Figure 11.2.1
The series-shunt feedback amplifier shown in Fig. 11.2.1 utilizes a differential amplifier with a gain $\mu=10^{3}$, input resistance $R_{i d}=100 \mathrm{k} \Omega$, and output resistance $r_{o}=1 \mathrm{k} \Omega$. The feedback amplifier is fed with a signal source $V_{S}$ having a source resistance $R_{S}=10 \mathrm{k} \Omega$, and a load resistance $R_{L}=$ $1 \mathrm{k} \Omega$ is connected to the output terminal.
(a) If it is required to obtain a closed-loop gain $A_{f} \equiv V_{o} / V_{S}$ that is ideally $10 \mathrm{~V} / \mathrm{V}$, what is the required feedback factor $\beta$ ? If $R_{1}=1 \mathrm{k} \Omega$, what value must $R_{2}$ have?
(b) Give the complete $A$ circuit and use it to determine $A, R_{i}$, and $R_{o}$.
(c) Find the actual value of $A_{f}$ realized.
(d) Find the input resistance $R_{\text {in }}$ of the feedback amplifier.
(e) Find the output resistance $R_{\text {out }}$ of the feedback amplifier.
(f) If, due to temperature variation, the gain $\mu$ changes by $10 \%$, what percentage change do you expect $A_{f}$ to have?
(g) If the high-frequency response of the amplifier $\mu$ is characterized by a drop of $20 \mathrm{~dB} /$ decade with a 3-dB frequency of $10 \mathrm{kHz}$, what do you expect the 3-dB frequency of the gain $A_{f}$ to be? At what frequency will the magnitude of $A_{f}$ become unity?

Question

Figure 11.2.1
The series-shunt feedback amplifier shown in Fig. 11.2.1 utilizes a differential amplifier with a gain $\mu=10^{3}$, input resistance $R_{i d}=100 \mathrm{k} \Omega$, and output resistance $r_{o}=1 \mathrm{k} \Omega$. The feedback amplifier is fed with a signal source $V_{S}$ having a source resistance $R_{S}=10 \mathrm{k} \Omega$, and a load resistance $R_{L}=$ $1 \mathrm{k} \Omega$ is connected to the output terminal.
(a) If it is required to obtain a closed-loop gain $A_{f} \equiv V_{o} / V_{S}$ that is ideally $10 \mathrm{~V} / \mathrm{V}$, what is the required feedback factor $\beta$ ? If $R_{1}=1 \mathrm{k} \Omega$, what value must $R_{2}$ have?
(b) Give the complete $A$ circuit and use it to determine $A, R_{i}$, and $R_{o}$.
(c) Find the actual value of $A_{f}$ realized.
(d) Find the input resistance $R_{\text {in }}$ of the feedback amplifier.
(e) Find the output resistance $R_{\text {out }}$ of the feedback amplifier. 
(f) If, due to temperature variation, the gain $\mu$ changes by $10 \%$, what percentage change do you expect $A_{f}$ to have?
(g) If the high-frequency response of the amplifier $\mu$ is characterized by a drop of $20 \mathrm{~dB} /$ decade with a 3-dB frequency of $10 \mathrm{kHz}$, what do you expect the 3-dB frequency of the gain $A_{f}$ to be? At what frequency will the magnitude of $A_{f}$ become unity?

Quizplus · Accepted Answer

The Answer is $Figure 11.2.1 (a) For A_{f} to be ideally 10 \mathrm{~V} / \mathrm{V} , \beta=\frac{1}{A_{f}}=0.1 \mathrm{~V} / \mathrm{V} Figure 11.2.2 From the feedback network shown in Fig. 11.2.2, \begin{aligned} \beta & =\frac{R_{1}}{R_{1}+R_{2}} \ 0.1 & =\frac{1}{1+R_{2}} \ \Rightarrow R_{2} & =9 \mathrm{k} \Omega \end{aligned} (b) The A circuit is shown in Fig. 11.2.3 (refer to Figure below). Figure 11.2.3 \begin{aligned} A & =\frac{V_{o}}{V_{i}}=\frac{V_{1}}{V_{i}} \frac{V_{o}}{V_{1}} \ & =\frac{R_{i d}}{R_{S}+R_{i d}+\left(R_{1} \| R_{2} ight)} \mu \frac{R_{L} \|\left(R_{1}+R_{2} ight)}{\left[R_{L} \|\left(R_{1}+R_{2} ight) ight]+r_{o}} \ & =\frac{100}{10+100+(1 \| 9)} imes 10^{3} imes \frac{1 \|(1+9)}{[1 \|(1+9)]+1} \ & =429.4 \mathrm{~V} / \mathrm{V} \ R_{i} & =R_{S}+R_{i d}+\left(R_{1} \| R_{2} ight) \ & =10+100+(1 \| 9) \ & =110.9 \mathrm{k} \Omega \ R_{O} & =R_{L}\left\|\left(R_{1}+R_{2} ight) ight\| r_{o} \ & =1\|(1+9)\| 1 \ & =476.2 \Omega \end{aligned} (c) \begin{aligned} A_{f} & =\frac{A}{1+A \beta} \ & =\frac{429.4}{1+429.4 imes 0.1} \ & =9.77 \mathrm{~V} / \mathrm{V} \end{aligned} (d) \begin{aligned} R_{i f} & =(1+A \beta) R_{i} \ & =(1+429.4 imes 0.1) imes 110.9 \ & =4.873 \mathrm{M} \Omega \ R_{ ext {in }} & =R_{i f}-R_{S}=4873-10 \ & =4.863 \mathrm{M} \Omega \end{aligned} (e) \begin{aligned} R_{o f} & =\frac{R_{o}}{1+A \beta} \ & =\frac{476.2}{43.94}=10.84 \Omega \end{aligned} Since \begin{aligned} \frac{1}{R_{ ext {of }}} & =\frac{1}{R_{L}}+\frac{1}{R_{\mathrm{out}}} \ \Rightarrow R_{\mathrm{out}} & =10.96 \Omega \end{aligned} (f) \begin{aligned} \frac{ riangle \mu}{\mu} & =10 \% \ \Rightarrow \frac{ riangle A}{A} & =10 \% \ \Rightarrow \frac{ riangle A_{f}}{A_{f}} & =\frac{ riangle A / A}{1+A \beta} \ & =\frac{10}{43.94}=0.23 \% \end{aligned} (g) \begin{aligned} f_{H f} & =(1+A \beta) f_{H} \ & =43.94 imes 10 \ & =439.4 \mathrm{kHz} \end{aligned} \left|A_{f} ight| reaches unity at \begin{aligned} f_{t} & =\left|A_{f} ight| imes f_{H f}=9.77 imes 439.4 \ & =4.29 \mathrm{MHz} \end{aligned}$

Figure 11.2.1
(a) For

A_{f}

to be ideally

10 \mathrm{~V} / \mathrm{V}

,

\beta=\frac{1}{A_{f}}=0.1 \mathrm{~V} / \mathrm{V}

$Figure 11.2.1 (a) For A_{f} to be ideally 10 \mathrm{~V} / \mathrm{V} , \beta=\frac{1}{A_{f}}=0.1 \mathrm{~V} / \mathrm{V} Figure 11.2.2 From the feedback network shown in Fig. 11.2.2, \begin{aligned} \beta & =\frac{R_{1}}{R_{1}+R_{2}} \ 0.1 & =\frac{1}{1+R_{2}} \ \Rightarrow R_{2} & =9 \mathrm{k} \Omega \end{aligned} (b) The A circuit is shown in Fig. 11.2.3 (refer to Figure below). Figure 11.2.3 \begin{aligned} A & =\frac{V_{o}}{V_{i}}=\frac{V_{1}}{V_{i}} \frac{V_{o}}{V_{1}} \ & =\frac{R_{i d}}{R_{S}+R_{i d}+\left(R_{1} \| R_{2} ight)} \mu \frac{R_{L} \|\left(R_{1}+R_{2} ight)}{\left[R_{L} \|\left(R_{1}+R_{2} ight) ight]+r_{o}} \ & =\frac{100}{10+100+(1 \| 9)} imes 10^{3} imes \frac{1 \|(1+9)}{[1 \|(1+9)]+1} \ & =429.4 \mathrm{~V} / \mathrm{V} \ R_{i} & =R_{S}+R_{i d}+\left(R_{1} \| R_{2} ight) \ & =10+100+(1 \| 9) \ & =110.9 \mathrm{k} \Omega \ R_{O} & =R_{L}\left\|\left(R_{1}+R_{2} ight) ight\| r_{o} \ & =1\|(1+9)\| 1 \ & =476.2 \Omega \end{aligned} (c) \begin{aligned} A_{f} & =\frac{A}{1+A \beta} \ & =\frac{429.4}{1+429.4 imes 0.1} \ & =9.77 \mathrm{~V} / \mathrm{V} \end{aligned} (d) \begin{aligned} R_{i f} & =(1+A \beta) R_{i} \ & =(1+429.4 imes 0.1) imes 110.9 \ & =4.873 \mathrm{M} \Omega \ R_{ ext {in }} & =R_{i f}-R_{S}=4873-10 \ & =4.863 \mathrm{M} \Omega \end{aligned} (e) \begin{aligned} R_{o f} & =\frac{R_{o}}{1+A \beta} \ & =\frac{476.2}{43.94}=10.84 \Omega \end{aligned} Since \begin{aligned} \frac{1}{R_{ ext {of }}} & =\frac{1}{R_{L}}+\frac{1}{R_{\mathrm{out}}} \ \Rightarrow R_{\mathrm{out}} & =10.96 \Omega \end{aligned} (f) \begin{aligned} \frac{ riangle \mu}{\mu} & =10 \% \ \Rightarrow \frac{ riangle A}{A} & =10 \% \ \Rightarrow \frac{ riangle A_{f}}{A_{f}} & =\frac{ riangle A / A}{1+A \beta} \ & =\frac{10}{43.94}=0.23 \% \end{aligned} (g) \begin{aligned} f_{H f} & =(1+A \beta) f_{H} \ & =43.94 imes 10 \ & =439.4 \mathrm{kHz} \end{aligned} \left|A_{f} ight| reaches unity at \begin{aligned} f_{t} & =\left|A_{f} ight| imes f_{H f}=9.77 imes 439.4 \ & =4.29 \mathrm{MHz} \end{aligned}$

Figure 11.2.2
From the feedback network shown in Fig. 11.2.2,

\begin{aligned}\beta & =\frac{R_{1}}{R_{1}+R_{2}} \0.1 & =\frac{1}{1+R_{2}} \\Rightarrow R_{2} & =9 \mathrm{k} \Omega\end{aligned}

(b) The

A

circuit is shown in Fig. 11.2.3 (refer to Figure below).
$Figure 11.2.1 (a) For A_{f} to be ideally 10 \mathrm{~V} / \mathrm{V} , \beta=\frac{1}{A_{f}}=0.1 \mathrm{~V} / \mathrm{V} Figure 11.2.2 From the feedback network shown in Fig. 11.2.2, \begin{aligned} \beta & =\frac{R_{1}}{R_{1}+R_{2}} \ 0.1 & =\frac{1}{1+R_{2}} \ \Rightarrow R_{2} & =9 \mathrm{k} \Omega \end{aligned} (b) The A circuit is shown in Fig. 11.2.3 (refer to Figure below). Figure 11.2.3 \begin{aligned} A & =\frac{V_{o}}{V_{i}}=\frac{V_{1}}{V_{i}} \frac{V_{o}}{V_{1}} \ & =\frac{R_{i d}}{R_{S}+R_{i d}+\left(R_{1} \| R_{2} ight)} \mu \frac{R_{L} \|\left(R_{1}+R_{2} ight)}{\left[R_{L} \|\left(R_{1}+R_{2} ight) ight]+r_{o}} \ & =\frac{100}{10+100+(1 \| 9)} imes 10^{3} imes \frac{1 \|(1+9)}{[1 \|(1+9)]+1} \ & =429.4 \mathrm{~V} / \mathrm{V} \ R_{i} & =R_{S}+R_{i d}+\left(R_{1} \| R_{2} ight) \ & =10+100+(1 \| 9) \ & =110.9 \mathrm{k} \Omega \ R_{O} & =R_{L}\left\|\left(R_{1}+R_{2} ight) ight\| r_{o} \ & =1\|(1+9)\| 1 \ & =476.2 \Omega \end{aligned} (c) \begin{aligned} A_{f} & =\frac{A}{1+A \beta} \ & =\frac{429.4}{1+429.4 imes 0.1} \ & =9.77 \mathrm{~V} / \mathrm{V} \end{aligned} (d) \begin{aligned} R_{i f} & =(1+A \beta) R_{i} \ & =(1+429.4 imes 0.1) imes 110.9 \ & =4.873 \mathrm{M} \Omega \ R_{ ext {in }} & =R_{i f}-R_{S}=4873-10 \ & =4.863 \mathrm{M} \Omega \end{aligned} (e) \begin{aligned} R_{o f} & =\frac{R_{o}}{1+A \beta} \ & =\frac{476.2}{43.94}=10.84 \Omega \end{aligned} Since \begin{aligned} \frac{1}{R_{ ext {of }}} & =\frac{1}{R_{L}}+\frac{1}{R_{\mathrm{out}}} \ \Rightarrow R_{\mathrm{out}} & =10.96 \Omega \end{aligned} (f) \begin{aligned} \frac{ riangle \mu}{\mu} & =10 \% \ \Rightarrow \frac{ riangle A}{A} & =10 \% \ \Rightarrow \frac{ riangle A_{f}}{A_{f}} & =\frac{ riangle A / A}{1+A \beta} \ & =\frac{10}{43.94}=0.23 \% \end{aligned} (g) \begin{aligned} f_{H f} & =(1+A \beta) f_{H} \ & =43.94 imes 10 \ & =439.4 \mathrm{kHz} \end{aligned} \left|A_{f} ight| reaches unity at \begin{aligned} f_{t} & =\left|A_{f} ight| imes f_{H f}=9.77 imes 439.4 \ & =4.29 \mathrm{MHz} \end{aligned}$ Figure 11.2.3

\begin{aligned}A & =\frac{V_{o}}{V_{i}}=\frac{V_{1}}{V_{i}} \frac{V_{o}}{V_{1}} \& =\frac{R_{i d}}{R_{S}+R_{i d}+\left(R_{1} \| R_{2} ight)} \mu \frac{R_{L} \|\left(R_{1}+R_{2} ight)}{\left[R_{L} \|\left(R_{1}+R_{2} ight) ight]+r_{o}} \& =\frac{100}{10+100+(1 \| 9)} imes 10^{3} imes \frac{1 \|(1+9)}{[1 \|(1+9)]+1} \& =429.4 \mathrm{~V} / \mathrm{V} \R_{i} & =R_{S}+R_{i d}+\left(R_{1} \| R_{2} ight) \& =10+100+(1 \| 9) \& =110.9 \mathrm{k} \Omega \R_{O} & =R_{L}\left\|\left(R_{1}+R_{2} ight) ight\| r_{o} \& =1\|(1+9)\| 1 \& =476.2 \Omega\end{aligned}

(c)

\begin{aligned}A_{f} & =\frac{A}{1+A \beta} \& =\frac{429.4}{1+429.4 imes 0.1} \& =9.77 \mathrm{~V} / \mathrm{V}\end{aligned}

(d)

\begin{aligned}R_{i f} & =(1+A \beta) R_{i} \& =(1+429.4 imes 0.1) imes 110.9 \& =4.873 \mathrm{M} \Omega \R_{ ext {in }} & =R_{i f}-R_{S}=4873-10 \& =4.863 \mathrm{M} \Omega\end{aligned}

(e)

\begin{aligned}R_{o f} & =\frac{R_{o}}{1+A \beta} \& =\frac{476.2}{43.94}=10.84 \Omega\end{aligned}

Since

\begin{aligned}\frac{1}{R_{ ext {of }}} & =\frac{1}{R_{L}}+\frac{1}{R_{\mathrm{out}}} \\Rightarrow R_{\mathrm{out}} & =10.96 \Omega\end{aligned}

(f)

\begin{aligned}\frac{ riangle \mu}{\mu} & =10 \% \\Rightarrow \frac{ riangle A}{A} & =10 \% \\Rightarrow \frac{ riangle A_{f}}{A_{f}} & =\frac{ riangle A / A}{1+A \beta} \& =\frac{10}{43.94}=0.23 \%\end{aligned}

(g)

\begin{aligned}f_{H f} & =(1+A \beta) f_{H} \& =43.94 imes 10 \& =439.4 \mathrm{kHz}\end{aligned}

\left|A_{f} ight|

reaches unity at

\begin{aligned}f_{t} & =\left|A_{f} ight| imes f_{H f}=9.77 imes 439.4 \& =4.29 \mathrm{MHz}\end{aligned}

Figure 112 μ=103\mu=10^{3}μ=103 , Input Resistance Rid=100kΩR_{i d}=100 \mathrm{k} \OmegaRid​=100kΩ , and Output Resistance

Figure 112 $\mu=10^{3}$ , Input Resistance $R_{i d}=100 \mathrm{k} \Omega$ , and Output Resistance