Small Signal Amplifiers Questions and Answers

Answer: A

If both ends of a resistor move together, no signal current flows in it and it is effectively absent from the signal path, however low its resistance. The capacitor merely couples the driving voltage, reflection is a different rule, and no phase cancellation is involved. Making a component invisible by removing the voltage across it rather than by increasing its value is an idea worth recognising, since it appears in current sources and active loads as well.

Answer: B

The load and emitter resistor combine in parallel against the intrinsic resistance:

\[ \begin{aligned} R &= 1000 \| 220 = 180 \ \Omega \\ A_{v} &= \frac{180}{180+13} \approx 0.93 \end{aligned} \]

so about 0.93, noticeably below the unloaded 0.99 because the effective emitter resistance has fallen. Even a follower's gain drops when the load approaches its own emitter resistance, which is a useful reminder that unity is a limit rather than a guarantee.

Answer: C

Setting the collector resistor ten times the emitter resistor gives a gain of ten set by two resistors, holding across devices and temperatures - subject to a small correction for the intrinsic emitter resistance. Choosing an operating current would give the right answer only at one temperature, and the remaining options do not produce a gain of ten at all. A specification that reduces to a ratio of passive components is as reliable as those components, which is the whole appeal of feedback.

Answer: D

The denominator is the external resistor plus a term proportional to temperature and inversely to current, so the ratio is not quite exact and drifts a little. It does not touch the collector resistor, introduces no phase shift at these frequencies, and does not depend on beta. Making the external resistance large compared with the intrinsic term is the remedy, and it is why heavy degeneration is more accurate than light. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: A

Both are true and the second is the reason: beta appears nowhere in the gain expression, which is why gain is far more predictable than input impedance. Beta does affect how much drive current the stage needs and how heavily it loads its source, but not the voltage ratio itself. Knowing which specifications inherit the unreliable parameter and which do not is what makes a data sheet's spread tolerable. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: B

Both statements are true but the second does not explain the first: the accuracy comes from the gain becoming a ratio of two resistors rather than a function of the operating current, and the reduction in gain is the price rather than the mechanism. A circuit could reduce gain with a simple attenuator and gain no accuracy at all. Distinguishing the mechanism from the cost is exactly what this item tests. The shared properties explain why these schemes form a family, and the differences explain why more than one exists. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: C

The reason is true and the assertion is false: the follower's value lies in transforming impedance - high looking in, low looking out - which lets a delicate source drive a demanding load. Voltage gain is only one service a stage can provide. Judging a circuit by the one specification it does not offer is a mistake worth guarding against generally, and buffers are where it is made most often.

Answer: D

The first two are correct and the third is the reverse of the truth: the input resistance is beta times the emitter resistance and inherits beta's full spread. That one parameter depends on beta and another does not is the most useful distinction in the model, because it predicts which of a stage's specifications will hold when the device is changed and which will not.

Answer: A

The pairing is A-P, B-Q, C-R, and all three follow from one rule: whatever is in the emitter circuit appears at the base multiplied by beta, and appears at the emitter itself unmultiplied. Deriving the three from a single reflection rule is far more reliable than remembering three separate results, and it extends to the output impedances too.

Answer: B

The first three are correctly paired and the fourth is not, so three match. A stiffer divider is a lower resistance in parallel with the input, so it lowers the input impedance rather than raising it. Each of these changes moves two specifications at once, and being able to name both directions is what turns a list of formulae into a working feel for the circuit.