Question

Consider the circuit shown in the figure below:

For the transistor $T_1$, the value of ${\mu }_n{C}_{ox}=50\frac{mA}{V^2}and{V}_t=1V$.

Then the value of $\left(\frac{W}{L}\right)$ for the transistor is
(Assuming, there is no channel length modulation)

• A. 0.04

Answer 1

The correct option is A 0.04

In the transistor $V_{GS}=V_{DS}$

Since, the gate and drain terminals are shorted,

The transistor will always be in saturation mode.

Thus, $I_D=\frac{{\mu }_n{C}_{ox}}{2}\left(\frac{W}{L}\right)(V_{GS}-V_t{)}^2$

now, $I_D=\frac{V_0}{R}=\frac{3}{3}\times {10}^{-3}=1mA$

thus, $1\times {10}^{-3}=\frac{50\times {10}^{-3}}{2}\left(\frac{W}{L}\right)\times (2-1{)}^2$

$\left(\frac{W}{L}\right)=\frac{1}{25}=0.04$