Analog Electronics Test 5

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Analog Electronics Test 5

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Question 1
1 / -0

An amplifier with mid-band gain |A| = 600 has negative feedback \(\left| \beta \right| = \frac{1}{{100}}.\) If the upper cut-off without feedback were at 60 kHz, then with feedback it would become _______ kHz.

A
420-420

B
4204-4201

C
4207-4206

D
4209-4202
Solution
Concept:
With negative Feedback the effect on frequency is:
- Upper cut-off frequency increase by (1 + AB)
- Lower cut-off frequency decrease by (1 + AB)
Calculation:
Given:
\(\left| \beta \right| = \frac{1}{{100}}.\)
∴ The upper cut-off frequency (f'u) with feedback will be:
f'_u = f_U(1 + AB)
\(= 60\left( {1 + \frac{{600}}{{100}}} \right) \)
f'u = 420 kHz
Question 2
1 / -0

An amplifier has an open loop gain of 2000 ±20. Negative feedback is provided such that the gain variation remains with 0.2%. What is the amount of Feedback β_F?

A
\(\frac{1}{{500}}\)

B
\(\frac{1}{{100}}\)

C
\(\frac{1}{{10}}\)

D
\(\frac{9}{{1000}}\)

Solution

Suppose there is a small change in the internal resistance of an amplifier
∴ Fractional change of gain with feedback is given by
\(\begin{array}{l} {A_F} = \frac{A}{{1 + A\beta }}\\ \frac{{\partial {A_F}}}{{\partial A}} = \frac{{1 + A\beta - A\beta }}{{{{\left( {1 + A\beta } \right)}^2}}}\\\end{array}\)
\(\frac{{\partial {A_F}}}{{{A_F}}} = \frac{1}{{{{\left( {1 + A\beta } \right)}^2}}} \cdot \frac{{\partial A}}{{{A_F}}}\\ \frac{{\partial {A_F}}}{{{A_F}}} = \frac{1}{{{{\left( {1 + A\beta } \right)}^2}}} \cdot \frac{{\partial A}}{A} \times \left( {1 + A\beta } \right)\\ \frac{{\partial {A_F}}}{{{A_F}}} = \frac{1}{{\left( {1 + A\beta } \right)}}\frac{{\partial A}}{A}\\ \frac{{\partial {A_F}}}{{{A_F}}} = \frac{1}{{1 + A\beta }} \cdot \frac{{\partial A}}{A}\\ \frac{{0.2}}{{100}} = \frac{{20}}{{2000}}\left[ {\frac{1}{{1 + 2000\beta }}} \right]\\ \beta = \frac{1}{{500}}\)