Question

The transistor circuit as shown in figure below has $V_T=1Vand{\mu }_n{C}_{ox}\left(\frac{\omega }{L}\right)=2mA/V^2.$ The drain to source voltage $\left(V_{DS}\right)$ will be _____V.

• A. 2.925
• B. 2.86
• C. 5.16
• D. 2.7375

Answer 1

The correct option is D 2.7375

Let us assume that transistor is operating in saturation region.

$I_D={\mu }_n{C}_{ox}\left(\frac{\omega }{2L}\right)(V_{GS}-V_T{)}^2$

$V_G=1.8V,$ $V_s=0.5I_D$
$V_{GS}=1.8-0.5I_D$
$\Rightarrow$ $V_{GS}-V_T=(0.8-0.5I_D)$
$I_D=1\times (0.8-0.5I_D{)}^2$
$I_D=0.64+0.25I_D^2-0.8I_D$

$\frac{1}{4}{I}_D^2-1.8I_D+0.64=0$
On solving, $I_D=0.3750mA,6.824mA$
$I_D=6.824\left(notpossible\right),So,I_D=0.375mA$
$V_S=0.5I_D=0.1875V$
$V_D=V_{DD}-I_D{R}_D=3.3-1\times 0.3750=2.925V$
$V_{DS}=2.925-0.1875=2.7375V$

For saturation region,
$V_{DS}\ge (V_{GS}-V_T)$
$2.7375>(1.8-0.1875-1)$ True

So, given transistor is operating in saturation region.