Frequency Response of Amplifiers Questions and Answers

Answer: D

The upper device is a common-base stage, so its input is its emitter and its collector-base capacitance does not bridge input to output. Its capacitance is the same device property, its base is bypassed rather than driven, and it does take substantial voltage gain. The cascode works because neither device presents a multiplied capacitance: the lower has no gain and the upper has no bridging path, which is why the combination beats either configuration alone.

Answer: A

The arrangement needs a second transistor and enough supply voltage to bias both in their active regions, which is the whole of its cost. The gain is comparable or better, it remains beta-independent, and the input capacitance is far lower rather than higher. A technique whose only price is a component and some headroom is unusually cheap, which is why the cascode appears wherever both gain and bandwidth are wanted from one stage.

Answer: B

The pairing is A-P, B-Q, C-R, and it explains the whole bandwidth ranking of the three configurations from one mechanism. Common emitter suffers because it has both a bridging capacitance and a large inverting gain; the other two each lack one of those two ingredients. Since the Miller penalty needs both, removing either is enough, and the cascode removes them from different devices in the same stage.

Answer: C

Bandwidth is conventionally the interval between the two frequencies at which the gain has fallen three decibels from its passband value. The ninety-degree phase point sits well beyond the break, rise time gives the bandwidth only through a known factor and only for a single-pole response, and distortion is unrelated to the roll-off. A definition fixed by convention rather than by physics still has to be applied exactly, or two measurements of the same amplifier will not agree.

Answer: D

A large input drives the stage into non-linearity, and the resulting compression looks like a fall in gain, so the measured roll-off would be an artefact of the drive level rather than a property of the amplifier. Break frequencies are set by resistances and capacitances and do not move with amplitude, and neither the capacitors nor the bias point are the issue at reasonable levels. A measurement must not disturb the quantity it is measuring, and here that means staying inside the small-signal condition.

Answer: A

A gap of four orders of magnitude between device and stage is the signature of Miller multiplication working into a high source impedance, and the remedy is a stiffer drive or a cascode. Coupling capacitors affect the low end, decoupling affects stability and hum, and bias resistors set the operating point. When a measured limit is far below a device limit, the circuit rather than the device is responsible, and the Miller effect is the first suspect.

Answer: B

A break twenty times too high at the low end points at a bypass capacitor too small for the very low resistance it works into, which is where the largest capacitance is needed. Too large a coupling capacitor would push the break lower rather than higher, the collector resistor does not set a low-frequency break, and Miller capacitance acts at the high end. Since the highest low-frequency break dominates, the capacitor working into the smallest resistance is the one to check.

Answer: C

A peak before the roll-off is the signature of overcorrection: the load impedance is rising faster than the shunt capacitance is reducing the gain. A small coupling capacitor or an open bypass would affect the low end, and a high source impedance would reduce the bandwidth without peaking it. Any compensation capable of undercorrecting can also overcorrect, and reading the sign of the error off the response shape is what tells you which way to adjust it.

Answer: D

A shortfall in the flat part of the response, rather than at either end, points at interstage loading: the first stage's effective load is its own collector resistor in parallel with the second stage's input impedance. Compounding roll-offs affect the band edges and not the passband, a drifted bias would distort or clip, and a small coupling capacitor would show at the low end only. Where in the response a defect appears is what localises its cause, and the passband points at loading.

Answer: A

A probe presents tens of picofarads to earth, which shunts the signal exactly as any other stray capacitance would and lowers the break frequency at the point being measured. Its resistive loading is negligible at megohms, and it adds no significant series inductance. The instrument disturbing what it measures is a general hazard, and it is why a compensated low-capacitance probe is used for high-frequency work and why a measurement should be repeated at a different point when the result looks suspicious.