Small Signal Amplifiers Questions and Answers

Q: A stage has \( R_{C} = 4.7 \ \mathrm{k\Omega} \), \( I_{C} = 1 \ \mathrm{mA} \) and a bypassed emitter. Its gain is about

Dividing the load by the intrinsic emitter resistance:\[ A_{v} = -\frac{4700}{26} \approx -180 \]giving about -180. The figure depends on the operating current through that 26 ohms, so it moves with temperature and with any drift in the bias - which is exactly the unpredictability that degeneration is added to remove.

Q: With \( R_{E} = 470 \ \Omega \) left unbypassed, that gain becomes about

The external resistance now dominates the denominator:\[ A_{v} \approx -\frac{4700}{470+26} \approx -9.5 \]so about -9.5, close to the simple ratio of ten. Nearly twenty times the gain has been traded for a figure set by two resistors, which will hold across devices and temperatures. Whether that trade is worth making depends entirely on how much gain the design has to spare.

Q: What is the input impedance of a common-emitter stage with a bypassed emitter?

Looking into the base gives beta times the emitter resistance, and the bias divider appears in parallel with it - often dominating, since a stiff divider is deliberately made low. The third option applies when the emitter resistor is unbypassed. Remembering that the bias network is part of the signal path is what stops input impedance being overestimated by a wide margin.

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.

11.

A stage has \( R_{C} = 4.7 \ \mathrm{k\Omega} \), \( I_{C} = 1 \ \mathrm{mA} \) and a bypassed emitter. Its gain is about

A.-4700
B.-26
C.-180
D.-47, ignoring the emitter resistance entirely

Answer: C

Dividing the load by the intrinsic emitter resistance:

\[ A_{v} = -\frac{4700}{26} \approx -180 \]

giving about -180. The figure depends on the operating current through that 26 ohms, so it moves with temperature and with any drift in the bias - which is exactly the unpredictability that degeneration is added to remove.

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12.

With \( R_{E} = 470 \ \Omega \) left unbypassed, that gain becomes about

A.-180, unchanged because the emitter resistor is a direct-current element
B.-100, a little more than half the bypassed figure
C.-1
D.-9.5

Answer: D

The external resistance now dominates the denominator:

\[ A_{v} \approx -\frac{4700}{470+26} \approx -9.5 \]

so about -9.5, close to the simple ratio of ten. Nearly twenty times the gain has been traded for a figure set by two resistors, which will hold across devices and temperatures. Whether that trade is worth making depends entirely on how much gain the design has to spare.

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13.

What is the input impedance of a common-emitter stage with a bypassed emitter?

A.Beta times the intrinsic emitter resistance, in parallel with the divider
B.The collector load divided by the voltage gain of the stage
C.The emitter resistor multiplied by one plus the current gain
D.The divider's Thevenin resistance taken by itself alone

Answer: A

Looking into the base gives beta times the emitter resistance, and the bias divider appears in parallel with it - often dominating, since a stiff divider is deliberately made low. The third option applies when the emitter resistor is unbypassed. Remembering that the bias network is part of the signal path is what stops input impedance being overestimated by a wide margin.

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14.

How does leaving the emitter resistor unbypassed change the input impedance?

A.It becomes equal to the collector load divided by the gain
B.It rises, because the emitter resistance is now larger
C.It falls, because the resistor appears in parallel at the input
D.It is unchanged, since the resistor is outside the base circuit

Answer: B

The reflected quantity is now the external resistor plus the intrinsic one, so the input impedance rises by roughly the same factor as the gain falls. It appears in series with the emitter rather than in parallel with the input, and it is certainly inside the reflected path. That gain and input impedance move in opposite directions under series feedback is a general result, and it is often the reason degeneration is added in the first place. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

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15.

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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16.

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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17.

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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18.

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.

Discuss in Forum

19.

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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20.

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.

Discuss in Forum

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