Power Amplifiers Questions and Answers

Answer: A

Limiting senses the output current and robs the drive when it becomes excessive, so a shorted loudspeaker lead does not destroy the output devices. An open load draws no current and is harmless, ripple is a supply matter, and reverse voltage is handled by clamp diodes. Protecting against the one fault a user is most likely to cause, by hand and repeatedly, is worth the components it costs.

Answer: B

A fixed current limit permits the same current at forty volts across the device as at four, and the safe operating area does not. The remedy is V-I limiting, which reduces the permitted current as the voltage across the device rises. Response is fast, normal output is unaffected, and the circuit resets automatically. A protection scheme has to follow the shape of the limit it is protecting, and a straight line across a curved boundary leaves part of it unguarded. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: C

V-I limiting senses the collector-emitter voltage as well as the current and tightens the limit as the voltage grows, so the permitted operating points follow the safe operating area's own shape. Automatic reset is common to both schemes, the devices are not switched off, and relay disconnection is a separate protection layer. Making a protection boundary match a device boundary is the whole idea, and the alternative is to be over-cautious in normal use or unsafe in a fault.

Answer: D

If an output device fails short, the full supply rail appears across the loudspeaker and destroys it, so a detector opens a relay when direct voltage is sensed. Current limiting protects the amplifier rather than the load, impedance matching is not needed in a voltage-source amplifier, and rail switching is a different technique. The relay protects the expensive item from the cheap one's failure, which is the usual direction of this kind of protection. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: A

A transformer presents the load reflected by the square of the turns ratio, so a few-ohm loudspeaker can be made to look like the several kilohms a valve or a high-voltage stage wants. Isolation from direct voltage is a genuine secondary benefit, and neither filtering nor storage is the point. Where a device's optimum load and the actual load differ by orders of magnitude, an impedance transformation is unavoidable, and a transformer is the passive way to get one. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

Low-voltage, high-current transistors can drive a few-ohm load directly, and a complementary pair does it without even a phase splitter, so the transformer's cost, weight and distortion became unnecessary. Transformers handle large currents perfectly well, delay is not the issue, and solid-state stages want a lower load impedance rather than a higher one. A component becomes obsolete when the problem it solved stops existing, which is what changed here. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: C

A class D stage switches its devices fully on or fully off at a high rate, encoding the signal in the duty cycle and recovering it with a low-pass filter. A switch dissipates little because it never carries current and voltage together, which is where the efficiency comes from. Conduction angle is not how the switching classes are described, and neither deep class A nor a transformer is involved. Moving from a linear device to a switch is the same step that transformed power supplies.

Answer: D

Dissipation is the product of current and voltage, and a device that is either fully on, with little voltage across it, or fully off, with no current through it, has a small product in both states. The filter passes energy to the load rather than returning it, temperature is a consequence, and conduction angle is not the relevant description. Reducing the overlap of current and voltage is the general route to efficiency, and switching reduces it as far as it can go.

Answer: A

Real switches take time to change state, dissipating during each transition, and they have a finite resistance when on, so the two losses rise with frequency and with current respectively. Filter saturation is a design fault rather than a limit, conduction angle does not apply, and supply ripple affects the output rather than the efficiency. Every idealisation fails somewhere, and locating where is what turns a theoretical efficiency into a predicted one. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

The filter passes the audio and rejects the switching frequency and its harmonics, recovering the original waveform from the pulse train. Current limiting, impedance matching and feedback are separate functions. Its cut-off must sit well above the audio band and well below the switching frequency, which is why raising the switching rate eases the filter's design at the cost of more switching loss - a trade that has moved steadily as devices have got faster. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.