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transistor Dc analysis

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AMS012

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Voltage across EB is 0.65V (PNP). I through 10k across EB is 65mA. Writing the KCL at node E gives (9-Ve)/10k =65mA + (β+1) Ib. At B gives 65mA + Ib = (Vb - Vc)/10k. At C gives (Vb-Vc)/10k + β Ib = Vc/10k. And one more equation (9-Ve)/10k = Vc/10k and Ve -0.65 =Vb. Solving these equations will get you the answers.
 

keith1200rs

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If you recognise it as a Vbe multiplier, ignoring the base current Vcb = Vbe = 0.65V. Different ratios of the base-emitter and base-collector resistors is used to give different multipliers. E.g. 10k from base to emitter and 20k from base to collector will give Vcb = 2xVbe.

Keith.
 

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