Question

Consider the circuit shown in the figure below:


For the transistor T1, the value of
${\mu }_n{C}_{ox}=50mA/V^{2}and{V}_t=1V.\text{Then value of}\left(\frac{W}{L}\right)$ for the transistor is
(Assuming, there is no channel length modulation)

• A. 0.005
• B. 0.04
• C. 0.02
• D. 0.01

Answer 1

The correct option is B 0.04

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

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