Small Signal Amplifiers Questions and Answers

Answer: A

A common-base input is naturally tens of ohms, which matches such a line without any transformer or padding. A common-emitter input is a few kilohms and a follower's is far higher still. Matching is about choosing the configuration whose natural impedance suits the source, and it is one of the few situations where common base is the obvious first choice rather than a specialist's option.

Answer: B

For the same voltage across it, the base supplies only a beta-th of the current, so the apparent impedance is one plus beta times larger. The other options are gain expressions rather than reflection factors. The rule works because impedance is voltage over current and only the current has been scaled, which is the same reasoning that governs a transformer.

Answer: C

Looking the other way the same rule inverts: a source impedance at the base appears at the emitter divided by one plus beta, which is why a follower's output impedance is so low. That the factor is the same in both directions, multiplying one way and dividing the other, makes the pair easy to hold once the reason is clear rather than the formula memorised.

Answer: D

The reflected source adds to the intrinsic emitter resistance:

\[ r_{out} \approx 26 + \frac{4700}{151} \approx 57 \ \Omega \]

giving about 57 ohms. Both terms matter here, which is why omitting either gives a noticeably wrong answer. A follower divides a source impedance by beta but cannot go below the emitter resistance at the operating current, and that floor is worth remembering. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

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.