Question

Consider the circuit shown in the figure below:

The MOSFET is biased in saturation region having the minimum value of $V_{DD}$ for which the MOSFET will remain in saturation region is _____ V
$V_T=2.5Vand\left(\frac{{\mu }_n{C}_{ox}W}{L}\right)=0.5mA/V$

• A. 2

Answer 1

The correct option is A 2

For the transistor

$V_s=V_B=V_A$

due to virtual ground,

thus, $V_s=0V$

Hence, $I_D=\frac{0-(-10)}{10\times {10}^5}=1mA$

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

$\therefore V_{GS}-V_T=\sqrt{\frac{I_D}{\frac{{\mu }_n{C}_{ox}W}{2L}}}$

$V_{GS}-V_T=\sqrt{\frac{1\times {10}^{-3}}{\frac{0.5\times {10}^{-3}}{2}}}$

$V_{GS}-V_T=2V$

For the MOSFET to be in saturation region

$V_{DS}\ge V_{GS}-V_T$

$\therefore$ at the edge of saturation

$V_{DS}=V_{GS}-V_T=2V$

$\because V_s=0$

$\therefore V_D=V_G-V_T$

$\Rightarrow V_{DD}=2V$