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Turns Ratio

The current i...

Question

The current in an enhancement mode NMOS transistor biased in saturation mode was measured to be $1 m A$ at a drain-source voltage of 5V. When the drain to source voltage was increased to $6 V$ while keeping gate-source voltage same, the drain current increased to $1.02 m A$ . Assume the drain to source saturation voltages is much smaller than the applied drain –source voltage, The channel length modulation parameter $λ$ (in $V^{- 1})$ is
0.02

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Solution

The correct option is A 0.02
$l a m b d a = \frac{Δ I_{D}}{I_{D} Δ V_{D S}} = \frac{0.02 m A}{1 m A \times 1 V} = 0.02$

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