The feedforward transconductances $g_{m f 0}$ and $g_{m f 1}$ will be used to move the zeros to well beyond the unity-gain frequency. To simplify the design equations, let $g_{m f 0}=g_{m 0}$ and $g_{m f 1}=g_{m 1}$, based on (9.73)-(9.76) and the assumption that $C_0$ and $C_1$ are small compared to $C_{m 1}$ and $C_{m 2}$.

When $g_{m f 0}=g_{m f 1}=0$, the coefficients $a_i$ of the denominator of the transfer function are given by (9.57). With nonzero $g_{m f 0}$ and $g_{m f 1}$, however, the coefficients of $s$ and $s^2$ in the denominator of the transfer function change and are given by (9.75). From (9.75c), $b_3=a_3$. Also, as noted in the text following (9.76), the term added to $a_1$ in (9.75a) is small compared to $a_1$, so $b_1 \approx a_1$ and $p_1$ is given by (9.59). Hence, poles $p_2$ and $p_3$ are changed due to the added term that includes $g_{m f 1}$ in $b_2$ in (9.75b). Assuming $C_1 \ll C_{m 1}$, (9.75b) reduces to

$$
b_2 \approx a_2+g_{m f 1} R_0 R_1 R_2 C_{m 1} C_{m 2}
$$


Substituting the approximate expression for $a_2$ in (9.62) and using $g_{m f 1}=g_{m 1}$, this equation becomes

$$
b_2 \approx g_{m 2} R_0 R_1 R_2 C_{m 1} C_{m 2}
$$


Following the analysis from (9.60) to (9.67), we find

$$
\begin{aligned}
& p_2 \approx-\frac{b_1}{b_2} \approx-\frac{g_{m 1}}{C_{m 1}} \\
& p_3 \approx \frac{b_1}{b_3} \frac{1}{p_2} \approx-\frac{g_{m 2}}{C_2}
\end{aligned}
$$

To satisfy $\left|p_2\right| \ll\left|p_3\right|$, let $\left|p_3\right|=10\left|p_2\right|$. Substituting (9.77) in this equality and rearranging yields

$$
C_{m 1}=10 \frac{g_{m 1}}{g_{m 2}} C_2
$$


To ensure that $C_{m 1}$ is not much larger than $C_2=5 \mathrm{pF}$ we need $g_{m 1} / g_{m 2} \ll 1$ in (9.78). Here, we chose $g_{m 1} / g_{m 2}=0.2$. Substituting this value into (9.78) gives

$$
C_{m 1}=10(0.2)(5 \mathrm{pF})=10 \mathrm{pF}
$$


With widely spaced poles, placing $\left|p_2\right|$ at the unity-gain frequency gives a $45^{\circ}$ phase margin. Since $\mid$ gain $\mid \times$ frequency is constant for frequencies between $\left|p_1\right|$ and $\left|p_2\right|$, we can write

$$
\left|a_0\right| \cdot\left|p_1\right|=1 \cdot\left|p_2\right|
$$

where

$$
\left|a_0\right|=g_{m 0} R_0 g_{m 1} R_1 g_{m 2} R_2
$$

is the low-frequency gain. Substitution of (9.59), (9.77a), and (9.80) into (9.79) gives

$$
\frac{g_{m 0}}{C_{m 2}}=\frac{g_{m 1}}{C_{m 1}}
$$


If the first two gain stages are made identical to reduce the circuit-design effort, $g_{m 0}=g_{m 1}$, and the last equation reduces to

$$
C_{m 2}=C_{m 1}=10 \mathrm{pF}
$$


Now the transconductances can be found from the low-frequency gain requirement and (9.80),

$$
\left|a_0\right|=g_{m 0} R_0 g_{m 1} R_1 g_{m 2} R_2=\frac{g_{m 1}^3}{0.2}(5 \mathrm{k} \Omega)^3=20,000=86 \mathrm{~dB}
$$

since $g_{m 0}=g_{m 1}=0.2 g_{m 2}$ has been selected. Solving gives $g_{m 1}=g_{m 0}=g_{m f 1}=g_{m f 0}= 3.2 \mathrm{~mA} / \mathrm{V}$ and $g_{m 2}=g_{m 1} / 0.2=16 \mathrm{~mA} / \mathrm{V}$.

SPICE simulation of this op amp gives a dc gain of 86.3 dB and a phase margin of 52 degrees with a unity-gain frequency of 40 MHz . These values are close enough to the specifications to illustrate the usefulness of the calculations. The pole locations are $\left|p_1\right| / 2 \pi= 2.3 \mathrm{kHz},\left|p_2\right| / 2 \pi=59 \mathrm{MHz}$, and $\left|p_3\right| / 2 \pi=464 \mathrm{MHz}$. The zero locations are complex with a magnitude much larger than the unity-gain frequency, at $z_{1,2} / 2 \pi=-345 \mathrm{MHz} \pm j 1.58 \mathrm{GHz}$. Running simulations with slight changes to the compensation capacitors, we find that using $C_{m 1}=10.4 \mathrm{pF}$ and $C_{m 2}=8.3 \mathrm{pF}$ gives a phase margin of 47 degrees with a unity-gain frequency of 45 MHz .