SOLUTION

1. To meet the slew rate, $I_5 \geq 50 \mu \mathrm{~A}$. For maximum $P_{\text {diss }}, I_5 \leq 200 \mu \mathrm{~A}$.

2. An $f_{-3 \mathrm{~dB}}$ of 100 kHz implies that $R_{\text {out }} \leq 318 \mathrm{k} \Omega . R_{\text {out }}$ can be expressed as $$

R_{\mathrm{out}}=\frac{2}{\left(\lambda_N+\lambda_P\right) I_5} \leq 318 \mathrm{k} \Omega

$$



which gives $I_5 \geq 70 \mu \mathrm{~A}$. Therefore, we will pick $I_5=100 \mu \mathrm{~A}$.

3. The maximum input common-mode voltage gives



$$

V_{S G 3}=V_{D D}-V_{I C}(\max )+V_{T N 1}=2.5-2+0.7=1.2 \mathrm{~V}

$$





Therefore, we can write



$$

V_{S G 3}=1.2 \mathrm{~V}=\sqrt{\frac{2 \cdot 50 \mu \mathrm{~A}}{\left(50 \mu \mathrm{~A} / \mathrm{V}^2\right)\left(W_3 / L_3\right)}}+0.7

$$





Solving for $W_3 / L_3$ gives



$$

\frac{W_3}{L_3}=\frac{W_4}{L_4}=\frac{2}{(0.5)^2}=8

$$

4. The small-signal gain specification gives



$$

100 \mathrm{~V} / \mathrm{V}=g_{m 1} R_{\mathrm{out}}=\frac{g_{m 1}}{g_{d s 2}+g_{d s 4}}=\frac{\sqrt{\left(2 \cdot 110 \mu \mathrm{~A} / \mathrm{V}^2\right)\left(W_1 / L_1\right)}}{(0.04+0.05) \sqrt{50 \mu \mathrm{~A}}}=23.31 \sqrt{W_1 / L_1}

$$





Solving for $W_1 / L_1$ gives



$$

\frac{W_1}{L_1}=\frac{W_2}{L_2}=18.4

$$



5. Using the minimum input common-mode voltage gives



$$

\begin{aligned}

V_{D S}(\mathrm{sat}) & =V_{I C}(\mathrm{~min})-V_{S S}-V_{G S 1}=-1.5+2.5-\sqrt{\frac{2 \cdot 50 \mu \mathrm{~A}}{110 \mu \mathrm{~A} / \mathrm{V}^2(18.4)}}-0.7 \\

& =0.3-0.222-0.0777 \mathrm{~V}

\end{aligned}

$$





This value of $V_{D S 5}$ (sat) gives a $W_5 / L_5$ of



$$

\frac{W_5}{L_5}=\frac{2 I_5}{K_N^{\prime} V_{D S}(\mathrm{sat})^2}=301

$$

We should probably increase $W_1 / L_1$ to reduce $V_{G S 1}$, to allow for a variation in $V_{T N}$. Therefore, select $W_1 / L_1\left(W_2 / L_2\right)=40$, which gives $W_5 / L_5=82$. The small-signal gain will increase to $147 \mathrm{~V} / \mathrm{V}$, which should be okay.