Feedback Amplifiers Questions and Answers

Q: By what factor does negative feedback extend an amplifier's bandwidth?

Bandwidth rises by one plus the loop gain, the same factor by which the gain falls, so their product is unchanged. That the two move by exactly the same factor is what makes the gain-bandwidth product a conserved quantity and what makes the trade so clean. The fraction and the open-loop gain appear only through the loop gain. One factor governing every benefit and the single cost is why feedback is described by one number. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Q: Why does feedback reduce distortion generated inside the amplifier?

Distortion arising within the loop is fed back, compared with the input and largely cancelled, so it is divided by one plus the loop gain just as any other internal error is. The reduced drive is a consequence of the same mechanism rather than a separate one, no filtering occurs, and bias stability is a different matter. Any imperfection that arises inside a loop is reduced by the loop gain, and any that arises at the input is not - which is the distinction the next question turns on.

Q: Why does feedback not reduce noise arriving at the amplifier's input?

Noise entering at the input is indistinguishable from signal, so the loop amplifies both and the ratio between them is unchanged. Bandwidth, location relative to the network and correlation are all beside the point. Feedback improves the signal-to-noise ratio only for noise generated inside the loop, such as a later stage's hum; the input stage's own noise is untouched, which is why low-noise design still begins with the first device. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Q: Which imperfection does feedback reduce most usefully in a power amplifier?

Crossover distortion arises in the output stage, well inside the loop, so it is divided by one plus the loop gain and a modest output stage can be made to sound clean. Input noise and offset enter at the input and are amplified with the signal. The feedback resistors set the closed-loop gain, so their tolerance appears in the answer directly and is not reduced at all. Where an imperfection enters decides whether feedback helps with it, and that is the question to ask first.

Q: An amplifier's open-loop distortion is \( 5 \) per cent. With a loop gain of \( 49 \) the closed-loop figure is about

Internal distortion is divided by one plus the loop gain:\[ \frac{5\%}{50} = 0.1\% \]giving about 0.1 per cent. A loop gain of forty-nine has turned an unusable amplifier into a respectable one, using nothing but gain that was surplus to requirements. That is the trade in its most persuasive form, and it is why open-loop gain is designed to be as large as stability allows.

Q: Why does feedback reduce an amplifier's drift with temperature?

A drift in an internal gain is an error inside the loop and is reduced by one plus the loop gain, exactly as a distortion or a gain change is. The devices still warm, the network's coefficient matters for the closed-loop gain rather than the drift reduction, and bias stabilisation is a separate local mechanism. Feedback does not care what caused an internal error, only where it entered, which is why one mechanism addresses drift, distortion and device spread together.

Q: Statement I: Feedback reduces distortion generated inside the loop. Statement II: Feedback reduces noise arriving at the amplifier's input. In the light of the above statements, choose the correct answer.

The first is true and the second is false: an error inside the loop is compared against the input and divided by one plus the loop gain, while noise at the input is indistinguishable from signal and is amplified with it. Where an imperfection enters is the whole question, and it decides whether feedback is a remedy or an irrelevance. This is why a feedback amplifier can have superb distortion figures and an unremarkable noise figure.

Q: What condition makes a feedback amplifier oscillate?

Oscillation requires the loop gain to be one in magnitude and the phase shift to have reached inversion, so the intended negative feedback has become positive. A large loop gain at low frequency is exactly what is wanted, ninety degrees is only halfway, and a large difference between the two gains is normal. Both conditions must hold at the same frequency, which is why a Bode plot is read for magnitude and phase together rather than either alone.

21.

By what factor does negative feedback extend an amplifier's bandwidth?

A.One plus the loop gain
B.The feedback fraction
C.The open-loop gain itself
D.The square root of the loop gain

Answer: A

Bandwidth rises by one plus the loop gain, the same factor by which the gain falls, so their product is unchanged. That the two move by exactly the same factor is what makes the gain-bandwidth product a conserved quantity and what makes the trade so clean. The fraction and the open-loop gain appear only through the loop gain. One factor governing every benefit and the single cost is why feedback is described by one number. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

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22.

An amplifier of gain \( 1000 \) and bandwidth \( 20 \ \mathrm{kHz} \) is given feedback to a closed-loop gain of \( 100 \). Its bandwidth becomes about

A.2 kHz
B.200 kHz
C.20 kHz
D.2 MHz

Answer: B

The gain has fallen tenfold, so the bandwidth rises tenfold:

\[ \begin{aligned} B_{f} &= 20\mathrm{k} \times \frac{1000}{100} \\ &= 200 \ \mathrm{kHz} \end{aligned} \]

giving about 200 kHz, with the gain-bandwidth product held at twenty megahertz throughout. That the product is the same before and after is worth checking on every such calculation, since it catches a direction error at once.

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23.

Why does feedback reduce distortion generated inside the amplifier?

A.The harmonics are filtered by the feedback network
B.The operating point is stabilised against drift
C.The distortion appears in the loop and is divided down
D.The devices are driven at a smaller amplitude

Answer: C

Distortion arising within the loop is fed back, compared with the input and largely cancelled, so it is divided by one plus the loop gain just as any other internal error is. The reduced drive is a consequence of the same mechanism rather than a separate one, no filtering occurs, and bias stability is a different matter. Any imperfection that arises inside a loop is reduced by the loop gain, and any that arises at the input is not - which is the distinction the next question turns on.

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24.

Why does feedback not reduce noise arriving at the amplifier's input?

A.Its spectrum is too wide for the loop to follow
B.It is generated outside the feedback network
C.It has no correlation with the output signal
D.It is amplified along with the signal

Answer: D

Noise entering at the input is indistinguishable from signal, so the loop amplifies both and the ratio between them is unchanged. Bandwidth, location relative to the network and correlation are all beside the point. Feedback improves the signal-to-noise ratio only for noise generated inside the loop, such as a later stage's hum; the input stage's own noise is untouched, which is why low-noise design still begins with the first device. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

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25.

Which imperfection does feedback reduce most usefully in a power amplifier?

A.Crossover distortion in the output stage
B.Thermal noise in the input stage's first device
C.The offset voltage of the input differential pair
D.The tolerance of the feedback resistors themselves

Answer: A

Crossover distortion arises in the output stage, well inside the loop, so it is divided by one plus the loop gain and a modest output stage can be made to sound clean. Input noise and offset enter at the input and are amplified with the signal. The feedback resistors set the closed-loop gain, so their tolerance appears in the answer directly and is not reduced at all. Where an imperfection enters decides whether feedback helps with it, and that is the question to ask first.

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26.

Why does the feedback network's own tolerance matter so much?

A.Its capacitance sets the loop's stability margin
B.The closed-loop gain is set by it alone
C.It carries the full output current of the amplifier
D.Its noise is amplified by the open-loop gain

Answer: B

With a large loop gain the closed-loop gain is the reciprocal of the feedback fraction, so the network's component tolerances appear directly in the answer with no reduction whatever. It carries a small fraction of the output current, its noise is not amplified by the full open-loop gain, and stray capacitance affects stability separately. Feedback transfers the accuracy requirement from the devices to a handful of passive components, and those components must then be worth trusting.

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27.

An amplifier's open-loop distortion is \( 5 \) per cent. With a loop gain of \( 49 \) the closed-loop figure is about

A.0.5 per cent
B.5 per cent
C.0.1 per cent
D.1 per cent

Answer: C

Internal distortion is divided by one plus the loop gain:

\[ \frac{5\%}{50} = 0.1\% \]

giving about 0.1 per cent. A loop gain of forty-nine has turned an unusable amplifier into a respectable one, using nothing but gain that was surplus to requirements. That is the trade in its most persuasive form, and it is why open-loop gain is designed to be as large as stability allows.

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28.

Why does feedback reduce an amplifier's drift with temperature?

A.The devices are held at a constant temperature
B.The feedback network has a low temperature coefficient
C.The operating currents are stabilised by the bias network
D.Drift within the loop is divided down like any error

Answer: D

A drift in an internal gain is an error inside the loop and is reduced by one plus the loop gain, exactly as a distortion or a gain change is. The devices still warm, the network's coefficient matters for the closed-loop gain rather than the drift reduction, and bias stabilisation is a separate local mechanism. Feedback does not care what caused an internal error, only where it entered, which is why one mechanism addresses drift, distortion and device spread together.

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29.

Statement I: Feedback reduces distortion generated inside the loop.

Statement II: Feedback reduces noise arriving at the amplifier's input.

In the light of the above statements, choose the correct answer.

A.Statement I is correct but Statement II is incorrect
B.Both Statement I and Statement II are correct
C.Both Statement I and Statement II are incorrect
D.Statement I is incorrect but Statement II is correct

Answer: A

The first is true and the second is false: an error inside the loop is compared against the input and divided by one plus the loop gain, while noise at the input is indistinguishable from signal and is amplified with it. Where an imperfection enters is the whole question, and it decides whether feedback is a remedy or an irrelevance. This is why a feedback amplifier can have superb distortion figures and an unremarkable noise figure.

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30.

What condition makes a feedback amplifier oscillate?

A.Open-loop gain exceeding the closed-loop gain by 60 dB
B.Loop gain of unity with 180 degrees of extra phase
C.Loop gain above unity at any frequency at all
D.Phase shift of 90 degrees at the unity-gain frequency

Answer: B

Oscillation requires the loop gain to be one in magnitude and the phase shift to have reached inversion, so the intended negative feedback has become positive. A large loop gain at low frequency is exactly what is wanted, ninety degrees is only halfway, and a large difference between the two gains is normal. Both conditions must hold at the same frequency, which is why a Bode plot is read for magnitude and phase together rather than either alone.

Discuss in Forum

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