Small Signal Amplifiers Questions and Answers

Q: What is an emitter follower's voltage gain?

The gain is the emitter load divided by that load plus the intrinsic emitter resistance, so just below one and non-inverting. It cannot exceed unity, which rules out the second option immediately. There is no inversion, so the fourth is wrong in sign as well. A stage whose gain can be stated as a number rather than an expression is usually one dominated by feedback, and a follower has as much of that as a circuit can have. The shared properties explain why these schemes form a family, and the differences explain why more than one exists. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Q: A follower has \( R_{E} = 1 \ \mathrm{k\Omega} \) at \( 2 \ \mathrm{mA} \). Its voltage gain is about

The intrinsic resistance is thirteen ohms at two milliamps:\[ A_{v} = \frac{1000}{1000+13} \approx 0.987 \]giving about 0.987. A value above one is impossible for a follower, which rules out 1.013 without any calculation. Checking a computed gain against the physical bound the topology imposes is a quick and reliable sanity test.

Q: Why does a common-base stage have such a wide bandwidth?

With the base common, the collector-base capacitance no longer bridges input and output across an inverting gain, so there is no Miller multiplication. The capacitance itself is the same device property in every configuration. The low input impedance helps but is not the mechanism, and the current gain is unrelated. Bandwidth is nearly always about which capacitance sees which gain, rather than about the size of the capacitance. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Q: What is a common-base stage's input impedance?

Driving the emitter directly presents the intrinsic emitter resistance, tens of ohms at typical currents, with no beta multiplication because the base is not the input. The second option is the common-emitter figure. A very low input impedance is a good match to a transmission line or to a current-source drive, which is why the configuration survives in high-frequency and cascode arrangements. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Q: Which configuration would you choose to drive a 50 ohm line directly?

A common-base input is naturally tens of ohms, which matches such a line without any transformer or padding. A common-emitter input is a few kilohms and a follower's is far higher still. Matching is about choosing the configuration whose natural impedance suits the source, and it is one of the few situations where common base is the obvious first choice rather than a specialist's option.

Q: By what factor does an emitter impedance appear larger at the base?

For the same voltage across it, the base supplies only a beta-th of the current, so the apparent impedance is one plus beta times larger. The other options are gain expressions rather than reflection factors. The rule works because impedance is voltage over current and only the current has been scaled, which is the same reasoning that governs a transformer.

Q: By what factor does a base impedance appear smaller at the emitter?

Looking the other way the same rule inverts: a source impedance at the base appears at the emitter divided by one plus beta, which is why a follower's output impedance is so low. That the factor is the same in both directions, multiplying one way and dividing the other, makes the pair easy to hold once the reason is clear rather than the formula memorised.

31.

What is the output impedance of a common-emitter stage?

A.The emitter resistor in parallel with the collector resistor
B.The intrinsic emitter resistance taken by itself alone
C.Essentially the collector load resistor
D.Beta times the intrinsic emitter resistance of the device

Answer: C

The device's own output resistance is far higher than a practical collector resistor, so the load resistor dominates and is essentially the output impedance. The other options belong to the input side or to the emitter circuit. A high output impedance is why a common-emitter stage cannot drive a low-impedance load directly, and why a follower so often appears after it.

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32.

What is an emitter follower's voltage gain?

A.Beta times the emitter resistance over the load
B.The emitter resistance divided by the collector resistance
C.Minus the load over the emitter resistance
D.Just below unity

Answer: D

The gain is the emitter load divided by that load plus the intrinsic emitter resistance, so just below one and non-inverting. It cannot exceed unity, which rules out the second option immediately. There is no inversion, so the fourth is wrong in sign as well. A stage whose gain can be stated as a number rather than an expression is usually one dominated by feedback, and a follower has as much of that as a circuit can have. The shared properties explain why these schemes form a family, and the differences explain why more than one exists. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

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33.

Why is an emitter follower's gain so insensitive to the device?

A.Its feedback is nearly complete
B.Its emitter resistance is negligible at any useful current
C.Its collector is bypassed and takes no part in the signal path
D.Its base current is too small to affect the emitter voltage

Answer: A

The whole output voltage is fed back in series with the input, so the feedback factor is essentially one and the gain is pinned just below unity whatever the device does. The emitter resistance is small but not negligible, and the collector's role and the base current are beside the point. Complete feedback is what makes a gain of one the most reliable gain in electronics, and it is why followers appear wherever fidelity matters more than amplification.

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34.

A follower has \( R_{E} = 1 \ \mathrm{k\Omega} \) at \( 2 \ \mathrm{mA} \). Its voltage gain is about

A.0.013
B.0.987
C.1.013
D.0.5, since the two resistances are comparable

Answer: B

The intrinsic resistance is thirteen ohms at two milliamps:

\[ A_{v} = \frac{1000}{1000+13} \approx 0.987 \]

giving about 0.987. A value above one is impossible for a follower, which rules out 1.013 without any calculation. Checking a computed gain against the physical bound the topology imposes is a quick and reliable sanity test.

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35.

Why does a common-base stage have such a wide bandwidth?

A.Its input impedance is low enough to swamp the device's capacitances
B.Its current gain of just under unity limits the phase shift available
C.Its input and output are not bridged by a gain
D.Its collector-base capacitance is smaller than in other configurations

Answer: C

With the base common, the collector-base capacitance no longer bridges input and output across an inverting gain, so there is no Miller multiplication. The capacitance itself is the same device property in every configuration. The low input impedance helps but is not the mechanism, and the current gain is unrelated. Bandwidth is nearly always about which capacitance sees which gain, rather than about the size of the capacitance. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

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36.

What is a common-base stage's input impedance?

A.Beta times the intrinsic emitter resistance of the device
B.The collector load divided by the voltage gain of the stage
C.The emitter resistor in parallel with the base divider network
D.The intrinsic emitter resistance

Answer: D

Driving the emitter directly presents the intrinsic emitter resistance, tens of ohms at typical currents, with no beta multiplication because the base is not the input. The second option is the common-emitter figure. A very low input impedance is a good match to a transmission line or to a current-source drive, which is why the configuration survives in high-frequency and cascode arrangements. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

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37.

Which configuration would you choose to drive a 50 ohm line directly?

A.Common base
B.Common emitter with a bypassed emitter resistor
C.Common collector with a large emitter resistor
D.Common emitter with heavy degeneration

Answer: A

A common-base input is naturally tens of ohms, which matches such a line without any transformer or padding. A common-emitter input is a few kilohms and a follower's is far higher still. Matching is about choosing the configuration whose natural impedance suits the source, and it is one of the few situations where common base is the obvious first choice rather than a specialist's option.

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38.

By what factor does an emitter impedance appear larger at the base?

A.The square of the current gain
B.One plus the current gain
C.The transconductance times the load
D.The ratio of collector to emitter resistance

Answer: B

For the same voltage across it, the base supplies only a beta-th of the current, so the apparent impedance is one plus beta times larger. The other options are gain expressions rather than reflection factors. The rule works because impedance is voltage over current and only the current has been scaled, which is the same reasoning that governs a transformer.

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39.

By what factor does a base impedance appear smaller at the emitter?

A.The ratio of the two supply rails in the circuit
B.The square root of the current gain
C.One plus the current gain
D.The transconductance times the source resistance

Answer: C

Looking the other way the same rule inverts: a source impedance at the base appears at the emitter divided by one plus beta, which is why a follower's output impedance is so low. That the factor is the same in both directions, multiplying one way and dividing the other, makes the pair easy to hold once the reason is clear rather than the formula memorised.

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40.

A source of \( 4.7 \ \mathrm{k\Omega} \) drives a follower with \( \beta = 150 \) at \( 1 \ \mathrm{mA} \). The output impedance is about

A.4.7 kilohms
B.31 ohms, taking only the reflected source resistance
C.705 kilohms
D.57 ohms

Answer: D

The reflected source adds to the intrinsic emitter resistance:

\[ r_{out} \approx 26 + \frac{4700}{151} \approx 57 \ \Omega \]

giving about 57 ohms. Both terms matter here, which is why omitting either gives a noticeably wrong answer. A follower divides a source impedance by beta but cannot go below the emitter resistance at the operating current, and that floor is worth remembering. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

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