Ignoring early effect, if R1 is the total resistance connected to the collector; what is the output pole of a simple C.B. stage?(a) 1/[R1 * (Ccs + Cµ)](b) 1/[R1* (Ccs + 2*Cµ)](c) 1/[R1 * (2*Ccs + Cµ)](d) 1/[R1 * 2*(Ccs + Cµ)]The question was asked during an interview for a job.The origin of the question is Effect of Various Capacitors on Frequency Response topic in division Transistor Frequency Response of Analog Circuits

Ignoring early effect, if R1 is the total resistance connected to the

1.	Ignoring early effect, if R1 is the total resistance connected to the collector; what is the output pole of a simple C.B. stage?(a) 1/[R1 * (Ccs + Cµ)](b) 1/[R1* (Ccs + 2Cµ)](c) 1/[R1 (2Ccs + Cµ)](d) 1/[R1 2*(Ccs + Cµ)]The question was asked during an interview for a job.The origin of the question is Effect of Various Capacitors on Frequency Response topic in division Transistor Frequency Response of Analog Circuits
Answer» Right answer is (a) 1/[R1 * (Ccs + Cµ)] The explanation: The output POLE is calculated, approximately, by the inverse product of the total resistance and the capacitance connected at the output node. We find that the total resistance connected to the output node is R1 while the total capacitance is Ccs + Cµ. In ABSENCE of early EFFECT, 1/[R1 * (Ccs + Cµ)] becomes the output pole.

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