Frequency Response of Amplifiers Questions and Answers

Answer: C

The diode's capacitance sits at the summing node, so it works against the feedback resistance to set the dominant pole - which is why receivers use small-area diodes and reverse bias them hard to reduce it. It does not reduce the photocurrent, is at the input rather than the output, and affects noise through a separate mechanism. That the detector rather than the amplifier often sets the speed is worth knowing before a faster amplifier is specified.

Answer: D

The dominant pole is set by the feedback resistance and the input capacitance:

\[ \begin{aligned} f &= \frac{1}{2\pi (10^{4})(5 \times 10^{-12})} \\ &\approx 3.2 \ \mathrm{MHz} \end{aligned} \]

giving about 3.2 MHz. Reducing the feedback resistor to one kilohm would give ten times the bandwidth and a tenth of the transimpedance, which is the trade in its plainest form. The same relation appears wherever a resistance and a capacitance set a limit.

Answer: A

Halving the capacitance halves the time constant and so doubles the bandwidth, while the feedback resistor and hence the transimpedance stay where they are. Halving the resistor would double the bandwidth too but halve the transimpedance, which the question forbids. Extra open-loop gain and a higher supply do not move the dominant pole. When a specification must be held while another is improved, the change has to act on a component that appears in only one of them.

Answer: B

Both are true and the second is the reason: the break frequency is set by the product of the source resistance and the multiplied capacitance, so reducing either raises it. That is why a follower placed ahead of a common-emitter stage can transform its bandwidth without any change to the stage itself. When a limit is a product of two quantities, either factor is a legitimate target and the cheaper one wins.

Answer: C

The reason is true but the assertion is false: each stage may be wide, but their roll-offs compound and the combined 3 dB point falls below any single stage's. What extra stages buy is gain, at the price of bandwidth rather than in addition to it. A statement containing always is worth testing against the compounding effect, which is exactly the consideration this pair separates.

Answer: D

The input coupling capacitor is in series with the signal, so it governs only the low-frequency end; making it larger or smaller leaves the high end untouched. The other three all appear in the high-frequency time constants. Sorting the components by whether they sit in series or in shunt with the signal path settles which end each affects, and it is the single most useful habit in reading a response.

Answer: A

Raising the collector load increases the gain and lowers the high-frequency break in the same proportion, so the bandwidth falls rather than widening. The other three all attack the limitation directly: the drive impedance, the Miller multiplication, or the load impedance's frequency dependence. A change that moves a conserved product's two factors in opposite directions cannot improve both, and recognising that saves proposing it.

Answer: B

This is the Miller effect, and it is usually what limits a common-emitter stage's bandwidth. The Early effect is the base-narrowing that tilts a bipolar device's output characteristics, the Hall effect is the transverse voltage in a magnetic field, and the Zener effect is a tunnelling breakdown mechanism. Effects named after people carry no clue in the name, so each has to be attached firmly to its own mechanism.

Answer: C

A single pole falls at six decibels per octave, equivalently twenty per decade. Three decibels is the drop at the break itself rather than a rate, and twelve per octave describes two poles. Twenty is right only when the unit is a decade rather than an octave. Two ways of stating one slope, plus one number that is a point rather than a slope, is the confusion this topic turns on most often.

Answer: D

Midband is the flat region between the two roll-offs, where the coupling capacitors are effectively short circuits and the shunt capacitances effectively open ones. The geometric mean happens to lie in that region but does not define it, one kilohertz is a common measurement frequency rather than a definition, and no averaging is involved. The whole value of the midband idea is that both sets of capacitances can be ignored there, which is exactly the assumption every small-signal gain calculation makes.