Design of a Folded-Cascode Op Amp

Follow the procedure of Table 6.5-1 to design the folded-cascode op amp of Fig. 6.5-9(b) when the slew rate is $10 \mathrm{~V} / \mu \mathrm{s}$, the load capacitor is 10 pF , the maximum and minimum output voltages are 2 V and 0.5 V for a 2.5 V power supply, the $G B$ is 10 MHz , the minimum input common-mode voltage is +1 V , and the maximum input common-mode voltage is 2.5 V . The differential voltage gain should be greater than $3000 \mathrm{~V} / \mathrm{V}$ and the power dissipation should be less than 5 mW . Use $K_N^{\prime}=120 \mu \mathrm{~A} / \mathrm{V}^2, K_P^{\prime}=25 \mu \mathrm{~A} / \mathrm{V}^2, V_{T N}=\left|V_{T P}\right|=0.5 \mathrm{~V}, \lambda_N= 0.06 \mathrm{~V}^{-1}$, and $\lambda_P=0.08 \mathrm{~V}^{-1}$. Let $L=0.5 \mu \mathrm{~m}$.

\begin{table}

\captionsetup{labelformat=empty}

\caption{Table 6.5-1 Design Approach for the Folded-Cascode Op Amp}

\begin{tabular}{|l|l|l|l|}

\hline Step & Relationship & Design Equation/Constraint & Comments \\

\hline 1 & Slew rate & $I_3=S R \cdot C_L$ & \\

\hline 2 & Bias currents in output cascodes & $I_4=I_5=1.2 I_3$ to $1.5 I_3$ & Avoid zero current in cascodes \\

\hline 3 & Maximum output voltage, $v_{\text {out }}$ (max) & $S_5=\frac{2 I_5}{K_P{ }^{\prime} V_{S D 5}^2}, S_7=\frac{2 I_7}{K_P{ }^{\prime} V_{S D 7}^2},\left(S_4=S_5\right.$ and $\left.S_6=S_7\right)$ & $$

\begin{aligned}

& V_{S D 5}(\mathrm{sat})=V_{S D 7}(\mathrm{sat}) \\

& =0.5\left[V_{D D}-V_{\text {out }}(\max )\right]

\end{aligned}

$$ \\

\hline 4 & Minimum output voltage, $v_{\text {out }}(\mathrm{min})$ & $S_{11}=\frac{2 I_{11}}{K_N{ }^{\prime} V_{D S 11}^2}, S_9=\frac{2 I_9}{K_N{ }^{\prime} V_{D S 9}^2},\left(S_{10}=S_{11}\right.$ and $\left.S_8=S_9\right)$ & $$

\begin{aligned}

& V_{D S 9}(\mathrm{sat})=V_{D S 11}(\mathrm{sat}) \\

& =0.5\left[V_{\text {out }}(\mathrm{min})-V_{S S}\right]

\end{aligned}

$$ \\

\hline 5 & $G B=\frac{g_{m 1}}{C_L}$ & $S_1=S_2=\frac{g_{m 1}^2}{K_N{ }^{\prime} I_3}=\frac{G B^2 C_L^2}{K_N{ }^{\prime} I_3}$ & \\

\hline 6 & Minimum input CM & $S_3=\frac{2 I_3}{K_N{ }^{\prime}\left(V_{\text {in }}(\min )-V_{S S}-\sqrt{\left(I_3 K_N{ }^{\prime} S_1\right)}-V_{T 1}\right)^2}$ & \\

\hline 7 & Maximum input CM & $S_4=S_5=\frac{2 I_4}{K_P{ }^{\prime}\left(V_{D D}-V_{i n}(\max )+V_{T 1}\right)^2} 2$ & $S_4$ and $S_5$ must meet or exceed value in step 3 \\

\hline 8 & Differential-voltage gain & $\frac{v_{\text {out }}}{v_{\text {in }}}=\left(\frac{g_{m 1}}{2}+\frac{g_{m 2}}{2(1+k)}\right) R_{\text {out }}=\left(\frac{2+k}{2+2 k}\right) g_{m l} R_{\text {out }}$ & $k=\frac{R_{I I}\left(g_{d s 2}+g_{d s 4}\right)}{g_{m 7} r_{d s 7}}$ \\

\hline 9 & Power dissipation & $P_{\text {diss }}=\left(V_{D D}-V_{S S}\right)\left(I_3+I_{10}+I_{11}\right)$ & \\

\hline

\end{tabular}

\end{table}