Chapter11

Microelectronic Circuit Design

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CHAPTER 11 11.1 v O = v S 1 M 1 M Ω+ 5 k 1000 () 1 k 1 k 0.5 | A v = v O v S = 990 or 59.9 dB i S = v S 1 M 5 k and i O = 990 v S 1 k A i = i O i S = 990 1000 10 6 = 9.9 x 10 5 or 120 dB A P = A v A i = 990 9.9 x 10 5 = 9.8 x 10 8 or 89.9 dB v S = v O A V = 5 V 990 = 5.05 mV 11.2 v O = v S 5 k 5 k 5 k 31.6 1 k 1 k 1 k and A v = v O v S = 7.91 or 18.0 dB v s = v O A v = 10 V 7.91 = 1.27 V Since R out has the same value as R L , the power dissipated in R out is also 0.5W. The power dissipated in R id will be P I = V id 2 2 R id = V S 2 2 2 R id = V S 2 8 R id where V S = V O 7.91 and V O = 2 0.5 W 1000 = 31.6 V P I = 4 2 8 5000 = 0.4 mW . The total power dissipated in the amplifier is P = 500mW + 0.4mW = 500 mW. 11.3 0.99 mV 1 mV R id R id + 50 k R id 4.95 M 11.4 I o = 2 100 W 50 = 2 A and I o 2 R out 2 5W or R out 2.5 11.5 v id = v O A = 10 V 10 5 = 0.1 mV 10 V A 10 6 V requires A 10 7 or 140 dB 11.6 v id = v O A = 15 V 10 6 = 15 µ V 15 V A 10 6 V requires A 15 x 10 6 or 144 dB i + = 15 V 1 M = 15 pA 11-1
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11.7 a () A v =− R 2 R 1 220 k 4.7 k 46.8 | 20log 46.8 = 33.4 dB R in = R 1 = 4.7 k R out = 0 b A v R 2 R 1 2.2 M 47 k 46.8 20log 46.8 = 33.4 dB R in = R 1 = 47 k R out = 11.8 a A v R 2 R 1 120 k 12 k 10.0 20log 10.0 = 20.0 dB R in = R 1 = 12 k R out = b A v R 2 R 1 330 k 140 k 2.36 20log 2.36 = 7.46 dB R in = R 1 = 140 k R out = c A v R 2 R 1 240 k 4.3 k 55.8 20log 55.8 = 34.9 dB R in = R 1 = 4.3 k R out = 11.9 A v R 2 R 1 8200 750 10.9 V O 10.9 0.05 V 0.547 V v o t 0.547sin 4638 t V I s = V s R 1 = 0.05 V 750 =66 .7 µ A i s t = 66.7sin 4638 t A 11.10 a A v 110 k 22 k 5 v o = A v v s = 0 b V O = A v V S = -5 0.22 V 1.10 V b v o 5 0.15 V [] sin2500 π t 0.75sin2500 t V d v o 1.10 + 0.75sin2500 t V e I S = 0.22 V 22 k = 10.0 A i s = 0.15 V 22 k sin2500 t = 6.82sin2500 t A i S = 10.0 - 6.82sin2500 t A f i O =- i S I O 10.0 A i o 6.82sin2500 t A i O = -10.0 + 6.82sin2500 t A g v - = 0 11.11 v O i TH R 11.12 R in = R 1 = 1.5 k A v R 2 R 1 10 40 20 100 R 2 = 100 R 1 = 150 k The resistors exist as standard values. 11-2
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11.13 R in = R 1 = 30 k | A v =− R 2 R 1 10 26 20 20 R 2 = 20 R 1 = 600 k Using the closest values from Appendix A, R 1 30.1 k and R 2 604 k The values for the final design are A v 604 k 30.1 k 20.1 and R in = 30.1 k 11.14 R in = R 1 = 100 k A v R 2 R 1 10 12 20 15.8 R 2 = 15.8 R 1 =1.58 M Using the closest 1% values from Appendix A, R 1 100 k and R 2 1.00 M Ω+ 576 k The values for the final design are A v 1.576M 100 k 15.8 and R in = 100 k If we used 5% values, we could select 100 k and 1.6 M . 11.15 A v = 1 + R 2 R 1 = 1 + 750 k 8.2 k = 92.5 20log 92.5 () = 39.3 dB R in =∞ R out = 0 11.16 a A v = 1 + R 2 R 1 = 1 + 120 k 24 k = 6.00 20log 6.00 = 15.6 dB R in R out = b A v = 1 + R 2 R 1 = 1 + 300 k 15 k = 21.0 20log 21.0 = 26.4 dB R in R out = c A v = 1 + R 2 R 1 = 1 + 360 k 4.3 k = 84.7 20log 84.7 = 38.6 dB R in R out = 11.17 A v = 1 + R 2 R 1 = 1 + 8200 910 = 10.0 | V O = 10.0 0.05 V = 0.500 V v o t = 0.500sin9125 t V 11-3
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11.18 a () A v = 1 + 110 k 22 k = +6 v o = 6 v s = 0 b V O = A v V S = 60 .33 V =+ 1.98 V b v o = .18 V [] sin3250 π t = 1.08sin3250 t V d v o = 1.98 1.08sin3250 t V e i S = 0 f i O v O R 1 + R 2 I O = 1.98 V 132 k = 15.0 µ A i o = 8.18sin3250 t A i O = 15.0 + 8.18sin3250 t A g v - = 0.33 0.18sin3250 t V 11.19 a A v nom = 1 + R 2 R 1 = 1 + 47 k 0.18 k = 262 | 20log 262 =48 .4 dB R in = 10 k Ω+∞=∞ R out = b A v max = 1 + 47 k 1.1 0.18 k 0.9 = 320 A v min = 1 + 47 k 0.9 0.18 k 1.1 = 215 A v max A v nom A v nom = 320 262
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Chapter11 - CHAPTER 11 11.1 v O = vS iS = v 1M 1k (1000)1k...

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