Frequency Response of Amplifiers Questions and Answers

Q: Why does an overshooting step response indicate a peaked frequency response?

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.

Q: Why does an emitter follower have a wide bandwidth?

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.

Q: Why is a common-base stage's input capacitance small?

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.

Q: What limits an emitter follower's bandwidth if not the Miller effect?

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.

Q: What is the usual remedy for a follower oscillating into a capacitive load?

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.

Q: A follower's output looks inductive at \( 500 \ \mathrm{nH} \) into a \( 100 \ \mathrm{pF} \) load. The resonance is near

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.

31.

Why does an overshooting step response indicate a peaked frequency response?

A.The stage's phase shift exceeds ninety degrees
B.Excess gain near the band edge appears as overshoot
C.A slow roll-off always produces a slow edge
D.The low-frequency break has been set too high

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.

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32.

What does droop on a long pulse indicate?

A.The stage's gain is lower than designed
B.The coupling capacitors are too large in value
C.The low-frequency response falls off too early
D.The high-frequency response has been peaked

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.

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33.

Why is a square wave a useful test of an amplifier's response?

A.It contains only odd harmonics of its fundamental
B.Its amplitude is easy to measure accurately
C.It has the same spectrum as a step function
D.Its edges and its flats probe both ends at once

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.

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34.

Which capacitance dominates a common-emitter stage's input at high frequency?

A.The Miller-multiplied collector-base capacitance
B.The base-emitter diffusion capacitance alone
C.The collector-substrate capacitance of the device
D.The stray capacitance of the wiring to earth

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.

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35.

Why does an emitter follower have a wide bandwidth?

A.Its output impedance is low enough to drive any load
B.Its voltage gain is unity, so nothing is multiplied
C.Its collector-base capacitance is grounded at the collector
D.Its input impedance is high and works into little 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.

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36.

Why is a common-base stage's input capacitance small?

A.Its collector-base capacitance is reverse biased harder
B.Its transconductance is higher than other configurations
C.Its input is the emitter, not the base
D.Its base is bypassed to earth for signals

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.

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37.

What limits an emitter follower's bandwidth if not the Miller effect?

A.The collector load resistor and the stray capacitance
B.The emitter resistor and the output capacitance
C.The device's saturation voltage at high frequency
D.The base-emitter capacitance and the source impedance

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.

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38.

Why can a follower become unstable when driving a capacitive load?

A.The load and the reflected source inductance can resonate
B.The voltage gain rises above unity at high frequencies
C.The base-emitter capacitance shunts the input entirely
D.The output impedance falls to zero at high frequencies

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.

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39.

What is the usual remedy for a follower oscillating into a capacitive load?

A.A higher operating current in the device
B.A small series resistor at the output
C.A larger emitter resistor to earth
D.A capacitor from base to earth

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.

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40.

A follower's output looks inductive at \( 500 \ \mathrm{nH} \) into a \( 100 \ \mathrm{pF} \) load. The resonance is near

A.220 MHz
B.710 kHz
C.22 MHz
D.2.2 MHz

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.

Discuss in Forum

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