Ch11s - Chapter 11 Problem Solutions 11.1 (a) ( ) 0.7 3 0.1...

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Unformatted text preview: Chapter 11 Problem Solutions 11.1 (a) ( ) 0.7 3 0.1 23 K 3 1.5 0.05 30 K E E C C R R R R = = = = (b) ( ) ( ) 2 2 2 6 2 6 7 6 CE C C E C v i R R i = + = (c) ( ) 2 2 max 0.7 V cm CB CE v v v = = So ( ) 2 2 0.7 6 76 69.74 A C C i i = = ( ) ( ) ( ) ( ) ( ) max 0.7 3 2 0.06974 max 0.908 V 23 CM CM v v = = ( ) ( ) min 3 V min 2.3 V CM S CM v V v = = 11.2 180, 85 dB 180 17,783 0.01012 d d B d cm cm cm A C M R R A C M RR A A A = = = = = = Assume the common-mode gain is negative. ( ) ( ) ( ) 180 0.01012 180 2sin mV 0.01012 2sin V 0.36sin 0.02024sin d d cm cm d c m v A v A v v v v t t v t t = + = = = Ideal Output: ( ) 0.360sin V v t = Actual Output: ( ) 0.340sin V v t = 11.3 a. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 2 4 4 2 2 4 4 10 2 0.7 1.01 mA 8.5 1.01 1.01 mA 2 2 1 1 1 1 1 1 100 1.01 0.50 mA 101 2 0 0.7 5 4.3 V 5 0.5 2 0.7 4.7 V C C C C CE CE CE CE I I I I I I I V V V V = = = = + + + = = = = = b. ( ) ( ) ( ) ( ) 4 4 4 4 2 4 2 1 2 1 1 For 2.5 V 0.7 2.5 1.8 V 5 1.8 1.6 mA 2 1 1 1 2 2 1.6 3.23 mA 100 3.23 mA 10 2 0.7 2.66 k 3.23 CE C C C C C C C V V I I I I I I I R R = = + = = = + + = = = = = 11.4 a. Neglecting base currents 1 3 1 1 1 1 30 0.7 400 A R 73.25 k 0.4 10 V V 9.3 V 15 9.3 28.5 k 0.2 CE C C C I I R V R R = = = = = = = = b. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 10 100 0.026 13 k 0.2 50 125 k 0.4 We have 100 28.5 62 2 2 13 10 1 2 1 1 100 28.5 1 0.113 2 125 101 13 10 1 13 10 62 20log 0.113 C d d B C cm B B cm dB r r Q R A A r R R A r r R r R A C M RR C = = = = = = = + + = + + + + = = + + + = 54.8 dB dB M RR = c. ( ) ( ) ( ) ( ) ( ) 2 2 13 10 46 k 1 2 1 2 1 13 10 2 101 125 12.6 M 2 id B id icm B icm R r R R R r R r R = + = + = = + + + = + + = 11.5 (a) ( ) max CM CB v V = so that ( ) ( ) ( ) ( ) 0.5 max 5 5 8 2 2 Q CM C I v R = = ( ) max 3 V CM v = (b) ( ) ( ) 2 0.25 0.018 0.08654 mA 2 2 0.026 2 0.08654 8 0.692 V CQ d d m T C C I V V I g V V I R = = = = = = = (c) ( ) ( ) 2 0.25 0.010 0.04808 mA 0.026 2 0.04808 8 0.385 V C I V = = = = 11.6 ( ) ( ) ( ) ( ) 1 4 1 4 1 1 4 1 1 1 2 so 1.2 2 6 0.1 3 0.7 3 53 0.1 3 1 For 1 1 40 0.05 C C C CM C C C C P I I V V I I I I I mA R R k v V V V V R R k + = + = = = = = = + = = = = One-sided output...
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This homework help was uploaded on 02/10/2008 for the course ECE 138 taught by Professor N/a during the Spring '08 term at CSU Fresno.

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Ch11s - Chapter 11 Problem Solutions 11.1 (a) ( ) 0.7 3 0.1...

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