Differential Amplifiers Questions and Answers

Answer: C

The first three are correctly paired and the fourth is inverted: the rejection ratio is the differential gain over the common-mode gain, so a good stage has a large one rather than a small one. Getting the ratio the right way up matters because the inverted form would be a number far below one and would look like a fault. A figure of merit should grow as the device improves, which is a quick check on whether a ratio has been written the right way round. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: D

Taking the ratio of the two gains and converting to decibels:

\[ \begin{aligned} \text{CMRR} &= \tfrac{500}{0.01} = 50000 \\ &= 20\log_{10}50000 \approx 94 \ \mathrm{dB} \end{aligned} \]

giving about 94 dB. The offset voltage is a distractor here, since it plays no part in the rejection calculation. Spotting which of a table's figures the question does not need is half the work in a question of this shape.

Answer: A

This is the tail, which is why the arrangement is called a long-tailed pair. A cascode sits above the pair to improve gain and bandwidth, an active load replaces the collector resistors, and a bootstrap raises an apparent impedance by driving both ends of a component together. Four techniques that all appear in a good input stage, each addressing a different specification, and naming them apart is what makes a schematic readable. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: B

It converts to single-ended, taking the current change from one side and adding it to the other so that the full differential signal appears at one node. A differential or balanced output is what the pair already provides, and common-mode is what it rejects. Getting a single-ended output without the usual halving, and a large increase in impedance at the same time, is why the arrangement is universal in operational amplifier input stages. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: C

Offset voltage shifts the output but does not change how well a shared signal is rejected - it is a separate specification arising from the same mismatch that does affect rejection through the loads. The other three all bear on the ratio directly. Two specifications arising from one physical cause are still two specifications, and improving one does not automatically improve the other.

Answer: D

Increasing the collector load raises both gains in nearly the same proportion, so the ratio barely moves - what it buys is differential gain. The other three all act on the ratio itself, by raising the tail impedance, restoring the cancellation, or reducing the imbalance that converts common-mode into differential. A change affecting both terms of a ratio equally cannot improve it, which is worth checking before any such proposal is adopted. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: A

The pair proper uses two matched devices sharing a tail, though a practical stage adds a third for the current source and often two more for an active load. One device cannot resolve a difference, and four or more describes a more elaborate arrangement. That the essential circuit is two devices and a shared tail, with everything else an improvement to one specification or another, is a useful way to read any input stage schematic. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

The stage's offset, drift and part of its rejection all depend on how closely the two devices match in junction area, doping and temperature. Different gains or currents would unbalance it, and opposite polarities describe a complementary output stage rather than a differential input. That the circuit's accuracy rests on matching rather than on any absolute value is why it belongs to integrated design, where matching is cheap and absolute values are not. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: C

With equal inputs and matched devices each carries half the tail current, so the two collector voltages are equal and the differential output is zero. Complete steering to one side is what a large differential input produces, and neither the loads nor the supplies divide it at rest. That the resting condition is a clean half-and-half split is what makes the stage's balance point so well defined.

Answer: D

Beyond a difference of a hundred millivolts or so the pair is fully steered: one device takes the entire tail current and the other is off. The tail current is fixed by its source and cannot double, and both devices cannot carry it all or be off together. That full steering happens at such a small input is why a differential pair makes a good comparator and why an amplifier must keep its input difference far smaller.