Figure 118$\left(R_{1}, R_{2}\right)$ Each of the Four MOS Transistors Is Operating at with Negative

Question

Figure 118$\left(R_{1}, R_{2}\right)$  Each of the Four MOS Transistors Is Operating at with Negative

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The Answer is $Figure 11.8.1 Figure 11.8.2 Figure 11.8.3 (a) From the \beta circuit, shown in Fig. 11.8.2, \beta \equiv \frac{V_{f}}{V_{o}}=\frac{R_{1}}{R_{1}+R_{2}} The ideal value of the closed-loop gain is \begin{aligned} \left.A_{f} ight|_{ ext {ideal }} & \left.\equiv \frac{V_{o}}{V_{s}} ight|_{ ext {ideal }}=\frac{1}{\beta}=1+\frac{R_{2}}{R_{1}} \ \beta & =\frac{200}{200+800}=0.2 \mathrm{~V} / \mathrm{V} \ \frac{V_{o}}{V_{s}} & =1+\frac{800}{200}=5 \mathrm{~V} / \mathrm{V} \end{aligned} (b) The A circuit is shown in Fig. 11.8.3 above \begin{aligned} A \equiv \frac{V_{o}}{V_{i}} & =g_{m 1,2}\left[r_{o 2}\left\|r_{o 4} ight\|\left(R_{1}+R_{2} ight) ight] \ R_{O} & =r_{o 2}\left\|r_{o 4} ight\|\left(R_{1}+R_{2} ight) \ g_{m 1,2} & =\frac{2 I_{D 1,2}}{V_{O V}} \ & =\frac{2 imes 0.1}{0.2}=1 \mathrm{~mA} / \mathrm{V} \ r_{O 2} & =r_{o 4}=\frac{\left|V_{A} ight|}{I_{D 2,4}}=\frac{20}{0.1}=200 \mathrm{k} \Omega \end{aligned} Thus, \begin{aligned} R_{O} & =200\|200\| 1000=90.9 \mathrm{k} \Omega \ A & =1 imes 90.9=90.9 \mathrm{~V} / \mathrm{V} \end{aligned} (c) \begin{aligned} A_{f} \equiv \frac{V_{o}}{V_{S}} & =\frac{A}{1+A \beta} \ & =\frac{90.9}{1+90.9 imes 0.2}=\frac{90.9}{19.18} \ & =4.74 \mathrm{~V} / \mathrm{V} \end{aligned} which is only 5.2 \% lower than the ideal value found in (a). (d) \begin{aligned} R_{o f} & =\frac{R_{O}}{1+A \beta}=\frac{90.9 \mathrm{k} \Omega}{19.18}=4.74 \mathrm{k} \Omega \ R_{ ext {out }} & =R_{o f}=4.74 \mathrm{k} \Omega \end{aligned}$

Figure 11.8.1
$Figure 11.8.1 Figure 11.8.2 Figure 11.8.3 (a) From the \beta circuit, shown in Fig. 11.8.2, \beta \equiv \frac{V_{f}}{V_{o}}=\frac{R_{1}}{R_{1}+R_{2}} The ideal value of the closed-loop gain is \begin{aligned} \left.A_{f} ight|_{ ext {ideal }} & \left.\equiv \frac{V_{o}}{V_{s}} ight|_{ ext {ideal }}=\frac{1}{\beta}=1+\frac{R_{2}}{R_{1}} \ \beta & =\frac{200}{200+800}=0.2 \mathrm{~V} / \mathrm{V} \ \frac{V_{o}}{V_{s}} & =1+\frac{800}{200}=5 \mathrm{~V} / \mathrm{V} \end{aligned} (b) The A circuit is shown in Fig. 11.8.3 above \begin{aligned} A \equiv \frac{V_{o}}{V_{i}} & =g_{m 1,2}\left[r_{o 2}\left\|r_{o 4} ight\|\left(R_{1}+R_{2} ight) ight] \ R_{O} & =r_{o 2}\left\|r_{o 4} ight\|\left(R_{1}+R_{2} ight) \ g_{m 1,2} & =\frac{2 I_{D 1,2}}{V_{O V}} \ & =\frac{2 imes 0.1}{0.2}=1 \mathrm{~mA} / \mathrm{V} \ r_{O 2} & =r_{o 4}=\frac{\left|V_{A} ight|}{I_{D 2,4}}=\frac{20}{0.1}=200 \mathrm{k} \Omega \end{aligned} Thus, \begin{aligned} R_{O} & =200\|200\| 1000=90.9 \mathrm{k} \Omega \ A & =1 imes 90.9=90.9 \mathrm{~V} / \mathrm{V} \end{aligned} (c) \begin{aligned} A_{f} \equiv \frac{V_{o}}{V_{S}} & =\frac{A}{1+A \beta} \ & =\frac{90.9}{1+90.9 imes 0.2}=\frac{90.9}{19.18} \ & =4.74 \mathrm{~V} / \mathrm{V} \end{aligned} which is only 5.2 \% lower than the ideal value found in (a). (d) \begin{aligned} R_{o f} & =\frac{R_{O}}{1+A \beta}=\frac{90.9 \mathrm{k} \Omega}{19.18}=4.74 \mathrm{k} \Omega \ R_{ ext {out }} & =R_{o f}=4.74 \mathrm{k} \Omega \end{aligned}$
Figure 11.8.2
$Figure 11.8.1 Figure 11.8.2 Figure 11.8.3 (a) From the \beta circuit, shown in Fig. 11.8.2, \beta \equiv \frac{V_{f}}{V_{o}}=\frac{R_{1}}{R_{1}+R_{2}} The ideal value of the closed-loop gain is \begin{aligned} \left.A_{f} ight|_{ ext {ideal }} & \left.\equiv \frac{V_{o}}{V_{s}} ight|_{ ext {ideal }}=\frac{1}{\beta}=1+\frac{R_{2}}{R_{1}} \ \beta & =\frac{200}{200+800}=0.2 \mathrm{~V} / \mathrm{V} \ \frac{V_{o}}{V_{s}} & =1+\frac{800}{200}=5 \mathrm{~V} / \mathrm{V} \end{aligned} (b) The A circuit is shown in Fig. 11.8.3 above \begin{aligned} A \equiv \frac{V_{o}}{V_{i}} & =g_{m 1,2}\left[r_{o 2}\left\|r_{o 4} ight\|\left(R_{1}+R_{2} ight) ight] \ R_{O} & =r_{o 2}\left\|r_{o 4} ight\|\left(R_{1}+R_{2} ight) \ g_{m 1,2} & =\frac{2 I_{D 1,2}}{V_{O V}} \ & =\frac{2 imes 0.1}{0.2}=1 \mathrm{~mA} / \mathrm{V} \ r_{O 2} & =r_{o 4}=\frac{\left|V_{A} ight|}{I_{D 2,4}}=\frac{20}{0.1}=200 \mathrm{k} \Omega \end{aligned} Thus, \begin{aligned} R_{O} & =200\|200\| 1000=90.9 \mathrm{k} \Omega \ A & =1 imes 90.9=90.9 \mathrm{~V} / \mathrm{V} \end{aligned} (c) \begin{aligned} A_{f} \equiv \frac{V_{o}}{V_{S}} & =\frac{A}{1+A \beta} \ & =\frac{90.9}{1+90.9 imes 0.2}=\frac{90.9}{19.18} \ & =4.74 \mathrm{~V} / \mathrm{V} \end{aligned} which is only 5.2 \% lower than the ideal value found in (a). (d) \begin{aligned} R_{o f} & =\frac{R_{O}}{1+A \beta}=\frac{90.9 \mathrm{k} \Omega}{19.18}=4.74 \mathrm{k} \Omega \ R_{ ext {out }} & =R_{o f}=4.74 \mathrm{k} \Omega \end{aligned}$

Figure 11.8.3
(a) From the

\beta

circuit, shown in Fig. 11.8.2,

\beta \equiv \frac{V_{f}}{V_{o}}=\frac{R_{1}}{R_{1}+R_{2}}

The ideal value of the closed-loop gain is

\begin{aligned}\left.A_{f} ight|_{ ext {ideal }} & \left.\equiv \frac{V_{o}}{V_{s}} ight|_{ ext {ideal }}=\frac{1}{\beta}=1+\frac{R_{2}}{R_{1}} \\beta & =\frac{200}{200+800}=0.2 \mathrm{~V} / \mathrm{V} \\frac{V_{o}}{V_{s}} & =1+\frac{800}{200}=5 \mathrm{~V} / \mathrm{V}\end{aligned}

(b) The

A

circuit is shown in Fig. 11.8.3 above

\begin{aligned}A \equiv \frac{V_{o}}{V_{i}} & =g_{m 1,2}\left[r_{o 2}\left\|r_{o 4} ight\|\left(R_{1}+R_{2} ight) ight] \R_{O} & =r_{o 2}\left\|r_{o 4} ight\|\left(R_{1}+R_{2} ight) \g_{m 1,2} & =\frac{2 I_{D 1,2}}{V_{O V}} \& =\frac{2 imes 0.1}{0.2}=1 \mathrm{~mA} / \mathrm{V} \r_{O 2} & =r_{o 4}=\frac{\left|V_{A} ight|}{I_{D 2,4}}=\frac{20}{0.1}=200 \mathrm{k} \Omega\end{aligned}

Thus,

\begin{aligned}R_{O} & =200\|200\| 1000=90.9 \mathrm{k} \Omega \A & =1 imes 90.9=90.9 \mathrm{~V} / \mathrm{V}\end{aligned}

(c)

\begin{aligned}A_{f} \equiv \frac{V_{o}}{V_{S}} & =\frac{A}{1+A \beta} \& =\frac{90.9}{1+90.9 imes 0.2}=\frac{90.9}{19.18} \& =4.74 \mathrm{~V} / \mathrm{V}\end{aligned}

which is only

5.2 \%

lower than the ideal value found in (a).
(d)

\begin{aligned}R_{o f} & =\frac{R_{O}}{1+A \beta}=\frac{90.9 \mathrm{k} \Omega}{19.18}=4.74 \mathrm{k} \Omega \R_{ ext {out }} & =R_{o f}=4.74 \mathrm{k} \Omega\end{aligned}

Figure 118 $\left(R_{1}, R_{2}\right)$ Each of the Four MOS Transistors Is Operating at with Negative

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Figure 118 (R1,R2)\left(R_{1}, R_{2}\right)(R1​,R2​) Each of the Four MOS Transistors Is Operating at with Negative

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Figure 118 $\left(R_{1}, R_{2}\right)$ Each of the Four MOS Transistors Is Operating at with Negative

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