Frequency Response of Amplifiers Questions and Answers

Q: What is shunt peaking?

Shunt peaking places a small inductor in series with the collector resistor, so the load impedance rises with frequency just as the shunt capacitance starts to reduce the gain, and the two effects partly cancel to extend the bandwidth. A capacitor across the load would make matters worse, and the other options describe different techniques. Compensating a falling response with a rising impedance is a general idea, and it costs only a component and some care over the peaking that results.

Q: What is the penalty of over-generous shunt peaking?

Too large an inductor makes the load impedance rise more than the capacitance is reducing the gain, so the response peaks before falling - and a peaked response overshoots on a step input. The passband gain and the low-frequency break are unaffected, and the Miller capacitance is a property of the input side. Any compensation that can undercorrect can also overcorrect, and the flat-response condition is a particular value rather than a direction.

Q: Which device parameter caps the bandwidth a stage can reach however well it is designed?

The transition frequency is where the device's own current gain falls to unity, so no circuit arrangement can produce useful gain beyond it. Low-frequency beta, saturation voltage and the junction drop all matter elsewhere. Circuit techniques such as the cascode close the gap between what a stage achieves and what the device permits, but they cannot move the device limit itself, and knowing which limit is which prevents effort being spent on the wrong one.

Q: A stage gives a gain of \( 100 \) with a bandwidth of \( 200 \ \mathrm{kHz} \). At a gain of \( 10 \) its bandwidth is about

Holding the product constant:\[ \begin{aligned} GB &= 100 \times 200\mathrm{k} = 20 \ \mathrm{MHz} \\ B &= \tfrac{20\mathrm{M}}{10} = 2 \ \mathrm{MHz} \end{aligned} \]giving about 2 MHz. Twenty megahertz is the product itself rather than the new bandwidth. Reducing the gain by ten buys ten times the bandwidth, which is the trade every feedback amplifier makes and the reason a high open-loop gain is worth having.

Q: Why does a cascade's bandwidth fall below that of its individual stages?

Two identical stages each 3 dB down at some frequency are 6 dB down together, so the combined 3 dB point sits at a lower frequency than either stage's own. Coupling capacitors set low-frequency poles rather than high-frequency ones, loading reduces gain rather than bandwidth directly, and the product applies per stage. Roll-offs multiply as ratios and add as decibels, and that compounding is what shrinks a cascade's bandwidth even when nothing is loading anything.

21.

What is shunt peaking?

A.An inductor in series with the collector load
B.A capacitor across the collector load resistor
C.An inductor in series with the base lead
D.A resistor in parallel with the load capacitance

Answer: A

Shunt peaking places a small inductor in series with the collector resistor, so the load impedance rises with frequency just as the shunt capacitance starts to reduce the gain, and the two effects partly cancel to extend the bandwidth. A capacitor across the load would make matters worse, and the other options describe different techniques. Compensating a falling response with a rising impedance is a general idea, and it costs only a component and some care over the peaking that results.

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22.

What is the penalty of over-generous shunt peaking?

A.An increase in the Miller capacitance seen
B.A peak in the response before the roll-off
C.A loss of gain across the whole passband
D.A rise in the stage's low-frequency break

Answer: B

Too large an inductor makes the load impedance rise more than the capacitance is reducing the gain, so the response peaks before falling - and a peaked response overshoots on a step input. The passband gain and the low-frequency break are unaffected, and the Miller capacitance is a property of the input side. Any compensation that can undercorrect can also overcorrect, and the flat-response condition is a particular value rather than a direction.

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23.

Which device parameter caps the bandwidth a stage can reach however well it is designed?

A.The collector-emitter saturation voltage
B.The base-emitter forward drop
C.The transition frequency
D.The current gain at low frequency

Answer: C

The transition frequency is where the device's own current gain falls to unity, so no circuit arrangement can produce useful gain beyond it. Low-frequency beta, saturation voltage and the junction drop all matter elsewhere. Circuit techniques such as the cascode close the gap between what a stage achieves and what the device permits, but they cannot move the device limit itself, and knowing which limit is which prevents effort being spent on the wrong one.

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24.

What does the gain-bandwidth product of a stage stay roughly constant under?

A.Changes to the device's operating current
B.Changes to the supply voltage available
C.Changes to the ambient temperature
D.Changes to the collector load resistance

Answer: D

Increasing the load resistance raises the gain and lowers the break frequency in the same proportion, so their product is unchanged and gain can be traded for bandwidth freely. Operating current changes the transconductance and so the product itself, and supply and temperature act on other things. A conserved product is the most useful kind of relation in design, because it states exactly what one improvement will cost elsewhere.

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25.

A stage gives a gain of \( 100 \) with a bandwidth of \( 200 \ \mathrm{kHz} \). At a gain of \( 10 \) its bandwidth is about

A.2 MHz
B.200 kHz
C.20 kHz
D.20 MHz

Answer: A

Holding the product constant:

\[ \begin{aligned} GB &= 100 \times 200\mathrm{k} = 20 \ \mathrm{MHz} \\ B &= \tfrac{20\mathrm{M}}{10} = 2 \ \mathrm{MHz} \end{aligned} \]

giving about 2 MHz. Twenty megahertz is the product itself rather than the new bandwidth. Reducing the gain by ten buys ten times the bandwidth, which is the trade every feedback amplifier makes and the reason a high open-loop gain is worth having.

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26.

Why is a frequency response plotted on logarithmic axes?

A.The gain is only meaningful in decibels above the break
B.Roll-offs then appear as straight lines
C.Logarithmic axes remove the phase information
D.Linear axes cannot show a gain above one thousand

Answer: B

A single pole falls at a constant number of decibels per octave, so on logarithmic axes it becomes a straight line and the whole response can be sketched from its break points and slopes alone. Phase is plotted separately either way, linear axes can represent any number, and gain is meaningful in any units. A coordinate system that turns the behaviour of interest into straight lines is worth adopting for that reason by itself, and it is why Bode's construction survived the arrival of computers.

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27.

At what rate does a two-pole roll-off fall?

A.Three decibels per octave
B.Twenty decibels per octave
C.Twelve decibels per octave
D.Six decibels per octave

Answer: C

Each pole contributes six decibels per octave, so two together give twelve decibels per octave, equivalently forty per decade. Six per octave is one pole, three decibels is the drop at a break rather than a slope, and twenty per decade is one pole expressed in the other unit. Slopes add when poles are cascaded, which is why counting poles is the quickest way to predict a response's steepness and its phase shift together.

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28.

What total phase shift do two poles approach well beyond both breaks?

A.Ninety degrees
B.Forty-five degrees
C.Three hundred and sixty degrees
D.One hundred and eighty degrees

Answer: D

Each pole approaches ninety degrees, so two together approach one hundred and eighty degrees - which is exactly the condition that turns negative feedback into positive and makes an amplifier oscillate. That is why a feedback amplifier's stability depends on how many poles lie below the frequency at which the loop gain falls to unity. The phase count and the slope count are the same count, which is what makes a Bode plot a stability tool rather than merely a description.

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29.

Why does a cascade's bandwidth fall below that of its individual stages?

A.Each stage's roll-off adds to the others
B.Each coupling capacitor adds a high-frequency pole
C.The loading between stages reduces the load resistance
D.The gain-bandwidth product applies to the cascade as a whole

Answer: A

Two identical stages each 3 dB down at some frequency are 6 dB down together, so the combined 3 dB point sits at a lower frequency than either stage's own. Coupling capacitors set low-frequency poles rather than high-frequency ones, loading reduces gain rather than bandwidth directly, and the product applies per stage. Roll-offs multiply as ratios and add as decibels, and that compounding is what shrinks a cascade's bandwidth even when nothing is loading anything.

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30.

Two identical stages each of \( 1 \ \mathrm{MHz} \) bandwidth are cascaded. The overall bandwidth is about

A.2 MHz
B.640 kHz
C.1 MHz
D.500 kHz

Answer: B

For n identical single-pole stages the combined 3 dB point shrinks by a known factor:

\[ B_{n} = B \sqrt{2^{1/n}-1} = 10^{6}(0.64) \]

giving about 640 kHz. Half a megahertz would be the answer if bandwidths simply halved, which they do not. The shrinkage is modest for two stages and becomes severe for five or six, which is why a long cascade of narrow stages is a poor way to obtain both gain and bandwidth.

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