Small Signal Amplifiers Questions and Answers

Answer: A

Multiplying amplitude by gain and noting the sign:

\[ v_{o} = -180 \times 5 \ \mathrm{mV} = -0.9 \ \mathrm{V} \]

so 0.9 volts, inverted. The magnitude alone is not the whole answer in a common-emitter stage, where the inversion matters as soon as the stage is placed in a feedback loop or cascaded. Carrying the sign through rather than taking magnitudes is a habit worth keeping.

Answer: B

Five millivolts is a fifth of the twenty-six millivolt thermal voltage, so the exponential is well approximated by its tangent and the small-signal analysis holds. The output amplitude is irrelevant to the question, which concerns the excursion at the device. A one-millivolt threshold is arbitrary. Judging smallness against the thermal voltage rather than against a fixed figure is what makes the criterion meaningful. 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: C

At twice the thermal voltage the exponential curvature is plainly significant and the output acquires substantial harmonic content, so the small-signal model no longer predicts the waveform. The gain does not rise usefully, the capacitors are unaffected by amplitude, and the input impedance change is a second-order effect. Knowing where a model stops being valid is more useful than knowing the model, because that boundary is where real circuits get designed.

Answer: D

Higher current means a smaller intrinsic emitter resistance and so a larger gain from the same collector resistor. It uses more of the supply in the collector resistor's drop, demands a larger bypass capacitor because the emitter resistance is lower, and reduces the input impedance. One change improving one specification while worsening three is the normal case, and the design question is only which of the four matters most. 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: A

Lower current raises the intrinsic emitter resistance and hence the reflected input impedance at the base, at the cost of gain, linearity and bandwidth, since the transition frequency also falls at low current. Every one of those moves the wrong way. That the same parameter is the right lever in opposite directions for different specifications is why the operating current is chosen last, once the binding requirement is known. Checking the units of each quantity catches a wrong formula before the arithmetic begins. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

The transition frequency is set by the transconductance against the junction capacitances, and transconductance falls with current while the depletion capacitance does not, so the ratio worsens. Base resistance and the Early voltage do not behave that way, and the beta peak is a separate effect that also contributes. A figure of merit that is a ratio degrades whenever only its numerator falls, which is the general shape of this argument.

Answer: C

A piezoelectric source is a very high impedance, so the stage must not load it: a bootstrapped divider keeps the bias stiff for direct current while presenting a very high impedance to the signal. Common base is the lowest input impedance available, a stiff ordinary divider would shunt the source, and a high current lowers the input impedance further. Matching the input impedance to the source is the first question with any sensor, and it often decides the topology outright.

Answer: D

A loudspeaker is a low impedance and a high-gain stage has a high output impedance, so a follower's impedance transformation is exactly what is needed, at the cost of no further voltage gain. Another common-emitter stage would have a high output impedance too, and common base is an input configuration. Interfaces are where impedance rather than gain decides the choice, and a follower exists for precisely those places. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: A

The usual cause is that each stage's effective load is its own collector resistor in parallel with the next stage's input impedance, which can halve the gain or worse. Supply and bias faults would show as wrong quiescent voltages, and coupling capacitors affect only the low-frequency end. Computing the interstage loading before suspecting anything else is the right first step, because it is a design effect rather than a fault. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: B

All three are correct, and they are all the same mechanism seen from three sides: series feedback divides the gain, multiplies the impedance and divides the non-linearity by the same factor. That one component buys two improvements for one sacrifice is why degeneration is so nearly universal in discrete amplifier design, and why the sacrifice is usually made willingly.