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hw10_sol - ECE 442 HW10 Solutions 1 a From the rearranged...

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ECE 442 HW10 Solutions 1. a) From the rearranged small-signal schematic shown in Fig. 1a, we can write, noting v π = - v e : Figure 1. KCL @ emitter node: v e r π || 1 sC π + v e - v sig R sig + g m v e = 0 v e = v sig R sig ( 1 r π || 1 sC π + 1 R sig + g m ) = r π || 1 sC π || R sig || 1 g m R sig v sig KVL @ collector (output) impedance: v out = g m v e · ( R 0 L || 1 sC μ ) = g m R 0 L r π || 1 sC π || R sig || 1 g m R sig v sig A v = v out v sig = g m ( R 0 L || 1 μ ) × r π || 1 sC π || R sig || 1 g m R sig Using the equality R || 1 sC = R 1 + sCR , we can rewrite the voltage gain as: A v = g m R 0 L 1 + μ R 0 L × r π || R sig || 1 gm R sig 1 + π ( r π || R sig || 1 g m ) 1
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From here, we can identify: A vo = g m R 0 L × r π || R sig || 1 g m R sig ω 1 = 1 C π ( r π || R sig || 1 g m ) , ω 2 = 1 C μ R 0 L b) Refer to Fig. 1b. At midband, we have (replacing capacitors with open circuits), knowing that for CB amp R i = r π || 1 g m : A 1 = v π v sig = - R i R i + R sig = - r π || 1 g m r π || 1 g m + R sig × R sig R sig = - r π || 1 g m || R sig R sig A 2 = v out v π = - g m R 0 L A vo = A 1 × A 2 Resistance ’seen’ by capacitor C π , replacing capacitor C μ
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hw10_sol - ECE 442 HW10 Solutions 1 a From the rearranged...

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