Small Signal Amplifiers Questions and Answers

Answer: C

Including the intrinsic 26 ohms in the denominator:

\[ A_{v} \approx -\frac{3300}{330+26} \approx -9.3 \]

so about -9, slightly below the bare ratio of ten. The uncorrected ratio is a defensible first estimate, and recognising it as an approximation rather than the answer is what the question is really asking.

Answer: D

This is the intrinsic emitter resistance, sometimes written as little r-e, and it is a property of the operating current rather than of any component. An external emitter resistor adds to it, reflection is what happens to it at the base, and differential mode belongs to a pair of devices. Distinguishing a resistance that exists because of the physics from one that exists because a component was fitted is essential to reading these expressions.

Answer: A

The current gain does not appear in the voltage gain expression at all - the gain is transconductance times load, or equivalently a resistance ratio. Beta affects the input impedance and the drive required rather than the voltage ratio. Knowing which parameters a specification is immune to is as useful as knowing which it depends on, because it tells you what you can change freely.

Answer: B

Distortion appears when the excursion is large enough that the exponential characteristic can no longer be replaced by its tangent, which is precisely the boundary of the small-signal assumption. Capacitor break points change the amplitude and phase without distorting the waveform, and loading and dissipation are separate concerns. A model's failure and the onset of distortion are the same event, which is why the small-signal condition matters practically.

Answer: C

Series feedback makes the stage's gain depend on a resistor ratio rather than on the device's curved characteristic, so the non-linearity is divided down along with the gain. The amplitude at the base is actually reduced as a consequence, which is part of the same effect rather than a separate one. It does not change the current or filter anything. Feedback trades gain for every kind of accuracy at once, and linearity is one of the benefits.

Answer: D

At useful frequencies shot noise in the junction currents sets the floor, with thermal noise in the base spreading resistance contributing as well. Flicker noise matters only at low frequencies, divider thermal noise can be made small by design, and quantisation noise belongs to digital coding entirely. Identifying which mechanism dominates decides where design effort belongs, and in a bipolar front end it is the operating current. 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

Raising the current increases the shot noise but lowers the equivalent input noise resistance, so the optimum depends on the source impedance the stage works from. Neither extreme is universally better. Noise is certainly current-dependent. A parameter with an optimum rather than a direction is one where the source impedance has to be known before the design can be settled, which is why low-noise design begins with the source. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: B

Referring the noise to the input removes the gain from the comparison, so two amplifiers of different gain can be judged on the same footing. Distortion, bias and bandwidth are all specified separately. Referring an imperfection to the input is a general technique, and it works because everything at the input is multiplied by the same gain as the signal.

Answer: C

A low source resistance generates little thermal noise of its own, so the amplifier's fixed input noise voltage becomes a large fraction of the total and the ratio degrades. Loading and capacitor reactance affect the signal rather than the noise ratio. Noise figure is always a comparison against what the source itself contributes, which is why the same amplifier can look excellent on one source and poor on another. Framing the question as which quantity the change acts upon makes the answer follow without calculation. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

Referring the output figure back through the gain:

\[ v_{n} = \frac{1 \ \mathrm{mV}}{100} = 10 \ \mathrm{\mu V} \]

giving 10 microvolts. Multiplying rather than dividing gives 100 millivolts, which would be an absurdly noisy amplifier. Referring a quantity to the input always divides by the gain, and checking the answer for plausibility catches the direction error at once.