Question

A MOSFET is operated in saturation region and channel length modulation is present. Then the drain to source conductane $g_{ds}$ in terms of channel modulation parameter $\left(\lambda \right)$ is

• A. $g_{ds}=\frac{I_D}{1+\lambda V_{DS}}$
• B. $g_{ds}=\frac{I_D}{\frac{1}{\lambda }+V_{DS}}$
• C. $g_{ds}=\frac{I_D}{\frac{2}{\lambda }+V_{DS}}$
• D. $g_{ds}=\frac{2I_D}{\frac{1}{\lambda }+V_{DS}}$

Answer 1

The correct option is B $g_{ds}=\frac{I_D}{\frac{1}{\lambda }+V_{DS}}$

$\text{Given MOSFET is operated in saturation region and channel length modulation is present.}$
$\therefore \text{Drain~current},I_D={\mu }_n{C}_{ox}\frac{W}{2L}(V_{GS}-V_r{)}^2[1+\lambda V_{DS}]...(i)$
Drain to source conductance.
$g_{ds}=\frac{\alpha I_D}{\alpha V_{DS}}={\mu }_n{C}_{ox}\frac{W}{2L}(V_{GS}-V_T{)}^2(\lambda )$
From equation (i), we can write.
$\frac{I_D}{1+\lambda V_{DS}}={\mu }_n{C}_{ox}\frac{W}{2L}(V_{GS}-V_T{)}^2$
We can re-write equation (ii) as,
$g_{ds}=\frac{I_D}{1+\lambda V_{DS}}\lambda$
$g_{ds}=\frac{I_D}{\frac{1}{\lambda }+V_{DS}}$