Chapter10

Microelectronic Circuit Design

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CHAPTER 10 10.1 A/C temperature Automobile coolant temperature gasoline level oil pressure sound intensity inside temperature Battery charge level Battery voltage Fluid level Computer display hue contrast brightness Electrical variables voltage amplitude voltage phase current amplitude current phase power power factor spectrum Fan speed Humidity Lawn mower speed Light intensity Oven temperature Refrigerator temperature Sewing machine speed Stereo volume Stove temperature Time TV picture brightness TV sound level Wind velocity 10.2 ( a ) 20 log (120) = 41.6 dB | 20 log (60) = 35.6 dB | 20 log (50000) = 94.0 dB 20 log(100000) = 100 dB | 20 log(0.90) = 0.915 dB ( b ) 20 log (600) = 55.6 dB | 20 log (3000) = 69.5 dB | 20 log (10 6 ) = 120 dB 20 log(200000) = 106 dB | 20 log(0.95) = 0.446 dB ( c ) 10 log (2x10 9 ) = 93.0 dB | 10 log (4x10 5 ) = 56.0 dB 10 log (6x10 8 ) = 87.8 dB | 10 log(10 10 ) = 100 dB 10.3 (a) -4 -2 0 2 4 0 0.5 1 1.5 2 2.5 3 3.5 4 x10 -3 v S v O 10-1
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b () 500 Hz: 1 0 o | 1500 Hz: 0.333 0 o | 2500 Hz: 0.200 0 o c 500 Hz: 2 30 o | 1500 Hz: 1 30 o | 2500 Hz: 1 30 o d 30 o | 1500 Hz: 3 30 o | 2500 Hz: 5 30 o e Yes 10.4 V s = 0.0025 V | P O = 40 W V o = 2 P O R L = 240 ( ) 8 ( ) = 25.3 V A v = 25.3 .0025 = 10100 20 log 10100 = 80.1 dB I s = 0.0025 V 5 k Ω+ 50 k = 45.45 nA I o = V o 8 = 25.3 V 8 = 3.162 A A i = 3.162 A 45.45 nA = 6.96 x 10 7 20 log 3.48 x 10 7 = 157 dB A p = 40 W .0025 V 45.45 nA 2 = 7.04 x 10 11 10 log 7.04 x 10 11 = 118 dB 10.5 V s = 0.01 V P O = 20 mW V o = 2 P O R L = 2.02 ( ) 8 ( ) = 0.566 V A v = 0.566 .01 = 56.6 20 log 56.6 = 35.0 dB I s = 0.01 V 2 k 50 k = 192 nA I o = V o 8 = 0.566 V 8 = 70.8 mA A i = 70.8 mA 192 nA = 3.68 x 10 5 20 log 3.68 x 10 5 = 111 dB A p = 0.02 W .01 V 192 nA 2 = 2.08 x 10 7 10 log 2.08 x 10 7 = 73.2 dB 10.6 + - R th v th a v th = v oc = 0.768 2 = 1.09 V v o = R L R th + R L v th R th = R L v th v o v o = 430 0.768 0.721 0.721 = 28.0 b v th = v oc = 0.760 2 = 1.08 V v o = R L R th + R L v th R th = R L v th v o v o = 1040 0.760 0.740 0.740 = 28.1 c 1.09 V and 1.08 V 9% error and 8% error 10-2
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10.7 G4 laptop – 1 V, 28 . 10.8 a () V o = 2 R L P O = 28 20 = 17.9 V P i = V i 2 2 R i = 1 2 40066 = 25.0 µ W | I i = 1 V 20000 Ω+ 32 = 49.9 A I o = 17.9 V 8 = 2.24 A A v = V o V i = 17.9 V 1 V = 17.9 A P = 20 W 25 W = 8.00 x 10 5 A i = 2.24 A 49.9 A = 4.49 x 10 4 b V o = 17.9 V ; recommend ± 20- V supplies 10.9 V = 2 PR I = 2 P R The 24- case represents a good trade off between voltage and current. R ( ) V (V) I (mA) 8 1.27 158 24 2.19 91.3 1000 14.1 14.1 10.10 In the dc steady state, the internal circuit voltages cannot exceed the power supply limits. (a) +15 V (b) -9 V 10.11 (a) For V B = 0.6 V , V O =+ 8 V A v = dv O dv I v I = 0.6 V = 12 4 0.5 0.7 =− 40 A v = 32 dB A V = 180 o V M 0.100 V for linear operation (b) v I t = 0.6 + 0.1sin1000 t V v O t = 8 4sin1000 t V 10-3
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10.12 (a) For V B = 0.5 V , V O =+ 12 V | dv O dv I is different for positive and negative values of V M sin1000 t . Thus, the gain is different for positive and negative signal excursions and the output will always be a distorted sine wave. This is not a useful choice of bias point for the amplifier. (b) For V B = 1.1 V V O 2 V and dv O dv I = 0. The gain is zero for this bias point. Thus this is also not a useful choice of bias point for the amplifier. 10.13 a () For V B = 0.8 V V O 3 V A v = dv O dv I v I = 0.8 V = 4 2 0.7 0.9 =− 10 A v = 20 dB A V = 180 o V M 0.100 V for linear operation b For V B = 0.2 V V O 14 V A v = dv O dv I v I = 0.8 V = 0 The output signal will be distorted regardless of the value of V M .
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This note was uploaded on 03/05/2008 for the course EE 307 taught by Professor Braun during the Spring '08 term at Cal Poly.

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Chapter10 - CHAPTER 10 10.1 A/C temperature Automobile...

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