In a common emitter configuration with suitable bias, it is given that RL is the load resistance and RBE is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by: β is current gain, IB, IC and IE are respectively base, collector and emitter currents. (a) βRLRBE, (Δ IC)/(Δ IB), β²RLRBE (b) βRLRBE, (Δ IE)/(Δ IB), β²RLRBE (c) β²RLRBE, (Δ IC)/(Δ IE), β²RLRBE (d) β²RLRBE, (Δ IC)/(Δ IB), βRLRBE

Answer: (a). Current gain β = (Δ IC)/(Δ IB) Voltage gain AV = βRLRBE Power gain AP = β AV = β²RLRBE So the answer is (a).

In a common emitter configuration with suitable bias, it is given… | JEE Main 2018 (15 Apr, Shift 1)

Q90JEE Main 2018 · 15 APR · SHIFT 1Easy

Junction Transistor

In a common emitter configuration with suitable bias, it is given that $R_L$ is the load resistance and $R_{BE}$ is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by:

$\beta$ is current gain, $\mathrm{I}_B, \mathrm{I}_C$ and $\mathrm{I}_E$ are respectively base, collector and emitter currents.

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Answer: (a)

Current gain $\beta = \frac{\Delta I_C}{\Delta I_B}$
Voltage gain $A_V = \beta\frac{R_L}{R_{BE}}$
Power gain $A_P = \beta A_V = \beta^2\frac{R_L}{R_{BE}}$
So the answer is (a).

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Junction Transistor question from JEE Main 2018 (15 Apr, Shift 1)

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