A MOSFET carries drain current of 1 mA with $V_{D S} = 0.5$ V in saturation. If the channel length modulation parameter $λ = 0.1 V^{- 1}$ , then the output impedance of the device will be

10.5 k Ω

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1.42 k Ω

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10 k Ω

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1 k Ω

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Solution

The correct option is A $10.5 k Ω$
$I_{D} = k_{n} (V_{G S} - V_{t h})^{2} (1 + λ V_{D S})$
$r_{o} = \frac{Δ V_{D S}}{Δ I_{D}} = \frac{1}{(\partial I_{D} / \partial V_{D S})}$
$\frac{\partial I_{D}}{\partial V_{D S}} = k_{n} (V_{G S} - V_{t h})^{2} λ$
at $V_{G S} = 0.5 I_{D} = 1 m A$
so, $1 m A = k_{n} (V_{G S} - V_{t h})^{2} (1 + 0.05)$
$k_{n} (V_{G S} - V_{t h})^{2} = \frac{1}{1.05} m ℧$
$r_{0} = \frac{1.05}{0.1} k Ω = 10.5 k Ω$

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