An npn transistor operates as a common emitter amplifier with a power gain of \(10^{6}\). The input circuit resistance is \(100 \Omega\) and the output load resistance is \(10 \mathrm{k} \Omega\). The common emitter current gain ' \(\beta\) ' will be ⋯------. (Round off to the Nearest Integer)

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Semiconductors

An npn transistor operates as a common emitter amplifier with a power gain of

1 0^{6}

. The input circuit resistance is

100Ω

and the output load resistance is

10 k Ω

. The common emitter current gain '

β

' will be ⋯------. (Round off to the Nearest Integer)

Answer: 100

A common emitter amplifier has current gain

β

and voltage gain

β \frac{R _{L}}{R _{in}}

, so its power gain is the product

A_{P} = β^{2} \frac{R _{L}}{R _{in}}

Substituting

A_{P} = 1 0^{6}

R_{L} = 10 k Ω = 1 0^{4} Ω

and

R_{in} = 100 Ω

1 0^{6} = β^{2} \times \frac{1 0 ^{4}}{100} = 100 β^{2}

β^{2} = 1 0^{4}

and the common emitter current gain is

β = 100

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer