Since $(\lambda L)_1=(\lambda L)_2$, if we take $L_1=L_2=0.2 \mu$ m we ensure $r_{d s 2}=R_L=r_{d s 1}$ and using (3.20),

$$
\mathrm{r}_{\mathrm{in}} \cong 2 / \mathrm{g}_{\mathrm{m} 1}
$$


Hence, to ensure $\mathrm{r}_{\text {in }}=50 \Omega$,

$$
\mathrm{g}_{\mathrm{m} 1} \cong 2 /(50 \Omega)=40 \mathrm{~mA} / \mathrm{V}
$$


If we take $\mathrm{V}_{\text {eff1 }}=200 \mathrm{mV}$,

$$
I_{D 2}=I_{D 1}=g_{m 1} V_{e f f 1} / 2=(40 m A / V)(200 m V) / 2=4 m A
$$


This requires the current mirror to provide a gain of $4 \mathrm{~mA} / 100 \mu \mathrm{~A}=40$.

$$
(\mathrm{W} / \mathrm{L})_2=40(\mathrm{~W} / \mathrm{L})_3=80 \mu \mathrm{~m} / 0.2 \mu \mathrm{~m}
$$


Finally,

$$
(\mathrm{W} / \mathrm{L})_1=\frac{\mathrm{g}_{\mathrm{m} 1}^2}{2 \mu_{\mathrm{n}} \mathrm{C}_{\mathrm{ox}} \mathrm{I}_{\mathrm{D} 1}}=148 \mu \mathrm{~m} / 0.2 \mu \mathrm{~m}
$$