1.) $\mathrm{p}_2=2 \mathrm{~GB} \Rightarrow \mathrm{~g}_{\mathrm{m} 6} / \mathrm{C}_{\mathrm{L}}=2 \mathrm{~g}_{\mathrm{m} 1} / \mathrm{C}_{\mathrm{c}}$ and $\mathrm{z}=10 \mathrm{~GB} \Rightarrow \mathrm{~g}_{\mathrm{m} 6}=10 \mathrm{~g}_{\mathrm{m} 1} . \therefore \mathrm{C}_{\mathrm{c}}=\mathrm{C}_{\mathrm{L}} / 5=2 \mathrm{pF}$
2.) $\mathrm{I}=\mathrm{C}_{\mathrm{c}} \cdot \mathrm{SR}=\left(2 \times 10^{-12}\right) \cdot 10^7=20 \mu \mathrm{~A} \quad \therefore \quad \mathrm{I}=20 \mu \mathrm{~A}$
3.) $\mathrm{GB}=\mathrm{g}_{\mathrm{m} 1} / \mathrm{C}_{\mathrm{c}} \Rightarrow \mathrm{g}_{\mathrm{m} 1}=20 \pi \times 10^6 .2 \times 10^{-12}=40 \pi \times 10^{-6}=125.67 \mu \mathrm{~S}$

$$
\frac{\mathrm{W}_1}{\mathrm{~L}_1}=\frac{\mathrm{W}_2}{\mathrm{~L}_2}=\frac{\mathrm{g}_{\mathrm{m} 1} 2^2}{2 \mathrm{~K}_{\mathrm{N}}(\mathrm{I} / 2)}=\frac{\left(125.67 \times 10^{-6}\right)^2}{2 \cdot 24 \times 10^{-6} \cdot 10 \times 10^{-6}}=32.9 \Rightarrow \mathrm{~W} 1=\mathrm{W}_2=33 \mu \mathrm{~m}
$$

4.) $\mathrm{V}_{\mathrm{ic}}(\min )=\mathrm{V}_{\mathrm{DS} 5}($ sat. $)+\mathrm{V}_{\mathrm{GS} 1}(10 \mu \mathrm{~A})=1 \mathrm{~V} \rightarrow \mathrm{~V}_{\mathrm{DS} 5}($ sat. $)=1-\sqrt{\frac{2 \cdot 10}{24 \cdot 33}}-0.75=0.0908$

$$
\mathrm{V}_{\mathrm{DS} 5}(\mathrm{sat})=0.0908=\sqrt{\frac{2 \cdot \mathrm{I}}{\mathrm{~K}_{\mathrm{N}} \mathrm{~S}_5}} \rightarrow \mathrm{~W}_5=\frac{2 \cdot 20}{24 \cdot(0.0908)^2}=201.9 \mu \mathrm{~m} \quad \mathrm{~W}_5=202 \mu \mathrm{~m}
$$

5.) $\mathrm{V}_{\mathrm{ic}}(\max )=\mathrm{V}_{\mathrm{DD}}-\mathrm{V}_{\mathrm{SD} 11}(\mathrm{sat})+\mathrm{V}_{\mathrm{TN}}=1.5-\mathrm{V}_{\mathrm{SD} 11}(\mathrm{sat})+0.75=2 \mathrm{~V} \rightarrow \mathrm{~V}_{\mathrm{SD} 11}(\mathrm{sat})=0.25 \mathrm{~V}$

$$
\mathrm{V}_{\mathrm{SD} 11}(\mathrm{sat}) \leq \sqrt{\frac{2 \cdot 1.5 \mathrm{I}}{\mathrm{~K}_{\mathrm{P}} \cdot \mathrm{~S}_{11}}} \rightarrow \mathrm{~S}_{11}=\mathrm{W}_{11} \geq \frac{2 \cdot 30}{(0.25)^2 \cdot 8}=120 \rightarrow \underline{\underline{\mathrm{~W}}}_{11}=\underline{\underline{\mathrm{W}_{12}}} \underline{\underline{120 \mu \mathrm{~m}}}
$$

6.) Choose $S_3\left(S_4\right)$ by satisfying $V_{i c}(\max )$ specification then check mirror pole.

$$
\begin{aligned}
& \mathrm{V}_{\mathrm{ic}}(\max ) \geq \mathrm{V}_{\mathrm{GS} 3}(20 \mu \mathrm{~A})+\mathrm{V}_{\mathrm{TN}} \rightarrow \mathrm{~V}_{\mathrm{GS} 3}(20 \mu \mathrm{~A})=1.25 \mathrm{~V} \geq \sqrt{\frac{2 \cdot \mathrm{I}}{\mathrm{~K}_{\mathrm{N}} \cdot \mathrm{~S}_3}}+0.75 \mathrm{~V} \\
& \mathrm{~S}_3=\mathrm{S}_4=\frac{2 \cdot 20}{(0.5)^2 \cdot 24}=6.67 \Rightarrow \mathrm{~W}_3=\mathrm{W}_4=7 \mu \mathrm{~m}
\end{aligned}
$$

7.) Check mirror pole ( $\mathrm{p}_3=\mathrm{gm}_3 / \mathrm{C}_{\text {Mirror }}$ ).

$$
\mathrm{p}_3=\frac{\mathrm{g}_{\mathrm{m} 3}}{\mathrm{C}_{\text {Mirror }}}=\frac{\mathrm{g}_{\mathrm{m} 3}}{2 \cdot 0.667 \cdot \mathrm{~W}_3 \cdot \mathrm{~L}_3 \cdot \mathrm{C}_{\mathrm{ox}}}=\frac{\sqrt{2 \cdot 24 \cdot 6.67 \cdot 20} \times 10^{-6}}{2 \cdot 0.667 \cdot 6.67 \cdot 0.5 \times 10^{-15}}=17.98 \times 10^9
$$

which is much greater than $10 \mathrm{~GB}\left(0.0628 \times 10^9\right)$. Therefore, $\mathrm{W}_3$ and $\mathrm{W}_4$ are OK .
8.) $\mathrm{g}_{\mathrm{m} 6}=10 \mathrm{~g}_{\mathrm{m} 1}=1256.7 \mu \mathrm{~S}$
a.) $\mathrm{g}_{\mathrm{m} 6}=\sqrt{2 \mathrm{~K}_{\mathrm{N}} \mathrm{S}_6 10 \mathrm{I}} \Rightarrow \mathrm{W}_6=164.5 \mu \mathrm{~m}$
b.) $\mathrm{V}_{\text {out }}(\min )=0.5 \Rightarrow \mathrm{~V}_{\mathrm{DS} 6}(\mathrm{sat})=0.5=\sqrt{\frac{2 \cdot 10 \mathrm{I}}{\mathrm{K}_{\mathrm{N}} \mathrm{S}_6}} \Rightarrow \mathrm{~W}_6=66.67 \mu \mathrm{~m}$

Therefore, use $\mathrm{W}_6=165 \mu \mathrm{~m}$
Note: For proper mirroring, $S_4=\frac{I_4}{I_6} S_6=8.25 \mu \mathrm{~m}$ which is close enough to $7 \mu \mathrm{~m}$.
9.) Use the $\mathrm{V}_{\text {out }}$ (max) specification to design $\mathrm{W}_7$.

$$
\begin{aligned}
& \mathrm{V}_{\text {out }}(\max )=0.25 \mathrm{~V} \geq \mathrm{V}_{\mathrm{DS} 7}(\mathrm{sat})=\sqrt{\frac{2 \cdot 200 \mu \mathrm{~A}}{8 \times 10^{-6} \cdot \mathrm{~S}_7}} \\
& \therefore \mathrm{~S}_7 \geq \frac{400 \mu \mathrm{~A}}{8 \times 10^{-6}(0.25)^2} \Rightarrow \mathrm{~W}_7=800 \mu \mathrm{~m}
\end{aligned}
$$

10.) Now to achieve the proper currents from the current source I gives,

$$
\mathrm{S}_9=\mathrm{S}_{10}=\frac{\mathrm{S}_7}{10}=80 \rightarrow \mathrm{~W}_9=\mathrm{W}_{10}=80 \mu \mathrm{~m}
$$

and

$$
\mathrm{S}_{11}=\mathrm{S}_{12}=\frac{1.5 \cdot \mathrm{~S}_7}{10}=120 \rightarrow \mathrm{~W}_{11}=\mathrm{W}_{12}=120 \mu \mathrm{~m} . \text { We saw in step } 5 \text { that } \mathrm{W}_{11}
$$

and $\mathrm{W}_{12}$ had to be greater than $120 \mu \mathrm{~m}$ to satisfy $\mathrm{V}_{\mathrm{ic}}(\max ) . \therefore \quad \mathrm{W}_{11}=\mathrm{W}_{12}=120 \mu \mathrm{~m}$