Small Signal Amplifiers Questions and Answers

Q: A coupling capacitor works into \( 10 \ \mathrm{k\Omega} \) total. For a \( 20 \ \mathrm{Hz} \) break it must be at least about

Rearranging the break frequency relation:\[ \begin{aligned} C &= \frac{1}{2\pi f R} \\ &= \frac{1}{2\pi(20)(10^{4})} \\ &\approx 0.8 \ \mathrm{\mu F} \end{aligned} \]giving about 0.8 microfarads. The others misplace a decade. Because the capacitance needed is inversely proportional to both frequency and resistance, low-frequency response into a low impedance is what drives capacitor values up. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Q: An emitter bypass capacitor works into about \( 26 \ \Omega \). For the same \( 20 \ \mathrm{Hz} \) break it must be at least about

The same relation with a far smaller resistance:\[ C = \frac{1}{2\pi(20)(26)} \approx 306 \ \mathrm{\mu F} \]so about 306 microfarads, nearly four hundred times the coupling capacitor's value for the same frequency. That contrast is why the emitter bypass is invariably the largest capacitor in a small-signal stage and usually the one that sets the low-frequency limit.

Q: Which capacitor usually dominates a stage's low-frequency roll-off?

The emitter bypass works into the smallest resistance by far, so for any given capacitance its break frequency is the highest of the three signal capacitors. Coupling capacitors see kilohms and so break much lower. Supply decoupling is not in the signal path at all. When several break points contribute, the highest of them determines where the response starts to fall, which makes it the one worth calculating first. Checking the units of each quantity catches a wrong formula before the arithmetic begins. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Q: Why are direct-coupled stages used in integrated circuits?

A microfarad occupies an impossible area in silicon, so integrated designs couple directly and pay for it with the need to handle direct offsets and drift throughout. High-frequency response is unaffected by coupling method, matching is a separate advantage, and bias is still needed. What a technology cannot make cheaply shapes its circuits more than what it can, and the absence of large capacitors is the clearest example in analogue integration. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Q: A stage has \( R_{C} = 2.2 \ \mathrm{k\Omega} \), \( R_{E} = 220 \ \Omega \) unbypassed, at \( 1 \ \mathrm{mA} \). Its gain is about

Including the intrinsic resistance in the denominator:\[ A_{v} \approx -\frac{2200}{220+26} \approx -9 \]giving about -9, a little below the naive ratio of ten because the 26 ohms is not quite negligible. Neglecting it would be a ten per cent error here and much worse at lower currents, which is why the intrinsic term is worth carrying in any stage with a modest emitter resistor. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Q: That stage's input impedance at the base with \( \beta = 120 \) is about

The whole emitter resistance reflects to the base:\[ r_{in} = 120 \times (220+26) \approx 30 \ \mathrm{k\Omega} \]so about 30 kilohms, before the bias divider is placed in parallel with it. Degeneration has raised the input impedance by the same factor of nine that it lowered the gain, which is the reciprocal relation series feedback always produces.

Q: A bias divider of \( 47 \ \mathrm{k} \) and \( 12 \ \mathrm{k} \) is placed across that input. The stage's input impedance becomes about

Three resistances now appear in parallel:\[ \begin{aligned} r_{in} &= 30 \ \mathrm{k} \| 47 \ \mathrm{k} \| 12 \ \mathrm{k} \\ &\approx 6.4 \ \mathrm{k\Omega} \end{aligned} \]giving about 6.4 kilohms - the divider has cut the figure by nearly five to one. That the bias network usually dominates a stage's input impedance is the single most common surprise in these calculations, and it follows from the divider being made deliberately stiff.

Q: Why does a stiff bias divider hurt the input impedance so much?

The two requirements pull directly against each other: a stiff divider is by definition a low resistance, and it sits in parallel with the signal input. Base current and reflected resistance are unaffected by the divider's value. Loading the previous stage is the consequence rather than the mechanism. When two specifications depend oppositely on one component value, the choice is a compromise and stating both sides is what makes it deliberate. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

31.

A coupling capacitor works into \( 10 \ \mathrm{k\Omega} \) total. For a \( 20 \ \mathrm{Hz} \) break it must be at least about

A.0.08 microfarads
B.80 microfarads
C.0.8 microfarads
D.8 microfarads

Answer: C

Rearranging the break frequency relation:

\[ \begin{aligned} C &= \frac{1}{2\pi f R} \\ &= \frac{1}{2\pi(20)(10^{4})} \\ &\approx 0.8 \ \mathrm{\mu F} \end{aligned} \]

giving about 0.8 microfarads. The others misplace a decade. Because the capacitance needed is inversely proportional to both frequency and resistance, low-frequency response into a low impedance is what drives capacitor values up. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

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32.

An emitter bypass capacitor works into about \( 26 \ \Omega \). For the same \( 20 \ \mathrm{Hz} \) break it must be at least about

A.0.8 microfarads
B.31 microfarads
C.3 microfarads
D.306 microfarads

Answer: D

The same relation with a far smaller resistance:

\[ C = \frac{1}{2\pi(20)(26)} \approx 306 \ \mathrm{\mu F} \]

so about 306 microfarads, nearly four hundred times the coupling capacitor's value for the same frequency. That contrast is why the emitter bypass is invariably the largest capacitor in a small-signal stage and usually the one that sets the low-frequency limit.

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33.

Which capacitor usually dominates a stage's low-frequency roll-off?

A.The emitter bypass capacitor
B.The input coupling capacitor
C.The output coupling capacitor into the load
D.The supply decoupling capacitor

Answer: A

The emitter bypass works into the smallest resistance by far, so for any given capacitance its break frequency is the highest of the three signal capacitors. Coupling capacitors see kilohms and so break much lower. Supply decoupling is not in the signal path at all. When several break points contribute, the highest of them determines where the response starts to fall, which makes it the one worth calculating first. Checking the units of each quantity catches a wrong formula before the arithmetic begins. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

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34.

What happens between a stage's bypass break frequency and its coupling break frequency?

A.The phase shift reaches ninety degrees and stays there
B.The gain falls to the degenerated value and then holds
C.The gain falls continuously at six decibels per octave
D.The gain rises as the emitter resistance becomes negligible

Answer: B

Below the bypass break the capacitor no longer shorts the emitter resistor, so the gain settles at the degenerated ratio and stays there until the coupling capacitor starts to act. It is a step in the response rather than a continuous fall. That an amplifier can have a shelf in its low-frequency response, rather than a simple roll-off, surprises people who expect one slope, and it follows directly from the two break points being far apart. Framing the question as which quantity the change acts upon makes the answer follow without calculation. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

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35.

Why are direct-coupled stages used in integrated circuits?

A.Integrated devices match too closely for capacitors to be needed
B.Direct coupling removes the need for any bias network at all
C.Large capacitors cannot be fabricated on a chip
D.Direct coupling gives a wider high-frequency response

Answer: C

A microfarad occupies an impossible area in silicon, so integrated designs couple directly and pay for it with the need to handle direct offsets and drift throughout. High-frequency response is unaffected by coupling method, matching is a separate advantage, and bias is still needed. What a technology cannot make cheaply shapes its circuits more than what it can, and the absence of large capacitors is the clearest example in analogue integration. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Discuss in Forum

36.

A stage has \( R_{C} = 2.2 \ \mathrm{k\Omega} \), \( R_{E} = 220 \ \Omega \) unbypassed, at \( 1 \ \mathrm{mA} \). Its gain is about

A.-85, the value with the emitter fully bypassed
B.-10
C.-100
D.-9

Answer: D

Including the intrinsic resistance in the denominator:

\[ A_{v} \approx -\frac{2200}{220+26} \approx -9 \]

giving about -9, a little below the naive ratio of ten because the 26 ohms is not quite negligible. Neglecting it would be a ten per cent error here and much worse at lower currents, which is why the intrinsic term is worth carrying in any stage with a modest emitter resistor. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Discuss in Forum

37.

That stage's input impedance at the base with \( \beta = 120 \) is about

A.30 kilohms
B.3.1 kilohms
C.26 kilohms, using the emitter resistor alone
D.300 kilohms

Answer: A

The whole emitter resistance reflects to the base:

\[ r_{in} = 120 \times (220+26) \approx 30 \ \mathrm{k\Omega} \]

so about 30 kilohms, before the bias divider is placed in parallel with it. Degeneration has raised the input impedance by the same factor of nine that it lowered the gain, which is the reciprocal relation series feedback always produces.

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38.

A bias divider of \( 47 \ \mathrm{k} \) and \( 12 \ \mathrm{k} \) is placed across that input. The stage's input impedance becomes about

A.39 kilohms
B.6.4 kilohms
C.30 kilohms
D.9.6 kilohms, the divider's parallel value alone

Answer: B

Three resistances now appear in parallel:

\[ \begin{aligned} r_{in} &= 30 \ \mathrm{k} \| 47 \ \mathrm{k} \| 12 \ \mathrm{k} \\ &\approx 6.4 \ \mathrm{k\Omega} \end{aligned} \]

giving about 6.4 kilohms - the divider has cut the figure by nearly five to one. That the bias network usually dominates a stage's input impedance is the single most common surprise in these calculations, and it follows from the divider being made deliberately stiff.

Discuss in Forum

39.

Why does a stiff bias divider hurt the input impedance so much?

A.Stiffness reduces the reflected emitter resistance at the base
B.Stiffness increases the loading the previous stage experiences
C.Stiffness means low resistance, which appears in parallel
D.Stiffness raises the base current the stage has to supply

Answer: C

The two requirements pull directly against each other: a stiff divider is by definition a low resistance, and it sits in parallel with the signal input. Base current and reflected resistance are unaffected by the divider's value. Loading the previous stage is the consequence rather than the mechanism. When two specifications depend oppositely on one component value, the choice is a compromise and stating both sides is what makes it deliberate. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Discuss in Forum

40.

How can a stage have both a stiff divider and a high input impedance?

A.Increase the divider resistances and accept beta dependence
B.Bypass the divider to ground with a large capacitor
C.Insert a coupling capacitor between the divider and the base
D.Bootstrap the divider from the emitter

Answer: D

Bootstrapping drives the divider's midpoint from the emitter through a capacitor, so it sees almost no signal voltage across it and its apparent impedance rises enormously while its direct resistance stays low. Raising the resistances sacrifices stiffness, bypassing to ground would make the loading worse, and a coupling capacitor would block the bias. Feeding a component with the signal it is trying to shunt is a neat way to escape a compromise, and it recurs throughout amplifier design.

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