Bipolar Junction Transistors Questions and Answers

Q: An emitter resistor of \( 100 \ \Omega \) is now added without bypassing. The gain becomes about

The emitter resistance adds to the intrinsic 26 ohms:\[ A_{v} \approx -\frac{5000}{100+26} \approx -40 \]so about -40, five times less than before. What has been bought is predictability: the gain now depends on two resistors rather than on the operating current, so it barely moves with temperature or device spread. Trading gain for stability through local feedback is the single most common move in analogue design.

Q: Why does an unbypassed emitter resistor stabilise the voltage gain?

With the emitter resistance dominant, the gain is set by the collector and emitter resistors, both of which are far more predictable than the operating current or beta. The transconductance is unchanged, bias stability is a related but separate benefit, and the input impedance rises rather than falls. Whenever a performance figure can be made a ratio of passive components, it becomes as good as those components, which is the whole appeal of feedback.

Q: What does the hybrid-pi model represent?

The hybrid-pi model is a linearised picture valid for small excursions about a chosen bias point, with a transconductance, an input resistance and an output resistance. Switching, thermal and breakdown behaviour all need other models. Replacing a non-linear device by a linear model about one point is the standard way to make amplifier analysis tractable, and the model's validity is exactly as wide as the signal is small. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Q: Which hybrid-pi element accounts for the Early effect?

The finite slope of the output characteristics appears as a resistance from collector to emitter, and it caps the gain a stage can reach even with an infinite load. Transconductance follows from the operating current, input resistance from beta and transconductance together, and base spreading resistance is a physical bulk resistance. Every element of a small-signal model corresponds to one physical effect, and tracing that correspondence is what makes the model memorable.

Q: What does the base spreading resistance limit?

Base spreading resistance works with the junction capacitances to form a low-pass network, so it limits bandwidth and adds noise. It has little effect on the direct current gain, and neither voltage rating nor saturation voltage depends on it. It is also why a thin base, which helps transistor action, hurts the high-frequency noise figure - a genuine conflict the device designer has to resolve rather than avoid.

Q: What does the transition frequency of a transistor state?

The transition frequency is where the common-emitter current gain has fallen to unity, and it equals the low-frequency beta times the beta cut-off frequency. The 3 dB point of the voltage gain depends on the circuit rather than the device. The remaining options describe nothing standard. A gain-bandwidth product expressed as a single frequency lets two devices be compared regardless of the gain each is used at.

Q: Why does a transistor's beta fall at high frequencies?

At high frequencies the base-emitter capacitance takes an increasing share of the drive current, so less of it produces collector current. Base thickness is fixed, leakage is a temperature effect, and the Early effect is a voltage effect. A gain that falls with frequency is nearly always a capacitance diverting something, and finding which capacitance and where it goes is the whole of high-frequency analysis.

Q: What is the Miller effect in a common-emitter stage?

Because the collector swings in antiphase to the base, the collector-base capacitance appears at the input multiplied by one plus the voltage gain, which usually dominates the stage's bandwidth. The emitter resistance is indeed multiplied at the base, but that is the impedance-reflection rule rather than the Miller effect. A small capacitance across an inverting gain becomes a large one, which is why the cascode arrangement, with no voltage gain at the lower device, exists at all.

31.

An emitter resistor of \( 100 \ \Omega \) is now added without bypassing. The gain becomes about

A.-400
B.-40
C.-200
D.-2

Answer: B

The emitter resistance adds to the intrinsic 26 ohms:

\[ A_{v} \approx -\frac{5000}{100+26} \approx -40 \]

so about -40, five times less than before. What has been bought is predictability: the gain now depends on two resistors rather than on the operating current, so it barely moves with temperature or device spread. Trading gain for stability through local feedback is the single most common move in analogue design.

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32.

Why does an unbypassed emitter resistor stabilise the voltage gain?

A.The collector current no longer depends on beta
B.The input impedance falls in the same proportion
C.The gain becomes a ratio of two resistors
D.The transconductance rises to compensate for it

Answer: C

With the emitter resistance dominant, the gain is set by the collector and emitter resistors, both of which are far more predictable than the operating current or beta. The transconductance is unchanged, bias stability is a related but separate benefit, and the input impedance rises rather than falls. Whenever a performance figure can be made a ratio of passive components, it becomes as good as those components, which is the whole appeal of feedback.

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33.

What does the hybrid-pi model represent?

A.The large-signal switching behaviour of the device
B.The thermal behaviour of the junction and package
C.The breakdown behaviour under excessive voltage
D.The small-signal behaviour about an operating point

Answer: D

The hybrid-pi model is a linearised picture valid for small excursions about a chosen bias point, with a transconductance, an input resistance and an output resistance. Switching, thermal and breakdown behaviour all need other models. Replacing a non-linear device by a linear model about one point is the standard way to make amplifier analysis tractable, and the model's validity is exactly as wide as the signal is small. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Discuss in Forum

34.

Which hybrid-pi element accounts for the Early effect?

A.The output resistance
B.The transconductance
C.The input resistance
D.The base spreading resistance

Answer: A

The finite slope of the output characteristics appears as a resistance from collector to emitter, and it caps the gain a stage can reach even with an infinite load. Transconductance follows from the operating current, input resistance from beta and transconductance together, and base spreading resistance is a physical bulk resistance. Every element of a small-signal model corresponds to one physical effect, and tracing that correspondence is what makes the model memorable.

Discuss in Forum

35.

What does the base spreading resistance limit?

A.The saturation voltage under heavy drive
B.The high-frequency performance of the stage
C.The direct current gain the device achieves
D.The maximum collector voltage permitted

Answer: B

Base spreading resistance works with the junction capacitances to form a low-pass network, so it limits bandwidth and adds noise. It has little effect on the direct current gain, and neither voltage rating nor saturation voltage depends on it. It is also why a thin base, which helps transistor action, hurts the high-frequency noise figure - a genuine conflict the device designer has to resolve rather than avoid.

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36.

What does the transition frequency of a transistor state?

A.The highest frequency at which the device may be biased
B.The frequency at which the input impedance becomes real
C.The frequency at which the current gain falls to one
D.The frequency at which the voltage gain falls by 3 dB

Answer: C

The transition frequency is where the common-emitter current gain has fallen to unity, and it equals the low-frequency beta times the beta cut-off frequency. The 3 dB point of the voltage gain depends on the circuit rather than the device. The remaining options describe nothing standard. A gain-bandwidth product expressed as a single frequency lets two devices be compared regardless of the gain each is used at.

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37.

A transistor has \( \beta = 100 \) and \( f_{T} = 300 \ \mathrm{MHz} \). Its beta cut-off frequency is

A.300 MHz
B.30 MHz
C.30 kHz
D.3 MHz

Answer: D

Dividing the transition frequency by the low-frequency gain:

\[ \begin{aligned} f_{\beta} &= \frac{300 \ \mathrm{MHz}}{100} \\ &= 3 \ \mathrm{MHz} \end{aligned} \]

giving 3 MHz, beyond which the current gain starts to fall. The other options misplace a decade. The reciprocal relation between gain and bandwidth is what the transition frequency encodes, and it means a stage run at lower gain automatically has more bandwidth available.

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38.

Why does a transistor's beta fall at high frequencies?

A.Junction capacitances shunt the drive current away
B.The base becomes too thin to carry the current
C.Leakage current grows with the signal frequency
D.The Early effect becomes stronger as frequency rises

Answer: A

At high frequencies the base-emitter capacitance takes an increasing share of the drive current, so less of it produces collector current. Base thickness is fixed, leakage is a temperature effect, and the Early effect is a voltage effect. A gain that falls with frequency is nearly always a capacitance diverting something, and finding which capacitance and where it goes is the whole of high-frequency analysis.

Discuss in Forum

39.

What is the Miller effect in a common-emitter stage?

A.The output resistance appears magnified by the load
B.The collector-base capacitance appears magnified at the input
C.The emitter resistance appears magnified at the base
D.The base current appears magnified at the collector

Answer: B

Because the collector swings in antiphase to the base, the collector-base capacitance appears at the input multiplied by one plus the voltage gain, which usually dominates the stage's bandwidth. The emitter resistance is indeed multiplied at the base, but that is the impedance-reflection rule rather than the Miller effect. A small capacitance across an inverting gain becomes a large one, which is why the cascode arrangement, with no voltage gain at the lower device, exists at all.

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40.

Why is a transistor driven hard into saturation when used as a switch?

A.To reduce the leakage current when it is off
B.To speed up the transition into the off state
C.To keep its dissipation low while conducting
D.To make its current gain as large as possible

Answer: C

In saturation the collector-emitter voltage is a few tenths of a volt, so the product of current and voltage is small and the device stays cool. Beta is irrelevant there, since the current is set by the external circuit. Off-state leakage is a cut-off property, and heavy drive actually slows turn-off because the stored charge must be removed. A switch wants low voltage when on and low current when off, and saturation delivers the first of those. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

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