Feedback Amplifiers Questions and Answers

Answer: A

Applying the feedback equation:

\[ A_{f} = \frac{200}{1 + 10} \approx 18.2 \]

giving 18.2, some nine per cent below the ideal twenty. A loop gain of only ten leaves a noticeable dependence on the open-loop gain, which is why a loop gain of a hundred or more is preferred where accuracy matters. The size of the departure from the ideal is itself a measure of how much feedback is present.

Answer: B

Series feedback multiplies the input impedance by one plus the loop gain:

\[ Z_{if} = 2\mathrm{k} \times 11 = 22 \ \mathrm{k\Omega} \]

giving 22 kilohms. Two hundred ohms would be the result of dividing, which is what shunt feedback does. The same factor of eleven appears in the gain reduction, the impedance rise and the distortion improvement, which is why computing the loop gain first answers every question about the circuit. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: C

Voltage sampling divides the output impedance by the same factor:

\[ Z_{of} = \frac{500}{11} \approx 45 \ \Omega \]

giving about 45 ohms. Multiplying instead gives 5.5 kilohms, which is what current sampling would produce. One factor, applied as a multiplication at one port and a division at the other, covers the whole impedance behaviour of a topology once its two words are read.

Answer: D

Loop gain in decibels is the difference between the two:

\[ 80 - 40 = 40 \ \mathrm{dB} \]

which is 40 dB, a factor of a hundred, so every internal imperfection is divided by about a hundred. Reading the loop gain straight off a Bode plot as the vertical gap between the open-loop curve and the closed-loop line is one of the construction's most useful properties.

Answer: A

Each pole contributes the arctangent of the crossover frequency over its own:

\[ \begin{aligned} \phi &= \arctan 8 + \arctan 2 \\ &= 83^\circ + 63^\circ = 146^\circ \end{aligned} \]

leaving a margin of about thirty-four degrees - below the forty-five usually specified, so the step response would overshoot noticeably. Compensation is needed, either to bring the crossover down or to add a zero near it. Summing each pole's phase contribution at the crossover frequency is the whole calculation.

Answer: B

At the lower crossover the first pole gives forty-five degrees and the second only fourteen:

\[ \begin{aligned} \phi &= 45^\circ + 14^\circ = 59^\circ \\ \text{margin} &= 180 - 59 = 121^\circ \end{aligned} \]

so about one hundred and twenty degrees, a very comfortable margin bought by giving up a factor of eight in bandwidth. Moving the crossover down to the first pole's own frequency is the standard dominant-pole answer, and this is the price it charges.

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.

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.

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.

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.