Small Signal Amplifiers Questions and Answers

Answer: C

The first is true and the second is false: input impedance is beta times the emitter resistance and so inherits the full threefold device spread, while the gain contains no beta at all. Keeping this pair straight predicts exactly which of a stage's measured specifications will move when the transistor is changed, which is the practical use of the distinction.

Answer: D

Bypassing the emitter resistor lowers the input impedance, because it removes the resistance that was being reflected to the base - the reflected quantity falls to the intrinsic value alone. The other three all raise it. That the same component raises or lowers a specification depending on whether a capacitor is fitted across it is worth pausing over, since it is the commonest source of confusion in reading these stages. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: A

The thermal voltage is about 26 millivolts near 300 kelvin, and it appears in the diode equation, in the transconductance and in the intrinsic emitter resistance. The 0.7 volt figure is a junction's forward drop, and sixty millivolts is the change in that drop per decade of current - a related figure but not the same one. Three small voltages in one topic, all easily confused, so it is worth attaching each to the expression it appears in. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: B

The exponential law gives about sixty millivolts per decade, which is why the drop is nearly constant over a wide range and the 0.7 volt approximation works so well. The thermal voltage is twenty-six millivolts, and two millivolts is the change per degree of temperature. Each of these figures answers a different question, and matching the figure to the question is most of using them correctly.

Answer: C

The product being roughly constant means gain can be traded for bandwidth and back again, which is why a stage run at lower gain reaches higher frequencies. The other options do not describe any standard figure. A conserved product is the most useful kind of relation in design, because it says exactly what one improvement will cost elsewhere.

Answer: D

Above the transition frequency the device provides no current gain at all, so it cannot amplify and the model describes something that no longer behaves as an amplifier. The transconductance does not reverse sign, measurement difficulty is not a model failure, and the device stays biased in the active region. A model breaking down at the point where the device stops doing its job is not a coincidence, since both follow from the same capacitances.

Answer: A

Converting from decibels for a voltage ratio:

\[ A_{v} = 10^{40/20} = 100 \]

giving 100. Ten thousand would follow from dividing by ten rather than twenty, which is the power convention rather than the voltage one. Whether to divide by ten or twenty depends on whether the quantity is a power or an amplitude, and getting that wrong squares the answer.

Answer: B

Decibel gains add:

\[ 40 + 40 = 80 \ \mathrm{dB} \]

which is 80 dB, a voltage ratio of ten thousand. Forty decibels is the answer only if the two gains are read as a ratio and multiplied instead, and 1600 dB comes from multiplying the decibel figures themselves, which is never a meaningful operation. Loading between the stages will reduce the real figure, often by six decibels a stage. That gains multiply as ratios and add as decibels is exactly why the logarithmic unit is used for cascades in the first place.

Answer: C

Adding the reduced figures:

\[ 34 + 34 = 68 \ \mathrm{dB} \]

so 68 dB, twelve decibels below the unloaded estimate - a factor of four in voltage. Six decibels lost per stage is exactly what halving each effective load produces, which is the common case when two identical stages are coupled. Small per-stage losses compound quickly in a cascade.

Answer: D

A single pole falls at a constant number of decibels per octave, so on logarithmic axes it is a straight line and the whole response can be sketched from its break points. Gain is meaningful in any units, phase is plotted separately either way, and linear axes can represent any number. A coordinate system that turns the behaviour of interest into straight lines is worth adopting for that reason alone.