SOLUTION

Assuming the conditions for a two-stage op amp necessary to achieve a $60^{\circ}$ phase margin and that the RHP zero is at least $10 G B$ gives



$$

C_c=0.2 C_L=2 \mathrm{pF}

$$





The slew rate is directly related to the current in M5 and gives



$$

I_5=C_c \cdot S R=2 \times 10^{-11} \cdot 10^7=20 \mu \mathrm{~A}

$$





We also know the input transconductances from $G B$ and $C_c$. They are given as



$$

g_{m 1}=g_{m 2}=G B \cdot C_c=20 \pi \times 10^6 \cdot 2 \times 10^{-12}=125.67 \mu \mathrm{~S}

$$





Knowing the current flow in M1 and M2 gives the $W / L$ ratios as



$$

\frac{W_1}{L_1}=\frac{W_2}{L_2}=\frac{g_{m 1}^2}{2 K_N^{\prime}\left(I_1 / 2\right)}=\frac{\left(125.67 \times 10^{-6}\right)^2}{2 \cdot 110 \times 10^{-6} \cdot 10 \times 10^{-6}}=7.18

$$





Next, we find the $W / L$ of M5 that will satisfy the $V_{\text {icm }}(\min )$ specification.



$$

V_{i c m}(\min )=V_{D S 5}(\text { sat })+V_{G S 1}(10 \mu \mathrm{~A})=1 \mathrm{~V}

$$

This gives



$$

V_{D S 5}(\mathrm{sat})=1-\sqrt{\frac{2 \cdot 10}{110 \cdot 7.18}}-0.75=1-0.159-0.75=0.0909 \mathrm{~V}

$$





Therefore,



$$

V_{D S 5}(\text { sat })=0.0909=\sqrt{\frac{2 I_5}{K_N^{\prime}\left(W_5 / L_5\right)}} \rightarrow \frac{W_5}{L_5}=\frac{2 \cdot 20}{110 \cdot(0.0909)^2}=44

$$





The design of M3 and M4 is accomplished from the upper input common-mode voltage and is



$$

V_{i c m}(\max )=V_{D D}-V_{S D 3}(\mathrm{sat})+V_{T N}=2-V_{S D 3}(\mathrm{sat})+0.75=2.5 \mathrm{~V}

$$





Solving for $V_{S D 3}$ (sat) gives 0.25 V . Let us assume that the currents in M6 and M7 are $20 \mu \mathrm{~A}$. This gives a current of $30 \mu \mathrm{~A}$ in M3 and M4. Knowing the current in M3 (M4) gives



$$

V_{S D 3}(\text { sat }) \leq \sqrt{\frac{2 \cdot 30}{50 \cdot\left(W_3 / L_3\right)}} \rightarrow \frac{W_3}{L_3}=\frac{W_4}{L_4} \geq \frac{2 \cdot 30}{(0.25)^2 \cdot 50}=19.2

$$





Next, using the $V_{S D}(\mathrm{sat})=V_{O N}$ of M3 and M4, design M10 through M12. Let us assume that $I_{10}=I_5=20 \mu \mathrm{~A}$, which gives $W_{10} / L_{10}=44 . R_1$ is designed as $R_1=0.25 \mathrm{~V} / 20 \mu \mathrm{~A}= 12.5 \mathrm{k} \Omega$. The $W / L$ ratios of M11 and M12 can be expressed as

$$

\frac{W_{11}}{L_{11}}=\frac{W_{12}}{L_{12}}=\frac{2 \cdot I_{11}}{K_P^{\prime} V_{S D 11}^2(\mathrm{sat})}=\frac{2 \cdot 20}{50 \cdot(0.25)^2}=12.8

$$





It turns out that since the source-gate voltages and currents of M6 and M7 are the same as M11 and M12, the $W / L$ values are equal. Thus,



$$

\frac{W_6}{L_6}=\frac{W_7}{L_7}=12.8

$$





M8 and M9 should be as small as possible to reduce the parasitic (mirror) pole. However, the voltage drop across M4, M6, and M8 must be less than the power supply. Using this to design the gate-source voltage of M8 gives



$$

V_{G S 8}=V_{D D}-2 V_{O N}=2 \mathrm{~V}-2 \cdot 0.25=1.5 \mathrm{~V}

$$





Thus,



$$

\frac{W_8}{L_8}=\frac{W_9}{L_9}=\frac{2 \cdot I_8}{K_N^{\prime} \cdot V_{D S 8}^2(\mathrm{sat})}=\frac{2 \cdot 20}{110 \cdot(0.8)^2}=0.57 \approx 1

$$





This value suggests that it might be possible to make M8 and M9 a cascode current mirror, which would boost the gain (see Problem 7.6-9). Because M8 and M9 are small, the mirror

pole will be insignificant. The next poles of interest would be those at the sources of M6 and M7. These poles are given as



$$

p_6 \approx \frac{g_{m 6}}{C_{G S 6}}=\frac{\sqrt{2 K_p^{\prime} \cdot\left(W_6 / L_6\right) \cdot I_6}}{\left(\frac{2}{3}\right) \cdot W_6 \cdot L_6 \cdot C_{\mathrm{ox}}}=\frac{\sqrt{2 \cdot 50 \cdot 12.8 \cdot 20 \times 10^{-6}}}{\left(\frac{2}{3}\right) \cdot 12.8 \cdot 1 \cdot 2.47 \times 10^{-15}}=7.59 \times 10^9 \mathrm{rad} / \mathrm{s}

$$





We have assumed that the channel length is $1 \mu \mathrm{~m}$ in the above calculation. This value is about 100 times greater than $G B$ so we should be okay even though we have neglected the drainground capacitances of M1 and M3.



Finally, the $W / L$ ratios of the second stage must be designed. We can either use the relationship for a $60^{\circ}$ phase margin of $g_{m 14}=10 g_{m 1}=1256.7 \mu \mathrm{~S}$ or consider proper mirroring between M9 and M14. Combining the equations for saturation region, we get



$$

\frac{W}{L}=\frac{g_m}{K_N^{\prime} V_{D S}(\mathrm{sat})}

$$





Substituting $1256.7 \mu \mathrm{~S}$ for $g_{m 1}$ and 0.5 V for $V_{D S 14}$ gives $W_{14} / L_{14}=22.85$. The current corresponding to this $g_m$ and $W / L$ is $I_{14}=314 \mu \mathrm{~A}$. The $W / L$ of M13 is designed by the necessary current ratio desired between the two transistors and is



$$

\frac{W_{13}}{L_{13}}=\frac{I_{13}}{I_{12}} I_{12}=\frac{314}{20} \cdot 12.8=201

$$

Now, we must check to make sure that the $V_{\text {out }}(\max )$ is satisfied. The saturation voltage of M13 is



$$

V_{S D 13}(\mathrm{sat})=\sqrt{\frac{2 \cdot I_{13}}{K_P^{\prime}\left(W_{13} / L_{13}\right)}}=\sqrt{\frac{2 \cdot 314}{50 \cdot 201}}=0.25 \mathrm{~V}

$$



which exactly meets the specification. For proper mirroring, the $W / L$ ratio of M9 should be



$$

\frac{W_9}{L_9}=\frac{I_9}{I_{14}} \frac{W_{14}}{L_{14}}=1.42

$$





Since $W_9 / L_9$ was selected as 1 , this is close enough.

Let us check to see what gain is achieved at low frequencies. The small-signal voltage gain can be written as



$$

\frac{v_{\text {out }}}{v_{\text {in }}}=\left(\frac{g_{m 1}}{g_{d s 7}+g_{d s 9}}\right)\left(\frac{g_{m 14}}{g_{d s 13}+g_{d s 14}}\right)

$$



$g_{d s 7}=1 \mu \mathrm{~S}, g_{d s 8}=0.8 \mu \mathrm{~S}, g_{d s 13}=15.7 \mu \mathrm{~S}$, and $g_{d s 14}=12.56 \mu \mathrm{~S}$. Putting these values in the above equation gives



$$

\frac{v_{\text {out }}}{v_{\text {in }}}=\left(\frac{125.6}{1.8}\right)\left(\frac{1256.7}{28.26}\right)=69.78 \cdot 44.47=3103 \mathrm{~V} / \mathrm{V}

$$





The power dissipation, including $I_{\text {BIAS }}$ of $20 \mu \mathrm{~A}$, is $708 \mu \mathrm{~W}$. The minimum power supply possible with no regard to the input common-mode voltage range is $V_T+3 V_{O N}$. With $V_T=0.7 \mathrm{~V}$ and $V_{O N} \approx 0.25 \mathrm{~V}$, this op amp should be capable of operating with a power supply of 1.5 V .