Small Signal Amplifiers Questions and Answers

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.

Q: Why is a differential pair preferred as an integrated input stage?

Two matched devices sharing a temperature drift together, so the difference between their inputs is amplified while the common drift cancels. Its gain per device is no higher than a single stage's, it needs a tail current source, and single-rail operation is unrelated. Arranging a circuit so that the imperfection appears as a common-mode quantity and the signal as a differential one is the foundation of every precision analogue front end.

91.

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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92.

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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93.

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.

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94.

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.

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95.

Why does an active load occupy less silicon than an equivalent resistor?

A.An active load dissipates less power than a resistor would
B.A resistor cannot be made above a few kilohms in silicon
C.A high resistance needs a great deal of area
D.Transistors are always smaller than resistors on a chip

Answer: C

Integrated resistors are long strips of lightly doped material, so a hundred kilohms occupies an area a transistor delivering the same impedance does not need. Transistors are not universally smaller, dissipation is comparable at the same current, and large resistors can be made, just wastefully. What is cheap in a technology shapes its circuits, and area is the currency in integration.

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96.

What penalty does an active load bring?

A.A much lower gain than a resistor would give
B.A dependence of the gain on beta
C.An inability to work at low frequencies
D.Reduced output voltage headroom

Answer: D

A current source needs a volt or so across it to stay in its useful region, so the output cannot swing as close to the supply rail as it could with a resistor. The gain rises rather than falls, does not become beta-dependent, and the arrangement works down to direct current. Trading headroom for gain is the exchange, and it is affordable precisely because integrated stages are usually followed by more gain rather than by a load.

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97.

Why is a differential pair preferred as an integrated input stage?

A.Matched devices cancel their own drift
B.It offers a higher voltage gain than a single stage can
C.It needs no bias current source of its own
D.It works from a single supply rail rather than two

Answer: A

Two matched devices sharing a temperature drift together, so the difference between their inputs is amplified while the common drift cancels. Its gain per device is no higher than a single stage's, it needs a tail current source, and single-rail operation is unrelated. Arranging a circuit so that the imperfection appears as a common-mode quantity and the signal as a differential one is the foundation of every precision analogue front end.

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98.

A stage's gain rises from \( 40 \) to \( 2000 \) when its collector resistor is replaced by a current source. The device's output resistance is about

A.Twice the old load resistance
B.Fifty times the old load
C.Equal to the old load resistance
D.One fiftieth of the old load

Answer: B

Gain is proportional to the effective load, so the ratio of gains gives the ratio of loads:

\[ \frac{2000}{40} = 50 \]

so the device's output resistance is about fifty times the old collector resistor, which is now the only thing limiting the gain. Reading a hidden parameter from the change a circuit modification produces is often quicker than measuring it directly.

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99.

A stage of gain \( -150 \) drives an identical stage. The cascade's gain is about

A.-150
B.300
C.11000
D.22500

Answer: C

Each stage's load is halved by the next stage's input impedance if the two are equal, so each gain falls to about seventy-five:

\[ (-75) \times (-75) \approx 5600 \]

and with lighter loading the figure rises towards about eleven thousand, the product of one loaded and one unloaded stage. The naive 22500 assumes no loading at all. Two inversions give a positive result, which is worth noting before the magnitude is considered.

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100.

Why does the last stage of a cascade often show the full unloaded gain?

A.Its operating current is usually higher than the others
B.Its emitter resistor is always fully bypassed
C.Its input impedance is higher than the preceding stages
D.Its load is the external one rather than another stage

Answer: D

Only the stages that drive another stage suffer interstage loading, so the final one sees whatever external load the design intends, which may be much lighter. Operating current, bypassing and input impedance vary by design rather than by position. Recognising that loading is a property of the interface rather than of the stage explains why the stages of a cascade often differ in measured gain despite being identical on paper.

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