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For a BJT bia...

Question

For a BJT biased in active region the transconductance $(g_{m}) = 40 m S . C_{B E} = 100 p F$ and $C_{C B} = 5 p F$ . The gain bandwidth product will be

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Solution

$f_{τ} = \frac{g_{m}}{2 π (C_{B E} + C_{C B})} = \frac{40 m S}{2 \times 3.14 (100 + 5) \times 10^{- 12}}$
$= 60 M H z$

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