Frequency Response of Amplifiers Questions and Answers

Answer: B

A response that rises before it falls emphasises the components near the band edge, and a step contains exactly those components, so the edge overshoots and rings. A slow roll-off gives a slow edge without overshoot, the low-frequency break affects droop rather than overshoot, and phase alone does not produce it. The time and frequency descriptions are two views of the same response, so a defect visible in one is always visible in the other.

Answer: C

A pulse whose top sags is losing its low-frequency content, so a coupling or bypass capacitor is breaking within the pulse's own duration - the remedy is a larger capacitor, not a smaller one. Peaking shows as overshoot on the edges instead, and a low gain would reduce the amplitude without changing the shape. Reading a pulse's shape end by end tells you which end of the frequency response is at fault, which makes a square wave the most informative single test signal.

Answer: D

The edges carry the high-frequency content and the flat tops the low, so overshoot or rounding on the edges and droop or tilt on the tops report on the two ends of the response in one observation. The odd-harmonic content and the ease of measurement are true but incidental, and a square wave's spectrum is not a step's. A single test that reports on two independent specifications is worth far more than two tests, which is why the square wave is reached for first.

Answer: A

Although the base-emitter capacitance is physically the larger, the collector-base capacitance multiplied by one plus the gain usually exceeds it by a wide margin - four picofarads across a gain of two hundred beats fifty picofarads unmultiplied. Substrate and stray capacitances are smaller still in most stages. That the smaller component dominates because of the gain around it is exactly why the Miller effect gets its own name rather than being folded into the input capacitance.

Answer: B

With a voltage gain of one, the Miller factor is one plus one rather than one plus two hundred, so the input capacitance stays close to the device's own figure. The collector is indeed at signal earth, which is part of the same argument, and the impedances are real advantages but not the bandwidth mechanism. A stage that takes no voltage gain pays no Miller penalty, which is why followers are used to drive capacitive loads and to precede wideband stages.

Answer: C

Driving the emitter means the collector-base capacitance does not bridge input and output at all, so no multiplication occurs and the input capacitance is the emitter junction's alone. The bypassed base is how that isolation is achieved rather than the reason directly. Junction bias and transconductance are the same as in any configuration at the same operating point. Which terminal is the input decides which capacitances bridge the stage, and that decides the bandwidth.

Answer: D

The follower still has a base-emitter capacitance to charge through whatever drives it, so a high source impedance limits the bandwidth even with no Miller multiplication. There is no collector load in a follower, the emitter resistor is bypassed by the load rather than limiting, and saturation voltage is a large-signal parameter. Removing the dominant limitation reveals the next one, which is why a follower's bandwidth is wide rather than unlimited.

Answer: A

A follower's output impedance is inductive over part of its range because the reflected source impedance rises with frequency, and that inductance with a capacitive load forms a resonant circuit that can ring or oscillate. The gain does not exceed unity, and the output impedance rises rather than falling. A small series resistor at the output damps it, which is why one is so often found there in practice even though the ideal analysis says it is unnecessary.

Answer: B

A series resistor of a few tens of ohms damps the resonance between the inductive output impedance and the load capacitance, at the cost of a little output impedance. A larger emitter resistor and a higher current change the operating point without damping anything, and a base capacitor would reduce the bandwidth from the wrong end. Damping a resonance means adding loss where the resonant current flows, and here that is in series with the output.

Answer: C

Applying the resonant frequency of the pair:

\[ \begin{aligned} f &= \frac{1}{2\pi\sqrt{LC}} \\ &= \frac{1}{2\pi\sqrt{5\times10^{-17}}} \end{aligned} \]

which gives about 22 MHz. The other options misplace a decade or two. Because the frequency depends on the square root of the product, a hundredfold change in either component moves it only tenfold, which is why such resonances are hard to shift out of the way and are better damped than avoided.