1.) $S R=100 \mathrm{~V} / \mu \mathrm{s} \quad \rightarrow \quad I_{\text {out }}=C_L \cdot S R=5 \times 10^{-12} \cdot 10^8=500 \mu \mathrm{~A} \quad \rightarrow I_3=50 \mu \mathrm{~A}$
2.) $V_{\text {icm }}{ }^{+}=0.75 \mathrm{~V} \rightarrow V_{S G 4}=V_{D D}-V_{\text {icm }}{ }^{+}+V_{T N}=1.0-0.75+0.5=0.75 \mathrm{~V}$

$$
\begin{aligned}
& V_{O N 3}=0.75-0.5=0.25 \mathrm{~V} \quad \rightarrow \frac{W_4}{L_4}=\frac{2 I_4}{K_N\left(V_{O N 4}\right)^2}=\frac{50}{70(0.25)^2}=11.43=12 \\
& \therefore \underline{\underline{W}}_{\underline{4}}=\underline{\underline{W}}_{\underline{5}}=12 \mu \mathrm{~m} \quad \rightarrow \quad \underline{\underline{W}}_{\underline{6}}=\underline{\underline{W}}_{\underline{7}}=120 \mu \mathrm{~m}
\end{aligned}
$$

3.) $G B=200 \mathrm{MHz}$ or $G B=400 \pi \times 10^6 \mathrm{rads} / \mathrm{sec}$.

$$
\begin{aligned}
& G B=\frac{g_{m 1}}{C_L} 10 \quad \rightarrow \quad g_{m 1}=\frac{G B \cdot C_L}{10}=\frac{400 \pi \times 10^6 \cdot 5 \times 10^{-12}}{10}=628 \mu \mathrm{~S} \\
& \frac{W_1}{L_1}=\frac{g_{m 1}}{2 K_N I_1}=\frac{628^2}{50 \cdot 300}=26.32=27 \quad \therefore \quad \underline{\underline{W}}_1=W_2=27 \mu \mathrm{~m}
\end{aligned}
$$

4.) $V_{\text {icm }}{ }^{-}=-0.25 \mathrm{~V} \quad \rightarrow \quad V_{D S 3}=V_{\text {icm }}{ }^{-} V_{G S 1}-V_{S S}$

$$
\begin{aligned}
& V_{G S 1}=\sqrt{\frac{2 \cdot 25}{300 \cdot 27}}+0.5=0.5786 \mathrm{~V} \quad \rightarrow V_{D S 3}=-0.25-0.5786+1=0.1714 \mathrm{~V} \\
& \therefore \frac{W_3}{L_3}=\frac{2 I_3}{K_N\left(V_{O N 3}\right)^2}=\frac{2 \cdot 50}{300(0.1714)^2}=11.34=12 \quad \therefore \underline{W}_3=12 \mu \mathrm{~m}
\end{aligned}
$$

$$
\begin{aligned}
& \text { 5.) } V_{\text {out }}{ }^{+}=0.5 \mathrm{~V} \\
& \qquad V_{S D 6}=\sqrt{\frac{2 \cdot I_6}{K_N\left(W_6 / L_6\right)}}=\sqrt{\frac{2 \cdot 250}{70 \cdot 120}}=0.243 \mathrm{~V} \rightarrow V_{S D 8}=0.256 \mathrm{~V} \\
& \therefore \quad \frac{W_8}{L_8}=\frac{2 I_8}{K_P\left(V_{O N 8}\right)^2}=\frac{2 \cdot 250}{70(0.256)^2}=108.99=109 \quad \therefore \underline{W}_8=\underline{\underline{W}} \underline{\underline{9}}=109 \mu \mathrm{~m}
\end{aligned}
$$

6.) $V_{\text {out }}=-0.5 \mathrm{~V} \quad$ Let $V_{D S 10}=V_{D S 12}=0.25 \mathrm{~V}$

$$
\begin{array}{ll} 
& \frac{W_{12}}{L_{12}}=\frac{2 I_{12}}{K_N\left(V_{O N 12}\right)^2}=\frac{2 \cdot 250}{300(0.25)^2}=26.67=27 \\
\therefore & \underline{\underline{W}} 10=W_{11}=W_{12}=W_{13}=27 \mu \mathrm{~m}
\end{array}
$$