Figure 13.2.1
It is required to design the folded-cascode op amp shown in Fig. 13.2.1. Let $V_{D D}=V_{S S}=1 \mathrm{~V}$ and $V_{t n}=-V_{t p}=0.4 \mathrm{~V}$, and assume that for all transistors, $\left|V_{A}\right|=6 \mathrm{~V}$. Design so that the power dissipated in the circuit (with no input signal applied) is $0.6 \mathrm{~mW}$, and so that each of $Q_{1}$ and $Q_{2}$ is operating at a current four times that at which each of $Q_{3}$ and $Q_{4}$ is operating. Also, design so that all transistors operate at $\left|V_{O V}\right|=0.2 \mathrm{~V}$.
(a) Show that the current drawn from each of the two power supplies is $2 I_{B}$, and hence find $I_{B}$ and $I$ that result in the circuit operating at its specified power dissipation.
(b) Find the dc current at which each of $Q_{1}$ to $Q_{11}$ is operating. Present your results in a table.
(c) Find the input common-mode range.
(d) Find the required values of $V_{\mathrm{BIAS} 1}, V_{\mathrm{BIAS} 2}$, and $V_{\mathrm{BIAS} 3}$ that result in the maximum allowable value of $v_{O}$ to be as high as possible.
(e) Find the allowable range of $v_{O}$.
(f) Find the overall transconductance $G_{m}$.
(g) Find the output resistance $R_{O}$.
(h) Find the low-frequency voltage gain.
(i) If the amplifier at its output is modeled by a controlled current-source $G_{m} V_{i d}$ (where $V_{i d}$ is the differential input voltage) feeding the output resistance $R_{O}$ and the total capacitance at the output node $C_{L}$, find the value of $C_{L}$ that results in the amplifier having a unity-gain bandwidth of $100 \mathrm{MHz}$. Assume that the dominant pole is that formed at the output.

Question

Figure 13.2.1
It is required to design the folded-cascode op amp shown in Fig. 13.2.1. Let $V_{D D}=V_{S S}=1 \mathrm{~V}$ and $V_{t n}=-V_{t p}=0.4 \mathrm{~V}$, and assume that for all transistors, $\left|V_{A}\right|=6 \mathrm{~V}$. Design so that the power dissipated in the circuit (with no input signal applied) is $0.6 \mathrm{~mW}$, and so that each of $Q_{1}$ and $Q_{2}$ is operating at a current four times that at which each of $Q_{3}$ and $Q_{4}$ is operating. Also, design so that all transistors operate at $\left|V_{O V}\right|=0.2 \mathrm{~V}$.
(a) Show that the current drawn from each of the two power supplies is $2 I_{B}$, and hence find $I_{B}$ and $I$ that result in the circuit operating at its specified power dissipation.
(b) Find the dc current at which each of $Q_{1}$ to $Q_{11}$ is operating. Present your results in a table.
(c) Find the input common-mode range. 
(d) Find the required values of $V_{\mathrm{BIAS} 1}, V_{\mathrm{BIAS} 2}$, and $V_{\mathrm{BIAS} 3}$ that result in the maximum allowable value of $v_{O}$ to be as high as possible.
(e) Find the allowable range of $v_{O}$.
(f) Find the overall transconductance $G_{m}$.
(g) Find the output resistance $R_{O}$.
(h) Find the low-frequency voltage gain.
(i) If the amplifier at its output is modeled by a controlled current-source $G_{m} V_{i d}$ (where $V_{i d}$ is the differential input voltage) feeding the output resistance $R_{O}$ and the total capacitance at the output node $C_{L}$, find the value of $C_{L}$ that results in the amplifier having a unity-gain bandwidth of $100 \mathrm{MHz}$. Assume that the dominant pole is that formed at the output.

Quizplus · Accepted Answer

The Answer is $Figure 13.2.1 (a) I_{D D}=I_{D 9}+I_{D 10}=2 I_{B} Since each of Q_{1} and Q_{2} is conducting a current I / 2 , we can see from node equations at D_{1} and D_{2} that I_{D 3}=I_{D 4}=I_{B}-\frac{I}{2} Now, \begin{aligned} I_{S S} & =I_{D 11}+I_{D 7}+I_{D 8} \ & =I+\left(I_{B}-\frac{I}{2} ight)+\left(I_{B}-\frac{I}{2} ight) \ & =2 I_{B} \quad ext { Q.E.D } \ P_{D} & =V_{D D} imes 2 I_{B}+V_{S S} imes 2 I_{B} \ & =1 imes 2 I_{B}+1 imes 2 I_{B}=4 I_{B} \end{aligned} For P_{D}=0.6 \mathrm{~mW} , I_{B}=\frac{0.6}{4}=0.15 \mathrm{~mA}=150 \mu \mathrm{A} For I_{D 1,2}=4 I_{D 3,4} , \begin{aligned} \frac{I}{2} & =4\left(I_{B}-\frac{I}{2} ight) \ \Rightarrow I & =\frac{4 I_{B}}{2.5}=\frac{600}{2.5}=240 \mu \mathrm{A} \end{aligned} (b) (c) \begin{aligned} V_{I C M \max } & =V_{D 1 \max }+V_{t n} \ & =V_{D D}-\left|V_{O V 9} ight|+V_{t n} \ & =+1-0.2+0.4=+1.2 \mathrm{~V} \ V_{I C M \min } & =-V_{S S}+V_{O V 11}+V_{G S 1} \ & =-1+0.2+(0.4+0.2)=-0.2 \mathrm{~V} \end{aligned} Thus, -0.2 \mathrm{~V} \leq V_{I C M} \leq+1.2 \mathrm{~V} (d) To make v_{O \max } as high as possible, \begin{aligned} V_{\mathrm{BIAS} 1} & =V_{D D}-\left|V_{O V 9,10} ight|-V_{S G 3,4} \ & =+1-0.2-(0.4+0.2)=+0.2 \mathrm{~V} \ V_{\mathrm{BIAS} 2} & =V_{D D}-V_{S G 9,10} \ & =+1-(0.4+0.2)=+0.4 \mathrm{~V} \ V_{\mathrm{BIAS} 3} & =-V_{S S}+V_{G S 11} \ & =-1+(0.4+0.2)=-0.4 \mathrm{~V} \end{aligned} (e) \begin{aligned} v_{O \max } & =V_{\mathrm{BIAS} 1}+\left|V_{t p} ight| \ & =0.2+0.4=+0.6 \mathrm{~V} \ v_{O \min } & =-V_{S S}+V_{G S 7}+V_{G S 5}-V_{t n} \ & =-1+(0.4+0.2)+(0.4+0.2)-0.4 \ & =-0.2 \mathrm{~V} \end{aligned} Thus, -0.2 \mathrm{~V} \leq v_{O} \leq+0.6 \mathrm{~V} (f) \begin{aligned} G_{m} & =g_{m 1,2}=\frac{2 I_{D 1,2}}{V_{O V}} \ & =\frac{2 imes 0.12}{0.2}=1.2 \mathrm{~mA} / \mathrm{V} \end{aligned} (g) \begin{aligned} R_{O} & =R_{O 4} \| R_{o 6} \ R_{o 4} & =\left(g_{m 4} r_{o 4} ight)\left[r_{o 10} \| r_{o 2} ight] \ r_{o 10} & =\frac{\left|V_{A} ight|}{I_{B}}=\frac{6}{0.15}=40 \mathrm{k} \Omega \ r_{o 2} & =\frac{V_{A}}{I_{D 2}}=\frac{6}{0.12}=50 \mathrm{k} \Omega \ g_{m 4} & =\frac{2 I_{D 4}}{\left|V_{O V} ight|}=\frac{2 imes 0.03}{0.2}=0.3 \mathrm{~mA} / \mathrm{V} \ r_{o 4} & =\frac{\left|V_{A} ight|}{I_{D 4}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \ g_{m 4} r_{o 4} & =0.3 imes 200=60 \ R_{o 4} & =60(40 \| 50) \ & =1.33 \mathrm{M} \Omega \ R_{o 6} & =\left(g_{m 6} r_{o 6} ight) r_{o 8} \ g_{m 6} & =\frac{2 I_{D 6}}{V_{O V}}=\frac{2 imes 0.03}{0.2}=0.3 \mathrm{~mA} / \mathrm{V} \ r_{o 6} & =\frac{V_{A}}{I_{D 6}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \end{aligned} \begin{aligned} r_{o 8} & =\frac{V_{A}}{I_{D 8}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \ R_{o 6} & =(0.3 imes 200) imes 200=12 \mathrm{M} \Omega \ R_{O} & =R_{o 4}\left\|R_{o 6}=1.33 ight\| 12 \ & =1.2 \mathrm{M} \Omega \end{aligned} (h) \begin{aligned} A_{v} & =G_{m} R_{O}=1.2 imes 1.2 imes 10^{3}=1.44 imes 10^{3} \ & =1440 \mathrm{~V} / \mathrm{V} \end{aligned} (i) Figure 13.2.2 Figure 13.2.2 shows the output equivalent circuit of the amplifier. At high frequencies, \begin{aligned} & V_{o}=\frac{G_{m} V_{i d}}{s C_{L}} \ \left|\frac{V_{o}}{V_{i d}} ight|=1 ext { at } & \ & \omega=\omega_{t}=\frac{G_{m}}{C_{L}} \end{aligned} Thus, \begin{aligned} C_{L} & =\frac{G_{m}}{\omega_{t}} \ & =\frac{1.2 imes 10^{-3}}{2 \pi imes 100 imes 10^{6}}=1.9 \mathrm{pF} \end{aligned}$

Figure 13.2.1
(a)

I_{D D}=I_{D 9}+I_{D 10}=2 I_{B}

Since each of

Q_{1}

and

Q_{2}

is conducting a current

I / 2

, we can see from node equations at

D_{1}

and

D_{2}

that

I_{D 3}=I_{D 4}=I_{B}-\frac{I}{2}

Now,

\begin{aligned}I_{S S} & =I_{D 11}+I_{D 7}+I_{D 8} \& =I+\left(I_{B}-\frac{I}{2} ight)+\left(I_{B}-\frac{I}{2} ight) \& =2 I_{B} \quad ext { Q.E.D } \P_{D} & =V_{D D} imes 2 I_{B}+V_{S S} imes 2 I_{B} \& =1 imes 2 I_{B}+1 imes 2 I_{B}=4 I_{B}\end{aligned}

For

P_{D}=0.6 \mathrm{~mW}

,

I_{B}=\frac{0.6}{4}=0.15 \mathrm{~mA}=150 \mu \mathrm{A}

For

I_{D 1,2}=4 I_{D 3,4}

,

\begin{aligned}\frac{I}{2} & =4\left(I_{B}-\frac{I}{2} ight) \\Rightarrow I & =\frac{4 I_{B}}{2.5}=\frac{600}{2.5}=240 \mu \mathrm{A}\end{aligned}

(b)
$Figure 13.2.1 (a) I_{D D}=I_{D 9}+I_{D 10}=2 I_{B} Since each of Q_{1} and Q_{2} is conducting a current I / 2 , we can see from node equations at D_{1} and D_{2} that I_{D 3}=I_{D 4}=I_{B}-\frac{I}{2} Now, \begin{aligned} I_{S S} & =I_{D 11}+I_{D 7}+I_{D 8} \ & =I+\left(I_{B}-\frac{I}{2} ight)+\left(I_{B}-\frac{I}{2} ight) \ & =2 I_{B} \quad ext { Q.E.D } \ P_{D} & =V_{D D} imes 2 I_{B}+V_{S S} imes 2 I_{B} \ & =1 imes 2 I_{B}+1 imes 2 I_{B}=4 I_{B} \end{aligned} For P_{D}=0.6 \mathrm{~mW} , I_{B}=\frac{0.6}{4}=0.15 \mathrm{~mA}=150 \mu \mathrm{A} For I_{D 1,2}=4 I_{D 3,4} , \begin{aligned} \frac{I}{2} & =4\left(I_{B}-\frac{I}{2} ight) \ \Rightarrow I & =\frac{4 I_{B}}{2.5}=\frac{600}{2.5}=240 \mu \mathrm{A} \end{aligned} (b) (c) \begin{aligned} V_{I C M \max } & =V_{D 1 \max }+V_{t n} \ & =V_{D D}-\left|V_{O V 9} ight|+V_{t n} \ & =+1-0.2+0.4=+1.2 \mathrm{~V} \ V_{I C M \min } & =-V_{S S}+V_{O V 11}+V_{G S 1} \ & =-1+0.2+(0.4+0.2)=-0.2 \mathrm{~V} \end{aligned} Thus, -0.2 \mathrm{~V} \leq V_{I C M} \leq+1.2 \mathrm{~V} (d) To make v_{O \max } as high as possible, \begin{aligned} V_{\mathrm{BIAS} 1} & =V_{D D}-\left|V_{O V 9,10} ight|-V_{S G 3,4} \ & =+1-0.2-(0.4+0.2)=+0.2 \mathrm{~V} \ V_{\mathrm{BIAS} 2} & =V_{D D}-V_{S G 9,10} \ & =+1-(0.4+0.2)=+0.4 \mathrm{~V} \ V_{\mathrm{BIAS} 3} & =-V_{S S}+V_{G S 11} \ & =-1+(0.4+0.2)=-0.4 \mathrm{~V} \end{aligned} (e) \begin{aligned} v_{O \max } & =V_{\mathrm{BIAS} 1}+\left|V_{t p} ight| \ & =0.2+0.4=+0.6 \mathrm{~V} \ v_{O \min } & =-V_{S S}+V_{G S 7}+V_{G S 5}-V_{t n} \ & =-1+(0.4+0.2)+(0.4+0.2)-0.4 \ & =-0.2 \mathrm{~V} \end{aligned} Thus, -0.2 \mathrm{~V} \leq v_{O} \leq+0.6 \mathrm{~V} (f) \begin{aligned} G_{m} & =g_{m 1,2}=\frac{2 I_{D 1,2}}{V_{O V}} \ & =\frac{2 imes 0.12}{0.2}=1.2 \mathrm{~mA} / \mathrm{V} \end{aligned} (g) \begin{aligned} R_{O} & =R_{O 4} \| R_{o 6} \ R_{o 4} & =\left(g_{m 4} r_{o 4} ight)\left[r_{o 10} \| r_{o 2} ight] \ r_{o 10} & =\frac{\left|V_{A} ight|}{I_{B}}=\frac{6}{0.15}=40 \mathrm{k} \Omega \ r_{o 2} & =\frac{V_{A}}{I_{D 2}}=\frac{6}{0.12}=50 \mathrm{k} \Omega \ g_{m 4} & =\frac{2 I_{D 4}}{\left|V_{O V} ight|}=\frac{2 imes 0.03}{0.2}=0.3 \mathrm{~mA} / \mathrm{V} \ r_{o 4} & =\frac{\left|V_{A} ight|}{I_{D 4}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \ g_{m 4} r_{o 4} & =0.3 imes 200=60 \ R_{o 4} & =60(40 \| 50) \ & =1.33 \mathrm{M} \Omega \ R_{o 6} & =\left(g_{m 6} r_{o 6} ight) r_{o 8} \ g_{m 6} & =\frac{2 I_{D 6}}{V_{O V}}=\frac{2 imes 0.03}{0.2}=0.3 \mathrm{~mA} / \mathrm{V} \ r_{o 6} & =\frac{V_{A}}{I_{D 6}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \end{aligned} \begin{aligned} r_{o 8} & =\frac{V_{A}}{I_{D 8}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \ R_{o 6} & =(0.3 imes 200) imes 200=12 \mathrm{M} \Omega \ R_{O} & =R_{o 4}\left\|R_{o 6}=1.33 ight\| 12 \ & =1.2 \mathrm{M} \Omega \end{aligned} (h) \begin{aligned} A_{v} & =G_{m} R_{O}=1.2 imes 1.2 imes 10^{3}=1.44 imes 10^{3} \ & =1440 \mathrm{~V} / \mathrm{V} \end{aligned} (i) Figure 13.2.2 Figure 13.2.2 shows the output equivalent circuit of the amplifier. At high frequencies, \begin{aligned} & V_{o}=\frac{G_{m} V_{i d}}{s C_{L}} \ \left|\frac{V_{o}}{V_{i d}} ight|=1 ext { at } & \ & \omega=\omega_{t}=\frac{G_{m}}{C_{L}} \end{aligned} Thus, \begin{aligned} C_{L} & =\frac{G_{m}}{\omega_{t}} \ & =\frac{1.2 imes 10^{-3}}{2 \pi imes 100 imes 10^{6}}=1.9 \mathrm{pF} \end{aligned}$

(c)

\begin{aligned}V_{I C M \max } & =V_{D 1 \max }+V_{t n} \& =V_{D D}-\left|V_{O V 9} ight|+V_{t n} \& =+1-0.2+0.4=+1.2 \mathrm{~V} \V_{I C M \min } & =-V_{S S}+V_{O V 11}+V_{G S 1} \& =-1+0.2+(0.4+0.2)=-0.2 \mathrm{~V}\end{aligned}

Thus,

-0.2 \mathrm{~V} \leq V_{I C M} \leq+1.2 \mathrm{~V}

(d) To make

v_{O \max }

as high as possible,

\begin{aligned}V_{\mathrm{BIAS} 1} & =V_{D D}-\left|V_{O V 9,10} ight|-V_{S G 3,4} \& =+1-0.2-(0.4+0.2)=+0.2 \mathrm{~V} \V_{\mathrm{BIAS} 2} & =V_{D D}-V_{S G 9,10} \& =+1-(0.4+0.2)=+0.4 \mathrm{~V} \V_{\mathrm{BIAS} 3} & =-V_{S S}+V_{G S 11} \& =-1+(0.4+0.2)=-0.4 \mathrm{~V}\end{aligned}

(e)

\begin{aligned}v_{O \max } & =V_{\mathrm{BIAS} 1}+\left|V_{t p} ight| \& =0.2+0.4=+0.6 \mathrm{~V} \v_{O \min } & =-V_{S S}+V_{G S 7}+V_{G S 5}-V_{t n} \& =-1+(0.4+0.2)+(0.4+0.2)-0.4 \& =-0.2 \mathrm{~V}\end{aligned}

Thus,

-0.2 \mathrm{~V} \leq v_{O} \leq+0.6 \mathrm{~V}

(f)

\begin{aligned}G_{m} & =g_{m 1,2}=\frac{2 I_{D 1,2}}{V_{O V}} \& =\frac{2 imes 0.12}{0.2}=1.2 \mathrm{~mA} / \mathrm{V}\end{aligned}

(g)

\begin{aligned}R_{O} & =R_{O 4} \| R_{o 6} \R_{o 4} & =\left(g_{m 4} r_{o 4} ight)\left[r_{o 10} \| r_{o 2} ight] \r_{o 10} & =\frac{\left|V_{A} ight|}{I_{B}}=\frac{6}{0.15}=40 \mathrm{k} \Omega \r_{o 2} & =\frac{V_{A}}{I_{D 2}}=\frac{6}{0.12}=50 \mathrm{k} \Omega \g_{m 4} & =\frac{2 I_{D 4}}{\left|V_{O V} ight|}=\frac{2 imes 0.03}{0.2}=0.3 \mathrm{~mA} / \mathrm{V} \r_{o 4} & =\frac{\left|V_{A} ight|}{I_{D 4}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \g_{m 4} r_{o 4} & =0.3 imes 200=60 \R_{o 4} & =60(40 \| 50) \& =1.33 \mathrm{M} \Omega \R_{o 6} & =\left(g_{m 6} r_{o 6} ight) r_{o 8} \g_{m 6} & =\frac{2 I_{D 6}}{V_{O V}}=\frac{2 imes 0.03}{0.2}=0.3 \mathrm{~mA} / \mathrm{V} \r_{o 6} & =\frac{V_{A}}{I_{D 6}}=\frac{6}{0.03}=200 \mathrm{k} \Omega\end{aligned}

\begin{aligned}r_{o 8} & =\frac{V_{A}}{I_{D 8}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \R_{o 6} & =(0.3 imes 200) imes 200=12 \mathrm{M} \Omega \R_{O} & =R_{o 4}\left\|R_{o 6}=1.33 ight\| 12 \& =1.2 \mathrm{M} \Omega\end{aligned}

(h)

\begin{aligned}A_{v} & =G_{m} R_{O}=1.2 imes 1.2 imes 10^{3}=1.44 imes 10^{3} \& =1440 \mathrm{~V} / \mathrm{V}\end{aligned}

(i)
$Figure 13.2.1 (a) I_{D D}=I_{D 9}+I_{D 10}=2 I_{B} Since each of Q_{1} and Q_{2} is conducting a current I / 2 , we can see from node equations at D_{1} and D_{2} that I_{D 3}=I_{D 4}=I_{B}-\frac{I}{2} Now, \begin{aligned} I_{S S} & =I_{D 11}+I_{D 7}+I_{D 8} \ & =I+\left(I_{B}-\frac{I}{2} ight)+\left(I_{B}-\frac{I}{2} ight) \ & =2 I_{B} \quad ext { Q.E.D } \ P_{D} & =V_{D D} imes 2 I_{B}+V_{S S} imes 2 I_{B} \ & =1 imes 2 I_{B}+1 imes 2 I_{B}=4 I_{B} \end{aligned} For P_{D}=0.6 \mathrm{~mW} , I_{B}=\frac{0.6}{4}=0.15 \mathrm{~mA}=150 \mu \mathrm{A} For I_{D 1,2}=4 I_{D 3,4} , \begin{aligned} \frac{I}{2} & =4\left(I_{B}-\frac{I}{2} ight) \ \Rightarrow I & =\frac{4 I_{B}}{2.5}=\frac{600}{2.5}=240 \mu \mathrm{A} \end{aligned} (b) (c) \begin{aligned} V_{I C M \max } & =V_{D 1 \max }+V_{t n} \ & =V_{D D}-\left|V_{O V 9} ight|+V_{t n} \ & =+1-0.2+0.4=+1.2 \mathrm{~V} \ V_{I C M \min } & =-V_{S S}+V_{O V 11}+V_{G S 1} \ & =-1+0.2+(0.4+0.2)=-0.2 \mathrm{~V} \end{aligned} Thus, -0.2 \mathrm{~V} \leq V_{I C M} \leq+1.2 \mathrm{~V} (d) To make v_{O \max } as high as possible, \begin{aligned} V_{\mathrm{BIAS} 1} & =V_{D D}-\left|V_{O V 9,10} ight|-V_{S G 3,4} \ & =+1-0.2-(0.4+0.2)=+0.2 \mathrm{~V} \ V_{\mathrm{BIAS} 2} & =V_{D D}-V_{S G 9,10} \ & =+1-(0.4+0.2)=+0.4 \mathrm{~V} \ V_{\mathrm{BIAS} 3} & =-V_{S S}+V_{G S 11} \ & =-1+(0.4+0.2)=-0.4 \mathrm{~V} \end{aligned} (e) \begin{aligned} v_{O \max } & =V_{\mathrm{BIAS} 1}+\left|V_{t p} ight| \ & =0.2+0.4=+0.6 \mathrm{~V} \ v_{O \min } & =-V_{S S}+V_{G S 7}+V_{G S 5}-V_{t n} \ & =-1+(0.4+0.2)+(0.4+0.2)-0.4 \ & =-0.2 \mathrm{~V} \end{aligned} Thus, -0.2 \mathrm{~V} \leq v_{O} \leq+0.6 \mathrm{~V} (f) \begin{aligned} G_{m} & =g_{m 1,2}=\frac{2 I_{D 1,2}}{V_{O V}} \ & =\frac{2 imes 0.12}{0.2}=1.2 \mathrm{~mA} / \mathrm{V} \end{aligned} (g) \begin{aligned} R_{O} & =R_{O 4} \| R_{o 6} \ R_{o 4} & =\left(g_{m 4} r_{o 4} ight)\left[r_{o 10} \| r_{o 2} ight] \ r_{o 10} & =\frac{\left|V_{A} ight|}{I_{B}}=\frac{6}{0.15}=40 \mathrm{k} \Omega \ r_{o 2} & =\frac{V_{A}}{I_{D 2}}=\frac{6}{0.12}=50 \mathrm{k} \Omega \ g_{m 4} & =\frac{2 I_{D 4}}{\left|V_{O V} ight|}=\frac{2 imes 0.03}{0.2}=0.3 \mathrm{~mA} / \mathrm{V} \ r_{o 4} & =\frac{\left|V_{A} ight|}{I_{D 4}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \ g_{m 4} r_{o 4} & =0.3 imes 200=60 \ R_{o 4} & =60(40 \| 50) \ & =1.33 \mathrm{M} \Omega \ R_{o 6} & =\left(g_{m 6} r_{o 6} ight) r_{o 8} \ g_{m 6} & =\frac{2 I_{D 6}}{V_{O V}}=\frac{2 imes 0.03}{0.2}=0.3 \mathrm{~mA} / \mathrm{V} \ r_{o 6} & =\frac{V_{A}}{I_{D 6}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \end{aligned} \begin{aligned} r_{o 8} & =\frac{V_{A}}{I_{D 8}}=\frac{6}{0.03}=200 \mathrm{k} \Omega \ R_{o 6} & =(0.3 imes 200) imes 200=12 \mathrm{M} \Omega \ R_{O} & =R_{o 4}\left\|R_{o 6}=1.33 ight\| 12 \ & =1.2 \mathrm{M} \Omega \end{aligned} (h) \begin{aligned} A_{v} & =G_{m} R_{O}=1.2 imes 1.2 imes 10^{3}=1.44 imes 10^{3} \ & =1440 \mathrm{~V} / \mathrm{V} \end{aligned} (i) Figure 13.2.2 Figure 13.2.2 shows the output equivalent circuit of the amplifier. At high frequencies, \begin{aligned} & V_{o}=\frac{G_{m} V_{i d}}{s C_{L}} \ \left|\frac{V_{o}}{V_{i d}} ight|=1 ext { at } & \ & \omega=\omega_{t}=\frac{G_{m}}{C_{L}} \end{aligned} Thus, \begin{aligned} C_{L} & =\frac{G_{m}}{\omega_{t}} \ & =\frac{1.2 imes 10^{-3}}{2 \pi imes 100 imes 10^{6}}=1.9 \mathrm{pF} \end{aligned}$
Figure 13.2.2
Figure 13.2.2 shows the output equivalent circuit of the amplifier. At high frequencies,

\begin{aligned}& V_{o}=\frac{G_{m} V_{i d}}{s C_{L}} \\left|\frac{V_{o}}{V_{i d}} ight|=1 ext { at } & \& \omega=\omega_{t}=\frac{G_{m}}{C_{L}}\end{aligned}

Thus,

\begin{aligned}C_{L} & =\frac{G_{m}}{\omega_{t}} \& =\frac{1.2 imes 10^{-3}}{2 \pi imes 100 imes 10^{6}}=1.9 \mathrm{pF}\end{aligned}

Figure 132 VDD=VSS=1 VV_{D D}=V_{S S}=1 \mathrm{~V}VDD​=VSS​=1 V And Vtn=−Vtp=0.4 VV_{t n}=-V_{t p}=0.4 \mathrm{~V}Vtn​=−Vtp​=0.4 V , and Assume That for All Transistors

Figure 132 $V_{D D}=V_{S S}=1 \mathrm{~V}$ And $V_{t n}=-V_{t p}=0.4 \mathrm{~V}$ , and Assume That for All Transistors