Consider the MOS transistor circuit shown in the figure below:

Assuming both the MOS transistor are identical with $μ_{n} C_{o x} = 2.5 μ A / V^{2}, V_{T h} = 1 V a n d \frac{W}{L} = 3$
then the value of current $I_{D}$ is equal to _____ $μ A$

8.44

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Solution

The correct option is A 8.44
$V_{D} = V_{G}$

$V_{D} - V_{s} = V_{G} - V_{S}$

$V_{D S} = V_{G S}$

$V_{D S} \geq V_{G S} - V_{T}$

$∴$ MOSFETs are in the saturation region

$I_{D 1} = I_{D 2}$

$\frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{G S 1} - V_{T h}) = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} {(V_{G S 2} - V_{T h})}^{2}$
$V_{G S 1} = V_{G S 2}$ $B y K V L, - 5 + V_{G S 1} + V_{G S 2} = 0$

$V_{G S 1} + V_{G S 2} = 5$

$V_{G S 1} = V_{G S 2} = 2.5$

$I_{D} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} {(V_{G S 1} - V_{T h})}^{2}$

$= \frac{1}{2} \times 2.5 \times 10^{- 6} \times 3 {(2.5 - 1)}^{2}$

$I_{D} = 8.437 μ A = 8.44 μ A$

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