One of the practice questions for my upcoming test is this: enter image description here

After some googling, it seems like the correct answer is A. But why?

  • \$\begingroup\$ Hm, can't from the top of my head find a reason why that should be true. Can you link to the answers you've found via Google? Are there perhaps implicit restrictions on the type of signal you're amplifying? \$\endgroup\$ – Marcus Müller Dec 15 '18 at 18:37
  • \$\begingroup\$ For example, if the goal is to amplify a signal that should yield the minimum (maximum) output voltage, I'd argue that a good push-pull output stage would yield an efficiency darn close to 1, depending on the rdson of the low-side (high-side)switching transistor. \$\endgroup\$ – Marcus Müller Dec 15 '18 at 18:40
  • 1
    \$\begingroup\$ It depends on the application, when it's a digital output stage, the efficiency turn to nearly 1, but if it's an audio or RF amplifier (B-class), the maximum should be at \$\pi/4\$ or 0.785 \$\endgroup\$ – Tom Kuschel Dec 15 '18 at 18:55
  • 1
    \$\begingroup\$ For a sinewave output there is a voltage across the transistor when the sine is not at it's peak(assuming Vpeak = VCC), so naturally there will be losses in the transistor. \$\endgroup\$ – Linkyyy Dec 15 '18 at 19:00

It is good to review this.
@Barry gives the good answer. Am attempting a more graphic solution using LTSPICE. The assumption is that a sine wave is the signal source, whose peak voltage just grazes the DC power supply that powers the Class-B output device(s).
This is unrealistic, because any transistor has some overhead voltage drop: at least its saturation voltage. The answer assumes it is zero, and the transistor works linearly all the way through the sine wave peak.

In this simulation, a sine wave current source of one amp is fed to a load resistor of two ohms. The peak voltage at this load resistor reaches 2V. So a 2V DC power supply is required to power this Class-B simulation. It can only do the top half of the sine wave - only one quarter cycle is scanned from a 1Hz source.

LTspice circuit
Below is a plot of power vs. time. One trace shows power dissipated in R1, while the other shows power dissipated in I1:
Power in I1, Power in R1
The result of the two LTspice measurements shows the average power during the quarter-second run:

p_source: AVG((2-v(n002))*i(i1))=0.273242 FROM 0 TO 0.25
p_r1: AVG(v(n002)*v(n002)/2)=0.999993 FROM 0 TO 0.25

Efficiency for your question seems to be defined as p_r1/(p_r1 + p_source)
That is: 1/(1 + 0.273242) = 78.54%

Note that for every watt dissipated into the load resistor, 0.273W is dissipated in the pull-up transistor, but that only lasts for half a cycle. During the next half cycle, this top transistor (it might be a PNP) dissipates nothing.

During the bottom half cycle (while the PNP dissipates nothing), the NPN would take over, and dissipate a similar 0.273W. So the PNP and NPN share power dissipation, each taking half of the 0.273W.
Don't forget, this is a unrealistically ideal case.


Push-Pull is used for analog , digital, Class B, D,E, half-bridge switches and all sorts of applications.

In theory with sine wave and ideal saturation =0V say with MOSFETs, it is η=π/4 or 78.5% but 78% may be closer to what is possible but given no assumptions choose (c) (d) for sine as this is the typical signal not a square wave which in theory is 100% but a practical design goal might be 99% for Class D or E.

The problem is lack of assumptions or specs so the question is vague but assume by push-pull they imply Class B not Class D or E. Unless one states signal shape, and distortion allowed, efficiency must be based on some standard test method. like 1KHz sine 1% THD, so the question is very typical of an academic one.

The archaic nature of this academic question leads me to suggest they only want you to compute the integral of a sine current (I-peak) squared times a fixed resistance, which I leave for you to figure out.

  • \$\begingroup\$ I think you meant option c.. I know you wrote the answer also, but just to remove any confusion.. :) \$\endgroup\$ – Linkyyy Dec 15 '18 at 19:13

The answer can be found by integrating over one cycle the product of the output current and the voltage across the collector of the output transistor, which represents the loss in the amplifier. The output voltage times the output current is the output power. The resultant value, found by dividing the power loss by the sum of the power loss and the output power, is pi/4 for a maximum level sinusoidal signal. Efficiency decreases as the output power decreases. This result can be found in any book on power amplifiers. It is only true for push-pull (class B) amplifiers. For class A amplifiers, the maximum is 50%. As pointed out, these results are for ideal cases, not real amplifiers.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.