Feedback Amplifiers Questions and Answers

Q: Why does a compensated amplifier have less bandwidth at a given gain?

Compensation moves a pole downward, so the open-loop gain starts to fall at a lower frequency and the closed-loop bandwidth at any chosen gain is correspondingly less. The closed-loop gain and the fraction are the designer's choice rather than consequences, and input capacitance is not what changes. Buying stability with bandwidth is the standard exchange, and it is why an uncompensated device is faster but usable only above some minimum gain.

Q: Assertion (A): Assertion: Negative feedback improves an amplifier's gain accuracy. Reason (R): Reason: The closed-loop gain tends to the reciprocal of the feedback fraction. Select the correct answer from the options below.

Both are true and the second is the reason: with a large loop gain the amplifier's own gain drops out and the result is set by the passive network alone. The accuracy is then as good as those components, which can be made far better than any device parameter. Transferring an accuracy requirement from an unreliable element to a reliable one is what feedback achieves, and every other benefit follows from the same division.

Q: Assertion (A): Assertion: Negative feedback improves an amplifier's signal-to-noise ratio. Reason (R): Reason: Feedback divides internal errors by one plus the loop gain. Select the correct answer from the options below.

The reason is true but the assertion is false in general: noise entering at the input is amplified with the signal and the ratio is unchanged, and only noise generated inside the loop is reduced. Since the input stage's own noise usually dominates, feedback typically does nothing for the noise figure. Where an imperfection enters decides whether the loop can act on it, and the input is precisely where it cannot.

Q: Assertion (A): Assertion: An amplifier stable at a closed-loop gain of one hundred may oscillate at unity gain. Reason (R): Reason: Reducing the closed-loop gain raises the frequency at which the loop gain passes unity. Select the correct answer from the options below.

Both hold and the second is the mechanism: a smaller closed-loop gain means more feedback and a higher crossover frequency, where additional poles have shifted the phase towards inversion. That is exactly why data sheets state a minimum stable gain for uncompensated devices. Stability is a property of a loop rather than of an amplifier, and the closed-loop gain is part of the loop.

Q: Consider the following statements about negative feedback. 1. It divides internal distortion by one plus the loop gain 2. It multiplies the bandwidth by one plus the loop gain 3. It divides the input impedance by one plus the loop gain in every topology Which of the statements given above is or are correct?

The first two are correct and the third overgeneralises: shunt feedback divides the input impedance but series feedback multiplies it, so the direction depends on the topology. Distortion and bandwidth improve by the same factor whatever the topology, which is why those two can be stated unconditionally and the impedance effects cannot. Knowing which results are general and which are topology-dependent is most of using this subject. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Q: With \( A = 500 \) and \( \beta = 0.02 \), the gain once the loop is closed comes to

Computing the loop gain first, then applying the equation:\[ \begin{aligned} A\beta &= 500 \times 0.02 = 10 \\ A_{f} &= \tfrac{500}{11} \approx 45.5 \end{aligned} \]giving about 45.5, some nine per cent below the ideal fifty. A loop gain of ten is modest, so the departure from the ideal is visible; at a loop gain of a thousand it would be a tenth of a per cent. The size of that departure is itself the measure of how much feedback is present.

Q: An amplifier has open-loop input impedance \( 1 \ \mathrm{k\Omega} \) and a loop gain of \( 24 \). With series feedback the input impedance becomes

Series feedback multiplies by one plus the loop gain:\[ Z_{if} = 1000 \times 25 = 25 \ \mathrm{k\Omega} \]so 25 kilohms. Using the loop gain without the added one gives 24 kilohms, a four per cent error that matters little here but matters greatly at low loop gains. Carrying the one plus rather than dropping it is a habit worth keeping, since it is exactly the term that describes the no-feedback case.

61.

Why does a compensated amplifier have less bandwidth at a given gain?

A.Its feedback fraction has been made smaller
B.Its input capacitance has been increased
C.Its open-loop response begins falling earlier
D.Its closed-loop gain has been reduced as well

Answer: C

Compensation moves a pole downward, so the open-loop gain starts to fall at a lower frequency and the closed-loop bandwidth at any chosen gain is correspondingly less. The closed-loop gain and the fraction are the designer's choice rather than consequences, and input capacitance is not what changes. Buying stability with bandwidth is the standard exchange, and it is why an uncompensated device is faster but usable only above some minimum gain.

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62.

Assertion (A): Assertion: Negative feedback improves an amplifier's gain accuracy.

Reason (R): Reason: The closed-loop gain tends to the reciprocal of the feedback fraction.

Select the correct answer from the options below.

A.Both A and R are true, but R is not the correct explanation of A
B.A is true but R is false
C.A is false but R is true
D.Both A and R are true, and R is the correct explanation of A

Answer: D

Both are true and the second is the reason: with a large loop gain the amplifier's own gain drops out and the result is set by the passive network alone. The accuracy is then as good as those components, which can be made far better than any device parameter. Transferring an accuracy requirement from an unreliable element to a reliable one is what feedback achieves, and every other benefit follows from the same division.

Discuss in Forum

63.

Assertion (A): Assertion: Negative feedback improves an amplifier's signal-to-noise ratio.

Reason (R): Reason: Feedback divides internal errors by one plus the loop gain.

Select the correct answer from the options below.

A.A is false but R is true
B.Both A and R are true, and R is the correct explanation of A
C.Both A and R are true, but R is not the correct explanation of A
D.A is true but R is false

Answer: A

The reason is true but the assertion is false in general: noise entering at the input is amplified with the signal and the ratio is unchanged, and only noise generated inside the loop is reduced. Since the input stage's own noise usually dominates, feedback typically does nothing for the noise figure. Where an imperfection enters decides whether the loop can act on it, and the input is precisely where it cannot.

Discuss in Forum

64.

Assertion (A): Assertion: An amplifier stable at a closed-loop gain of one hundred may oscillate at unity gain.

Reason (R): Reason: Reducing the closed-loop gain raises the frequency at which the loop gain passes unity.

Select the correct answer from the options below.

A.A is false but R is true
B.Both A and R are true, and R is the correct explanation of A
C.Both A and R are true, but R is not the correct explanation of A
D.A is true but R is false

Answer: B

Both hold and the second is the mechanism: a smaller closed-loop gain means more feedback and a higher crossover frequency, where additional poles have shifted the phase towards inversion. That is exactly why data sheets state a minimum stable gain for uncompensated devices. Stability is a property of a loop rather than of an amplifier, and the closed-loop gain is part of the loop.

Discuss in Forum

65.

Consider the following statements about negative feedback.

1. It divides internal distortion by one plus the loop gain

2. It multiplies the bandwidth by one plus the loop gain

3. It divides the input impedance by one plus the loop gain in every topology

Which of the statements given above is or are correct?

A.2 and 3 only
B.1, 2 and 3
C.1 and 2 only
D.1 and 3 only

Answer: C

The first two are correct and the third overgeneralises: shunt feedback divides the input impedance but series feedback multiplies it, so the direction depends on the topology. Distortion and bandwidth improve by the same factor whatever the topology, which is why those two can be stated unconditionally and the impedance effects cannot. Knowing which results are general and which are topology-dependent is most of using this subject. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Discuss in Forum

66.

Consider the following pairs of quantity and how feedback changes it.

Closed-loop gain - Divided by one plus loop gain

Bandwidth - Multiplied by one plus loop gain

Internal distortion - Divided by one plus loop gain

Input-referred noise - Divided by one plus loop gain

How many of the pairs given above are correctly matched?

A.Only one
B.Only two
C.All four
D.Only three

Answer: D

The first three are correctly paired and the fourth is not, so three match. Input-referred noise is amplified with the signal and is therefore unchanged in ratio terms. That three quantities share one factor and a fourth is exempt is the pattern worth fixing: the factor applies to everything that arises inside the loop and to nothing that arrives at its input.

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67.

A feedback amplifier is specified as follows.

Parameter	Value
Open-loop gain	100 dB
Closed-loop gain	20 dB
Open-loop bandwidth	10 Hz

What is its approximate closed-loop bandwidth?

A.100 kHz
B.10 Hz
C.1 MHz
D.10 kHz

Answer: A

The loop gain is eighty decibels, a factor of ten thousand, and the bandwidth rises by that same factor:

\[ B_{f} = 10 \times 10^{4} = 100 \ \mathrm{kHz} \]

giving 100 kHz. A ten hertz open-loop bandwidth with a hundred decibels of gain is typical of an internally compensated operational amplifier, and this calculation is how its closed-loop bandwidth is predicted at any chosen gain.

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68.

With \( A = 500 \) and \( \beta = 0.02 \), the gain once the loop is closed comes to

A.500
B.45.5
C.50
D.10

Answer: B

Computing the loop gain first, then applying the equation:

\[ \begin{aligned} A\beta &= 500 \times 0.02 = 10 \\ A_{f} &= \tfrac{500}{11} \approx 45.5 \end{aligned} \]

giving about 45.5, some nine per cent below the ideal fifty. A loop gain of ten is modest, so the departure from the ideal is visible; at a loop gain of a thousand it would be a tenth of a per cent. The size of that departure is itself the measure of how much feedback is present.

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69.

Taking \( A = 10^{5} \) and \( \beta = 10^{-3} \) instead, the result settles near

A.100
B.100000
C.990
D.1000

Answer: C

The loop gain is a hundred, so the ideal thousand is approached closely:

\[ A_{f} = \frac{100000}{1 + 100} \approx 990 \]

giving about 990, one per cent short. That one per cent is the residual dependence on the amplifier's own gain, and it halves each time the loop gain doubles. Quoting the ideal figure and the departure from it separately is more informative than either alone.

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70.

An amplifier has open-loop input impedance \( 1 \ \mathrm{k\Omega} \) and a loop gain of \( 24 \). With series feedback the input impedance becomes

A.24 kilohms
B.1 kilohm
C.40 ohms
D.25 kilohms

Answer: D

Series feedback multiplies by one plus the loop gain:

\[ Z_{if} = 1000 \times 25 = 25 \ \mathrm{k\Omega} \]

so 25 kilohms. Using the loop gain without the added one gives 24 kilohms, a four per cent error that matters little here but matters greatly at low loop gains. Carrying the one plus rather than dropping it is a habit worth keeping, since it is exactly the term that describes the no-feedback case.

Discuss in Forum

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