Since we know what the gain is to be, let us assume that $C_o$ will be 100 fF . Thus, to get a $G B$ of $500 \mathrm{MHz}, R_1$ must be $3.2 \mathrm{k} \Omega$ and $R_2=32 \mathrm{k} \Omega$. Therefore, $R_3$ must be less than $300 \Omega . R_3$ is designed by $\Delta V / I$, where $\Delta V$ is the saturation voltage of M1-M4. Therefore, we can write



$$

R_3=\frac{\Delta V}{I}=\left(\frac{2}{I K^{\prime}(W / L)}\right)^{1 / 2}=300 \Omega \rightarrow 22.2 \times 10^{-6}=K^{\prime} I \frac{W}{L}

$$



or the product of $I$ and $(W / L)$ must be equal to 0.202 . At this point we have a problem because if $W / L$ is small to minimize $C_o$, the current will be too high. If we select $W / L=200 \mu \mathrm{~m} / 1 \mu \mathrm{~m}$ we will get a current of 1 mA . However, using this $W / L$ for M4 and M6 will give a value of $C_o$ that is greater than 100 fF . The solution to this problem is to select the $W / L$ of 200 for M1, M3, M5, and M7 and a smaller $W / L$ for M2, M4, M6, and M8. Let us choose a $W / L= 20 \mu \mathrm{~m} / 1 \mu \mathrm{~m}$ for M2, M4, M6, and M8, which gives a current in these transistors of $100 \mu \mathrm{~A}$. However, now we are multiplying the $R_2 / R_1$ ratio by $1 / 11$ according to Eq. (7.2-10). Thus, we will select $R_2 110$ times $R_1$ or $352 \mathrm{k} \Omega$.



Now select a $W / L$ for M12 of $20 \mu \mathrm{~m} / 1 \mu \mathrm{~m}$, which will now permit us to calculate $C_o$. We will assume zero bias on all voltage-dependent capacitors. Furthermore, we will assume the diffusion area as $2 \mu \mathrm{~m}$ times the $W . C_o$ can be written as



$$

C_o=C_{g d 4}+C_{b d 4}+C_{g d 6}+C_{b d 6}+C_{g s 12}

$$

The information required to calculate these capacitors is found from Table 3.2-1. The various capacitors are



$$

\begin{aligned}

C_{g d 4}= & C_{g d 6}=\mathrm{CGDO} \times 10 \mu \mathrm{~m}=\left(220 \times 10^{-12}\right)\left(20 \times 10^{-6}\right)=4.4 \mathrm{fF} \\

C_{b d 4}= & \mathrm{CJ} \times \mathrm{AD}_4+\mathrm{CJSW} \times \mathrm{PD}_4 \\

= & \left(770 \times 10^{-6}\right)\left(20 \times 10^{-12}\right)+\left(380 \times 10^{-12}\right)\left(44 \times 10^{-6}\right)=15.4 \mathrm{fF} \\

& +16.7 \mathrm{fF}=32.1 \mathrm{fF} \\

C_{b d 6}= & \left(560 \times 10^{-6}\right)\left(20 \times 10^{-12}\right)+\left(350 \times 10^{-12}\right)\left(44 \times 10^{-6}\right)=26.6 \mathrm{fF} \\

C_{g s 12}= & \left(220 \times 10^{-12}\right)\left(20 \times 10^{-6}\right)+(0.67)\left(20 \times 10^{-6} \cdot 10^{-6} \cdot 24.7 \times 10^{-4}\right)=37.3 \mathrm{fF}

\end{aligned}

$$

Therefore,



$$

C_o=4.4 \mathrm{fF}+32.1 \mathrm{fF}+4.4 \mathrm{fF}+26.6 \mathrm{fF}+37.3 \mathrm{fF}=105 \mathrm{fF}

$$





Note that if we had not reduced the $W / L$ of M2, M4, M6, and M8 then $C_o$ would have easily exceeded 100 fF . Since 105 fF is close to our original guess of 100 fF , let us keep the values of $R_1$ and $R_2$. If this value was significantly different, then we would adjust the values of $R_1$ and $R_2$ so that the $G B$ is 500 MHz . One must also check to make sure that the input pole is greater than 500 MHz (see Problem 7.2-6).



The design can be completed by assuming that $I_{\text {BIAS }}=100 \mu \mathrm{~A}$ and that the current in M9 through M12 is $100 \mu \mathrm{~A}$. Thus, $W_{13} / L_{13}=W_{14} / L_{14}=20 \mu \mathrm{~m} / 1 \mu \mathrm{~m}$ and $W_9 / L_9$ through $W_{12} / L_{12}$ are $20 \mu \mathrm{~m} / 1 \mu \mathrm{~m}$. Figure 7.2-13 shows the simulated frequency response of this design. The -3 dB frequency is seen to be close to 38 MHz and the $G B$ approximately 300 MHz . The gain loss of -2 dB is due to the source-follower loss and the fact that $R_1$ is larger than $3.2 \mathrm{k} \Omega$ because of $R_3$. A second pole occurs slightly above 1 GHz . To get these results, it was necessary to bias the input at -1.7 Vdc using $\pm 3 \mathrm{~V}$ power supplies. The difference between the desired -3 dB frequency and the simulated is probably due to the fact that $R_1$ is larger than $3.2 \mathrm{k} \Omega$ and that the actual capacitance may be slightly larger than that calculated in this example. The gain is more than doubled ( 26.4 dB ) and the new -3 dB frequency is 32 MHz . The new unity-gain bandwidth is 630 MHz ! This illustrates the inherent ability of current feedback to increase the value of $G B$ as the closed-loop gain increases.