Differential Amplifiers Questions and Answers

Q: What does a Widlar current source achieve in an input stage's bias?

Adding a small resistor in one emitter of a mirror makes the copied current far smaller than the reference, so microamps can be generated without the megohm resistor a plain mirror would need - which matters because large resistors occupy impossible area on a chip. Supply independence and temperature tracking are separate properties of other arrangements. A technique that exists because one component is expensive in a given technology is worth recognising as such.

Q: A pair's collector loads are \( 20 \ \mathrm{k\Omega} \) with a tail current of \( 200 \ \mathrm{\mu A} \). Its single-ended gain is about

Each device carries a hundred microamps:\[ \begin{aligned} g_{m} &= \tfrac{100\mathrm{\mu}}{26\mathrm{m}} \approx 3.85 \ \mathrm{mS} \\ A_{v} &\approx \tfrac{3.85\mathrm{m} \times 2\times10^{4}}{1} \approx 77 \end{aligned} \]Taking the single-ended halving into account gives about 77. Halving the tail current before computing the transconductance is the step that matters, and forgetting it doubles the answer.

Q: A pair's offset drifts \( 5 \ \mathrm{\mu V} \) per degree over \( 40 \ \mathrm{^\circ C} \). The total drift is

Multiplying the coefficient by the temperature span:\[ 5 \times 40 = 200 \ \mathrm{\mu V} \]giving 200 microvolts, which in a stage of gain a thousand becomes two hundred millivolts at the output. Since a trim cancels the offset only at the temperature it was set, this drift is what actually limits a precision measurement across a working day.

Q: Why is a chopper-stabilised amplifier used for very small direct signals?

Chopping translates a slowly varying signal up to the chopping frequency, where offset drift and one-over-f noise have fallen away, then recovers it by synchronous detection. Gain, bias current and bandwidth are not what it addresses - bandwidth is if anything reduced. It is the same technique used to escape flicker noise in any low-frequency measurement, applied here to escape the drift that limits a direct-coupled stage.

Q: What does an auto-zeroing amplifier do?

Auto-zeroing disconnects the input at intervals, measures the offset that remains, and stores a correction that is applied while the amplifier is in use. Gain control, input coupling and common-mode nulling are other functions. Because the measurement is repeated, the correction follows the drift rather than being fixed at one temperature - which is precisely the limitation of a one-off trim that this technique escapes.

Q: Which of the following is not a use of a differential pair? Rectifying an alternating supply Amplifying a small difference with high rejection Comparing two voltages and switching Multiplying two signals in a mixer

Rectification needs a diode's asymmetric conduction and has nothing to do with a pair, which is symmetrical by construction. The other three are its standard roles as amplifier, comparator and multiplier. That one arrangement serves three quite different purposes according to how it is driven and where the output is taken is what makes it the most reused block in analogue design.

Q: Two differential pairs steered by a third, forming a four-quadrant multiplier, make up a ______ cell.

This is the Gilbert cell. Widlar names a current source that produces a small current from a modest resistor, Wilson names an improved current mirror, and Darlington names a cascaded pair that multiplies current gain. Four names attached to four distinct integrated building blocks, and using the right one makes a description intelligible to whoever reads it next.

Q: Why does a Wilson mirror copy a current more accurately than a simple one?

The third device senses the output current and adjusts the drive, so the base currents that a simple two-device mirror loses are largely compensated. Matching is still between devices, the current is a design choice, and it needs more headroom rather than less. Adding feedback around a copying arrangement to cancel its principal error is the same idea that makes an operational amplifier accurate, applied here to a current rather than a voltage.

Q: What sets the accuracy of a simple two-device current mirror?

Two error sources dominate: how well the two devices match, and the two base currents drawn from the reference branch, which make the copied current slightly smaller. Supply and load affect the reference current and the compliance rather than the copying accuracy, and the resistor's coefficient affects the reference. Knowing which two mechanisms limit an arrangement is what tells you which improvement - better matching or a Wilson connection - is worth making.

Q: Why is a current mirror's output impedance important as an active load?

The stage's gain is transconductance times load impedance, so a mirror's hundreds of kilohms let the pair approach its intrinsic gain limit where a practical resistor could not. The tail current, the common-mode range and the slew rate are set elsewhere. That the same component supplies both the conversion to single-ended output and a large increase in gain is why it displaced resistive loading entirely.

91.

What does a Widlar current source achieve in an input stage's bias?

A.A current independent of the supply voltage
B.A current that tracks the temperature exactly
C.A very small current from a modest resistor
D.A very large current from a small resistor

Answer: C

Adding a small resistor in one emitter of a mirror makes the copied current far smaller than the reference, so microamps can be generated without the megohm resistor a plain mirror would need - which matters because large resistors occupy impossible area on a chip. Supply independence and temperature tracking are separate properties of other arrangements. A technique that exists because one component is expensive in a given technology is worth recognising as such.

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92.

A pair's collector loads are \( 20 \ \mathrm{k\Omega} \) with a tail current of \( 200 \ \mathrm{\mu A} \). Its single-ended gain is about

A.154
B.40
C.308
D.77

Answer: D

Each device carries a hundred microamps:

\[ \begin{aligned} g_{m} &= \tfrac{100\mathrm{\mu}}{26\mathrm{m}} \approx 3.85 \ \mathrm{mS} \\ A_{v} &\approx \tfrac{3.85\mathrm{m} \times 2\times10^{4}}{1} \approx 77 \end{aligned} \]

Taking the single-ended halving into account gives about 77. Halving the tail current before computing the transconductance is the step that matters, and forgetting it doubles the answer.

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93.

A pair's offset drifts \( 5 \ \mathrm{\mu V} \) per degree over \( 40 \ \mathrm{^\circ C} \). The total drift is

A.200 microvolts
B.20 microvolts
C.2 millivolts
D.45 microvolts

Answer: A

Multiplying the coefficient by the temperature span:

\[ 5 \times 40 = 200 \ \mathrm{\mu V} \]

giving 200 microvolts, which in a stage of gain a thousand becomes two hundred millivolts at the output. Since a trim cancels the offset only at the temperature it was set, this drift is what actually limits a precision measurement across a working day.

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94.

Why is a chopper-stabilised amplifier used for very small direct signals?

A.It has a wider bandwidth than a plain amplifier
B.It moves the signal above the flicker noise and drift
C.It has a higher differential gain than a plain pair
D.It needs no input bias current at all

Answer: B

Chopping translates a slowly varying signal up to the chopping frequency, where offset drift and one-over-f noise have fallen away, then recovers it by synchronous detection. Gain, bias current and bandwidth are not what it addresses - bandwidth is if anything reduced. It is the same technique used to escape flicker noise in any low-frequency measurement, applied here to escape the drift that limits a direct-coupled stage.

Discuss in Forum

95.

What does an auto-zeroing amplifier do?

A.Removes the direct component from its input
B.Nulls the common-mode voltage at its inputs
C.Measures and subtracts its own offset periodically
D.Adjusts its gain to keep the output constant

Answer: C

Auto-zeroing disconnects the input at intervals, measures the offset that remains, and stores a correction that is applied while the amplifier is in use. Gain control, input coupling and common-mode nulling are other functions. Because the measurement is repeated, the correction follows the drift rather than being fixed at one temperature - which is precisely the limitation of a one-off trim that this technique escapes.

Discuss in Forum

96.

Which of the following is not a use of a differential pair?

Rectifying an alternating supply

Amplifying a small difference with high rejection

Comparing two voltages and switching

Multiplying two signals in a mixer

A.Amplifying a small difference with high rejection
B.Comparing two voltages and switching
C.Multiplying two signals in a mixer
D.Rectifying an alternating supply

Answer: D

Rectification needs a diode's asymmetric conduction and has nothing to do with a pair, which is symmetrical by construction. The other three are its standard roles as amplifier, comparator and multiplier. That one arrangement serves three quite different purposes according to how it is driven and where the output is taken is what makes it the most reused block in analogue design.

Discuss in Forum

97.

Two differential pairs steered by a third, forming a four-quadrant multiplier, make up a ______ cell.

A.Gilbert
B.Widlar
C.Wilson
D.Darlington

Answer: A

This is the Gilbert cell. Widlar names a current source that produces a small current from a modest resistor, Wilson names an improved current mirror, and Darlington names a cascaded pair that multiplies current gain. Four names attached to four distinct integrated building blocks, and using the right one makes a description intelligible to whoever reads it next.

Discuss in Forum

98.

Why does a Wilson mirror copy a current more accurately than a simple one?

A.It needs less voltage headroom to work
B.Feedback cancels most of the base current error
C.It uses matched resistors rather than devices
D.It operates at a much higher current

Answer: B

The third device senses the output current and adjusts the drive, so the base currents that a simple two-device mirror loses are largely compensated. Matching is still between devices, the current is a design choice, and it needs more headroom rather than less. Adding feedback around a copying arrangement to cancel its principal error is the same idea that makes an operational amplifier accurate, applied here to a current rather than a voltage.

Discuss in Forum

99.

What sets the accuracy of a simple two-device current mirror?

A.The load impedance the output drives
B.The temperature coefficient of the reference resistor
C.Device matching and the base currents lost
D.The supply voltage feeding the reference

Answer: C

Two error sources dominate: how well the two devices match, and the two base currents drawn from the reference branch, which make the copied current slightly smaller. Supply and load affect the reference current and the compliance rather than the copying accuracy, and the resistor's coefficient affects the reference. Knowing which two mechanisms limit an arrangement is what tells you which improvement - better matching or a Wilson connection - is worth making.

Discuss in Forum

100.

Why is a current mirror's output impedance important as an active load?

A.It determines the tail current the pair draws
B.It fixes the input common-mode range
C.It limits the stage's slew rate
D.It sets the gain the differential stage reaches

Answer: D

The stage's gain is transconductance times load impedance, so a mirror's hundreds of kilohms let the pair approach its intrinsic gain limit where a practical resistor could not. The tail current, the common-mode range and the slew rate are set elsewhere. That the same component supplies both the conversion to single-ended output and a large increase in gain is why it displaced resistive loading entirely.

Discuss in Forum

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