An npn transistor operates as a common emitter amplifier with a power gain of 106. The input circuit resistance is 100 Ω and the output load resistance is 10 K Ω. The common emitter current gain ' β ' will be ________. (Round off to the Nearest Integer). Numerical answer type. Work it out, then open the answer.

Answer: 100. Power gain of CE amplifier: Ap = β² (Ro)/(Ri) 10⁶ = β² × (10⁴)/(100) = 100 β² β² = 10⁴ ⇒ β = 100 So β = 100.

An npn transistor operates as a common emitter amplifier with a power… | JEE Main 2021 (18 Mar, Shift 1)

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Junction Transistor

An npn transistor operates as a common emitter amplifier with a power gain of 106. The input circuit resistance is 100 $\Omega$ and the output load resistance is 10 K $\Omega$ . The common emitter current gain ' $\beta$ ' will be ________. (Round off to the Nearest Integer).

Numerical answer type. Work it out, then open the answer.

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Answer: 100

Power gain of CE amplifier: $A_p = \beta^2 \dfrac{R_o}{R_i}$
$10^6 = \beta^2 \times \dfrac{10^4}{100} = 100\,\beta^2$
$\beta^2 = 10^4 \Rightarrow \beta = 100$
So $\beta = 100$ .

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