Feedback Amplifiers Questions and Answers

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.

Q: The same amplifier's \( 100 \ \Omega \) open-loop output impedance, with voltage sampling, becomes

Voltage sampling divides by the same factor:\[ Z_{of} = \frac{100}{25} = 4 \ \Omega \]giving 4 ohms, which is why a feedback power amplifier can present a fraction of an ohm to a loudspeaker despite output devices that are far from ideal. Multiplying instead gives 2500 ohms, the current-sampling result. One factor applied as a multiplication at one port and a division at the other covers the whole impedance behaviour.

Q: An amplifier's open-loop gain is \( 60 \ \mathrm{dB} \) and its closed-loop gain \( 26 \ \mathrm{dB} \). Its loop gain is about

The difference in decibels converts to a ratio:\[ \begin{aligned} 60 - 26 &= 34 \ \mathrm{dB} \\ 10^{34/20} &\approx 50 \end{aligned} \]giving about 50. Thirty-four is the answer in decibels rather than as a ratio, and the distinction matters because the feedback equation needs the ratio. Reading the vertical gap on a Bode plot gives the loop gain in decibels directly, and converting it is the only extra step.

Q: With that loop gain, an open-loop distortion of \( 2 \) per cent becomes about

Internal distortion is divided by one plus the loop gain:\[ \frac{2\%}{51} \approx 0.04\% \]giving about 0.04 per cent, a fiftyfold improvement obtained from gain that was surplus to requirements. This calculation is why audio power amplifiers with unremarkable output stages can meet demanding distortion specifications, and why open-loop gain is designed to be as large as stability permits.

Q: An amplifier of \( 1 \ \mathrm{MHz} \) gain-bandwidth product is used at a closed-loop gain of \( 5 \). Its bandwidth is

Dividing the product by the closed-loop gain:\[ B = \frac{10^{6}}{5} = 200 \ \mathrm{kHz} \]giving 200 kHz. Multiplying gives five megahertz, more bandwidth than the amplifier possesses at any gain, which is a useful check on the direction. This one division is the calculation behind every operational amplifier selection, and it is why the product rather than either factor is the figure quoted.

51.

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.

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52.

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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53.

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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54.

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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55.

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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56.

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.

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57.

The same amplifier's \( 100 \ \Omega \) open-loop output impedance, with voltage sampling, becomes

A.4 ohms
B.2500 ohms
C.100 ohms
D.40 ohms

Answer: A

Voltage sampling divides by the same factor:

\[ Z_{of} = \frac{100}{25} = 4 \ \Omega \]

giving 4 ohms, which is why a feedback power amplifier can present a fraction of an ohm to a loudspeaker despite output devices that are far from ideal. Multiplying instead gives 2500 ohms, the current-sampling result. One factor applied as a multiplication at one port and a division at the other covers the whole impedance behaviour.

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58.

An amplifier's open-loop gain is \( 60 \ \mathrm{dB} \) and its closed-loop gain \( 26 \ \mathrm{dB} \). Its loop gain is about

A.2.3
B.50
C.34
D.86

Answer: B

The difference in decibels converts to a ratio:

\[ \begin{aligned} 60 - 26 &= 34 \ \mathrm{dB} \\ 10^{34/20} &\approx 50 \end{aligned} \]

giving about 50. Thirty-four is the answer in decibels rather than as a ratio, and the distinction matters because the feedback equation needs the ratio. Reading the vertical gap on a Bode plot gives the loop gain in decibels directly, and converting it is the only extra step.

Discuss in Forum

59.

With that loop gain, an open-loop distortion of \( 2 \) per cent becomes about

A.2 per cent
B.0.02 per cent
C.0.04 per cent
D.0.4 per cent

Answer: C

Internal distortion is divided by one plus the loop gain:

\[ \frac{2\%}{51} \approx 0.04\% \]

giving about 0.04 per cent, a fiftyfold improvement obtained from gain that was surplus to requirements. This calculation is why audio power amplifiers with unremarkable output stages can meet demanding distortion specifications, and why open-loop gain is designed to be as large as stability permits.

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60.

An amplifier of \( 1 \ \mathrm{MHz} \) gain-bandwidth product is used at a closed-loop gain of \( 5 \). Its bandwidth is

A.5 MHz
B.1 MHz
C.20 kHz
D.200 kHz

Answer: D

Dividing the product by the closed-loop gain:

\[ B = \frac{10^{6}}{5} = 200 \ \mathrm{kHz} \]

giving 200 kHz. Multiplying gives five megahertz, more bandwidth than the amplifier possesses at any gain, which is a useful check on the direction. This one division is the calculation behind every operational amplifier selection, and it is why the product rather than either factor is the figure quoted.

Discuss in Forum

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