If early effect is included, and R1 is the total resistance connected at the collector. What is the output pole of a simple C.B. stage?(a) 1/[(R1 || ro) * 2(Ccs + Cµ)](b) 1/[(R1 || ro) * (Ccs + Cµ)](c) 1/[(R1 || ro) * (2*Ccs + Cµ)](d) 1/[(R1 || ro) * 2*(Ccs + 2*Cµ)]I had been asked this question in an online interview.Enquiry is from Effect of Various Capacitors on Frequency Response in portion Transistor Frequency Response of Analog Circuits

If early effect is included, and R1 is the total resistance connected

1.	If early effect is included, and R1 is the total resistance connected at the collector. What is the output pole of a simple C.B. stage?(a) 1/[(R1 \|\| ro) * 2(Ccs + Cµ)](b) 1/[(R1 \|\| ro) * (Ccs + Cµ)](c) 1/[(R1 \|\| ro) * (2Ccs + Cµ)](d) 1/[(R1 \|\| ro) 2(Ccs + 2Cµ)]I had been asked this question in an online interview.Enquiry is from Effect of Various Capacitors on Frequency Response in portion Transistor Frequency Response of Analog Circuits
Answer» The correct OPTION is (b) 1/[(R1 \|\| RO) * (Ccs + Cµ)] 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 in parallel with ro, due to early effect, while the total capacitance is C2 ie Ccs + Cµ. Thus, the correct option is 1/[(R1 \|\| ro) * (Ccs + Cµ)].

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