The first step is to calculate the minimum value of the compensation capacitor $C_c$, which is



$$

C_c>(2.2 / 10)(10 \mathrm{pF})=2.2 \mathrm{pF}

$$





Choose $C_c$ as 3 pF . Using the slew-rate specification and $C_c$ calculate $I_5$.



$$

I_5=\left(3 \times 10^{-12}\right)\left(10 \times 10^6\right)=30 \mu \mathrm{~A}

$$





Next calculate $(W / L)_3$ using ICMR requirements. Using Eq. (6.3-12) we have



$$

(W / L)_3=\frac{30 \times 10^{-6}}{\left(50 \times 10^{-6}\right)[2.5-2-0.85+0.55]^2}=15

$$





Therefore,



$$

(W / L)_3=(W / L)_4=15

$$





Now we can check the value of the mirror pole, $p_3$, to make sure that it is in fact greater than $10 G B$. Assume the $C_{\mathrm{ox}}=2.47 \mathrm{fF} / \mu \mathrm{m}^2$. The mirror pole can be found as



$$

p_3 \approx \frac{-g_{m 3}}{2 C_{g s 3}}=\frac{-\sqrt{2 K_p^{\prime} S_3 I_3}}{2(0.667) W_3 L_3 C_{\mathrm{ox}}}=2.81 \times 10^9 \mathrm{rad} / \mathrm{s}

$$



or 448 MHz . Thus, $p_3$ and $z_3$ are not of concern in this design because $p_3 \gg 10 G B$.

The next step in the design is to calculate $g_{m 1}$ using Eq. (6.3-13).



$$

g_{m 1}=\left(5 \times 10^6\right)(2 \pi)\left(3 \times 10^{-12}\right)=94.25 \mu \mathrm{~S}

$$





Therefore, $(W / L)_1$ is



$$

(W / L)_1=(W / L)_2=\frac{g_{m 1}^2}{2 K_N^{\prime} I_1}=\frac{(94.25)^2}{2 \cdot 110 \cdot 15}=2.79 \approx 3.0

$$





Next, calculate $V_{D S 5}$ using Eq. (6.3-15).



$$

V_{D S 5}=(-1)-(-2.5)-\sqrt{\frac{30 \times 10^{-6}}{110 \times 10^{-6} \cdot 3}}-0.85=0.35 \mathrm{~V}

$$





Using $V_{D S 5}$ calculate $(W / L)_5$ from Eq. (6.3-16).



$$

(W / L)_5=\frac{2\left(30 \times 10^{-6}\right)}{\left(110 \times 10^{-6}\right)(0.35)^2}=4.49 \approx 4.5

$$





From Eq. (6.2-20), we know that



$$

g_{m 6} \geq 10 g_{m 1} \geq 942.5 \mu \mathrm{~S}

$$

Assuming that $g_{m 6}=942.5 \mu \mathrm{~S}$ and calculating $g_{m 4}$ as $150 \mu \mathrm{~S}$, we use Eq. (6.3-18) to get



$$

S_6=S_4 \frac{g_{m 6}}{g_{m 4}}=15 \cdot \frac{942.5}{150}=94.25 \approx 94

$$





Calculate $I_6$ using Eq. (6.3-19).



$$

I_6=\frac{\left(942.5 \times 10^{-6}\right)^2}{(2)\left(50 \times 10^{-6}\right)(94)}=94.5 \mu \mathrm{~A} \approx 95 \mu \mathrm{~A}

$$





Designing $S_6$ by using Eq. (6.3-20) gives $S_6 \approx 15$. Since the $W / L$ ratio of 94 from above is greater, the maximum output voltage specification will be met.



Finally, calculate $(W / L)_7$ using Eq. (6.3-21).



$$

(W / L)_7=4.5\left(\frac{95 \times 10^{-6}}{30 \times 10^{-6}}\right)=14.25 \approx 14

$$





Let us check the $V_{\text {out }}(\min )$ specification although the $W / L$ of M7 is large enough that this is probably not necessary. The value of $V_{\text {out }}(\mathrm{min})$ is



$$

V_{\min }(\text { out })=V_{D S 7}(\mathrm{sat})=\sqrt{\frac{2 \cdot 95}{110 \cdot 14}}=0.351 \mathrm{~V}

$$



which is less than required. At this point, the first-cut design is complete.

The power dissipation can be calculated as



$$

P_{\text {diss }}=5 \mathrm{~V} \cdot(30 \mu \mathrm{~A}+95 \mu \mathrm{~A})=0.625 \mathrm{~mW}

$$





Now check to see that the gain specification has been met.



$$

A_v=\frac{(2)\left(92.45 \times 10^{-6}\right)\left(942.5 \times 10^{-6}\right)}{30 \times 10^{-6}(0.04+0.05) 95 \times 10^{-6}(0.04+0.05)}=7696 \mathrm{~V} / \mathrm{V}

$$



which meets specifications. If more gain were desired, an easy way to achieve it would be to increase the $W$ and $L$ values by a factor of 2 , which because of the decreased value of $\lambda$ would increase the gain by a factor of 20 .