Given
(a)

$$
\begin{aligned}
& I_{\mathrm{IN}}=100 \mu \mathrm{~A} \\
& I_{\text {out }}=10 \mu \mathrm{~A}
\end{aligned}
$$

(b)

$$
V_{o v 1}=0.2 \mathrm{~V}
$$

(c)
$M_2$ must operate in active region if the voltage from the drain of $M_2$ to ground is atleast 0.2 V
(d)

$$
\begin{aligned}
& R_o=50 \mathrm{M} \Omega \\
& L_{\text {dram }}=1 \mu \mathrm{~m}
\end{aligned}
$$


Ignoring the body effect and $X_d=L_d=0$
We know that $V_{\text {ov1 }}=\sqrt{\frac{2 I_{D 1}}{k^t\left(\frac{W}{L}\right)_1}}$
From above equation $I_{\mathrm{IN}}=I_{D 1}$

$$
V_{o v 1}=\sqrt{\frac{2 I_{\mathrm{NN}}}{k^{\prime}\left(\frac{W}{L}\right)_1}}
$$


Squaring on both sides of above equation, we get:

$$
V_{\text {ov1 }}^2=\frac{2 I_{\mathrm{WN}}}{k^{\prime}\left(\frac{W}{L}\right)_1}
$$

$$
\left(\frac{W}{L}\right)_1=\frac{2 I_{\mathrm{nN}}}{k^{\prime} V_{\mathrm{ovl}}^2}
$$


The capacitance per unit area of the $80 \AA$ gate oxide is

$$
\begin{aligned}
C_{a x} & =\frac{\varepsilon_{a x}}{t_{a x}} \\
C_{a x} & =\frac{3.9 \times 8.86 \times 10^{-14} \mathrm{~F} / \mathrm{cm}}{80 \times 10^{-3} \mathrm{~cm}} \\
C_{a x} & =4.32 \times 10^{-7} \frac{\mathrm{~F}}{\mathrm{~cm}^2}
\end{aligned}
$$


We have $\mu_n=450 \mathrm{~cm}^2 / \mathrm{V}-\mathrm{s}$

$$
\begin{aligned}
& \mu_n C_{o x}=450 \times 4.32 \times 10^{-7} \\
& \mu_n C_{o x}=194 \times 10^{-6} \mathrm{~A} / \mathrm{V}^2 \\
& k^{\prime}=\mu_n C_{o x}=194 \mu \mathrm{~A} / \mathrm{V}^2
\end{aligned}
$$


Substituting these values in equation (1), we get:

$$
\begin{aligned}
& \left(\frac{W}{L}\right)_1=\frac{2 \times 100 \mu}{194 \mu \times 0.2^2} \\
& \therefore\left(\frac{W}{L}\right)_1=25.8
\end{aligned}
$$


Writing KVL around the gate-source loop gives

$$
V_{G S 1}-V_{G S 2}-I_{\text {out }} R_2=0
$$

If we ignore the body effect, the threshold components of the gate source voltages cancel and equation (2) simplifies to

$$
I_{\text {our }} R_2+V_{\text {ov } 2}-V_{\text {ov } 1}=0
$$


If the transistors operate in strong inversion and $V_A \rightarrow \infty$,

$$
I_{\text {out }} R_2+\sqrt{\frac{2 I_{\text {out }}}{k^t\left(\frac{W}{L}\right)_2}}-V_{\text {ovl }}=0
$$


This quadratic equation can be solved for $I_{\text {out }}$

$$
\sqrt{I_{\mathrm{our}}}=-\frac{-\sqrt{\frac{2}{k^{\prime}\left(\frac{W}{L}\right)_2}} \pm \sqrt{\frac{2}{k^{\prime}\left(\frac{W}{L}\right)_2}+4 R_2 V_{o v 1}}}{2 R_2}
$$


Where $V_{\text {ov } 1}=\sqrt{\frac{2 I_{\mathrm{DN}}}{k^{\prime}\left(\frac{W}{L}\right)_1}}$
When $M_2$ operates in the active region, which means that $V_{G S 2}>V_t$.

$$
\sqrt{I_{\mathrm{OUT}}}=\sqrt{\frac{k^{\prime}\left(\frac{W}{L}\right)_2}{2}}\left(V_{G S 2}-V_t\right)
$$


So $\sqrt{I_{\text {our }}}>0$ and the potential solution where the second term in the numerator of (3) is subtracted from the first, cannot occur in practice. Therefore $$
\sqrt{I_{\text {out }}}=\frac{-\sqrt{\frac{2}{k^{\prime}\left(\frac{W}{L}\right)_2}}+\sqrt{\frac{2}{k^{\prime}\left(\frac{W}{L}\right)_2}+4 R_2 V_{\text {ov1 }}}}{2 R_2}
$$


Multiply both sides of above equation by the term $\sqrt{2 k^{\prime}\left(\frac{W}{L}\right)_2}$, we get:

$$
\sqrt{2 I_{\text {our }} k^{\prime}\left(\frac{W}{L}\right)_2}=\frac{-\sqrt{\frac{2}{k^{\prime}\left(\frac{W}{L}\right)_2}}+\sqrt{\frac{2}{k^{\prime}\left(\frac{W}{L}\right)_2}+4 R_2 V_{\text {ovl }}}}{2 R_2} \sqrt{2 k^{\prime}\left(\frac{W}{L}\right)_2}
$$


We know that $g_{m 2}=\sqrt{2 I_{\text {our }} k^{\prime}\left(\frac{W}{L}\right)_2}$

$$
g_{m 2} R_2=-1+\sqrt{1+2 k^{\prime}\left(\frac{W}{L}\right)_2 R_2 V_{o v 1}}
$$


We know that

$$
\begin{aligned}
& R_o=r_{o 2}\left(1+g_{m 2} R_2\right) \\
& g_{m 2} R_2=\frac{R_o}{r_{o 2}}-1
\end{aligned}
$$

We have $V_A=L_{\text {eff }}\left(\frac{d X_d}{d V_{D S}}\right)^{-1}$

$$
\begin{aligned}
& V_A=1 \mu \times(0.02 \mu)^{-1} \\
& V_A=50 \mathrm{~V} \\
& r_{o 2}=\frac{V_A}{I_{\mathrm{OUT}}} \\
& r_{o 2}=\frac{50}{10 \times 10^{-6}} \\
& r_{o 2}=5 \times 10^6 \\
& r_{o 2}=5 \mathrm{M} \Omega
\end{aligned}
$$


Substituting this value in equation (5), we get:

$$
\begin{aligned}
& g_{m 2} R_2=\frac{50 \mathrm{M}}{5 \mathrm{M}}-1 \\
& g_{m 2} R_2=10-1 \\
& g_{m 2} R_2=9
\end{aligned}
$$

$\begin{aligned} & \left(\frac{W}{L}\right)_2=\frac{99}{2 \times 16.38 \times 10^3 \times 0.2 \times 194 \times 10^{-6}} \\ & \left(\frac{W}{L}\right)_2=\frac{99}{1.27} \\ & \therefore\left(\frac{W}{L}\right)_2=77.9\end{aligned}$