Differential Amplifiers Questions and Answers

Answer: C

A current source presents hundreds of kilohms of dynamic impedance where a practical resistor might present a few kilohms, and the common-mode gain falls in proportion. Its standing current is whatever the design chooses, the differential gain is untouched, and it does nothing about device matching. Improving the numerator of a ratio while leaving the denominator alone is the cleanest possible improvement, and this is a textbook case of it. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: D

Taking the output from one collector loses the cancellation that a differential output provides, since the common-mode shift appears at both collectors equally and only the difference between them is free of it. The differential gain is indeed halved, which reduces the ratio further, but the loss of cancellation is the larger effect. An active load restores much of what is lost, which is why integrated stages use one rather than a plain resistor. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: A

A current mirror as the collector load takes the current change from one side and adds it to the other, so the full differential signal appears at one node and the stage's gain is not halved. The tail current, the bias and any swing limiting come from elsewhere. Getting a single-ended output without the usual penalty is what makes the arrangement universal in operational amplifier input stages, and it also raises the gain by presenting a very high impedance. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: B

The gain is the transconductance times the load impedance, and a current mirror's output impedance is hundreds of kilohms rather than the few kilohms a practical resistor allows, so the stage approaches the device's intrinsic gain limit. Tail current, base current and input capacitance are unaffected. One component supplying both the conversion to single-ended and a large increase in gain is why the arrangement displaced resistive loading entirely. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: C

With the external load impedance no longer dominant, the gain settles at the transconductance times the device's own output resistance, a few thousand. The tail impedance governs common-mode rejection rather than differential gain, current gain does not appear in the expression, and the supply limits swing. Removing whichever element dominates a limit reveals the limit beneath it, and here what is revealed is the device's intrinsic gain. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

Offset voltage is the small difference that must be applied between the inputs to bring the output to zero, and it arises from mismatch between the two devices and their loads. The other options describe common-mode conditions and input ranges. Because it is referred to the input, it is multiplied by the full gain on its way to the output - which is why a millivolt of offset matters so much more than a millivolt of anything measured later. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: A

Differences in junction area, doping and load resistance mean the two sides need slightly different base voltages to carry equal currents, and that difference is the offset. Tail impedance governs common-mode rejection, and neither supply nor source impedance produces it. Because it is a matching problem rather than an absolute one, offset is far smaller in an integrated pair made side by side on one chip than in two separately packaged devices. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

Bias current is the average of the two base currents, and it flows through whatever resistance is connected to each input, developing a voltage there. The difference between them is the offset current, a separate and usually much smaller specification. Tail and collector currents are internal. Because the bias current develops a voltage across the source resistance, matching the resistances at the two inputs cancels most of its effect. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: C

If both inputs see the same resistance, equal bias currents produce equal voltages that cancel in the difference, leaving only the mismatch between them to matter - and that mismatch is the offset current. Neither depends on signal presence or supply, and both are quoted at a stated temperature. Splitting a specification into a part that can be cancelled by design and a part that cannot is what makes it useful, and it tells the designer to match the input resistances.

Answer: D

The bias current develops a voltage across the unbalanced resistance:

\[ V = 80 \times 10^{-9} \times 10^{5} = 8 \ \mathrm{mV} \]

giving 8 millivolts, far larger than the pair's own offset voltage of perhaps one. Putting an equal resistance at the other input would cancel almost all of it, leaving only the offset current's contribution. A fault introduced by the surrounding circuit can easily dominate the device's own specification, and this is the commonest example.