Question

Consider the circuit shown in the figure below.

The value of $V_T=2Vand\frac{{\mu }_n{C}_{ox}W}{2L}=0.2\mu A/V^2$. Then the value of power dissipated is equal to _____$mW.$

• A. 2.35

Answer 1

The correct option is A 2.35
$V_s=I_D{R}_s=0.623\times {10}^{-3}\times 10\times {10}^3=6.23V$

$V_{DS}=10-6.23=3.77V$

$\therefore$ Power dissipated, $P=V_{DS}\times I_D=3.77\times 0.623\times {10}^{-3}=2.35mW$