Given

$$
\begin{aligned}
& I_{\mathrm{out}}=0.1 \mu \mathrm{~A} \\
& I_{\mathrm{nN}}=1 \mu \mathrm{~A}
\end{aligned}
$$


From the circuit if the $I_{M V}$ is small and positive, the voltage drop on $R$ is small And $M_1$ operates in the active region. Assume that $M_2$ also operates in the active region.
By applying KVL around the gate source loop we get

$$
V_{G S 1}-I_{\mathbb{N}} R-V_{G S 2}=0
$$


Since the sources of $M_1, M_2$ are connected to gether, the threshold cancel and we
get $V_{o v 2}=V_{o v 1}-I_{\mathbb{N}} R$

Here both the transistors are operate in weak inversion.
We know that drain current under weak inversion is

$$
\begin{aligned}
& I_D=\frac{W}{L} I_t e^{\left(\frac{V_{G S}-V_t}{n V_r}\right)}\left[1-e^{\left(-\frac{V_{D S}}{V_r}\right)}\right] \\
& \text { but } V_{G S 2}-V_t=I_{D V} R \\
& \text { Then } I_{\text {OVT }} \approx \frac{W}{L} I_t e^{\left(\frac{V_{G 2}-V_t}{n V_r}\right)}
\end{aligned}
$$

$$
\begin{gathered}
I_{O O T} \approx I_{\mathbb{N}} e^{\left(-\frac{I_N R}{n V_T}\right)} \\
\frac{I_{O O T}}{I_{M V}} \approx e^{\left(-\frac{I_R R}{n V_r}\right)} \\
\ln \left(\frac{I_{O O T}}{I_{M V}}\right)=\left(-\frac{I_{M N} R}{n V_T}\right)
\end{gathered}
$$


Then the resistance is $R=-\frac{n V_T}{I_{D V}} \ln \left(\frac{I_{D O T}}{I_W}\right)$

$$
R=\frac{n V_T}{I_{W V}} \ln \left(\frac{I_{W V}}{I_{O T S}}\right) \ldots
$$


By substituting the given parameters in the above expression we get

$$
\begin{aligned}
& R=\frac{(1.5)\left(26 \times 10^{-3} \mathrm{~V}\right)}{1 \times 10^{-6} \mathrm{~A}} \times \ln 10 \quad(\because n=1.5) \\
& R=39 \times 10^3 \times 2.30 \\
& R=90 \mathrm{k} \Omega
\end{aligned}
$$


To keep the transistor in week inversion $V_{G S}-V_t<0$.
Input transistor conduct more current.

$$
\begin{aligned}
& \text { i.e. } I_{\mathrm{IN}}=\frac{W}{L} I_t e^{\left[\frac{V_{G S}-V_t}{n V_T}\right]} \\
& \frac{I_{\mathrm{IN}}}{\left(\frac{W}{L}\right) I_t}=e^{\left[\frac{V_{G S}-V_T}{n V_T}\right]} \\
& \ln \left[\frac{I_{\mathrm{IN}}}{\left(\frac{W}{L}\right) I_t}\right]=\frac{V_{G S}-V_t}{n V_T} \\
& n V_T \ln \frac{I_{\mathrm{IN}}}{\frac{W}{L} I_t}=V_{G S 1}-V_t<0 \\
& I_{\mathrm{IN}}<\frac{W}{L} I_t \\
& \frac{W}{L}>\frac{I_{\mathrm{IN}}}{I_t} \ldots .(4) \\
& \frac{W}{L}>\frac{1 \times 10^{-6} \mathrm{~A}}{0.1 \times 10^{-6} \mathrm{~A}} \\
& \frac{W}{L}>10
\end{aligned}
$$