SOLUTION

Following the procedure outlined in Table 8.2-3, we calculate $I_6$ and $I_7$ as



$$

I_6=I_7=\frac{5 \times 10^{-12} \cdot 4}{50 \times 10^{-9}}=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=\frac{(0.2 \mathrm{pF})(4 \mathrm{~V})}{50 \mathrm{~ns}}=16 \mu \mathrm{~A} \rightarrow I_5=20 \mu \mathrm{~A}

$$





Step 5 gives $V_{S G 3}$ as 1.2 V , which results in



$$

\frac{W_3}{L_3}=\frac{W_4}{L_4}=\frac{20}{50(1.2-0.7)^2}=1.6 \rightarrow \frac{W_3}{L_3}=\frac{W_4}{L_4}=2

$$

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



$$

g_{m 1}=\frac{4000 \cdot 0.09 \cdot 10}{44.44}=81 \mu \mathrm{~S}

$$





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



$$

\frac{W_3}{L_3}=\frac{W_4}{L_4}=\frac{(81)^2}{110 \cdot 40}=1.49 \rightarrow \frac{W_1}{L_1}=\frac{W_2}{L_2}=2

$$

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}+0.4 \mathrm{fF}+119.5 \mathrm{fF}+20.4 \mathrm{fF}+15.3 \mathrm{fF} \\

& =156.5 \mathrm{fF}

\end{aligned}

$$



which is less than what was guessed.

Finally, the $W / L$ value of M5 is found by finding $V_{G S 1}$ as 1.00 V , which gives $V_{D S 5}(\mathrm{sat})=$ 0.25 V . This gives



$$

\frac{W_5}{L_5}=\frac{2 \cdot 20}{(0.25)^2 \cdot 110}=5.8 \approx 6

$$





As in the previous example, M5 and M7 cannot be connected gate-gate and source-source and a scheme like Fig. 8.2-8 must be used. The $W$ values are summarized below assuming that all channel lengths are $1 \mu \mathrm{~m}$.



$$

\begin{array}{lll}

W_1=W_2=2 \mu \mathrm{~m} & W_3=W_4=4 \mu \mathrm{~m} & W_5=6 \mu \mathrm{~m} \\

W_6=64 \mu \mathrm{~m} & W_7=29 \mu \mathrm{~m} &

\end{array}

$$