The slew rate will specify $I . \quad \therefore \quad I=C \cdot S R=10^{-11} \cdot 10^7=10^{-4}=100 \mu \mathrm{~A}$.
Use $G B$ to define $W_1$ and $W_2$.

$$
\begin{aligned}
& G B=\frac{g_{m 1}}{C} \quad \rightarrow \quad g_{m 1}=G B \cdot C=2 \pi \times 10^7 \cdot 10^{-11}=628 \mu \mathrm{~S} \\
& \therefore W_1=\frac{g_{m 1}{ }^2}{2 K_N(0.5 I)}=\frac{(628)^2}{2 \cdot 110 \cdot 50}=35.85 \quad \Rightarrow \quad \underline{W}_1=W_2=36 \mu \mathrm{~m}
\end{aligned}
$$

Design $W_{15}$ to give $V_T+2 V_{O N}$ bias for M 6 and $\mathrm{M} 7 . V_{O N}=0.5 \mathrm{~V}$ will meet the desired maximum output voltage specification. Therefore,

$$
\begin{array}{lr}
V_{S G 15}=V_{O N 15}+\left|V_T\right|=2(0.5 \mathrm{~V})+\left|V_T\right| & \rightarrow V_{O N 15}=1 \mathrm{~V}=\sqrt{\frac{2 I}{K_P W_{15}}} \\
\therefore W_{15}=\frac{2 I}{K_P V_{O N 15}{ }^2}=\frac{2 \cdot 100}{50 \cdot 1^2}=4 \mu \mathrm{~m} & \Rightarrow \quad \underline{\underline{W}} \underline{\underline{15}}=4 \mu \mathrm{~m}
\end{array}
$$

Design $W_3, W_4, W_6$ and $W_7$ to have a saturation voltage of 0.5 V with 1.5 I current.

$$
W_3=W_4=W_6=W_7=\frac{2(1.5 I)}{K_P V_{O N}^2}=\frac{2 \cdot 150}{50 \cdot 0.5^2}=24 \mu \mathrm{~m} \Rightarrow \underline{W}_3=\underline{W}_4=\underline{W}_6=\underline{W}_7=24 \mu \mathrm{~m}
$$

Next design $W_8, W_9, W_{10}$ and $W_{11}$ to meet the minimum output voltage specification. Note that we have not taken advantage of smallest minimum output voltage because a normal cascode current mirror is used which has a minimum voltage across it of $V_T+ 2 V_{O N}$. Therefore, setting $V_T+2 V_{O N}=1 \mathrm{~V}$ gives $V_{O N}=0.15 \mathrm{~V}$. Using worst case current, we choose 1.5I. Therefore,

$$
W_8=W_9=W_{10}=W_{11}=\frac{2(1.5 I)}{K_N V_{O N}^2}=\frac{2 \cdot 150}{110 \cdot 0.15^2}=121 \mu \mathrm{~m} \Rightarrow \underline{\underline{W}}_{\underline{8}}=W_{\underline{9}}=W_{\underline{10}}=W_{\underline{11}}=
$$

$\underline{121 \mu \mathrm{~m}}$

Check the maximum ICM voltage.

$$
V_{i c}(\max )=V_{D D}+V_{S D 3}(\mathrm{sat})+V_{T N}=3 \mathrm{~V}-0.5+0.7=3.2 \mathrm{~V} \text { which exceeds spec. }
$$

Use the minimum ICM voltage to design $W_5$.

$$
\begin{array}{ll} 
& V_{i c}(\mathrm{~min})=V_{S S}+V_{D S 5}(\mathrm{sat})+V_{G S 1}=-3+V_{D S 5}(\mathrm{sat})+\left(\sqrt{\frac{2 \cdot 50}{110 \cdot 36}}+0.7\right)=-1 \mathrm{~V} \\
\therefore & V_{D S 5}(\mathrm{sat})=1.141 \rightarrow W_5=\frac{2 I}{K_N V_{D S 5}(\mathrm{sat})^2}=1.39 \mu \mathrm{~m}=1.4 \mu \mathrm{~m} \\
& \text { Also, let } W_{12}=W_{13}=W_5 \quad \Rightarrow \quad W_{12}=W_{13}=W_5=1.4 \mu \mathrm{~m}
\end{array}
$$

$W_{14}$ is designed as

$$
W_{14}=W_3 \frac{I_{14}}{I_3}=24 \mu \mathrm{~m} \frac{I}{1.5 I}=16 \mu \mathrm{~m} \quad \Rightarrow \quad \underline{W}_{14}=16 \mu \mathrm{~m}
$$