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For a BJT bia...

Question

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

A

60 MHz

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B

1 MHz

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C

100 MHz

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D

1 MHz

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Solution

The correct option is A 60 MHz
$f_{T} = \frac{g_{m}}{2 π (C_{B E} + C_{C B})} = \frac{40 m S}{2 \times 3.14 (100 + 5) \times 10^{- 12}}$

$= 60 M H z$

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