SOLUTION

Following the procedure outlined in Table 8.2-2, we choose $m=1$ to get



$$

\left|p_I\right|=\left|p_{I I}\right|=\frac{10^9}{50 \sqrt{10}}=6.32 \times 10^6 \mathrm{rad} / \mathrm{s}

$$





This gives



$$

I_6=I_7=\frac{6.32 \times 10^6 \cdot 5 \times 10^{-12}}{0.04+0.05}=351 \mu \mathrm{~A} \quad \rightarrow \quad I_6=I_7=400 \mu \mathrm{~A}

$$





Therefore,



$$

\frac{W_6}{L_6}=\frac{2 \cdot 400}{(0.5)^2 \cdot 50}=64 \quad \text { and } \quad \frac{W_7}{L_7}=\frac{2 \cdot 400}{(0.5)^2 \cdot 110}=29

$$





Next, we guess $C_I=0.2 \mathrm{pF}$. This gives $I_5=32 \mu \mathrm{~A}$ and we will increase it to $40 \mu \mathrm{~A}$ for a margin of safety. Step 4 gives $V_{S G 3}$ as 1.2 V , which results in



$$

\frac{W_3}{L_3}=\frac{W_4}{L_4}=\frac{40}{50(1.2-0.7)^2}=3.2 \rightarrow \frac{W_3}{L_3}=\frac{W_4}{L_4}=4

$$





The desired gain is found to be 4000 , which gives an input transconductance of



$$

g_{m 1}=\frac{4000 \cdot 0.09 \cdot 20}{44.44}=162 \mu \mathrm{~S}

$$

This gives the $W / L$ ratios of M1 and M2 as



$$

\frac{W_1}{L_1}=\frac{W_2}{L_2}=\frac{(162)^2}{110 \cdot 40}=5.96 \rightarrow \frac{W_1}{L_1}=\frac{W_2}{L_2}=6

$$





To check the guess for $C_I$ we need to calculate it, which is done as



$$

\begin{aligned}

C_I & =C_{g d 2}+C_{g d 4}+C_{g s 6}+C_{b d 2}+C_{b d 4}=0.9 \mathrm{fF}+1.3 \mathrm{fF}+119.5 \mathrm{fF}+20.4 \mathrm{fF}+36.8 \mathrm{fF} \\

& =178.9 \mathrm{fF}

\end{aligned}

$$



which is less than what was guessed so we will make no changes.

Finally, the $W / L$ value of M5 is found by finding $V_{G S 1}$ as 0.946 V , which gives $V_{D S 5} (\mathrm{sat})=0.304 \mathrm{~V}$. This gives



$$

\frac{W_5}{L_5}=\frac{2 \cdot 40}{(0.304)^2 \cdot 110}=7.87 \approx 8

$$





Obviously, M5 and M7 cannot be connected gate-gate and source-source. The values of $I_5$ and $I_7$ must be derived separately as illustrated in Fig. 8.2-8. The $W$ values are summarized below, assuming that all channel lengths are $1 \mu \mathrm{~m}$.