SOLUTION

Following the approach outlined in Table 6.5-1 we obtain the following results:



$$

I_3=S R \cdot C_L=10 \times 10^6 \cdot 10^{-11}=100 \mu \mathrm{~A}

$$





Select $I_4=I_5=125 \mu \mathrm{~A}$.

Next, we see that the value of $0.5\left[V_{D D}-V_{\text {out }}(\min )\right]$ is $0.5 \mathrm{~V} / 2$ or 0.25 V . Thus,



$$

S_4=S_5=\frac{2 \cdot 125 \mu \mathrm{~A}}{25 \mu \mathrm{~A} / \mathrm{V}^2 \cdot(0.25 \mathrm{~V})^2}=\frac{2 \cdot 125 \cdot 16}{225}=160

$$

and assuming worst-case currents in M6 and M7 gives



$$

S_6=S_7=\frac{2 \cdot 125 \mu \mathrm{~A}}{25 \mu \mathrm{~A} / \mathrm{V}^2(0.25 \mathrm{~V})^2}=\frac{2 \cdot 125 \cdot 16}{25}=160

$$





The value of $0.5\left[V_{\text {our }}(\min )-\left|V_{S S}\right|\right]$ is also 0.25 V , which gives the value of $S_8, S_9, S_{10}$, and $S_{11}$ as



$$

S_8=S_9=S_{10}=S_{11}=\frac{2 \cdot I_8}{K_N^{\prime} V_{D S 8}^2}=\frac{2 \cdot 125}{120 \cdot(0.25)^2}=20

$$





In step 5, the value of $G B$ gives $S_1$ and $S_2$ as



$$

S_1=S_2=\frac{G B^2 \cdot C_L^2}{K_N^{\prime} I_3}=\frac{\left(20 \pi \times 10^6\right)^2\left(10^{-11}\right)^2}{120 \times 10^{-6} \cdot 100 \times 10^{-6}}=32.9 \approx 33

$$





The minimum input common-mode voltage defines $S_3$ as



$$

\begin{aligned}

S_3 & =\frac{2 I_3}{K_{N^{\prime}}\left(V_{i n}(\min )-V_{S S}-\sqrt{\frac{I_3}{K_N^{\prime} S_1}}-V_{T 1}\right)^2} \\

& =\frac{200 \times 10^{-6}}{110 \times 10^{-6}\left(-1.5+2.5-\sqrt{\frac{100}{120 \cdot 35.9}}-0.75\right)^2}=14.3 \approx 15

\end{aligned}

$$

We need to check that the values of $S_4$ and $S_5$ are large enough to satisfy the maximum input common-mode voltage. The maximum input common-mode voltage of 2.5 requires



$$

S_4=S_5 \geq \frac{2 I_4}{K_P^{\prime}\left[V_{D D}-V_{i n}(\max )+V_{T 1}\right]}=\frac{2 \cdot 125 \mu \mathrm{~A}}{25 \times 10^{-6} \mu \mathrm{~A} / \mathrm{V}^2[0.5 \mathrm{~V}]^2}=40

$$



which is much less than 160 . In fact, with $S_4=S_5=160$, the maximum input common-mode voltage is 2.75 V .



The power dissipation is found to be



$$

P_{\text {diss }}=2.5 \mathrm{~V}(125 \mu \mathrm{~A}+125 \mu \mathrm{~A}+125 \mu \mathrm{~A})=0.625 \mathrm{~mW}

$$





The small-signal voltage gain requires the following values to evaluate:



$$

S_4, S_5: g_m=\sqrt{2 \cdot 125 \cdot 25 \cdot 160}=1000 \mu \mathrm{~S} \text { and } g_{d s}=125 \times 10^{-6} \cdot 0.08=10 \mu \mathrm{~S}

$$

$$

\begin{aligned}

& S_6, S_7: g_m=\sqrt{2 \cdot 75 \cdot 25 \cdot 160}=774.6 \mu \mathrm{~S} \text { and } g_{d s}=75 \times 10^{-6} \cdot 0.08=6 \mu \mathrm{~S} \\

& S_8, S_8, S_{10}, S_{11}: g_m=\sqrt{2 \cdot 75 \cdot 120 \cdot 20}=600 \mu \mathrm{~S} \text { and } g_{d s}=75 \times 10^{-6} \cdot 0.06=4.5 \mu \mathrm{~S} \\

& S_1, S_2: g_{m I}=\sqrt{2 \cdot 50 \cdot 120 \cdot 33}=629 \mu \mathrm{~S} \text { and } g_{d s}=50 \times 10^{-6}(0.06)=3 \mu \mathrm{~S}

\end{aligned}

$$





Thus,



$$

\begin{aligned}

R_9 & \approx g_{m 9} r_{d s 9} r_{d s 11}=(600 \mu \mathrm{~S})\left(\frac{1}{4.5 \mu \mathrm{~S}}\right)\left(\frac{1}{4.5 \mu \mathrm{~S}}\right)=29.63 \mathrm{M} \Omega \\

R_{\text {out }} & \approx(29.63 \mathrm{M} \Omega) \|(774.6 \mu \mathrm{~S})\left(\frac{1}{6 \mu \mathrm{~S}}\right)\left(\frac{1}{10 \mu \mathrm{~S}+3 \mu \mathrm{~S}}\right)=7.44 \mathrm{M} \Omega \\

k & =\frac{R_9\left(g_{d s 2}+g_{d s}\right)}{g_{m 7} r_{d s 7}}=\frac{7.44 \mathrm{M} \Omega(3 \mu \mathrm{~S}+10 \mu \mathrm{~S})(6 \mu \mathrm{~S})}{774.6 \mu \mathrm{~S}}=0.75

\end{aligned}

$$





The small-signal differential-input voltage gain is



$$

A_{v d}=\left(\frac{2+k}{2+2 k}\right) g_{m l} R_{\text {out }}=\left(\frac{2+0.75}{2+1.5}\right) 0.629 \times 10^{-3} \cdot 7.44 \times 10^6=(0.786)(4680)=3678 \mathrm{~V} / \mathrm{V}

$$





The gain is slightly larger than required by the specifications but this should be okay.