Frequency Response of Amplifiers Questions and Answers

Answer: B

The probe capacitance works into the node's own resistance:

\[ \begin{aligned} f &= \frac{1}{2\pi (10^{4})(15 \times 10^{-12})} \\ &\approx 1.06 \ \mathrm{MHz} \end{aligned} \]

giving about 1.06 MHz - quite low enough to be mistaken for a property of the amplifier. At a high-impedance node the instrument can easily dominate the measurement, which is why probing at a low-impedance point is preferred whenever the choice exists.

Answer: C

The break frequency is inversely proportional to the product of resistance and capacitance, so the same few picofarads of stray matter far more where the resistance is megohms than where it is tens of ohms. The stray capacitance itself is a matter of geometry rather than impedance, amplitude does not enter, and the noise point is true but separate. Keeping high-impedance nodes few, short and shielded follows directly, and it is why layout matters more in some parts of a circuit than others.

Answer: D

Both are true and the second is the mechanism: the input capacitance is no longer multiplied by the stage gain, so the bandwidth is set by the device rather than by the Miller effect. The full voltage gain is still taken, but at the upper device where no capacitance bridges input to output. Separating where the gain is taken from where the capacitance is seen is the whole idea, and it recurs in every wideband topology.

Answer: A

The reason is true but the assertion is false: a coupling capacitor is in series with the signal, so its size affects only the low-frequency end and a larger one extends the response downward. The high end is set by shunt capacitances, which a bigger coupling capacitor does nothing about. Whether a capacitance is in series or in shunt decides which end of the response it governs, and applying a low-end remedy to a high-end problem is the mistake this item is built on.

Answer: B

The first two are correct and the third names the wrong gain: the Miller factor is one plus the voltage gain, not the current gain, since it arises from the antiphase voltage swing across the bridging capacitance. Getting that right matters because the two gains differ by a large factor in most stages, and using beta would give a quite wrong input capacitance. A statement with the correct mechanism and the wrong parameter is the most easily overlooked kind of error.

Answer: C

All three are correct. Roll-offs compound at both ends, narrowing the passband from above and below, while gains add in decibels because they multiply as ratios. The first two are the price of the third, and that trade is why a long chain of narrow stages is a poor way to get both gain and bandwidth - and why a feedback amplifier, which trades gain for bandwidth deliberately, so often beats it.

Answer: D

The first is true and the second is false: a capacitance in series with the signal passes high frequencies and blocks low ones, so it can only limit the response from below. High frequencies are limited by capacitances in shunt with the signal path, which divert current to earth. Series or shunt is the whole distinction, and once it is fixed the direction of every capacitor's effect follows without further thought.

Answer: A

Both are true, though for different reasons: a higher gain multiplies the capacitance more, and a higher source impedance makes the resulting break frequency lower for the same capacitance. The two combine, which is why a high-gain stage driven from a high impedance is the worst case and why the remedies attack either the gain, as in a cascode, or the drive impedance, as with a preceding follower.

Answer: B

The first three are correctly paired and the fourth is not, so three match. The collector load resistor works with the shunt capacitances to set the high-frequency break, so it affects that end very much - raising it increases the gain and lowers the bandwidth in the same proportion, which is the gain-bandwidth trade in its simplest form. A component that appears in a time constant always affects a break frequency, even when its obvious job is something else.

Answer: C

The passband runs from the highest low-frequency break to the lowest high-frequency break, so 120 Hz to 180 kHz. The 8 Hz break is already well below where the response has started to fall and so does not set the limit. Adding break frequencies has no meaning. Since it is always the innermost pair of breaks that bounds the useful band, sizing the emitter bypass generously is what would widen this stage's response at the low end.