Given constrains:
(a) Transistor $M_2$ must operate in the active region for values of $V_{\text {OOT }}$ to within 0.2 V of ground
(b) The output current must be $50 \mu \mathrm{~A}$
(c) The output current must change less than 1 percent for a change in output voltage of 1 V

$$
\begin{aligned}
& k^{\prime}=\mu_n C_{o x} \\
& k^{\prime}=\mu_n\left(\frac{\varepsilon_{o x}}{t_{o x}}\right) \\
& k^{\prime}=\frac{450 \times 3.9 \times 8.86 \times 10^{-14}}{80 \times 10^{-8}} \\
& k^{\prime}=194.366 \mu \mathrm{~A} / \mathrm{V}^2
\end{aligned}
$$


From the given constraints we can write

$$
I_{\text {OOT }}=50 \times 10^{-6}\left(1+\lambda \Delta V_{D S}\right)
$$


Considering the given condition $M_1$ and $M_2$ are identical, we have to minimize the total device area within the given constraints
From condition (c) we can write

$$
\lambda \Delta V_{D S} \leq 0.01
$$


Here given $\Delta V_{L S}=1 \mathrm{~V}$

$$
\begin{aligned}
& \lambda \leq \frac{0.01}{1} \\
& \lambda \leq 0.01 \mathrm{~V}^{-1}
\end{aligned}
$$


Drain current in active region is

$$
I_D=\frac{1}{2} k^{\prime} \frac{W}{L_{\text {eff }}}\left(V_{G S}-V_t\right)^2
$$


From condition (1) $\Rightarrow V_{\text {OTT }}=V_{D S 2}=V_{G S}-V_t=0.2 \mathrm{~V}$
From condition (2) $\Rightarrow I_D=50 \mu \mathrm{~A}$

$$
\begin{aligned}
& 50 \times 10^{-6}=\frac{1}{2}\left(194.366 \times 10^{-6}\right)\left(\frac{W}{L_{\text {eff }}}\right)(0.2)^2 \\
& \frac{W}{L_{\text {eff }}}=12.862
\end{aligned}
$$


We derived that

$$
\begin{aligned}
& \lambda \leq 0.01 \mathrm{~V}^{-1} \\
& \lambda \leq \frac{1}{100} \mathrm{~V}^{-1}
\end{aligned}
$$


We know that $\frac{d X_d}{d V_{D S}}=0.02 \mu \mathrm{~m} / \mathrm{V}$
We know that

$$
\begin{aligned}
& \lambda=\frac{1}{V_A}=\frac{1}{L_{e f f}}\left(\frac{d X_d}{d V_{D S}}\right) \\
& \frac{1}{L_{e f f}}\left(\frac{d X_d}{d V_{D B}}\right) \leq \frac{1}{100} \\
& \frac{1}{L_{e f f}} \leq \frac{1}{100 \times 0.02 \times 10^{-6}} \\
& \frac{1}{L_{e f f}} \leq \frac{1}{2 \times 10^{-6}} \\
& L_{e f f} \geq 2 \times 10^{-6} \\
& L_{e f f} \geq 2 \mu \mathrm{~m}
\end{aligned}
$$


We know that

$$
\begin{aligned}
& L_{\text {eff }}=L_{\text {draven }}-2 L_d-X_d \\
& \text { Here } X_d=0 \\
& L_{\text {drawn }}=L_{\text {eff }}+2 L_d \\
& L_{\text {drawn }}=2 \times 10^{-6}+2\left(0.09 \times 10^{-6}\right) \\
& L_{\text {drawn }}=2.18 \times 10^{-6} \\
& L_{\text {drawn }}=2.18 \mu \mathrm{~m}
\end{aligned}
$$


Device area $=$ total gate area

$$
\begin{aligned}
& W=\left(2.18 \times 10^{-6}\right)(12.862) \\
& W=28.039 \times 10^{-6} \\
& W=28.04 u \mathrm{~m}
\end{aligned}
$$