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Ignoring early effect, if C1 is the total capacitance tied to the emitter, what is the input pole of a simple C.B. stage?(a) 1/gm * C1(b) 2/gm * C1(c) gm * C1(d) gm * 2C1This question was posed to me during an interview for a job.My query is from Effect of Various Capacitors on Frequency Response topic in section Transistor Frequency Response of Analog Circuits

Answer» RIGHT choice is (a) 1/gm * C1

Best EXPLANATION: The resistance looking into the emitter of the B.J.T. is 1/gm. The CAPACITANCE CONNECTED to the input node is C1 (as mentioned). The inverse product of these two provides us the input pole of the C.B. stage.


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