Ignoring early effect, if R1 is the total resistance connected to the base and R2 is the total resistance connected at the collector, what could be the approximate input pole of a simple C.E. stage?(a) 1 / [R1 * (Cµ(2+gm*R2) + Cπ)](b) 1 / [R1 * (Cµ(1+2*gm*R2) + Cπ)](c) 1 / [R1 * (Cµ(1+gm*R2) + Cπ)](d) 1 / [R1 * (Cµ(1-gm*R2) + Cπ)]The question was asked in exam.My doubt is from Effect of Various Capacitors on Frequency Response topic in chapter 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 base and R2 is the total resistance connected at the collector, what could be the approximate input pole of a simple C.E. stage?(a) 1 / [R1 * (Cµ(2+gmR2) + Cπ)](b) 1 / [R1 (Cµ(1+2gmR2) + Cπ)](c) 1 / [R1 * (Cµ(1+gmR2) + Cπ)](d) 1 / [R1 (Cµ(1-gm*R2) + Cπ)]The question was asked in exam.My doubt is from Effect of Various Capacitors on Frequency Response topic in chapter Transistor Frequency Response of Analog Circuits
Answer» Correct answer is (C) 1 / [R1 * (Cµ(1+gmR2) + Cπ)] To EXPLAIN I would say: The INPUT pole can be approximately calculated by OBSERVING the input node. The input node is the node where the base of the B.J.T. is connected to the input voltage. The product of total resistance and capacitance connected at that particular node is R1 Cin and Cin is Cµ(1+gmR2) + Cπ- the inverse of this product gives us the input pole. Thus the correct OPTION is 1 / [R1 (Cµ(1+gm*R2) + Cπ)].

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