Differential Amplifiers Questions and Answers

Answer: C

Equal inputs should give equal collector currents, so an output offset means the two halves are not matched - different devices, different loads, or a fault in one side. A failed tail source would leave both collectors at the supply, an exceeded common-mode range needs the inputs moved away from earth, and oscillation would show as an alternating output. Reading the resting condition before applying any signal is the first step, and it separates matching faults from everything else.

Answer: D

Rejection is governed by the tail impedance while gain is governed by the loads and the transconductance, so a correct gain with poor rejection points at the tail rather than at anything in the signal path. Absolute load values affect gain, current gain affects neither much, and compensation affects stability. That two specifications depend on two different components is what lets one measurement localise a fault that the other would not reveal. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: A

Taking both inputs down eventually leaves too little voltage across the tail source for it to behave as a current source, and the stage collapses. Breakdown would be permanent, collector saturation limits the upper end of the range instead, and offset does not vary that way. That the two ends of the common-mode range fail by different mechanisms explains why they are rarely symmetrical, and why a data sheet quotes them separately. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

Full steering puts all the tail current into charging the compensation capacitor, and that fixed current produces a constant slope, which is a triangle. Clipping flattens the peaks instead, an exceeded common-mode range gives quite different symptoms, and instability produces ringing. A distortion whose shape identifies its mechanism is worth recognising, since it names the specification being exceeded and hence the remedy - more tail current or less compensation.

Answer: C

A cascode removes the Miller multiplication, widening the bandwidth, and raises the impedance at the output node, raising the gain - two improvements for the price of two devices and some headroom. A larger load raises gain and lowers bandwidth, more tail current raises gain and narrows the linear range, and degeneration reduces gain. A change that improves two specifications at once is rare enough to be worth recognising. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: D

Noise generated after the first stage is referred back to the input by dividing by the gain ahead of it, so with a first-stage gain of a hundred a later stage contributes a hundredth as much. Later stages do generate noise, feedback does not remove it, and the current is a design choice. This is the Friis cascade argument applied to an amplifier, and it is why the input pair's devices and operating current receive so much more attention than anything downstream. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: A

Both devices' noise appears at the output, so a pair is about three decibels noisier than a single stage at the same current per device. The tail source's noise is common to both sides and largely rejected, the loads are a design choice, and rejection does not convert noise. Three decibels is a real cost, paid for the rejection and the two inputs, and it is why a single-ended front end is still used where noise dominates every other consideration. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: B

The first two are correct and the third has the direction reversed: it is the later stages' noise that is divided by the input stage's gain, while the input stage's own noise is referred to the input undiminished. That asymmetry is why the first device dominates the noise figure of any chain. A statement with the right mechanism and the wrong direction is the most easily missed kind of error, and here it inverts the whole design priority.

Answer: C

The first three are correctly paired and the fourth is not, so three match. An active load improves the rejection of a single-ended output rather than reducing it, by restoring the cancellation that taking one collector would otherwise lose; its costs are headroom and the mirror's own matching. Naming what each improvement costs, as readily as what it buys, is what turns a list of techniques into a set of design choices.

Answer: D

A high-impedance source turns bias current into voltage error, so Stage B's picoamps beat Stage A's nanoamps by four orders of magnitude and that dominates everything else. Stage A's better offset and noise voltage would win with a low-impedance source. The two columns are a bipolar and a field effect input stage, and which is better depends entirely on the source impedance - which is why the source has to be named before the comparison can be decided.