Cwi = 120 pF Cwo = 10 pF Cbe = 40 pF Cbc = 10pF Cce = 14pF 15 V 1. 5 kΩ 4.7kΩ B = 100 1 μF - 4 V 5.6 n F HH 330 Ω om Vi

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answerhappygod
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Cwi = 120 pF Cwo = 10 pF Cbe = 40 pF Cbc = 10pF Cce = 14pF 15 V 1. 5 kΩ 4.7kΩ B = 100 1 μF - 4 V 5.6 n F HH 330 Ω om Vi

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Please solve all the questions with correct and neat answer. Iwill upvote for good work. Thank you
Cwi = 120 pF Cwo = 10 pF Cbe = 40 pF Cbc = 10pF Cce = 14pF 15 V 1. 5 kΩ 4.7kΩ B = 100 1 μF - 4 V 5.6 n F HH 330 Ω om Vi + Vs Vo 1 kQ
(a) By referring to Figure Q2(a), analyze; (i) (ii) (iii) (iv) (v) a.c. resistance diode, re a.c. equivalent circuit. midband gain, Avmid. low frequency response, fLE, and fic. high frequency response, fHi and fHo.
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