An amplifier with an open-loop gain of $1000 \mathrm{~V} / \mathrm{V}$ is connected in a feedback loop with the result that the closed-loop gain obtained is $10 \mathrm{~V} / \mathrm{V}$.
(a) Find the value of the amount of feedback, the loop gain, and the feedback factor.
(b) If the open-loop gain varies by $\pm 10 \%$, what is the corresponding variation of the closedloop gain?
(c) If the open-loop amplifier has 3-dB frequencies $f_{L}=1 \mathrm{kHz}$ and $f_{H}=100 \mathrm{kHz}$, and the gain rolls off at the low and high frequency ends with a slope of $20 \mathrm{~dB} / \mathrm{decade}$, find the $3-\mathrm{dB}$ frequencies of the closed-loop gain and sketch the Bode magnitude plot of both the open-loop and closed-loop gains (on the same axes).

Question

Quizplus · Accepted Answer

The Answer is

A=1000 \mathrm{~V} / \mathrm{V}, \quad A_{f}=10 \mathrm{~V} / \mathrm{V}

(a)

\begin{aligned}A_{f} & =\frac{A}{1+A \beta} \10 & =\frac{1000}{1+A \beta}\end{aligned}

Thus,
Amount of feedback

\equiv 1+A \beta=100

ext { Loop gain } \equiv A \beta=99

Feedback factor

\equiv \beta=\frac{99}{1000}=0.099 \mathrm{~V} / \mathrm{V}

(b)

\begin{aligned}\frac{\Delta A_{f}}{A_{f}} & =\frac{ riangle A / A}{1+A \beta} \& =\frac{ \pm 10 \%}{100}= \pm 0.1 \%\end{aligned}

(c)

\begin{aligned}f_{L} & =1 \mathrm{kHz}, \quad f_{H}=100 \mathrm{kHz} \f_{L f} & =\frac{f_{L}}{1+A \beta}=\frac{1000}{100}=10 \mathrm{~Hz} \f_{H f} & =(1+A \beta) f_{H}=100 imes 100=10 \mathrm{MHz}\end{aligned}

$A=1000 \mathrm{~V} / \mathrm{V}, \quad A_{f}=10 \mathrm{~V} / \mathrm{V} (a) \begin{aligned} A_{f} & =\frac{A}{1+A \beta} \ 10 & =\frac{1000}{1+A \beta} \end{aligned} Thus, Amount of feedback \equiv 1+A \beta=100 ext { Loop gain } \equiv A \beta=99 Feedback factor \equiv \beta=\frac{99}{1000}=0.099 \mathrm{~V} / \mathrm{V} (b) \begin{aligned} \frac{\Delta A_{f}}{A_{f}} & =\frac{ riangle A / A}{1+A \beta} \ & =\frac{ \pm 10 \%}{100}= \pm 0.1 \% \end{aligned} (c) \begin{aligned} f_{L} & =1 \mathrm{kHz}, \quad f_{H}=100 \mathrm{kHz} \ f_{L f} & =\frac{f_{L}}{1+A \beta}=\frac{1000}{100}=10 \mathrm{~Hz} \ f_{H f} & =(1+A \beta) f_{H}=100 imes 100=10 \mathrm{MHz} \end{aligned} Figure 11.1.1 Sketches of the Bode plots for |A| and \left|A_{f} ight| are shown in Fig. 11.1.1.$

Figure 11.1.1
Sketches of the Bode plots for

|A|

and

\left|A_{f} ight|

are shown in Fig. 11.1.1.

An Amplifier with an Open-Loop Gain Of 1000 V/V1000 \mathrm{~V} / \mathrm{V}1000 V/V

An Amplifier with an Open-Loop Gain Of $1000 \mathrm{~V} / \mathrm{V}$