Consider the transistor circuit shown in the figure below .

Assume the transistor is biased in saturation region with $g_{m} = 5 m A / V .$
Then the value of input resistance $R_{i n}$ is equal to

250 Ω

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728 Ω

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500 Ω

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200 Ω

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Solution

The correct option is D $200 Ω$

Applying KVL in loop (1), we get,
$V_{i n} + V_{g s} = 0$
$V_{i n} = - V_{g s}$
now $i_{i n} = g_{m} V_{g s}$
$∴ R_{i n} = \frac{V_{i n}}{i_{i n}} = \frac{V_{g s}}{g_{m} V_{g s}} = \frac{1}{g_{m}} = \frac{1}{5 \times 10^{- 3}} = 200 Ω$

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T h e t r a n s i s t o r i s b i a s e d s u c h t h a t t h e v a l u e o f g_{m} = 2 m A / V, t h e n t h e v a l u e o f o u t p u t r e s i s t a n c e R_{0} i s e q u a l t o . . . K Ω . [A s s u m e v a l u e o f λ = 0.1 V^{- 1}]

| V_{B E} |

The transistor is biased such that g_{m} = 1 mA/V. Then,

R_{i n}

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k Ω

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