${{V}_{th}}={{2}_{{}}}V$

${{R}_{D}}={{4}_{{}}}k$

When ${{I}_{D}}={{1.4}_{{}}}mA$

${{R}_{D}}={{1}_{{}}}k,{{I}_{D}}=??$

Let us assume the MOSFET works on saturation region

Then ${{I}_{D}}=\hbar {{\left( {{V}_{gs}}-{{V}_{th}} \right)}^{2}}\cdots \cdots \cdots \left( i \right)$

Also ${{I}_{D}}=\frac{10-{{V}_{D}}}{{{R}_{D}}}$

$\frac{10-{{V}_{D}}}{{{4}_{{}}}k}={{1.4}_{{}}}m$

${{V}_{D}}={{4.4}_{{}}}V$

Nearly no current flows through the gate hence;

${{V}_{D}}={{V}_{G}}={{4.4}_{{}}}V$

$\left( \therefore {{V}_{GS}}-{{V}_{th}}<{{V}_{DS}} \right)$

$\because$ Saturation

Using equation (i)

$1.4=\hbar {{\left( 4.4-2 \right)}^{2}}$

$\hbar ={{0.2430}_{{}}}{mA}/{{{V}^{2}}}\;\cdots \cdots \cdots \left( ii \right)$

$\therefore$ As

${{I}_{D}}=\frac{10-{{V}_{D}}}{{{R}_{D}}}$

The we can write:

$\frac{10-{{V}_{D}}}{{{1}_{{}}}k}=0.2430{{\left( {{V}_{0}}-2 \right)}^{2}}_{{}}m$

$10-{{V}_{D}}=0.2430{{\left( {{V}_{0}}-2 \right)}^{2}}$

$10-{{V}_{D}}=0.2430{{V}_{D}}^{2}-0.972{{V}_{D}}+0.972$

$0=0.2430{{V}_{D}}^{2}-0.972{{V}_{D}}+{{V}_{D}}-10+0.972$

$0=0.243{{V}_{D}}^{2}+0.028{{V}_{D}}-9.028$

${{V}_{D}}=6.037,-6.15$

(Negative value not allowed for saturation)

$\therefore {{I}_{D}}=\left( 10-6.037 \right)mA\simeq {{3.96}_{{}}}mA$

IC Fabrication Techniques