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## Q. 3.7

Show that if $\left|V_{GS}\right|\ll \left|V_{P}\right|$ then drain current can be approximated as $I_{DS}=I_{DSS}-g_{mo} V_{GS}$

## Verified Solution

We know that
$I_{DS}=I_{DSS}[1-\frac{V_{GS}}{V_P} ]^2$
Simplifying, we get$I_{DS}=I_{DSS}[1-2\frac{V_{GS}}{V_P}+(\frac{V_{GS}}{V_P} )^2]^2$
Neglecting the square term, we get
$I_{DS}=I_{DSS}-\frac{2I_{DSS}V_{GS}}{V_P}$
Putting $g_{mo}=-\frac{2I_{DSS}V_{GS}}{V_P}$
$I_{DS}=I_{DSS}-g_{mo}V_{GS}$ Ans.