Frequency Response of Amplifiers Questions and Answers

Answer: A

The low end is set by capacitors in series with the signal, or in parallel with a degeneration resistor, whose reactance grows as the frequency falls until they no longer pass or bypass. Junction and stray capacitances shunt the signal away and so act at the high end instead. The collector resistor sets gain rather than a break frequency. Whether a capacitance is in series or in shunt with the signal path decides which end of the response it governs, and that single distinction accounts for the whole shape of a Bode plot.

Answer: B

The high end is set by capacitances in parallel with the signal - the junction capacitances, the Miller-multiplied collector-base capacitance, and stray wiring capacitance - which divert an increasing share of the signal current as frequency rises. Series coupling capacitors and emitter bypasses act at the low end, and supply decoupling is not in the signal path at all. A shunt path that improves with frequency always produces a high-frequency roll-off, whatever the circuit.

Answer: C

Reactance is inversely proportional to frequency, so at low enough frequencies the capacitor's reactance is no longer small beside the series resistance and it drops an appreciable part of the signal. Leakage, charging behaviour and dielectric loss are all real properties but none produces the roll-off. The break occurs where reactance equals resistance, which is why the break frequency contains the product of the two and nothing else.

Answer: D

The break is defined where the reactance equals the resistance, which makes the total impedance root two times the resistance and the gain 3 dB down. That equality is what gives the familiar expression with two pi, the frequency, the resistance and the capacitance all in one product. Doubling or halving would put the break at a different frequency, and a squared relation is dimensionally impossible. Every break frequency in this topic comes from setting one impedance equal to another.

Answer: A

At the break the gain is 3 dB below its passband value, a factor of about 0.707 in voltage, and the phase has shifted forty-five degrees. Six decibels per octave is the slope beyond the break rather than the drop at it, twenty decibels per decade is the same slope expressed differently, and one decibel is not a standard figure here. Confusing the drop at the break with the slope after it is the commonest error in reading these responses, and the two are quite separate quantities.

Answer: B

At the break the phase is forty-five degrees, on its way to ninety degrees well beyond it. Zero applies well inside the passband and one hundred and eighty would require two poles, which is exactly why stability analysis counts poles so carefully. Attaching each phase figure to its position relative to the break, rather than learning the three as a bare list, is what makes them usable when a feedback loop's margin has to be assessed.

Answer: C

Setting reactance equal to resistance:

\[ \begin{aligned} f &= \frac{1}{2\pi R C} \\ &= \frac{1}{2\pi (15000)(10^{-6})} \\ &\approx 10.6 \ \mathrm{Hz} \end{aligned} \]

giving about 10.6 Hz, comfortably below the audio band. The other options misplace a decade or invert part of the product. Because the frequency is inversely proportional to both resistance and capacitance, a low-impedance point in the circuit always demands a much larger capacitor for the same break.

Answer: D

The capacitance needed scales inversely with the frequency wanted:

\[ C = \frac{1}{2\pi (20)(15000)} \approx 0.53 \ \mathrm{\mu F} \]

so about 0.53 microfarads. Two microfarads would push the break lower rather than higher, which is the direction error to watch for: a larger capacitor extends the response downwards. Reading off which way a component change moves a break frequency is worth doing before any arithmetic.

Answer: A

The resistance at the emitter is the degeneration resistor in parallel with the small intrinsic emitter resistance, often only tens of ohms, so the same capacitance gives a break hundreds of times higher. Its reactance at a given frequency is identical, being the same capacitance, and the parallel connection is why it sees that low resistance rather than being the reason directly. A time constant is a product, so a small resistance forces a large capacitance if a low break is wanted.

Answer: B

The same relation with the far smaller resistance:

\[ C = \frac{1}{2\pi (20)(30)} \approx 265 \ \mathrm{\mu F} \]

giving about 265 microfarads, five 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 decides where the response actually starts to fall.