An n-p-n transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is 100 Ω and the output load resistance is 10 k Ω. The common emitter current gain β is: (a) 104 (b) 102 (c) 6 × 102 (d) 60

Answer: (b). 60 dB power gain: 10log ₁₀ Ap = 60 ⇒ Ap = 10⁶ For CE amplifier Ap = β² (Ro)/(Ri) = β² × (10⁴)/(100) = 100 β² β² = 10⁴ ⇒ β = 10² So the answer is (b) 10².

An n-p-n transistor operates as a common emitter amplifier, with a… | JEE Main 2019 (10 Apr, Shift 1)

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Junction Transistor

An n-p-n 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 k $\Omega$ . The common emitter current gain $\beta$ is:

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Answer: (b)

60 dB power gain: $10\log_{10} A_p = 60 \Rightarrow A_p = 10^6$
For CE amplifier $A_p = \beta^2 \dfrac{R_o}{R_i} = \beta^2 \times \dfrac{10^4}{100} = 100\,\beta^2$
$\beta^2 = 10^4 \Rightarrow \beta = 10^2$
So the answer is (b) $10^2$ .

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