Power Amplifiers Questions and Answers

Answer: C

The driver must sit a couple of volts above the output device's base to keep it conducting, so a supply a few volts higher than the output rail lets the output reach much nearer to that rail. Output dissipation, supply rejection and coupling are not what it addresses. It is the same problem bootstrapping solves, taken by a different route, and the choice between them is one of cost against complexity.

Answer: D

With one rail the output stage idles at half the supply, so a capacitor is needed to keep that standing voltage out of the loudspeaker while passing the signal. Switching frequency belongs to class D, matching is unnecessary, and feedback is taken separately. The capacitor must be large, since it works into a few ohms, and it sets the amplifier's low-frequency limit - which is the principal reason split supplies displaced the arrangement. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: A

The break frequency relation with a very small resistance:

\[ C = \frac{1}{2\pi (20)(8)} \approx 1000 \ \mathrm{\mu F} \]

giving about 1000 microfarads, a physically large and expensive part that also distorts a little as its dielectric is worked. That cost, and the distortion, are exactly why split supplies and direct coupling took over.

Answer: B

With equal positive and negative rails the output idles at earth potential, so the loudspeaker connects directly and the large output capacitor with its cost, size and distortion disappears. Dissipation and efficiency are unchanged, and smoothing is still needed. Removing a component that was there only to correct a standing voltage is a clean improvement, and it is why nearly every modern amplifier uses two rails.

Answer: C

Without a coupling capacitor to block it, a shorted output device puts the full rail across the loudspeaker and burns out the voice coil in seconds, which is why a direct-voltage detector and a relay are standard. There is no coupling capacitor to fail, unequal rails shift the operating point without destroying the load, and oscillation is a separate hazard. Removing a component removes its function too, and the protection circuit is what replaces the capacitor's incidental safety.

Answer: D

The first two are correct and the third is the reverse: ballast resistors dissipate power and cost a little output swing, so they reduce efficiency slightly - what they buy is safe current sharing between paralleled devices. Three protection measures addressing three different failures, and each costs something. Naming the cost of a protective measure alongside its benefit is what prevents it being omitted as free or added as harmless.

Answer: A

The first three are correctly paired and the fourth is not, so three match. A bootstrap capacitor lifts the driver's supply with the output to recover swing near the rail; blocking direct voltage is the output coupling capacitor's job in a single-supply design. Two capacitors in one stage doing quite different jobs is easy to conflate, and tracing what each is connected between settles it immediately.

Answer: B

The sink term is the largest of the three, so improving it has the greatest effect on a sum of 2.5 degrees per watt. The interface is already the smallest term and more compound would make it worse rather than better, since the compound conducts less well than metal. Saturation voltage affects the power dissipated rather than the thermal path, and smoothing is unrelated. Attacking the largest term in a series chain is the rule, and checking which term that is comes first.

Answer: C

Raising the ambient raises the junction temperature directly, since the junction sits at ambient plus the rise. The other three all reduce either the rise or the starting point. Because the junction temperature is a sum of an ambient and a rise, both terms have to be considered - and equipment that works on a bench can fail in a warm rack for exactly this reason.

Answer: D

This is class B, whose half-cycle conduction gives it both its high efficiency and its crossover distortion. Class A conducts the whole cycle, class AB slightly more than half, and class D switches rather than conducting for any part of a cycle in the usual sense. Since one parameter defines the class and every other property follows from it, the conduction angle is what these letters really name.