chap05

# chap05 - Chapter 5 Solution 1(a(b(c Rin = 1.5 M Rout = 60 A...

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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 + -
(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 + - V S

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At node 1, ( V S V 1 )/10 k = [ V 1 /100 k] + [( V 1 V 0 )/100 k] 10 V S – 10 V 1 = V 1 + V 1 V 0 which leads to V 1 = (10 V 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) (a)
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, 1mA = 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. Since no current enters the op amp, the voltage at the input of the op amp is v s . Hence v s = v o 2 v 10 10 10 o = + s o v v = 2

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Chapter 5, Solution 11. 8 k v b = V 2 ) 3 ( 5 10 10 = + i o b a + 5 k 2 k 4 k + v o 10 k + 3 V At node a, 8 v v 2 v 3 o a a = 12 = 5v a – v o But v a = v b = 2V, 12 = 10 – v o v o = -2V –i o = mA 1 4 2 8 2 2 4 v 0 8 v v o o a = + + = + i o = -1mA Chapter 5, Solution 12. 4 k b a + 2 k 1 k + v o 4 k + 1.2V
At node b, v b = o o o v 3 2 v 3 2 v 2 4 4 = = + At node a, 4 v v 1 v 2 . 1 o a a = , but v a = v b = o v 3 2 4.8 - 4 x o o o v v 3 2 v 3 2 = v o = V 0570 . 2 7 8 . 4 x 3 = v a = v b = 7 6 . 9 v 3 2 o = i s = 7 2 . 1 1 v 2 . a = 1 p = v s i s = 1.2 = 7 2 . 1 -205.7 mW Chapter 5, Solution 13. By voltage division, i 1 i 2 90 k 10 k b a + 100 k 4 k 50 k + i o + v o 1 V v a = V 9 . 0 ) 1 ( 100 = 90 v b = 3 v v 150 o o = 50 But v a = v b 9 . 0

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