Power Amplifiers Questions and Answers

Q: Why is a single-supply amplifier's output coupled through a capacitor?

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.

Q: An output capacitor works into \( 8 \ \Omega \). For a \( 20 \ \mathrm{Hz} \) break it must be at least about

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.

Q: What advantage does a split supply give an audio output stage?

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.

Q: What must a direct-coupled amplifier's protection guard against?

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.

Q: Consider the following statements about output stage protection. 1. V-I limiting follows the shape of the safe operating area 2. An output relay protects the loudspeaker rather than the amplifier 3. Emitter ballast resistors improve the amplifier's efficiency Which of the statements given above is or are correct?

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.

Q: An amplifier's thermal chain is measured as follows. PathResistance Junction to case1.0 C/WCase to sink0.3 C/WSink to air1.2 C/W Which change would most reduce the junction temperature at a given power?

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.

Q: Which of the following would not reduce an output device's junction temperature? Raising the ambient temperature around the equipment Fitting a larger heatsink Applying thermal compound at the interface Reducing the power the device dissipates

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.

91.

What is a bootstrap capacitor used for in an output stage?

A.To filter the supply feeding the driver stage
B.To lift the driver's supply above the rail on peaks
C.To couple the signal into the output devices
D.To provide the quiescent bias for the drivers

Answer: B

As the output approaches the positive rail the upper driver runs out of headroom, and a capacitor from the output to the driver's collector load carries its supply upward with the signal, letting the output swing closer to the rail. Coupling, biasing and filtering are other functions. Feeding a stage from a supply that moves with its own output is the same bootstrapping idea used to raise an input impedance, applied here to recover swing rather than impedance. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

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92.

Why do some designs use a separate higher supply for the driver stage?

A.It improves the amplifier's supply rejection
B.It allows a smaller output coupling capacitor
C.It lets the output swing closer to its own rails
D.It reduces the dissipation in the output devices

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.

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93.

Why is a single-supply amplifier's output coupled through a capacitor?

A.The capacitor blocks the switching frequency
B.The capacitor matches the loudspeaker's impedance
C.The capacitor provides the feedback path
D.The output sits at half the supply rather than earth

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.

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94.

An output capacitor works into \( 8 \ \Omega \). For a \( 20 \ \mathrm{Hz} \) break it must be at least about

A.1000 microfarads
B.100 microfarads
C.10 microfarads
D.10000 microfarads

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.

Discuss in Forum

95.

What advantage does a split supply give an audio output stage?

A.The supply requires no smoothing capacitors
B.The output sits at earth and needs no coupling capacitor
C.The output devices dissipate less at a given power
D.The amplifier's efficiency exceeds class B's limit

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.

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96.

What must a direct-coupled amplifier's protection guard against?

A.The supply rails becoming unequal in value
B.The output stage oscillating at high frequency
C.A fault putting a rail voltage across the speaker
D.The coupling capacitor failing short circuit

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.

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97.

Consider the following statements about output stage protection.

1. V-I limiting follows the shape of the safe operating area

2. An output relay protects the loudspeaker rather than the amplifier

3. Emitter ballast resistors improve the amplifier's efficiency

Which of the statements given above is or are correct?

A.1 and 3 only
B.2 and 3 only
C.1, 2 and 3
D.1 and 2 only

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.

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98.

Consider the following pairs of component and its purpose in an output stage.

Bias spreader - Setting the quiescent current

Emitter ballast - Equalising current between devices

Output inductor - Isolating a capacitive load from the loop

Bootstrap capacitor - Blocking direct voltage from the load

How many of the pairs given above are correctly matched?

A.Only three
B.Only one
C.Only two
D.All four

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.

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99.

An amplifier's thermal chain is measured as follows.

Path	Resistance
Junction to case	1.0 C/W
Case to sink	0.3 C/W
Sink to air	1.2 C/W

Which change would most reduce the junction temperature at a given power?

A.A larger supply smoothing capacitor
B.A better heatsink
C.More thermal compound at the interface
D.A device with a lower saturation voltage

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.

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100.

Which of the following would not reduce an output device's junction temperature?

Raising the ambient temperature around the equipment

Fitting a larger heatsink

Applying thermal compound at the interface

Reducing the power the device dissipates

A.Applying thermal compound at the interface
B.Reducing the power the device dissipates
C.Raising the ambient temperature around the equipment
D.Fitting a larger heatsink

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.

Discuss in Forum

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