Ch09s - Chapter 9 Problem Solutions 9.1 (a) vO = Ad ( v2 -...

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Chapter 9 Problem Solutions 9.1 (a) () 21 33 11 01 0 5 0 0 Od dd vA v v AA −− =− = (b) 3 22 2 1 500 10 1 0.5 500 3 mV vv v = + = = (c) 1 5 500 1 500 495 0.990 V v = = (d) 0 O v = (e) 2 2 2 35 0 0 0 . 5 250 3 500 0.506 V v v v −= = 9.2 (a) ( ) 3 2 3 2 1 0.49975 10 3 1 2000 1.49925 10 51 0 1 . 4 9 9 2 0 0 7.49625 V I Oo d O v v v v ⎛⎞ == × ⎜⎟ + ⎝⎠ = × × = (b) 3 3 3 1.49925 10 0 0 d od od v v A A 9.3 ( ) 2 1 12 12 12 25 k 12 25 300 k v i R AR R R RR R = Ω Ω 9.3 (a) 2 3.00 V v = (b) 2.500 3.010 3.00 250 d od od v v A A = 9.4
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() 25 0.790 0.80 25 0.9875 25 24.6875 0.0125 1975 K i id I i i i ii i i R vv R R R RR R R ⎛⎞ = ⎜⎟ + ⎝⎠ = + += = = 9.5 200 10 20 and for each case 20 k =− v i A R 9.6 a. 1 100 10 10 10 k == Ω v i A b. 1 100 100 5 10 10 k Ω & v i A c. 100 5 10 10 10 10 20 K v i A R + =+= 9.7
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11 1 1 2 2 1 0.5 5 K 0.1 15 75 K I v IR R R R R R = = = = = 9.8 2 1 v R A R =− (a) 10 v A (b) 1 v A (c) 0.20 v A (d) 10 v A (e) 2 v A (f) 1 v A 9.9 2 1 v R A R (a) 12 20 K, 40 K RR == (b) 20 K, 200 K (c) 20 K, 1000 K (d) 80 K, 20 K 9.10 2 21 1 2 1 88 1 For 1, 15 A 66.7 k 533.3 k μ = = = v I R AR R R vi R R R 9.11 2 1 30 30 Set 1 M 33.3 k = v R A R 9.12 a. 22 2 1 1.05 1.105 0.95 0.95 0.905 1.05 Deviation in gain is +10.5% and 9.5% ⎛⎞ = = ⎜⎟ ⎝⎠ = v R A R b. 1 1 1.01 0.99 1.02 0.98 0.99 1.01 Deviation in gain 2% v A R R = = 9.13 (a)
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() 15 15 1 15 150sin O v l Ol O v A v vv v t m V ω == = =− (b) 21 1 2 10sin 37.5sin 47.5sin I O LL L OL O v ii t A R v t A R i it A ωμ μ = = 9.14 2 15 30 2.5% 29.25 30.75 v R A RR AA + ± ≤≤ 22 11 So 29.25 and 30.75 ++ We have 29.25 2 30.75 1 += + 12 Which yields 18.5 and 599.6 Rk R k = Ω For 25 , then 0.731 0.769 IO vm V v V =≤ 9.15 2 1 4 1 3 1 34 3 120 , 0.2 1.2 V 20 75 1.2 6 V 15 0.2 10 A 20 1.2 80 A 15 OI O OO O O R v R R v R v R ⎛⎞ = =+ ⎜⎟ ⎝⎠ = 1st op-amp: 90 A into output terminal 2nd op-amp: 80 A out of output terminal. 9.16 (a) 2 1 22 22 1 R R (b) From Eq. (9.23) 2 1 2 4 1 22 1 1 3 10 21.95 v od v R A R R AR A ⎡⎤ + ⎢⎥ ⎣⎦ (c)
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() Want 22 0.98 21.56 22 So 21.56 1 12 3 2 3 21.56 1 23 0.020408 1127 v od od od od A A A A A =− −= + += = = 9.17 (a) 2 1 2 1 3 1 1 11 100 1 1 25 15 51 0 3.9960 v od v R A R R AR A ⎡⎤ ⎛⎞ ++ ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ + × (b) 3.9960 1.00 3.9960 V OO vv (c) 4 3.9960 100% 0.10% 4 ×= (d) 21 1 1 3 1 3.9960 0 0.7992 mV Oo d o d O od vA v v A v v v A v + = −− = × = 9.18 1 1 3 1 5 0 1 mV d o d O od v v A v v v A v + = = × 9.19 33 2 14 2 1 v RR R A R + + a. 2 2 2 2 100 100 10 1 100 100 2 10 1 450 k 100 R R R R + + =+ b. 2 2 2 100 1 4.95 M 100 R R 9.20 a.
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33 2 14 2 1 2 42 23 4 44 1 500 k 80 1 500 Set 500 k 500 500 80 1 1 1 2 6.41 k v RR R A R R R R ⎛⎞ =− + + ⎜⎟ ⎝⎠ =+ + == Ω + = + b. () ( ) 12 63 22 4 324 3 For 0.05 V 0.05 0.1 A 500 k 0.1 10 500 10 0.05 0.05 7.80 A 6.41 0.1 7.80 7.90 A I X X v ii vi R v R iii i μ Ω =− − × × = =+= − − 9.21 (a) 2 11 1 500 1000 0.5 K v R A R = = = (b) 2 2 1 1 250 500 500 1250 1000 1 250 250 1.25 K v R A R R + −− −= + + = = 9.22 2 3 I I AI v R v R R v R vv i =
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() 423 4 5 645 22 2 3 3 3 235 AA A I I B AI I I BI II I vv v v iii RR R R v vvi Rv R v R v i R vv v RRR =+= − − = = ⎛⎞ =− = −− = ⎜⎟ ⎝⎠ =− = + = 0 06 5 38 From Figure 9.12 3 I I v v v R v R A − − 9.23 (a) 2 1 2 1 5 1 1 11 50 1 4.99985 10 15 0 10 21 0 v od v R A R R AR A ⎡⎤ ++ ⎢⎥ ⎣⎦ × (b) 3 4.99985 100 10 499.985 OO m V × (c) 0.5 0.499985 Error 100% 0.003% 0.5 9.24 a. From Equation (9.23) 2 1 2 1 3 1 1 100 1 0.9980 100 0 0 100 10 v od R A R R ( ) 00 Then 0.9980 2 1.9960 V vI vA v v =⋅= b. 0
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Ch09s - Chapter 9 Problem Solutions 9.1 (a) vO = Ad ( v2 -...

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