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Ignoring early effect, if R2 is the total resistance at the collector, what could be the approximate output pole of a simple C.E. stage?(a) 1 / [R2 * (Ccs + Cµ*(1 + 2/gm*R2))](b) 1 / [R2 * (Ccs – Cµ*(1 + 1/gm*R2))](c) 1 / [R2 * (Ccs + Cµ*(1 – 1/gm*R2))](d) 1 / [R2 * (Ccs + Cµ*(1 + 1/gm*R2))]The question was asked during a job interview.I need to ask this question from Effect of Various Capacitors on Frequency Response in section Transistor Frequency Response of Analog Circuits

Answer»

Correct option is (d) 1 / [R2 * (Ccs + Cµ*(1 + 1/gm*R2))]

The best explanation: The output pole can be APPROXIMATELY calculated by OBSERVING the output node. For a C.E. stage, the output node is the node where the Collector of the B.J.T. is connected to the output MEASURING device. The product of TOTAL resistance and CAPACITANCE connected at that particular node is R2 * Cout and Cout is (Ccs + Cµ*(1 + 1/gm*R2). The inverse of this product gives us the output pole. Thus the correct option is 1 / [R2 * (Ccs + Cµ*(1 + 1/gm*R2))].



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