Frequency Response of Amplifiers Questions and Answers

Answer: C

Coming down in frequency, the first break encountered is the highest one, and the response starts falling there - so the highest low-frequency break is what determines the useful lower limit. Neither a mean nor a sum has any meaning here. That the emitter bypass usually sets it, being the break into the smallest resistance, is why sizing that one capacitor correctly matters more than the others put together.

Answer: D

Because the collector swings in antiphase to the base, the capacitance carries far more current than the base voltage alone would drive through it, and appears at the input multiplied by one plus the voltage gain. A few picofarads across a gain of two hundred becomes hundreds of picofarads. It is not divided, not placed in series, and remains a capacitance. That a small component becomes the dominant one through the gain around it is the central fact of high-frequency amplifier design.

Answer: A

Multiplying by one plus the gain magnitude:

\[ C_{M} = (1+150) \times 4 = 604 \ \mathrm{pF} \]

giving about 604 picofarads, over a hundred times the device's own figure and certain to dominate the input capacitance. Dividing rather than multiplying gives 37 pF, and ignoring the gain leaves 4 pF. The multiplication is why a stage's measured input capacitance bears so little resemblance to its data sheet value.

Answer: B

The source resistance works into the multiplied capacitance:

\[ \begin{aligned} f &= \frac{1}{2\pi (1000)(604 \times 10^{-12})} \\ &\approx 264 \ \mathrm{kHz} \end{aligned} \]

so about 264 kHz - a severe limitation for a device whose transition frequency might be three hundred megahertz. The gap between what the device can do and what the stage does is almost entirely the Miller effect, which is why so much circuit design is aimed at it.

Answer: C

The break frequency is set by the source resistance working into the Miller capacitance, so a stiffer drive pushes the break upward without changing the capacitance at all. The gain does not rise, and the junction bias is unaffected by the source. This is why a wideband stage is preceded by a follower or a low-impedance driver, and why the same transistor can look fast or slow depending entirely on what feeds it.

Answer: D

With a common-base stage above it holding the collector nearly fixed, the lower device's voltage gain is about unity, so its collector-base capacitance is barely multiplied at all. The overall voltage gain is still taken, but at the upper device where the capacitance does not bridge input and output. Capacitances are not placed in series and the input is not where gain appears. Separating where the gain is taken from where the capacitance is seen is the whole trick.

Answer: A

Shunt peaking places a small inductor in series with the collector resistor, so the load impedance rises with frequency just as the shunt capacitance starts to reduce the gain, and the two effects partly cancel to extend the bandwidth. A capacitor across the load would make matters worse, and the other options describe different techniques. Compensating a falling response with a rising impedance is a general idea, and it costs only a component and some care over the peaking that results.

Answer: B

Too large an inductor makes the load impedance rise more than the capacitance is reducing the gain, so the response peaks before falling - and a peaked response overshoots on a step input. The passband gain and the low-frequency break are unaffected, and the Miller capacitance is a property of the input side. Any compensation that can undercorrect can also overcorrect, and the flat-response condition is a particular value rather than a direction.

Answer: C

The transition frequency is where the device's own current gain falls to unity, so no circuit arrangement can produce useful gain beyond it. Low-frequency beta, saturation voltage and the junction drop all matter elsewhere. Circuit techniques such as the cascode close the gap between what a stage achieves and what the device permits, but they cannot move the device limit itself, and knowing which limit is which prevents effort being spent on the wrong one.

Answer: D

Increasing the load resistance raises the gain and lowers the break frequency in the same proportion, so their product is unchanged and gain can be traded for bandwidth freely. Operating current changes the transconductance and so the product itself, and supply and temperature act on other things. A conserved product is the most useful kind of relation in design, because it states exactly what one improvement will cost elsewhere.