Small Signal Amplifiers Questions and Answers

Answer: A

The first stage's collector resistor works into the second stage's input impedance in parallel, so its effective load and hence its gain fall. A properly sized coupling capacitor is negligible in the passband, phase shifts do not reduce magnitude, and the bias network affects the loading rather than the gain directly. Loading between stages is the usual reason a cascade underperforms the arithmetic, and it is worth computing rather than assuming. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: B

The two appear in parallel to the signal:

\[ R_{L} = \frac{5 \times 5}{10} = 2.5 \ \mathrm{k\Omega} \]

so 2.5 kilohms, halving the first stage's gain. Adding them would apply to a series connection, which is not what a coupled load is. Recognising which combination applies is settled by asking whether the two elements share a voltage or a current.

Answer: C

A follower presents a high impedance to the first stage and a low one to the second, so neither loads the other - impedance transformation at the cost of no voltage gain. A larger capacitor extends the low-frequency response instead, and reducing the collector resistor would lower the gain further rather than help. Inserting a stage whose only job is impedance is worthwhile precisely when the interface, rather than the gain, is the problem. 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: D

Two identical stages each 3 dB down at some frequency are 6 dB down together, so the combined 3 dB point sits at a lower frequency. Coupling capacitors set low-frequency poles rather than high-frequency ones, the gain-bandwidth product applies per stage rather than to the cascade, and loading affects gain rather than bandwidth directly. Roll-offs multiply in magnitude and add in decibels, and that compounding is what shrinks a cascade's bandwidth.

Answer: A

A coupling capacitor lets the signal through while keeping one stage's quiescent voltages from disturbing the next, which is what allows each stage to be biased independently. Removing the emitter resistor is the bypass capacitor's job, the high-frequency limit comes from junction and stray capacitances, and supply decoupling is a third function. Three capacitors doing three quite different jobs in one stage is worth keeping straight when reading a schematic.

Answer: B

The capacitor works into the series combination of the driving source's resistance and the stage's own input impedance, and that sum sets the time constant. Taking either alone understates the resistance and so overstates the required capacitance. Identifying the complete path a capacitor charges through, rather than the nearest resistor, is what makes these calculations come out.

Answer: C

Rearranging the break frequency relation:

\[ \begin{aligned} C &= \frac{1}{2\pi f R} \\ &= \frac{1}{2\pi(20)(10^{4})} \\ &\approx 0.8 \ \mathrm{\mu F} \end{aligned} \]

giving about 0.8 microfarads. The others misplace a decade. Because the capacitance needed is inversely proportional to both frequency and resistance, low-frequency response into a low impedance is what drives capacitor values up. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: D

The same relation with a far smaller resistance:

\[ C = \frac{1}{2\pi(20)(26)} \approx 306 \ \mathrm{\mu F} \]

so about 306 microfarads, nearly four hundred times the coupling capacitor's value for the same frequency. That contrast is why the emitter bypass is invariably the largest capacitor in a small-signal stage and usually the one that sets the low-frequency limit.

Answer: A

The emitter bypass works into the smallest resistance by far, so for any given capacitance its break frequency is the highest of the three signal capacitors. Coupling capacitors see kilohms and so break much lower. Supply decoupling is not in the signal path at all. When several break points contribute, the highest of them determines where the response starts to fall, which makes it the one worth calculating first. 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

Below the bypass break the capacitor no longer shorts the emitter resistor, so the gain settles at the degenerated ratio and stays there until the coupling capacitor starts to act. It is a step in the response rather than a continuous fall. That an amplifier can have a shelf in its low-frequency response, rather than a simple roll-off, surprises people who expect one slope, and it follows directly from the two break points being far apart. 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.