Because the specifications above are similar to the two-stage design of Example 6.3-1, we can use the results of that example for the first two stages of our design. However, we must convert the results of Example 6.3-1 to a PMOS input stage. The results of doing this give $W_1 / L_1=W_2 / L_2=6 \mu \mathrm{~m} / 1 \mu \mathrm{~m}, W_3 / L_3=W_4 / L_4=7 \mu \mathrm{~m} / 1 \mu \mathrm{~m}, W_5 / L_5=11 \mu \mathrm{~m} / 1 \mu \mathrm{~m}, W_6 / L_6= 43 \mu \mathrm{~m} / 1 \mu \mathrm{~m}$, and $W_7 / L_7=34 \mu \mathrm{~m} / 1 \mu \mathrm{~m}$ (see Problem 6.3-6). Let the value of $I_{\text {BIAS }}$ be $30 \mu \mathrm{~A}$ and the values of $W_{12}$ and $W_{13}$ be $44 \mu \mathrm{~m}$.



The design of the two followers is next. Let us begin with the BJT follower designed to meet the slew-rate specification. The current required for a 100 pF capacitor is 1 mA . Therefore, we want $I_{11}$ to be 1 mA . This means that $W_{11}$ must be equal to $44 \mu \mathrm{~m}(1000 \mu \mathrm{~A} / 30 \mu \mathrm{~A})=1467 \mu \mathrm{~m}$. The 1 mA bias current through the BJT means that the output resistance will be $0.0258 \mathrm{~V} / 1 \mathrm{~mA}$ or $25.8 \Omega$, which is less than $100 \Omega$. The $1000 \mu \mathrm{~A}$ flowing in the BJT will require $10 \mu \mathrm{~A}$ from the MOS follower stage. Therefore, let us select a bias current of $100 \mu \mathrm{~A}$ for M8. If $W_{12}=44 \mu \mathrm{~m}$, then $W_8=44 \mu \mathrm{~m}(100 \mu \mathrm{~A} / 30 \mu \mathrm{~A})=146 \mu \mathrm{~m}$. If $1 / g_{m 10}$ is $25.8 \Omega$, then we can use Eq. (7.1-9) to design $g_{m 9}$ as



$$

g_{m 9}=\frac{1}{\left(R_{\text {out }}-\frac{1}{g_{m 10}}\right)\left(1+\beta_F\right)}=\frac{1}{(100-25.8)(101)}=133.4 \mu \mathrm{~S}

$$





Solving for the $W / L$ of M9 given $g_{m 9}$ gives a $W / L$ of 0.809 . Let us select $W / L$ to be 10 for M9 in order to make sure that the contribution of M9 to the output resistance is sufficiently small and to increase the gain closer to unity. This will also help the fact that Eq. (7.1-9) neglects the bulk contribution to this resistance [see Eq. (5.5-18)]. This gives a transconductance for M9 of $300 \mu \mathrm{~S}$.

To calculate the voltage gain of the MOS follower we need to find $g_{m b s 9}$. This value is given as



$$

g_{m b s 9}=\frac{g_{m 9} \gamma_N}{2 \sqrt{2 \phi_F+V_{B S 9}}}=\frac{300 \cdot 0.4}{2 \sqrt{0.7+2}}=36.5 \mu \mathrm{~S}

$$



where we have assumed that the value of $V_{B S 9}$ is approximately -2 V . Therefore,



$$

A_{\mathrm{MOS}}=\frac{300 \mu \mathrm{~S}}{300 \mu \mathrm{~S}+36.5 \mu \mathrm{~S}+4 \mu \mathrm{~S}+5 \mu \mathrm{~S}}=0.8683 \mathrm{~V} / \mathrm{V}

$$





The voltage gain of the BJT follower is



$$

A_{\mathrm{BJT}}=\frac{500}{25.8+500}=0.951 \mathrm{~V} / \mathrm{V}

$$





Thus, the gain of the op amp is



$$

A_{v d}(0)=(7777)(0.8683)(0.951)=6422 \mathrm{~V} / \mathrm{V}

$$



which meets the specification. The power dissipation of this amplifier is given as



$$

P_{\text {diss }}=5 \mathrm{~V}(30 \mu \mathrm{~A}+30 \mu \mathrm{~A}+95 \mu \mathrm{~A}+100 \mu \mathrm{~A}+1000 \mu \mathrm{~A})=6.27 \mathrm{~mW}

$$