An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB . The input circuit resistance is $100 \Omega$ and the output load resistance is $10 \mathrm{k} \Omega$. the common emitter current gain $\beta$ is:

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Semiconductors

An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB . The input circuit resistance is

100Ω

and the output load resistance is

10 k Ω

. the common emitter current gain

β

is:

Answer: (B)

A power gain of

60

dB means

60 = 10 lo g_{10} \frac{P}{P _{0}}

, so

\frac{P}{P _{0}} = 1 0^{6}

. Written through the current gain, the power gain is

β^{2} \frac{R _{o u t}}{R _{in}} = β^{2} \frac{1 0 ^{4}}{100} = 100 β^{2}

. Setting

100 β^{2} = 1 0^{6}

gives

β^{2} = 1 0^{4}

, so

β = 100

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option