For a transistor amplifier in common emitter configuration having load impedance of $1 \text{ k}\Omega$ ($h_{fe} = 50$ and $h_{oe} = 25\,\mu\mathrm{S} = 25\times10^{-6}\,\Omega^{-1}$) the current gain is

JEE Main · 2004 · 24 Apr 2004 · Offline · Semiconductors

For a transistor amplifier in common emitter configuration having load impedance of

1 k Ω

(

h_{f e} = 50

and

h_{oe} = 25 μ S = 25 \times 1 0^{- 6} Ω^{- 1}

) the current gain is

Answer: (D)

In the common-emitter mode the current gain is

A_{i} = \frac{- h _{f e}}{1 + h _{oe} R _{L}}

. With

h_{f e} = 50

h_{oe} = 25 \times 1 0^{- 6}

and

R_{L} = 1000 Ω

, the denominator is

1 + 25 \times 1 0^{- 6} \times 1000 = 1.025

, so

A_{i} = \frac{- 50}{1.025} \approx - 48.78

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option