Transistors

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Question

A pnp transistor is used in CE mode in an amplifier circuit. A change of 40 $μ A$ in the base current brings a change of 2 mA in collector current and 0.04 V in base emitter voltage. The input resistance $(R_{i n})$ in $k Ω$ is

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Solution

$G i v e n : Δ I_{B} = 40 μ A, Δ I_{C} = 2 m A, V_{B E} = 0.04 V I n p u t R e s i s \tan c e : R_{i n} = \frac{V_{B E}}{Δ I_{B}} = \frac{0.04}{40 \times 10^{- 6}} = 1000 Ω = 1 k Ω$

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