hw11_sol - ECE 442 HW11 Solutions 1 Following the method of...

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ECE 442 HW11 Solutions 1. Following the method of analysis with r o as shown in class, we first solve for the gain A vo and input impedance R io of CB without R sig . Referring to Fig. P1 below, we have: Figure P1. v o = - R C ( g m v π + v o - v i r o ) v π = - v i ⇒ - v o R C = - g m v i + v o r o - v i r o v o R C || r o = v i 1 g m || r o v i 1 g m A vo = v o v i g m ( R C || r o ) i i = v i r π + g m v i + v i - v o r o = v i ( 1 r π + g m + 1 r o - A vo r o ) = v i r π || 1 g m || r o || ( r o - A vo ) r o - A vo = - r o g m ( R C || r o ) = - R C + r o g m R C ≈ - r o g m R C R io = v i i i = r π || 1 g m || r o || r o - g m R C 1 g m || r o - g m R C = 1 g m ( 1 + R C r o ) Now we can calculate the gain A v with R sig (noting from Fig. 1 that adding R sig will not affect R i = R io ) : A v = v o v sig = v o v i × v i v sig = A vo × R i R i + R sig g m ( R C || r o ) × 1 g m ( 1 + R C r o ) 1 g m ( 1 + R C r o ) + R sig 1
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A v = g m R C 1 + R C r o + g m R sig g m R C 1 + R C r o To calculate R o = R C || R X , we short the independent source v sig , apply the test voltage v x at the output node, and solve for current i x , as shown in Fig. P1. Writing the KCL equation at the output node: v x + v π r o + g m v π = - v π r π || R sig v π = - v
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This note was uploaded on 02/21/2010 for the course ECE 442 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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hw11_sol - ECE 442 HW11 Solutions 1 Following the method of...

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