D) -13.46

Fig. P3.5.5

3. The input to the circuit in fig. P3.5.3 is

(A) 600 kW

(B) 450 kW

(C) 4.5 MW

(D) 6 MW

vi = 2 sin wt mV. The current io is 6. For the op-amp circuit shown in fig. P3.5.6 the

10 kW vi

voltage gain Av = vo vi is

1 kW io

R

vo

R

R

4 kW vi

(B) -7 sin wt m A

(C) -5 sin wt m A

(D) 0

R

R vo

Fig. P3.5.3

(A) -2 sin wt m A

R

Fig. P3.5.6

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183

Operational Amplifiers

(A) -8

3.5

(B) 8

(C) -10

20 kW

20 kW

+0.5 V

(D) 10

40 kW -1 V

7. For the op-amp shown in fig. P3.5.7 open loop

vo

60 kW +2 V

differential gain is Aod = 10 3. The output voltage vo for vi = 2 V is

Fig. P3.5.10 100 kW

vi

100 kW

(A) 2.67 V

(B) -2.67 V

(C) -6.67 V

(D) 6.67 V

vo

11. In the circuit of fig. P3.5.11 the voltage vi1 is (1 + 2 sin wt) mV and vi2 = -10 mV. The output voltage vo is

Fig. P3.5.7

20 kW

(A) -1.996

(B) -1.998

(C) -2.004

20 kW vi1

(D) -2.006

2 kW 1 kW vo

vi2

8. The op-amp of fig. P3.5.8 has a very poor open-loop

1 kW

voltage gain of 45 but is otherwise ideal. The closed-loop gain of amplifier is

Fig. P3.5.11

100 kW 2 kW vo

(A) -0.4(1 + sin wt) mV

(B) 0.4(1 + sin wt) mV

(C) 0.4(1 + 2 sin wt) mV

(D) -0.4(1 + 2 sin wt) mV

12. For the circuit in fig. P3.5.12 the output voltage is

vi

vo = 2.5 V in response to input voltage vi = 5 V. The

Fig. P3.5.8

finite open-loop differential gain of the op-amp is

(A) 20

(B) 4.5

(C) 4

(D) 5

vi

500 kW

9. For the circuit shown in fig. P3.5.9 the input voltage

Fig. P3.5.12

vi is 1.5 V. The current io is 10 kW vi

vo

1 kW

6 kW 8 kW io

(A) 5 ´ 10 4

(B) 250.5

(C) 2 ´ 10

(D) 501

4

13. vo = ?

vo

100 kW

5 kW 100 kW vo

20 kW

Fig. P3.5.9

+18 V 40 kW

(A) -1.5 mA

(B) 1.5 mA

(C) -0.75 mA

(D) 0.75 mA

+15 V

Fig. P3.5.13

10. In the circuit of fig. P3.5.10 the output voltage vo is

(A) 34 V

(B) -17 V

(C) 32 V

(D) -32 V

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3.5

Operational Amplifiers

184

18. For the circuit shown in fig. P3.5.18 the true

14. vo = ?

relation is

100 kW 20 kW vo

60 kW +10 V

vo1

10 kW

20 kW

1 kW

+15 V

R

30 kW R vi

Fig. P3.5.14

(A) -5.5 V

(B) 4.58 V

(C) 5.5 V

(D) -4.58 V

15. Av =

Fig. P3.5.18

vo =? vi R

R

vo2

(A) vo1 = vo2

(B) vo1 = -vo2

(C) vo = 2 vo2

(D) 2 vo1 = vo2

R

19. vo = ?

R vo

R

10 kW 10 kW

vi vo +6 V 48 kW

Fig. P3.5.15

(A) 5

(B) -5

(C) 6

(D) -6

5 kW

6 kW

Fig. P3.5.19

Statement for Q.16–17:

(A)

4 V 3

(B) -

2 V 3

(C)

2 V 3

(D) -

4 V 3

The circuit is as shown in fig. P3.5.16–17.

vo

50 kW

20. vo = ?

1 kW

vi

3 kW

Fig. P3.5.16–17

(B) -1

(C) ¥

(D) 50

vo

12 V

16. The ideal closed-loop voltage gain is (A) 1

4 kW

2 kW R 1 kW

Fig. P3.5.20

17. If open-loop gain is Aod = 999, then closed-loop gain is (A) -0.999

(B) 0.999

(A) -12 V

(B) 12 V

(C) 1.001

(D) -1.001

(C) -18 V

(D) 18 V

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185

Operational Amplifiers

21. vo = ?

3.5

25. Avd = 8 kW

vo =? ( v1 - v2 )

4 kW v1

vo 0.1 mA

+

2 kW

2V

6 kW 2 kW 4 kW

vo

Fig. P3.5.21

(A) -30V

(B) 18V

(C) -18V

(D) 28V

-

v2

22. vo = ? 10 kW

Fig. 3.5.25

vo 0.1 mA

20 kW 5V

(A) 8

(B) -6

(C) 6

(D) -8

26. vo = ?

Fig. P3.5.22

vo

(A) 4 V

(B) -4 V

(C) 5 V

(D) -5 V

23. io = ?

1 3 kW

2 kW

3 kW 4 kW

3 kW

io 12 V

2

6V

2m A 6 kW

Fig. 3.5.26 Fig. P3.5.23

(A) 12 mA

(B) 8.5 mA

(C) 6 mA

(D) 7.5 mA

(A) 6 V

(B) -6 V

(C) -10 V

(D) 10 V

27. Av =

24. vo = ?

vo =? vi vi

6 kW

2 kW 1 kW vo

vo

2.5 V

3 kW 8 kW 6 kW

1 kW

4 kW

Fig. P3.5.27

Fig. P3.5.24

(A) -7.5 V

(B) 7.5 V

(A) 15.8

(B) -10

(C) 8 V

(D) -8 V

(C) -17.4

(D) -8

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3.5

Operational Amplifiers

28. For the circuit shown in fig. P3.5.28 the input

186

31. io = ?

resistance is

6 kW

io

2 kW vo

6A 2 kW 4 kW is

Fig. S3.5.31 2 kW 10 kW

(A) -18 A

(B) 18 A

(C) -36 A

(D) 36 A

Statement for Q.32–33:

Fig. P3.5.28

(A) 38 kW

(B) 17 kW

(C) 25 kW

(D) 47 kW

Consider the circuit shown below

29. In the circuit of fig. P3.5.29 the op-amp slew rate is SR = 0.5 V ms. If the amplitude of input signal is 0.02

vi

3 kW

D1

6 kW

D2

2 kW vo

V, then the maximum frequency that may be used is 240 kW vi

10 kW

Fig. P3.5.32–33 vo

32. If vi = 2 V, then output vo is

Fig. P3.5.29

(A) 4 V

(B) -4 V

(C) 3 V

(D) -3 V

33. If vi = -2 V, then output vo is (A) 0.55 ´ 106 rad/s

(B) 0.55 rad/s

(C) 1.1 ´ 106 rad/s

(D) 1.1 rad/s

(A) -6 V

(B) 6 V

(C) -3 V

(D) 3 V

34. vo( t) = ? 30. In the circuit of fig. P3.5.30 the input offset voltage and input offset current are Vio = 4 mV and I io = 150 8 mF

nA. The total output offset voltage is 500 kW vi

5u(t) mA

250 W

vo 50 W

1 kW

5 kW vo

Fig. P3.5.34

(A) e

5 kW

Fig. P3.5.30

(A) 479 mV

(B) 234 mV

(C) 168 mV

(D) 116 mV

(C) e

-

t 10

-

t 1 .6

-

u( t) V

(B) -e

u( t) V

(D) -e

t 10

-

t 1 .6

u( t) V u( t) V

35. The circuit shown in fig. P3.5.35 is at steady state before the switch opens at t = 0. The voltage vC ( t) for t > 0 is www.nodia.co.in

187

Operational Amplifiers

3.5

vss

t=0

10 kW X

20 kW

R

vs

vo

20 kW 20 kW + 4 mF

5V

vC

Fig. P3.5.38

-

(C) (A) 10 - 5 e -12 .5t V (C) 5 + 5 e

-

t 12 .5

(B) 5 + 5 e -12 .5t V (D) 10 - 5 e

V

(B) -vs vss

(A) vs vss

Fig. P3.5.35

-

t 12 .5

vs vss

(D)

vs vss

39. If the input to the ideal comparator shown in fig.

V

P3.5.39 is a sinusoidal signal of 8 V (peak to peak)

36. The LED in the circuit of fig. P3.5.36 will be on if vi

without any DC component, then the output of the comparator has a duty cycle of

is Input

10 kW +10 V

Output

470W 10 kW

Vref = 2 V

vi

Fig. P3.5.39

Fig. P3.5.36

(A) > 10 V

(B) < 10 V

(C) > 5 V

(D) < 5 V

37. In the circuit of fig. P3.5.37 the CMRR of the op-amp is 60 dB. The magnitude of the vo is

(A)

1 2

(B)

1 3

(C)

1 6

(D)

1 12

40. In the op-amp circuit given in fig. P3.5.40 the load current iL is R1

2V

R1 vs

100 kW R

R

1 kW

R

R

R2

vo

1 kW

IL

R2 RL

100 kW

Fig. P3.5.40 Fig. P3.5.37

(A) 1 mV

(B) 100 mV

(C) 200 mV

(D) 2 mV

38. The analog multiplier X of fig. P.3.5.38 has the characteristics vp = v1 v2 . The output of this circuit is

(A) -

vs R2

(B)

vs R2

(C) -

vs RL

(D)

vs RL

41. In the circuit of fig. P3.5.41 output voltage is |vo| = 1

V for a certain set of w, R, an C. The |vo| will be 2 V if

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3.5

Operational Amplifiers

R1

(A)

R1 vi = sin wt V

(C)

vo

188

1 mF 2p 1 2p 6

(B) 2p mF mF

(D) 2 p 6 mF

R C

45. In the circuit shown in fig. P3.5.45 the op-amp is ideal. If bF = 60, then the total current supplied by the

Fig. P3.5.41

15 V source is

(A) w is doubled

(B) w is halved

(C) R is doubled

(D) None of the above

+15 V

47 kW

42. In the filter circuit of fig. P3.5.42. the 3 dB cutoff frequency is 50 nF

6 kW vo

3 kW

vo

Vz = 5 V

100 W

vi

Fig. P3.5.42

Fig. P3.5.45

(A) 10 kHz

(B) 1.59 kHz

(C) 354 Hz

(D) 689 Hz

(A) 123.1 mA

(B) 98.3 mA

(C) 49.4 mA

(D) 168 mA

43. The phase shift oscillator of fig. P3.5.43 operate at f = 80 kHz. The value of resistance RF is

46. In the circuit in fig. P3.5.46 both transistor Q1 and Q2 are identical. The output voltage at T = 300 K is

RF 100 pF

100 pF

R

100 pF

R

R1 vo

R

R2

v1

v2

333 kW

Fig. P3.5.43

(A) 148 kW

(B) 236 kW

20 kW

(C) 438 kW

(D) 814 kW

20 kW

44. The value of C required for sinusoidal oscillation of

vo

333 kW

frequency 1 kHz in the circuit of fig. P3.5.44 is 2.1 kW

1 kW

Fig. P3.5.46

C 1 kW

1 kW

Fig. P3.5.44

C

æv R ö (A) 2 log10 çç 2 1 ÷÷ è v1 R2 ø

æv R ö (B) log10 çç 2 1 ÷÷ è v1 R2 ø

æv R ö (C) 2.303 log10 çç 2 1 ÷÷ è v1 R2 ø

æv R ö (D) 4.605 log10 çç 2 1 ÷÷ è v1 R2 ø

47. In the op-amp series regulator circuit of fig. P8.3.47 Vz = 6.2 V, VBE = 0.7 V and b = 60. The output voltage vo is www.nodia.co.in

189

Operational Amplifiers

vo

+36 V

3.5

Gain of second stage Av2 = -

1 kW

150 = -6 25

Total gain Av = Av1 Av2 = 30, vo = 30 ´ 0.2 = 6 V 30 kW

5. (B) Let vx be the node voltage vx v v - vo + x + x =0 R 100 100 10 kW

æ 2 + 100 ö vo = vx ç ÷ R ø è

Þ

0 - vi 0 - vx R + = 0 Þ vx = vi , 100 R 100 100 ö vo R æ =ç2 + ÷ = -10 vi R ø 100 è

Þ

Fig. P3.5.47

(A) 35.8 V

(B) 24.8 V

(C) 29.8 V

(D) None of the above

2 R + 100 = -1000 , R = 450 kW 6. (A)

*******

0 - vi 0 - v1 + = 0, v1 = -vi R R R

R

v1 R

Solutions

vi

R

v2 R

R vo

1. (A) This is inverting amplifier 400 R Av = - F = = - 10 40 R1

Fig. S3.5.6

2. (A) The noninverting terminal is at ground level. Thus inverting terminal is also at virtual ground. There will not be any current in 60 kW. 400 Av = = - 10 40 3. (B) vo = -

10 (2 sin wt) mV = - 20 sin wt mV 1

v1 - 0 v1 - v2 v1 + + = 0, 3v1 = v2 , v2 = -3vi R R R v2 - v1 v2 v2 - vo + + =0 R R R vo -3vi + vi - 3vi - 3vi = vo Þ = -8 vi 7. (A) i1 =

vi - v1 100 k

10 kW i2

vi

100 kW

i1

1 kW io ii

iL

100 kW

vi

vo

i1

v1

vo

4 kW

Fig. S3.5.7 Fig. S3.5.3

i2 =

v iL = o = -5 sin wt m A 4k 2 sin wt = 2 sin wt m A i1 = ii = 1k

Þ

io = iL - i1 = -5 sin wt - 2 sin wt = - 7 sin wt m A 4. (A) Gain of first stage Av1 = -

v1 - vo , i1 = i2 , v1 - vo = vi - v1 100 k

50 = -5 10

2 v1 - vo = vi , vo = - Aod v1 v 2v v1 = - o = o - vo = vi Aod Aod 1 vo = vi æ 2 ö çç 1 + ÷ Aod ÷ø è

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Þ

vo = -

2 = -1996 . (1 + 2 ´ 10 -3)

3.5

Operational Amplifiers

8. (B) A closed loop gain ACL =

vo Aod = vi 1 + Aod b

2k b= = 0.2 8k + 2k ACL =

45 = 4.5 1 + ( 45)(0.2)

9. (C) i1 =

15 . = 0.25 mA, i1 = i2 6k i2

10 kW vi

190

15. (A) v+ = vi = vlet v1 be the node voltage of T network v- v- - v1 + =0 Þ v1 = 2 v- = 2 vi R R v1 - v- v1 v - vo + + 1 = 0 Þ 3v1 = v- + vo , R R R vo 6vi = vi + vo Þ =5 vi vo =1 vi

16. (A) v+ = vi , v- = vi = vo ,

6 kW 8 kW io

i1

iL

vo

5 kW

Fig. S3.5.9

17. (B) v+ = vi , v- = vo Aod ( vi - vo) = vo Aod = 999 vo Aod 999 = = = 0.999 vi 1 + Aod 1 + 999

vo = -10 ki2 = -2.5 V, i2 + io = iL 2.5 , io = -0.75 mA 0.25m + io = 5k

18. (B) At second stage input to both op-amp circuit is

10. (B) This is summing amplifier

Av = 1. Lower op-amp circuit is inverting amplifier R having gain Av = - = -1. Therefore vo1 = -vo2 . R

1 2 ö æ 0.5 vo = -80ç + ÷ = -2.67 V è 20 40 60 ø 20 11. (B) Output of first op-amp vo1 = vi1 2 = -10(1 + 2 sin wt) mV

same. The upper op-amp circuit is buffer having gain

19. (A) v+ =

6´ 6 2 = V, 48 + 6 3

10 ö 4 æ vo = ç 1 + ÷ v+ = V 10 ø 3 è

20. (A) Applying KVL to loop,

The second stage is summing amplifier æ -10 (1 + 2 sin wt) 10 ö vo = -20 ç ÷ mV 1 1 ø è = 0.4(1 + sin wt) mV 12. (B) v+ =

vo

R 1 kW

18 ´ 40 15 ´ 20 + = 17 V 20 + 40 20 + 40

i2

Fig. S3.5.20

12 = 3ki1 + 2 ki1

Þ

i1 = 2.4 mA , io = i1 = 2.4 mA

i2 = -i1 = -2.4 mA vo = i2 (1k) = -2.4 V

v+ v - 10 v+ - 15 + + + =0 30 60 20 1 3 11 v+ = + = 6 4 12

vo = va - io( 4k) = -2.4 - (2.4)( 4) = -12 V

100 ö 11 æ vo = v+ ç 1 + (1 + 5) = 5.5 V ÷= 20 ø 12 è

vo + 8 = -2, vo = -30 V 3

14. (C)

io

2 kW

vi A v , vo = od i 500 + 1 501

100 k ö æ vo = ç 1 + ÷v+ = 34 V 100 k ø è

4 kW

i1

12 V

(2.5)(501) = Aod (5), Aod = 250.5 13. (A) v+ =

va

3 kW

21. (A) v1 =

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vo( 4) 12( 8) , v+ = -2 V, v+ = v+ 4+8 4+8

191

Operational Amplifiers

22. (A) v+ = 5 V = v- ,

v+ - vo = 0.1 mA 10 k

v- = v+ , 2 kis = 4 ki1

12 = 3 mA 4k

i ö æ vs = 2 kis + 10 kç is + s ÷ Þ 2ø è i2

i1

2 mA

iL

vo

w £ Fig. S3.5.23

500 ö æ = ç1 + ÷ 4m = 404 mV 5 ø è Due to I io, vo = RF I io = (500 k)(150n) = 75 mV

vo = 7.5 V

Þ

Total offset voltage vo = 404 + 75 = 479 mV

25. (C) v1 + = v1 = v1 - , v2 + = v2 = v2 -

-vo v , io = - 6 + o 6k 3k -6( 6 k) io = - 6 + = -18 A. 3k

31. (A) 6 =

Current through 2 kW resistor v - v2 v - v2 i = 1 = 1 2k 2k ( v - v2 ) vo = i( 6 k + 2 k + 4 k) = 1 (12 k) 2k vo = 6 = Avd v1 - v2

32. (B) If vi > 0, then vo < 0, D1 blocks and D2 conducts Av = -

26. (C) v2 + = v2 - = 0 V, current through 6 V source i=

0.5 / m SR . ´ 106 rad/s = = 11 0.48 vom

æ R ö 30. (A) The offset due to Vio is vo = çç 1 + 1 ÷÷Vio R1 ø è

i2 = 3 + 2 = 5 mA, vo = -(5)( 3) = -15 V -15 , io = 7.5 mA i2 = io + iL , 5 = io + 6 vo ( 4) = 2.5 8+4

vs = 17k = Rin is

The maximum output voltage vom = 24 ´ 0.02 = 0.48 V

6 kW

24. (B)v+ = 2.5 V = v- ,

is 2

½R ½ 240 k 29. (C) Closed loop gain A =½ F½ = = 24 ½ R1½ 10 k

io 12 V

is = 2 i1

i2 = is + i1 , vs = 2 kis + 10 k( is + i1 ), i1 =

3 kW 4 kW

Þ

vs = 2 kis + 10 ki2

v+ - vo = 1 , 5 - vo = 1 , vo = 4 V 23. (D) v+ = v- = 0, i1 =

3.5

6 = 2 mA, vo = -2m( 3k + 2 k) = -10 V 3k

vo(1) vo v (2) vo(1) = , v- = i + 1+ 3 4 2+1 2+1 v v 2v v v+ = v- , o = o + i , o = -8 vi 4 3 3

27. (D) v+ =

6k = -2 3k

Þ

vo = ( -2)(2) = -4 V

33. (D) If vi < 0, then vo > 0, D2 blocks and D1 conduct Av = -

3k . )=3 V = -15 . , vo = ( -2)( -15 2k

34. (A) Voltage follower vo = v- = v+ v+ (0 + ) = 5m(250 ||1000) = 1 V, v+ ( ¥) = 0 t = 8m(1000 + 250) = 10 s 35. (A) vc (0 - ) = 5 V = vc (0 + ) = 5 V

28. (B) Since op-amp is ideal

For t > 0 the equivalent circuit is shown in fig. S3.5.35 20 kW

i1 10 V

4 kW is

4 mF

+ vC –

is 2 kW

i2 10 kW

Fig. S3.5.28

Fig. S3.5.35

t = 20 k ´ 4m = 0.08 s vc = 10 + (5 - 10) e www.nodia.co.in

-

t 0 .08

= 10 - 5 e -12 .5t V for t > 0

3.5

Operational Amplifiers

36. (C) v- =

192

Thus when w and R is changed, the transfer function

(10)(10 k) =5 V 10 k + 10 k

is unchanged.

When v+ > 5 V, output will be positive and LED will be 42. (B) Let R1 = 3 kW , R2 = 6 kW , C = 50 nF

on. Hence (C) is correct.

vi v - vo + i =0 R2 æ 1 ö R1 ||ç ÷ è sC ø

R R 37. (B) v+ = (2) = 1 V, v- = (2) = 1 V, vd = 0 2R 2R v + vR VCM VCM = + = 1, vo = F 1 CMRR 2 100 1 CMRR = 60 dB = 10 3 , vo = = 100 mV 1 10 3

éR ù vi ê 2 (1 + sR1 C) + 1ú = vo ë R1 û vi [R2 + R1 + sR1 R2 C ] = vo R1

38. (C) v+ = 0 = v- ,

vo R + R1 = 2 vi R1

Let output of analog multiplier be vp . vp vs =Þ vs = -vp , vp = vss vo R R v vs = -vss vo , vo = - s vss

Þ

39. (B) When vi > 2 V, output is positive. When vi < 2 V,

=

V

2V 2p

p

t

5p 6

6

vs - v- v- - vo = R1 R1

v+ v v - vo + + + + =0 R2 RL R2

v+ = -

Þ

1 ( 80 k)(2 p 6 )(100 p)

RF = 29 R

5p p TON 1 Duty cycle = = 6 6 = 2p T 3

Þ 0 = vs +

1 1 = = 159 . kHz 2 p( 3k ||6 k)50n 2 p(2 k)50n

Þ R=

Fig. S3.5.39

æ R 2 v- = vs + çç 2 + 2 RL è

1 2 p( R1 || R2 ) C

43. (B) The oscillation frequency is 1 1 f = Þ 80 k = 2 p 6 RC 2 p 6 R(100 p)

4V

40.(A)

Þ

= 8.12 kW

RF = ( 8.12 k)(29) = 236 kW

44. (A) This is Wien-bridge oscillator. The ratio R2 2.1k = = 2.1 is greater than 2. So there will be R1 1k oscillation

2 v1 = vs + vo

R2

Þ

æ R vo = çç 2 + 2 RL è

ö ÷÷ v+ ø

R1

ö ÷÷ v+ , v- = v+ ø R

R2 v+ RL

RL vs , R2

é sR1 R2 C ù ê1 + R + R ú ë 1 2 û

vo æ R ö = çç 1 + 2 ÷÷(1 + s( R1 || R2 ) C) vi è R1 ø

f3dB =

output is negative.

v v vi + i = o æ ö R2 R2 R1 çç ÷÷ è 1 + sR1 C ø

Þ

iL =

R

v+ , RL

iL = -

vs R2

| H( jw)| =

Frequency = 1 + ( wR 2 C) 2 1 + ( wRC) 2

C

Fig. S3.5.44

41. (D) This is a all pass circuit vo 1 - jwRC , = H ( jw) = vi 1 + jwRC

C

=1

C=

1 mF 2p

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1 2pRC

Þ

1 ´ 10 3 =

1 2 p(1k) C

193

Operational Amplifiers

45. (C) v+ = 5 V = v- = vE , The input current to the op-amp is zero. i+15V = iZ + iC = iZ + a F iE =

15 - 5 60 æ 5 ö + ç ÷ = 49.4 mA 47 k 61 è 100 ø

46. (B) vo =

333 ( vo1 - vo2 ) 20

æi vo1 = -vBE1 - Vt ln çç c1 è is

ö æi ÷÷, vo2 = -vBE 2 - Vt ln çç c 2 ø è is

æi vo1 - vo2 = -Vt ln çç c1 è ic 2

ö ÷÷ ø

ic1 =

v1 , R1

ic 2 =

ö æi ÷÷ = Vt ln çç c 2 ø è ic1

ö ÷÷ ø

v2 R2

æv R ö vo1 - vo2 = Vt ln çç 2 1 ÷÷, Vt = 0.0259 V è R2 v1 ø vo =

æv R ö 333 333 ( vo1 - vo2 ) = (0.0259) ln çç 2 1 ÷÷ 20 20 è v1 R2 ø

æv R ö æv R ö = 0.4329 lnçç 2 1 ÷÷ = 0.4329(2.3026) log10 çç 2 1 ÷÷ v R è 1 2ø è v1 R2 ø æv R ö = log10 çç 2 1 ÷÷ è v1 R2 ø 47. (B) v+ = v- , vZ =

10 vo v = o 10 + 30 4

vo = 4 vz = 6.2 ´ 4 = 24.8 V ************

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3.5