chap05

Chap05 - Chapter 5 Solution 1(a(b(c Rin = 1.5 M Rout = 60 A = 8x104 Therefore AdB = 20 log 8x104 = 98.0 dB Chapter 5 Solution 2 v0 = Avd = A(v2 v1

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Chapter 5, Solution 1. (a) R in = 1.5 M ( b ) R out = 60 (c) A = 8x10 4 Therefore A dB = 20 log 8x10 4 = 98.0 dB Chapter 5, Solution 2. v 0 = Av d = A(v 2 - v 1 ) = 10 5 (20-10) x 10 -6 = 0.1V Chapter 5, Solution 3. v 0 = Av d = A(v 2 - v 1 ) = 2 x 10 5 (30 + 20) x 10 -6 = 10V Chapter 5, Solution 4. v 0 = Av d = A(v 2 - v 1 ) v 2 - v 1 = V 20 10 x 2 4 A v 5 0 µ = = If v 1 and v 2 are in mV, then v 2 - v 1 = -20 mV = 0.02 1 - v 1 = -0.02 v 1 = 1.02 mV Chapter 5, Solution 5. + v 0 - - v d + R 0 R in I v i - + Av d
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-v i + Av d + (R i - R 0 ) I = 0 (1) But v d = R i I, -v i + (R i + R 0 + R i A) I = 0 v d = i 0 i i R ) A 1 ( R R v + + ( 2 ) -Av d - R 0 I + v 0 = 0 v 0 = Av d + R 0 I = (R 0 + R i A)I = i 0 i i 0 R ) A 1 ( R v ) A R R ( + + + 4 5 5 4 i 0 i 0 i 0 10 ) 10 1 ( 100 10 x 10 100 R ) A 1 ( R A R R v v + + + = + + + = () = + 4 5 9 10 10 1 10 = 001 , 100 000 , 100 0.9999990 Chapter 5, Solution 6. - v d + + v o - R 0 R in I v i + - Av d
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(R 0 + R i )R + v i + Av d = 0 But v d = R i I, v i + (R 0 + R i + R i A)I = 0 I = i 0 i R ) A 1 ( R v + + ( 1 ) -Av d - R 0 I + v o = 0 v o = Av d + R 0 I = (R 0 + R i A)I Substituting for I in (1), v 0 = + + + i 0 i 0 R ) A 1 ( R A R R v i = () () 6 5 3 5 6 10 x 2 x 10 x 2 1 50 10 10 x 2 x 10 x 2 50 + + + mV 10 x 2 x 001 , 200 10 x 2 x 000 , 200 6 6 v 0 = -0.999995 mV Chapter 5, Solution 7. 100 k 1 2 10 k - + + V d - + V out - R out = 100 R in AV d + -
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At node 1, (V S – V 1 )/10 k = [V 1 /100 k] + [(V 1 – V 0 )/100 k] 1 0 V S – 10 V 1 = V 1 + V 1 – V 0 which leads to V 1 = (10V S + V 0 )/12 At node 2, (V 1 – V 0 )/100 k = (V 0 – AV d )/100 But V d = V 1 and A = 100,000, V 1 – V 0 = 1000 (V 0 – 100,000V 1 ) 0= 1001V 0 – 100,000,001[(10V S + V 0 )/12] 0 = -83,333,334.17 V S - 8,332,333.42 V 0 which gives us (V 0 / V S ) = -10 (for all practical purposes) If V S = 1 mV, then V 0 = -10 mV Since V 0 = A V d = 100,000 V d , then V d = (V 0 /10 5 ) V = -100 nV Chapter 5, Solution 8. (a) If v a and v b are the voltages at the inverting and noninverting terminals of the op amp. v a = v b = 0 1mA = k 2 v 0 0 v 0 = -2V (b) 10 k 2V + - + v a - 10 k i a + v o - + v o - v a v b i a 2 k 2V - + 1V - + - + (b)
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Since v a = v b = 1V and i a = 0, no current flows through the 10 k resistor. From Fig. (b), -v a + 2 + v 0 = 0 v a = v a - 2 = 1 - 2 = -1V Chapter 5, Solution 9. (a) Let v a and v b be respectively the voltages at the inverting and noninverting terminals of the op amp v a = v b = 4V At the inverting terminal, 1 m A = k 2 0 v 4 v 0 = 2V Since v a = v b = 3V, -v b + 1 + v o = 0 v o = v b - 1 = 2V + v b - + v o - + - (b) 1V Chapter 5, Solution 10.
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Chap05 - Chapter 5 Solution 1(a(b(c Rin = 1.5 M Rout = 60 A = 8x104 Therefore AdB = 20 log 8x104 = 98.0 dB Chapter 5 Solution 2 v0 = Avd = A(v2 v1

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