# Ch11s - Chapter 11 Problem Solutions 11.1(a-0.7-3 RE = 0.1...

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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 76 CE C C E C v i R R i = + = (c) ( ) 2 2 max 0 0.7 V cm CB CE v v v = = So ( ) 2 2 0.7 6 76 69.74 A C C i i μ = = ( ) ( ) ( ) ( ) ( ) 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 dB d cm cm cm A C M RR A C M RR A A A = = = = = = Assume the common-mode gain is negative. ( ) ( )( ) 0 0 0 180 0.01012 180 2sin mV 0.01012 2sin V 0.36sin 0.02024sin d d cm cm d cm v A v A v v v v t t v t t ω ω ω ω = + = = = Ideal Output: ( ) 0 0.360sin V v t ω = Actual Output: ( ) 0 0.340sin V v t ω = 11.3 a.

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( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) 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 1 100 101 100 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 β β = = = = + + + ⎞⎛ = ⎟⎜ ⎠⎝ = − − = = − − = 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 1 101 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 β β = = − + = = = + &