Power Amplifiers Questions and Answers

Q: What is secondary breakdown?

Secondary breakdown is a thermal instability in which current crowds into a small area that then conducts still better, destroying the device before any average calculation would predict trouble. It is a runaway rather than an avalanche, and it is why the safe operating area is bounded by more than a simple power hyperbola. A failure driven by positive feedback needs its own boundary on a rating chart, since no average figure anticipates it. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Q: Why are MOSFETs less prone to secondary breakdown?

A hot region of a MOSFET channel conducts less well, so current is pushed towards cooler areas and the crowding self-corrects. Gate current and package construction are irrelevant, and the falling threshold works the other way but is dominated by the resistance effect in the conducting state. The same property that lets MOSFETs parallel without ballast protects them here, which is one mechanism explaining two advantages.

Q: A class B stage runs from \( \pm 24 \ \mathrm{V} \) into \( 8 \ \Omega \). Its maximum output power is about

The peak output approaches the supply rail:\[ \begin{aligned} P &= \frac{V_{p}^{2}}{2R} \\ &= \frac{24^{2}}{16} \\ &= 36 \ \mathrm{W} \end{aligned} \]giving about 36 W, a little less in practice once the saturation voltage is allowed for. The factor of two in the denominator converts a peak to a mean for a sine, and omitting it doubles the answer. Whether a voltage is a peak or a root-mean-square value decides that factor, so it is worth stating which is meant.

Q: An amplifier's output devices are rated \( 100 \ \mathrm{W} \) at \( 25 \ \mathrm{^\circ C} \) case, derating linearly to zero at \( 150 \ \mathrm{^\circ C} \). At a case temperature of \( 75 \ \mathrm{^\circ C} \) each may dissipate

Interpolating along the derating line:\[ P = 100 \times \frac{150-75}{150-25} = 60 \ \mathrm{W} \]giving 60 W. Fifty watts would be the midpoint of the temperature range rather than of the power range. A rating quoted at twenty-five degrees is nearly always a straight line to zero at the junction limit, and reading it that way rather than taking the headline figure is what keeps a design honest.

Q: A \( 0.22 \ \Omega \) emitter ballast resistor carries \( 2 \ \mathrm{A} \). It dissipates

Applying the resistive loss law:\[ P = I^{2}R = 4 \times 0.22 = 0.88 \ \mathrm{W} \]giving 0.88 W, so a one or two watt part is needed. These resistors are easy to under-rate because their resistance is small, but the current is large and the loss goes as its square. They also cost a little output swing, which is the price paid for safe current sharing.

Q: Why is an output stage's supply usually unregulated?

Music and speech draw current in bursts, so a regulator would have to be sized for the peaks while dissipating heavily at all times, and the feedback loop already rejects most of the supply variation. A regulator could deliver the peaks if sized for them, and bandwidth is unaffected. The fourth option is true and is part of why regulation is unnecessary, but the decisive argument is cost and dissipation. Leaving a supply unregulated because a loop downstream already handles the variation is a common and sensible economy.

Q: What does a supply's ripple appear as at an amplifier's output?

Ripple at twice the mains frequency reaches the output as hum, attenuated by the amplifier's supply rejection and by whatever loop gain is available at that frequency. It is not broadband noise, does not selectively distort peaks, and the operating point is held by the feedback. Since loop gain is largest at low frequencies, hum rejection is usually good - which is why a poorly performing amplifier's hum often points at a fault in the feedback path rather than at the supply.

31.

What is secondary breakdown?

A.The base-emitter junction breaking down under reverse bias
B.Leakage doubling for every ten degrees of temperature rise
C.Current concentrating into a hot spot
D.A second avalanche region above the first breakdown voltage

Answer: C

Secondary breakdown is a thermal instability in which current crowds into a small area that then conducts still better, destroying the device before any average calculation would predict trouble. It is a runaway rather than an avalanche, and it is why the safe operating area is bounded by more than a simple power hyperbola. A failure driven by positive feedback needs its own boundary on a rating chart, since no average figure anticipates it. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Discuss in Forum

32.

Why are MOSFETs less prone to secondary breakdown?

A.Their gates draw no current from the driver
B.Their threshold voltage falls as they warm
C.Their packages conduct heat better than bipolar ones
D.Their on-resistance rises with temperature

Answer: D

A hot region of a MOSFET channel conducts less well, so current is pushed towards cooler areas and the crowding self-corrects. Gate current and package construction are irrelevant, and the falling threshold works the other way but is dominated by the resistance effect in the conducting state. The same property that lets MOSFETs parallel without ballast protects them here, which is one mechanism explaining two advantages.

Discuss in Forum

33.

A class B stage runs from \( \pm 24 \ \mathrm{V} \) into \( 8 \ \Omega \). Its maximum output power is about

A.36 W
B.9 W
C.72 W
D.18 W

Answer: A

The peak output approaches the supply rail:

\[ \begin{aligned} P &= \frac{V_{p}^{2}}{2R} \\ &= \frac{24^{2}}{16} \\ &= 36 \ \mathrm{W} \end{aligned} \]

giving about 36 W, a little less in practice once the saturation voltage is allowed for. The factor of two in the denominator converts a peak to a mean for a sine, and omitting it doubles the answer. Whether a voltage is a peak or a root-mean-square value decides that factor, so it is worth stating which is meant.

Discuss in Forum

34.

The maximum each device in that stage must dissipate is about

A.3.6 W
B.7.3 W
C.36 W
D.18 W

Answer: B

Class B dissipation peaks at about two thirds of full swing:

\[ \begin{aligned} P_{d} &= \frac{V_{cc}^{2}}{\pi^{2}R} \\ &\approx \frac{576}{79} \\ &\approx 7.3 \ \mathrm{W} \end{aligned} \]

which is about 7.3 W per device, roughly a fifth of the output power. Sizing the heatsink from the full-output condition instead would badly understate the requirement, since dissipation there is much lower - which is the practical reason the peak's position matters. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Discuss in Forum

35.

A class A stage draws \( 1 \ \mathrm{A} \) from \( 30 \ \mathrm{V} \). Its maximum output into a resistive load is about

A.15 W
B.3.75 W
C.7.5 W
D.30 W

Answer: C

At best a quarter of the input power reaches a resistive load:

\[ P_{out} \le 0.25 \times 30 = 7.5 \ \mathrm{W} \]

so about 7.5 W, with the remaining 22.5 W dissipated. Transformer coupling would raise the ceiling to 15 W by removing the standing drop across the load resistor. A class A stage's heatsink is sized from its idle condition, which is also its worst case - the reverse of class B.

Discuss in Forum

36.

A class B stage delivers \( 9 \ \mathrm{W} \) into \( 8 \ \Omega \) from \( \pm 24 \ \mathrm{V} \). Its efficiency is about

A.78 per cent
B.25 per cent
C.50 per cent
D.39 per cent

Answer: D

At a twelve volt peak the supply delivers rather more than the load takes:

\[ \begin{aligned} P_{dc} &= \tfrac{2 V_{cc} V_{p}}{\pi R} \approx 22.9 \ \mathrm{W} \\ \eta &= \tfrac{9}{22.9} \approx 39\% \end{aligned} \]

giving about 39 per cent. The 78.5 per cent figure applies only at full swing, and at half swing the efficiency is half of it. Quoting a class's peak efficiency as though it applied at all levels is the standard overstatement.

Discuss in Forum

37.

An amplifier's output devices are rated \( 100 \ \mathrm{W} \) at \( 25 \ \mathrm{^\circ C} \) case, derating linearly to zero at \( 150 \ \mathrm{^\circ C} \). At a case temperature of \( 75 \ \mathrm{^\circ C} \) each may dissipate

A.60 W
B.50 W
C.100 W
D.33 W

Answer: A

Interpolating along the derating line:

\[ P = 100 \times \frac{150-75}{150-25} = 60 \ \mathrm{W} \]

giving 60 W. Fifty watts would be the midpoint of the temperature range rather than of the power range. A rating quoted at twenty-five degrees is nearly always a straight line to zero at the junction limit, and reading it that way rather than taking the headline figure is what keeps a design honest.

Discuss in Forum

38.

A \( 0.22 \ \Omega \) emitter ballast resistor carries \( 2 \ \mathrm{A} \). It dissipates

A.0.11 W
B.0.88 W
C.0.44 W
D.1.76 W

Answer: B

Applying the resistive loss law:

\[ P = I^{2}R = 4 \times 0.22 = 0.88 \ \mathrm{W} \]

giving 0.88 W, so a one or two watt part is needed. These resistors are easy to under-rate because their resistance is small, but the current is large and the loss goes as its square. They also cost a little output swing, which is the price paid for safe current sharing.

Discuss in Forum

39.

Why is an output stage's supply usually unregulated?

A.Regulation would reduce the amplifier's bandwidth
B.The output stage rejects supply variation anyway
C.Regulation would waste power and add cost
D.A regulated supply cannot deliver the peak currents

Answer: C

Music and speech draw current in bursts, so a regulator would have to be sized for the peaks while dissipating heavily at all times, and the feedback loop already rejects most of the supply variation. A regulator could deliver the peaks if sized for them, and bandwidth is unaffected. The fourth option is true and is part of why regulation is unnecessary, but the decisive argument is cost and dissipation. Leaving a supply unregulated because a loop downstream already handles the variation is a common and sensible economy.

Discuss in Forum

40.

What does a supply's ripple appear as at an amplifier's output?

A.Distortion of the waveform's peaks only
B.A rise in the noise floor across the band
C.A shift in the quiescent operating point
D.Hum, reduced by the loop's rejection

Answer: D

Ripple at twice the mains frequency reaches the output as hum, attenuated by the amplifier's supply rejection and by whatever loop gain is available at that frequency. It is not broadband noise, does not selectively distort peaks, and the operating point is held by the feedback. Since loop gain is largest at low frequencies, hum rejection is usually good - which is why a poorly performing amplifier's hum often points at a fault in the feedback path rather than at the supply.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Power Amplifiers Questions and Answers || Power Amplifiers MCQ Questions and Answers || Power Amplifiers GK Questions and Answers || Power Amplifiers GK MCQ Questions || Power Amplifiers Multiple Choice Questions and Answers || Power Amplifiers GK || GK on Power Amplifiers || Analog Electronics Questions and Answers || Analog Electronics MCQ Questions and Answers || Analog Electronics GK Questions and Answers || GK on Analog Electronics

Power Amplifiers Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications