Feedback Amplifiers Questions and Answers

Answer: A

The Nyquist test plots the loop gain in the complex plane and asks whether the locus encircles the point corresponding to unity gain and inverted phase. It settles cases that a Bode reading cannot, such as conditionally stable loops. Phase margin and roll-off rate are practical rules derived from it, and closed-loop flatness is a different specification. A general criterion and the rules of thumb drawn from it should not be confused, since the rules fail in exactly the awkward cases.

Answer: B

It deliberately moves one pole down so the gain has fallen to unity before any second pole contributes phase, guaranteeing a wide margin at the cost of bandwidth. Pole-zero cancellation is a different and more delicate approach, poles cannot be moved above the unity-gain frequency at will, and the low-frequency gain is left intact. Making one effect deliberately dominant so the others need not be controlled precisely is a robust strategy well beyond amplifiers.

Answer: C

A lag network adds a pole, pulling the gain down earlier so that unity gain is reached before the amplifier's own poles have shifted the phase - the mechanism of dominant-pole compensation. A lead network is the one that adds a zero and recovers phase. Coincident poles and zeros cancel and do nothing. Two networks with opposite effects, and choosing between them means deciding whether to reduce the crossover frequency or to improve the phase at it.

Answer: D

A lead network adds a zero whose ninety degrees of phase lead offsets a pole's lag, restoring margin without sacrificing crossover frequency. Adding a pole is what a lag network does. It does not raise the low-frequency gain, and input capacitance is unrelated. Because it buys stability without costing bandwidth, a lead network is preferred where speed matters - at the price of needing the zero placed accurately against a pole that may itself move.

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.