HW#5 Solution

2 again by kcl at the base we have v π r π v x v π

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Again, by KCL at the base, we have V π r π + V X + V π R B + ( V X + V π ) sC μ = 0 Z out = V X I X = R B + r π + r π R B C μ s (1 + β )(1 + R B C μ s ) . From the transfer function of Z out , we know the zero and pole locations are | w z | = R B + r π r π R B C μ , | w p | = 1 R B C μ , | w z | > | w p | . The DC value is ( R B + r π ) / (1 + β ). The Bode plot of | Z out | is shown in Figure 5. Figure 5 6. Since I C 3 = 0 . 75 I C 1 , we have I C 2 = 0 . 25 I C 1 . The DC voltage gain is A V ≈ - g m 1 g m 2 = - I C 1 I C 2 = - 4 . By Miller’s theorem, we know C in = (1 - A V ) C μ 1 = 5 C μ 1 ; C out = (1 - 1 /A V ) C μ 1 = 1 . 25 C μ 1 . The poles of the circuit are given as follows. | w p,X | = 1 ( R S || r π )( C π 1 + 5 C μ 1 ) | w p,Y | = g m 2 1 . 25 C μ 1 + C CS 1 + C π 2 + C CS 3 + C μ 3 . | w p,out = 1 R L ( C CS 2 + C μ 2 ) . Since | A V | is larger, the Miller’s effect is more significant. 3

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