Differential Amplifiers Questions and Answers

Answer: A

Two transistors formed side by side in the same process, at the same temperature, match far better than any two parts selected from a batch, and offset is entirely a matching problem. Uniform temperature is part of the same advantage rather than a separate one, current gain is not what matters, and trimming is done on precision parts but is not the general reason. That integrated analogue design rests on matching rather than on absolute values is the single most useful thing to know about it.

Answer: B

Drift is the temperature coefficient of the offset voltage, quoted in microvolts per degree, and it matters because a nulled offset does not stay nulled as the equipment warms. Supply sensitivity, warm-up and unit-to-unit spread are three other specifications. Since offset can be trimmed to zero at one temperature but drift cannot be trimmed away, drift is the figure that limits a precision measurement over a working day.

Answer: C

A trimming adjustment cancels the mismatch at the temperature it was set at, and as the devices heat their mismatch changes, so the cancellation decays. The trim's own drift is a smaller secondary effect, drift is a device property rather than a supply one, and it acts across the whole range. A correction applied at one operating point holds only near that point, which is a general limitation of trimming and the reason low-drift devices command a premium. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: D

A change in the supply moves both sides of the pair together, so it appears at the output as a common-mode signal and is rejected by the same mechanism that rejects a shared input. The tail source does help, by holding the current constant, but the cancellation at the two collectors is the principal mechanism. That a symmetrical arrangement rejects anything affecting both halves equally is why the pair is used wherever interference and supply noise are expected. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: A

The common-mode range is how far both inputs may be moved together while every device stays in its active region. Beyond it the tail source or an input device saturates and the stage stops working, often abruptly. The other options describe differential range, supply range and output swing. A specification bounding one of the two components the stage resolves its inputs into is easy to overlook, and exceeding it produces symptoms that look nothing like overload. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: B

Taking both inputs down eventually leaves the tail source without enough voltage across it to behave as a current source, and the pair's operation collapses. The upper end is limited instead by the input devices saturating against their collector loads. Breakdown and output swing are separate limits. That the two ends of the range are set by different mechanisms is why they are rarely symmetrical about the supply midpoint. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: C

Each device carries half the tail current, so its transconductance follows from that:

\[ \begin{aligned} g_{m} &= \tfrac{0.5 \ \mathrm{mA}}{26 \ \mathrm{mV}} \approx 19 \ \mathrm{mS} \\ A_{v} &\approx 19\mathrm{m} \times 10^{4} \approx 190 \end{aligned} \]

Taking the usual halving into account for a single-ended output gives about 100. Remembering that each device sees only half the tail current is the step this calculation turns on.

Answer: D

A double-ended output collects both collectors' opposite excursions:

\[ A_{v(dd)} = 2 \times 100 = 200 \]

giving about 200, twice the single-ended figure, and with the common-mode movement cancelled as well. That one wiring choice buys both twice the gain and better rejection is why a differential output is preferred wherever the following stage can accept one.

Answer: A

Common-mode gain is roughly the collector load over twice the tail impedance:

\[ A_{cm} \approx \frac{5000}{10^{6}} = 0.005 \]

giving about 0.005, so the stage attenuates a shared signal two hundredfold while amplifying a difference. The factor of two in the denominator comes from the two devices sharing the tail, and omitting it doubles the answer.

Answer: B

Taking the ratio of the two gains and converting:

\[ \begin{aligned} \text{CMRR} &= \tfrac{100}{0.005} = 20000 \\ &= 20\log_{10} 20000 \approx 86 \ \mathrm{dB} \end{aligned} \]

giving about 86 dB, respectable for a discrete pair though well short of what an integrated design with a good current source achieves. Quoting the ratio in decibels rather than as twenty thousand makes it comparable with other specifications.