Question

Consider the following circuit shown below, it is given
$V_1=1Vand{\mu }_n{C}_{ox}=120mA/V^2,\lambda =0.$ If the drain current is 0.1 mA, then value of ratio of $\frac{(W/L{)}_2}{(W/L{)}_1}$ will be (Also given $V_{GS1}=2V,V_{GS2}=1.5V)$.

• A. 1
• B. 3
• C. 4
• D. 2

Answer 1

The correct option is C 4

Given, $V_{GS1}=2V$,

As the drain current is same for both transistor when both transistors are operating in saturation

So

$\frac{1}{2}{\mu }_n{C}_{ox}{\left(\frac{W}{L}\right)}_1(V_{GS1}-V_T{)}^2=\frac{1}{2}{\mu }_n{C}_{ox}{\left(\frac{W}{L}\right)}_2(V_{GS2}-V_T{)}^2$

$\frac{(W/L{)}_2}{(W/L{)}_1}=\frac{(V_{GS1}-V_T{)}^2}{(V_{GS2}-V_T{)}^2}$

$=\frac{(2-1{)}^2}{(1.5-1{)}^2}=\frac{1}{(0.5{)}^2}=4$