Small Signal Amplifiers Questions and Answers

Q: Which of the following is not part of the hybrid-pi model? Pinch-off voltage Transconductance Input resistance from base to emitter Output resistance from collector to emitter

Pinch-off voltage belongs to a field effect transistor, describing where its channel closes; the hybrid-pi model of a bipolar device has no such element. The other three are its core elements, corresponding to the operating current, beta, and the Early effect respectively. Spotting a parameter that presupposes a structure the device does not have is a quick way to keep the two families' models apart. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Q: Reflecting an emitter impedance to the base multiplies it by ______.

The factor is one plus beta, because the base supplies only a beta-th of the current at the same voltage. The transconductance and the thermal voltage appear in the gain and in the intrinsic emitter resistance respectively, and the collector load appears in the gain. One rule, applied in two directions, accounts for every input and output impedance in the three configurations, which makes it the most economical thing to learn in the topic.

Q: Why is the two-port description useful even though it hides the physics?

A black box description makes no assumption about what is inside, so the same four parameters describe a transistor, a valve or a complete amplifier module. It has the same number of parameters as the hybrid-pi model, is equally restricted to small signals, and must be measured rather than derived. Generality bought at the cost of insight is a real trade, and it is why the two-port form survives in measurement and specification even where the physical model is preferred for design.

Q: Which two-port parameter corresponds to the current gain?

The forward current ratio is beta in the common-emitter case, and it is the parameter that carries the device's threefold spread. The input impedance corresponds to the reflected emitter resistance, the reverse voltage ratio to the Early effect seen backwards, and the output conductance to the slope of the output characteristics. Mapping each two-port parameter onto its physical counterpart is what makes the black box description intelligible.

Q: Why is the reverse voltage ratio usually neglected in hand analysis?

The reverse ratio is a few ten-thousandths, so the error from ignoring it is far smaller than the spread in beta and the tolerance of the resistors. It is measurable, does not vanish with bypassing, and applies at all frequencies. Neglecting a term because it is small compared with the uncertainties already present is legitimate, and stating that comparison is what distinguishes a justified approximation from a careless one.

Q: A stage has \( g_{m} = 80 \ \mathrm{mS} \) and a load of \( 1.5 \ \mathrm{k\Omega} \). Its gain magnitude is

Multiplying transconductance by load resistance:\[ |A_{v}| = 0.08 \times 1500 = 120 \]giving 120. The transconductance of eighty millisiemens corresponds to two milliamps of collector current, so the same answer follows from dividing 1500 by the thirteen-ohm intrinsic emitter resistance. Two routes to one answer is a useful check, and it also shows that the two forms of the gain expression really are the same statement.

Q: The same stage's output impedance, taking the device's own output resistance as \( 100 \ \mathrm{k\Omega} \), is about

The two resistances appear in parallel:\[ \begin{aligned} 1.5 \ \mathrm{k} \| 100 \ \mathrm{k} \\ &\approx 1.48 \ \mathrm{k\Omega} \end{aligned} \]so about 1.48 kilohms, within two per cent of the load resistor alone. That is why the device's output resistance is normally ignored in a resistively loaded stage, and why it becomes the dominant term as soon as the load is replaced by a current source.

Q: When does the device's own output resistance stop being negligible?

Two resistances in parallel are dominated by the smaller, so the device's output resistance matters only once the external load becomes comparable with it - which happens with a current-source load or an active load. Operating current, bypassing and frequency change other things. Knowing the condition under which a neglected term becomes significant is what lets the same analysis be reused in a quite different circuit without surprises.

Q: What is an active load?

An active load is a transistor, usually configured as a current source, standing in for the collector resistor so that the load impedance is very high. The gain then rises towards the device's intrinsic limit. It is not a signal-dependent impedance, not a matching resistor, and not a shaping capacitor. Replacing a passive component with a device that behaves better in the one respect that matters is a recurring move in integrated design, and this is its commonest form.

101.

Consider the following pairs of quantity and what determines it.

Voltage gain - Transconductance and load

Input impedance - Beta and the emitter resistance

Output impedance - Chiefly the collector load resistor

Transconductance - The device's current gain

How many of the pairs given above are correctly matched?

A.Only three
B.Only one
C.Only two
D.All four

Answer: A

The first three are correctly paired and the fourth is not, so three match. Transconductance is the collector current over the thermal voltage and has nothing to do with beta, which is precisely why the voltage gain is the predictable specification and the input impedance is not. Tracing each specification to the parameters it actually depends on is the quickest way to know which will survive a change of device. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

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102.

Which of the following is not part of the hybrid-pi model?

Pinch-off voltage

Transconductance

Input resistance from base to emitter

Output resistance from collector to emitter

A.Output resistance from collector to emitter
B.Pinch-off voltage
C.Transconductance
D.Input resistance from base to emitter

Answer: B

Pinch-off voltage belongs to a field effect transistor, describing where its channel closes; the hybrid-pi model of a bipolar device has no such element. The other three are its core elements, corresponding to the operating current, beta, and the Early effect respectively. Spotting a parameter that presupposes a structure the device does not have is a quick way to keep the two families' models apart. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Discuss in Forum

103.

Reflecting an emitter impedance to the base multiplies it by ______.

A.the thermal voltage
B.the collector load
C.one plus beta
D.the transconductance

Answer: C

The factor is one plus beta, because the base supplies only a beta-th of the current at the same voltage. The transconductance and the thermal voltage appear in the gain and in the intrinsic emitter resistance respectively, and the collector load appears in the gain. One rule, applied in two directions, accounts for every input and output impedance in the three configurations, which makes it the most economical thing to learn in the topic.

Discuss in Forum

104.

Why is the two-port description useful even though it hides the physics?

A.It contains fewer parameters than any physical model does
B.It remains valid for large signals as well as small ones
C.It can be derived from the device geometry without measurement
D.It applies to any device or network without modification

Answer: D

A black box description makes no assumption about what is inside, so the same four parameters describe a transistor, a valve or a complete amplifier module. It has the same number of parameters as the hybrid-pi model, is equally restricted to small signals, and must be measured rather than derived. Generality bought at the cost of insight is a real trade, and it is why the two-port form survives in measurement and specification even where the physical model is preferred for design.

Discuss in Forum

105.

Which two-port parameter corresponds to the current gain?

A.The forward current ratio
B.The input impedance at the driven port
C.The reverse voltage ratio between the two ports
D.The output conductance at the far port

Answer: A

The forward current ratio is beta in the common-emitter case, and it is the parameter that carries the device's threefold spread. The input impedance corresponds to the reflected emitter resistance, the reverse voltage ratio to the Early effect seen backwards, and the output conductance to the slope of the output characteristics. Mapping each two-port parameter onto its physical counterpart is what makes the black box description intelligible.

Discuss in Forum

106.

Why is the reverse voltage ratio usually neglected in hand analysis?

A.It applies only above the transition frequency
B.It is small enough to make little difference
C.It cannot be measured with ordinary equipment
D.It becomes zero once the emitter is properly bypassed

Answer: B

The reverse ratio is a few ten-thousandths, so the error from ignoring it is far smaller than the spread in beta and the tolerance of the resistors. It is measurable, does not vanish with bypassing, and applies at all frequencies. Neglecting a term because it is small compared with the uncertainties already present is legitimate, and stating that comparison is what distinguishes a justified approximation from a careless one.

Discuss in Forum

107.

A stage has \( g_{m} = 80 \ \mathrm{mS} \) and a load of \( 1.5 \ \mathrm{k\Omega} \). Its gain magnitude is

A.12
B.1500
C.120
D.80

Answer: C

Multiplying transconductance by load resistance:

\[ |A_{v}| = 0.08 \times 1500 = 120 \]

giving 120. The transconductance of eighty millisiemens corresponds to two milliamps of collector current, so the same answer follows from dividing 1500 by the thirteen-ohm intrinsic emitter resistance. Two routes to one answer is a useful check, and it also shows that the two forms of the gain expression really are the same statement.

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108.

The same stage's output impedance, taking the device's own output resistance as \( 100 \ \mathrm{k\Omega} \), is about

A.100 kilohms
B.1.5 kilohms exactly
C.51 kilohms
D.1.48 kilohms

Answer: D

The two resistances appear in parallel:

\[ \begin{aligned} 1.5 \ \mathrm{k} \| 100 \ \mathrm{k} \\ &\approx 1.48 \ \mathrm{k\Omega} \end{aligned} \]

so about 1.48 kilohms, within two per cent of the load resistor alone. That is why the device's output resistance is normally ignored in a resistively loaded stage, and why it becomes the dominant term as soon as the load is replaced by a current source.

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109.

When does the device's own output resistance stop being negligible?

A.When the collector load approaches or exceeds it
B.When the operating current is raised above a milliamp
C.When the emitter resistor is left unbypassed in the stage
D.When the signal frequency approaches the transition frequency

Answer: A

Two resistances in parallel are dominated by the smaller, so the device's output resistance matters only once the external load becomes comparable with it - which happens with a current-source load or an active load. Operating current, bypassing and frequency change other things. Knowing the condition under which a neglected term becomes significant is what lets the same analysis be reused in a quite different circuit without surprises.

Discuss in Forum

110.

What is an active load?

A.A capacitor placed across the collector to shape the response
B.A transistor used in place of the collector resistor
C.A load whose impedance changes with the signal amplitude
D.A resistor chosen to match the following stage's input

Answer: B

An active load is a transistor, usually configured as a current source, standing in for the collector resistor so that the load impedance is very high. The gain then rises towards the device's intrinsic limit. It is not a signal-dependent impedance, not a matching resistor, and not a shaping capacitor. Replacing a passive component with a device that behaves better in the one respect that matters is a recurring move in integrated design, and this is its commonest form.

Discuss in Forum

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