Frequency Response of Amplifiers Questions and Answers

Answer: B

A marginally stable loop peaks in the frequency domain and rings in the time domain, so a step overshoots and takes several cycles to settle. A rounded rise indicates a wide phase margin and a modest bandwidth, droop is a low-frequency matter, and the final amplitude is set by the closed-loop gain. The frequency and time descriptions are two views of one response, so a margin can be assessed from a step just as readily as from a swept measurement.

Answer: C

All three are correct, and between them they are the whole construction: the slope a pole adds, the phase a zero adds, and the worst-case error of the straight-line approximation. Because poles and zeros are exact mirrors of one another, learning one set of rules gives both. A method with a known bounded error is what allows a response to be sketched by hand and still trusted.

Answer: D

The pairing is A-P, B-Q, C-R. One test signal reports on three independent properties, each visible in a different part of the waveform: the edges carry the high-frequency content, the flats the low, and overshoot betrays excess gain near the band edge. That is why a square wave is reached for before a sweep generator when an amplifier's behaviour is unknown.

Answer: A

The first three are correctly paired and the fourth is not, so three match. A larger coupling capacitor extends the low-frequency response, since it is in series with the signal, and does nothing at the high end. Three of these techniques act on the high end and one on the low, and sorting them that way is what stops a low-end remedy being applied to a high-end problem.

Answer: B

Gain margin is measured where the phase reaches inversion and phase margin where the gain reaches unity, so the two report the loop's distance from oscillation at two separate frequencies, and a loop can be comfortable on one and marginal on the other. Both concern the high end, both are readable from a Bode plot, and neither is notably easier to measure. Two conditions must both be satisfied, which is why a single figure of merit for stability has never been agreed.

Answer: C

A follower presents a low output impedance, and since the Miller break is set by the product of source resistance and multiplied capacitance, a stiffer drive raises it directly. Another common-emitter stage would have a high output impedance itself, a common-base stage is driven at its emitter rather than its base, and a shunt-peaked stage addresses its own load rather than the next stage's input. Attacking whichever factor of a product is cheapest to change is the general strategy, and here it is the source impedance.

Answer: D

The Miller-multiplied capacitance at the input works against whatever resistance drives it, so the same stage is fast or slow according to its source. Its gain is set by transconductance and load rather than by the drive, no reflection of the collector load occurs, and the low-frequency break does depend on the source but is not what limits the bandwidth. That a stage has no bandwidth of its own, only a bandwidth in a context, is worth internalising before any figure is quoted.

Answer: A

Computing the two break frequencies:

\[ \begin{aligned} f_{1} &= \tfrac{1}{2\pi(10^{4})(600\mathrm{p})} \approx 27 \ \mathrm{kHz} \\ f_{2} &= \tfrac{1}{2\pi(60)(600\mathrm{p})} \approx 4.4 \ \mathrm{MHz} \end{aligned} \]

so 27 kHz to 4.4 MHz, a factor of over one hundred and sixty for the price of one transistor. A buffer that takes no voltage gain can still transform a stage's usefulness, which is the clearest possible argument for impedance transformation.

Answer: B

Removing the dominant limitation reveals the next one - the base-emitter capacitance, the collector load working into stray capacitance, and the follower's own break. The follower does add a pole but well above this range. The Miller capacitance is unchanged, which is the point of attacking the resistance instead. Improving a design means finding the successive dominant limitations in turn, and each one uncovered is progress rather than failure.

Answer: C

A capacitive load works against the collector resistance to form an additional shunt time constant, so the break falls and the bandwidth narrows. Gain in the passband is unchanged, the low end is unaffected, and the phase shift increases rather than reducing. Since a capacitive load is the normal case - the next stage's input, a cable, a probe - a stage's bandwidth should always be quoted with the load it was measured into.