The feedback amplifier shown in Fig. 11.3.1(a) employs the shunt-shunt feedback topology. To see this more clearly and to apply the feedback analysis method, convert the signal source $V_{S}$ together with the resistance $R_{1}$ to its Norton's equivalent circuit composed of current source $I_{S}=$ $V_{S} / R_{1}$ together with a parallel resistance equal to $R_{1}$. The amplifier $\mu$ has the equivalent circuit shown in Fig. 11.3.1(b). We wish to investigate the case for which $\mu=10^{3} \mathrm{~V} / \mathrm{V}, R_{i d}=1 \mathrm{M} \Omega$, $r_{o}=100 \Omega, R_{1}=1 \mathrm{k} \Omega$, and $R_{2}=100 \mathrm{k} \Omega$.
(a) Find the $A$ circuit and derive expressions for $A \equiv V_{o} / I_{S}, R_{i}$, and $R_{o}$. Evaluate these expressions to determine $A, R_{i}$, and $R_{o}$.
(b) Find the $\beta$ circuit and derive an expression for $\beta$. Find the value of $\beta$.
(c) Find an expression for $A \beta$ and its value.
(d) Find the value of the amount of feedback.
(e) Find the closed-loop gain and hence $V_{o} / V_{S}$.
(f) Find the value of $R_{\text {in }}$.
(g) Find the value of $R_{\text {out }}$.
(h) If $\mu$ changes by $-10 \%$, what is the resulting change in $V_{o} / V_{S}$ ?

Question

Quizplus · Accepted Answer

The Answer is $(a) Figure 11.3.2(a) shows the shunt-shunt feedback amplifier (transresistance amplifier) with V_{S} and R_{1} converted to the Norton form. The \beta circuit is shown in Fig. 11.3.2(b) and the A circuit is shown in Fig. 11.3.2(c). From the latter, \begin{aligned} R_{i} & =R_{1}\left\|R_{2} ight\| R_{i d} \ & =1\|100\| 1000=989.1 \Omega \ V_{1} & =-I_{i} R_{i} \ V_{o} & =\mu V_{1} \frac{R_{2}}{R_{2}+r_{o}} \end{aligned} Thus, \begin{aligned} A & \equiv \frac{V_{o}}{I_{i}}=-\mu R_{i} \frac{R_{2}}{R_{2}+r_{o}} \ A & =-\mu\left(R_{1}\left\|R_{2} ight\| R_{i d} ight) \frac{R_{2}}{R_{2}+r_{o}} \ A & =-10^{3} imes 0.9891 imes \frac{100}{100+0.1} \ & =-988.1 \mathrm{k} \Omega \ R_{O} & =R_{2} \| r_{o} \ R_{O} & =100 \mathrm{k} \Omega \| 100 \Omega=99.9 \Omega \end{aligned} (b) The \beta circuit is shown in Fig. 11.3.2(b). \begin{aligned} & \beta \equiv \frac{I_{f}}{V_{o}}=-\frac{1}{R_{2}} \ & \beta=-\frac{1}{100}=-0.01 \mathrm{~mA} / \mathrm{V} \end{aligned} (c) \begin{aligned} A \beta & =-\mu\left(R_{1}\left\|R_{2} ight\| R_{i d} ight) \frac{R_{2}}{R_{2}+r_{o}} imes \frac{-1}{R_{2}} \ A \beta & =\mu \frac{R_{1}\left\|R_{2} ight\| R_{i d}}{R_{2}+r_{o}} \ & =10^{3} imes \frac{989.1 \Omega}{100.1 \mathrm{k} \Omega} \ & =9.88 \end{aligned} (d) 1+A \beta=10.88 (e) \begin{gathered} A_{f} \equiv \frac{V_{o}}{I_{S}}=\frac{A}{1+A \beta}=-\frac{988.1 \mathrm{k} \Omega}{10.88}=-90.8 \mathrm{k} \Omega \ \frac{V_{o}}{V_{S}}=\frac{V_{o}}{I_{S} R_{1}}=-\frac{90.8 \mathrm{k} \Omega}{1 \mathrm{k} \Omega}=-90.8 \mathrm{~V} / \mathrm{V} \end{gathered} (f) R_{i f}=\frac{R_{i}}{1+A \beta}=\frac{989.1}{10.88}=90.9 \Omega But, \begin{aligned} \frac{1}{R_{ ext {if }}} & =\frac{1}{R_{ ext {in }}}+\frac{1}{R_{1}} \ \Rightarrow R_{ ext {in }} & =1 /\left(\frac{1}{90.9}-\frac{1}{1000} ight)=100 \Omega \end{aligned} (g) R_{\mathrm{out}}=R_{o f}=\frac{R_{o}}{1+A \beta}=\frac{99.9}{10.88}=9.2 \Omega (h) \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}{10.88}=-0.92 \% \ & \Rightarrow ext { Change in } \frac{V_{o}}{V_{S}} ext { is }-0.92 \% \end{aligned} Figure 11.4.3$

(a) Figure 11.3.2(a) shows the shunt-shunt feedback amplifier (transresistance amplifier) with

V_{S}

and

R_{1}

converted to the Norton form. The

\beta

circuit is shown in Fig. 11.3.2(b) and the

A

circuit is shown in Fig. 11.3.2(c). From the latter,

\begin{aligned}R_{i} & =R_{1}\left\|R_{2} ight\| R_{i d} \& =1\|100\| 1000=989.1 \Omega \V_{1} & =-I_{i} R_{i} \V_{o} & =\mu V_{1} \frac{R_{2}}{R_{2}+r_{o}}\end{aligned}

Thus,

\begin{aligned}A & \equiv \frac{V_{o}}{I_{i}}=-\mu R_{i} \frac{R_{2}}{R_{2}+r_{o}} \A & =-\mu\left(R_{1}\left\|R_{2} ight\| R_{i d} ight) \frac{R_{2}}{R_{2}+r_{o}} \A & =-10^{3} imes 0.9891 imes \frac{100}{100+0.1} \& =-988.1 \mathrm{k} \Omega \R_{O} & =R_{2} \| r_{o} \R_{O} & =100 \mathrm{k} \Omega \| 100 \Omega=99.9 \Omega\end{aligned}

(b) The

\beta

circuit is shown in Fig. 11.3.2(b).

\begin{aligned}& \beta \equiv \frac{I_{f}}{V_{o}}=-\frac{1}{R_{2}} \& \beta=-\frac{1}{100}=-0.01 \mathrm{~mA} / \mathrm{V}\end{aligned}

(c)

\begin{aligned}A \beta & =-\mu\left(R_{1}\left\|R_{2} ight\| R_{i d} ight) \frac{R_{2}}{R_{2}+r_{o}} imes \frac{-1}{R_{2}} \A \beta & =\mu \frac{R_{1}\left\|R_{2} ight\| R_{i d}}{R_{2}+r_{o}} \& =10^{3} imes \frac{989.1 \Omega}{100.1 \mathrm{k} \Omega} \& =9.88\end{aligned}

(d)

1+A \beta=10.88

(e)

\begin{gathered}A_{f} \equiv \frac{V_{o}}{I_{S}}=\frac{A}{1+A \beta}=-\frac{988.1 \mathrm{k} \Omega}{10.88}=-90.8 \mathrm{k} \Omega \\frac{V_{o}}{V_{S}}=\frac{V_{o}}{I_{S} R_{1}}=-\frac{90.8 \mathrm{k} \Omega}{1 \mathrm{k} \Omega}=-90.8 \mathrm{~V} / \mathrm{V}\end{gathered}

(f)

R_{i f}=\frac{R_{i}}{1+A \beta}=\frac{989.1}{10.88}=90.9 \Omega

But,

\begin{aligned}\frac{1}{R_{ ext {if }}} & =\frac{1}{R_{ ext {in }}}+\frac{1}{R_{1}} \\Rightarrow R_{ ext {in }} & =1 /\left(\frac{1}{90.9}-\frac{1}{1000} ight)=100 \Omega\end{aligned}

(g)

R_{\mathrm{out}}=R_{o f}=\frac{R_{o}}{1+A \beta}=\frac{99.9}{10.88}=9.2 \Omega

(h)

\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}{10.88}=-0.92 \% \& \Rightarrow ext { Change in } \frac{V_{o}}{V_{S}} ext { is }-0.92 \%\end{aligned}

$(a) Figure 11.3.2(a) shows the shunt-shunt feedback amplifier (transresistance amplifier) with V_{S} and R_{1} converted to the Norton form. The \beta circuit is shown in Fig. 11.3.2(b) and the A circuit is shown in Fig. 11.3.2(c). From the latter, \begin{aligned} R_{i} & =R_{1}\left\|R_{2} ight\| R_{i d} \ & =1\|100\| 1000=989.1 \Omega \ V_{1} & =-I_{i} R_{i} \ V_{o} & =\mu V_{1} \frac{R_{2}}{R_{2}+r_{o}} \end{aligned} Thus, \begin{aligned} A & \equiv \frac{V_{o}}{I_{i}}=-\mu R_{i} \frac{R_{2}}{R_{2}+r_{o}} \ A & =-\mu\left(R_{1}\left\|R_{2} ight\| R_{i d} ight) \frac{R_{2}}{R_{2}+r_{o}} \ A & =-10^{3} imes 0.9891 imes \frac{100}{100+0.1} \ & =-988.1 \mathrm{k} \Omega \ R_{O} & =R_{2} \| r_{o} \ R_{O} & =100 \mathrm{k} \Omega \| 100 \Omega=99.9 \Omega \end{aligned} (b) The \beta circuit is shown in Fig. 11.3.2(b). \begin{aligned} & \beta \equiv \frac{I_{f}}{V_{o}}=-\frac{1}{R_{2}} \ & \beta=-\frac{1}{100}=-0.01 \mathrm{~mA} / \mathrm{V} \end{aligned} (c) \begin{aligned} A \beta & =-\mu\left(R_{1}\left\|R_{2} ight\| R_{i d} ight) \frac{R_{2}}{R_{2}+r_{o}} imes \frac{-1}{R_{2}} \ A \beta & =\mu \frac{R_{1}\left\|R_{2} ight\| R_{i d}}{R_{2}+r_{o}} \ & =10^{3} imes \frac{989.1 \Omega}{100.1 \mathrm{k} \Omega} \ & =9.88 \end{aligned} (d) 1+A \beta=10.88 (e) \begin{gathered} A_{f} \equiv \frac{V_{o}}{I_{S}}=\frac{A}{1+A \beta}=-\frac{988.1 \mathrm{k} \Omega}{10.88}=-90.8 \mathrm{k} \Omega \ \frac{V_{o}}{V_{S}}=\frac{V_{o}}{I_{S} R_{1}}=-\frac{90.8 \mathrm{k} \Omega}{1 \mathrm{k} \Omega}=-90.8 \mathrm{~V} / \mathrm{V} \end{gathered} (f) R_{i f}=\frac{R_{i}}{1+A \beta}=\frac{989.1}{10.88}=90.9 \Omega But, \begin{aligned} \frac{1}{R_{ ext {if }}} & =\frac{1}{R_{ ext {in }}}+\frac{1}{R_{1}} \ \Rightarrow R_{ ext {in }} & =1 /\left(\frac{1}{90.9}-\frac{1}{1000} ight)=100 \Omega \end{aligned} (g) R_{\mathrm{out}}=R_{o f}=\frac{R_{o}}{1+A \beta}=\frac{99.9}{10.88}=9.2 \Omega (h) \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}{10.88}=-0.92 \% \ & \Rightarrow ext { Change in } \frac{V_{o}}{V_{S}} ext { is }-0.92 \% \end{aligned} Figure 11.4.3$

Figure 11.4.3

The Feedback Amplifier Shown in Fig VSV_{S}VS​ Together with the Resistance R1R_{1}R1​

The Feedback Amplifier Shown in Fig $V_{S}$ Together with the Resistance $R_{1}$