$$
\omega_{-3 \mathrm{~dB}} \cong 1 /\left(\mathrm{r}_{\mathrm{ds}, 1}| | \mathrm{r}_{\mathrm{ds}, 2}\right) \mathrm{C}_{\mathrm{L}}=2 \pi \cdot 5 \cdot 10^6 \mathrm{rad} / \mathrm{sec}
$$


Substituting the fixed $\mathrm{C}_{\mathrm{L}}=10 \mathrm{pF}$ into (4.159) yields,

$$
\left(\mathrm{r}_{\mathrm{ds}, 1} \| \mathrm{r}_{\mathrm{ds}, 2}\right)=3.2 \mathrm{k} \Omega
$$


Note that $\left(\mathrm{r}_{\mathrm{ds}, 1} \| \mathrm{r}_{\mathrm{ds}, 2}\right)=1 / \mathrm{I}_{\mathrm{D}}\left(\lambda_1+\lambda_2\right)$. To minimize $\mathrm{I}_{\mathrm{D}}$ and, hence, power consumption we require large $\lambda_1$ and $\lambda_2$ which in turn demands small $\mathrm{L}_1$ and $\mathrm{L}_2$. Therefore, we use the minimum possible in this CMOS process.

$$
L_1=L_2=0.18 \mu \mathrm{~m}
$$


Using the values for $\lambda \mathrm{L}$ from Table 1.5,

$$
\lambda_1=\lambda_2=(0.08 \mu \mathrm{~m} / \mathrm{V}) / 0.18 \mu \mathrm{~m}=0.44 \mathrm{~V}^{-1}
$$


From this, the required drain current is found.

$$
\mathrm{I}_{\mathrm{D}}=\frac{1}{\mathrm{r}_{\mathrm{ds}} \cdot \lambda} \cong 350 \mu \mathrm{~A}
$$


The result has been rounded to the nearest integer multiple of the bias current, $50 \mu \mathrm{~A}$, in order to simplify the current mirror design. To meet the gain requirement,

$$
\begin{gathered}
\left|A_0\right| \cong g_{m 1}\left(r_{d s, 1}| | r_{d s, 2}\right)=\frac{2 I_D}{V_{e f f, 1}} \cdot \frac{1}{\left(\lambda_1+\lambda_2\right) I_D}=\frac{2}{V_{e f f, 1}} \cdot \frac{1}{\left(\lambda_1+\lambda_2\right)}>20 \\
\Rightarrow V_{e f f, 1}<\frac{2}{20\left(0.44 V^{-1}+0.44 V^{-1}\right)}=114 \mathrm{mV}
\end{gathered}
$$

For some margin, we take $\mathrm{V}_{\text {eff, } 1}=100 \mathrm{mV}$ and the resulting transistor width is

$$
W_1=\frac{2 I_{\mathrm{D} 1} L_1}{\mu_{\mathrm{n}} C_{\mathrm{ox}} V_{\mathrm{eff}, 1}^2}=\frac{2 \cdot 355 \mu \mathrm{~A} \cdot 0.18 \mu \mathrm{~m}}{270 \mu \mathrm{~A} / \mathrm{V}^2(0.1 \mathrm{~V})^2} \cong 47 \mu \mathrm{~m}
$$


The width of $Q_2$ may be chosen the same, $W_2=47 \mu \mathrm{~m}$ and the size of $Q_3$ chosen to provide the correct current ratio in the current mirror formed by $Q_2$ and $Q_3$ :

$$
\begin{aligned}
& \frac{\left(\mathrm{W}_3 / \mathrm{L}_3\right)}{\left(\mathrm{W}_2 / \mathrm{L}_2\right)}=\frac{50 \mu \mathrm{~A}}{350 \mu \mathrm{~A}} \\
\Rightarrow \mathrm{~L}_3= & 0.18 \mu \mathrm{~m} \text { and } \mathrm{W}_3=6.7 \mu \mathrm{~m}
\end{aligned}
$$