Feedback Amplifiers Questions and Answers

Answer: A

Bandwidth rises by one plus the loop gain:

\[ B_{f} = 5 \times 2000 = 10 \ \mathrm{kHz} \]

giving 10 kHz. A five hertz open-loop bandwidth sounds useless until the loop gain is applied, which is exactly the situation inside every internally compensated operational amplifier: enormous gain over a tiny bandwidth, traded for modest gain over a useful one.

Answer: B

The network connects across the output, so it samples voltage, and it joins the signal at a single input node, so the return is in shunt - voltage shunt feedback. Both impedances are therefore lowered, which is the inverting operational amplifier configuration and the transimpedance amplifier. Reading the two words off the circuit's two connections is quicker and more reliable than recognising the arrangement by its shape. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

A network carrying the emitter current samples the output current, and a voltage developed in the emitter lead appears in series with the base-emitter input - current series feedback. Both impedances rise. This is what an unbypassed emitter resistor is, which is why degeneration raises the input impedance and the output impedance together while reducing the gain.

Answer: D

Peaking with ringing is the signature of a small phase margin, so the remedy is compensation: a dominant pole to bring the crossover down, or a lead network to recover phase at it. Raising the closed-loop gain would also help, by reducing the loop gain, but it changes the amplifier's job rather than fixing it. Coupling capacitors and decoupling act elsewhere. A frequency-domain symptom and a time-domain one pointing at the same cause is the normal case, since they are two views of one response.

Answer: A

At unity gain the loop gain is at its largest, so the crossover frequency is highest and the accumulated phase greatest - the margin has run out there but not at a gain of ten. A faulty resistor would misset the gain rather than produce this pattern, low-frequency gain is wanted, and coupling capacitors act at the other end. That stability depends on the closed-loop gain is why devices are specified as unity-gain stable or not. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

The closed-loop gain is set by the feedback network, so a five per cent error points straight at a resistor rather than at the amplifier. A low open-loop gain would produce a gain slightly low rather than high, and by far less than five per cent if the loop gain is large. Margin and supply affect other behaviour. Feedback transfers the accuracy requirement to the passive network, so it also transfers the blame for an accuracy fault. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: C

Extra phase shift anywhere in the loop brings the total closer to inversion at the crossover frequency and eats into the margin, and stray reactances in a long lead are as much part of the loop as any component. Its resistance is negligible beside the feedback resistor, and neither gain is affected. That layout is part of the circuit, not merely its packaging, is something feedback amplifiers demonstrate more sharply than most circuits. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: D

Positive feedback returns a signal that adds to the input, so the denominator of the feedback equation becomes one minus the loop gain and the gain rises rather than falling. The fraction's size, the inversion and any reactance are all incidental. Every consequence reverses: gain rises, bandwidth narrows, distortion grows and impedances move the other way, which is why the two are used for entirely different purposes. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: A

With unity loop gain and zero net phase, the denominator becomes zero and the circuit sustains an output with no input - it oscillates. That is the Barkhausen condition, a failure in an amplifier and the design target in an oscillator. Neither the gain nor the bandwidth behaves as the other options suggest. The same condition appearing as a fault in one topic and an objective in another is worth noticing, since the mathematics is identical. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

Oscillators need it to sustain their output, and a comparator uses a little of it to give hysteresis and so clean switching in the presence of noise. Output stages, transimpedance amplifiers and bias networks all rely on negative feedback. A mechanism that is a hazard in most circuits and the operating principle of a few is best learned in both roles at once, since the conditions are the same and only the intent differs. The shared properties explain why these belong to one family, and the differences explain why more than one exists.