In theory, it is widely known that the maximum efficiency of class B amplifier (consider \$V_p = V_{cc}\$) is:

$$\eta_{MAX} = \frac{\frac{V_p²}{\sqrt{2}² R_L}}{2 \cdot V_p \cdot \frac{V_p}{\pi R_L}} = \frac{\pi}{4}$$

Why in the load is used the RMS value of the signal while in the sources is used the maximum voltage (i.e., the source voltage) and the average current? Isn't the power equal to the sum of the product of RMS voltage and RMS current for each source? If I did that, I'd obtain:

$$\eta_{MAX} = \frac{\frac{V_p²}{\sqrt{2}² R_L}}{2 \cdot V_p \cdot \frac{V_p}{2 R_L}} = \frac{1}{2}$$

Recall that in a half wave signal (the current of the sources) the average value is \$V_p / \pi\$ and the RMS value is \$V_p / 2\$.

  • \$\begingroup\$ The power is only equal to Vrms*Irms if the current and voltage are in phase. I'm pretty sure they should be in phase here, but do make sure to remember that! \$\endgroup\$
    – Hearth
    Commented Jun 11, 2023 at 15:33
  • \$\begingroup\$ @Hearth Yes all these are only valid for a resistive load! For reactive loads, efficiency drops significantly... \$\endgroup\$
    – bobflux
    Commented Jun 11, 2023 at 17:43

2 Answers 2


Efficiency is typically defined as the ratio between the average power in the fundamental delivered to the load and the average power drawn from the supply. For both terms, we can form an expression for the instantaneous power and average over a period of the input signal.

In the case of the power delivered to the load, we know that for a resistive load $$P=V_{RMS}^2/R$$, which simplifies to the numerator of your first efficiency equation.

For the power drawn from the load, we notice that from the perspective of the voltage source, the amplifier does NOT look like a purely resistive load. That is, the voltage is constant (\$V_{CC}\$) and the current is time varying (half period current pulses). The instantaneous power is $$P=V_{CC}I(t)$$ where \$I(t)\$ is the characteristic current pulse, a half-sinusoid. If we average this over one period, we get $$ \frac{1}{T}\int_0^TV_{CC}I_0sin(\frac{2\pi}{T}t)dt=\frac{1}{T}\frac{2T}{\pi}V_{CC}I_0=\frac{2V_{CC}^2}{\pi{}R_L} $$ since the peak current from the supply, \$I_0\$, is simply the supply voltage over the load current. This is the denominator of your first expression, as expected.

To see why the second efficiency expression is not valid, think back to how RMS is derived as a measure of AC power. When both current and voltage waveforms are sinusoidal, then we can write instantaneous power as proportional to the product of a single sinusoid squared times the power factor. Then to find the average power, we take the average of a squared value.

However for the case when the voltage and current waveforms have fundamentally different shapes, we cannot write instantaneous power in terms of a quantity squared. We must express in terms of both voltage and current, like in the supply power case.


A simulation...
One cycle of a 1kHz wave, with a ridiculously large DC supply voltage (so as to minimize the transistor's base-emitter cross-over loss of about 0.6V). The transistors are LTspice default, which are close to ideal.
Class B LTspice demo circuit
Efficiency doesn't include power delivered by V2 source signal. This kind of calculation is sometimes called "Collector efficiency".

pin: AVG(i(v1)*v(vcc))=-629.766 FROM 0 TO 0.001
pout: AVG(i(r1)*v(load))=-498.728 FROM 0 TO 0.001
p_npn: AVG((v(vcc)-v(out))*ic(q1))=67.8811 FROM 0 TO 0.001
p_pnp: AVG(v(out)*ic(q2))=-67.7222 FROM 0 TO 0.001

DC power in = 629.766 W
Power delivered to R1 = 498.728 W
As a check, power dissipated by Q1 and by Q2 are also calculated, 67.8W each. Yes, 100% (nearly) 629.766W power pulled from V1 is accounted for.

TLDR: Efficiency = 498.728/629.766 = 79.193%, which is close to \$\pi/4\$

  • \$\begingroup\$ Interestingly, I always heard the figure 3/4 in school. Which is a reasonable enough approximation, but not mathematically accurate. \$\endgroup\$
    – Hearth
    Commented Jun 11, 2023 at 17:48
  • \$\begingroup\$ @Hearth, I too recall "3/4". Seems easier to recall than "PI-over-4" and is easier to approach. Your caution about real loads versus reactive loads is one to watch! \$\endgroup\$
    – glen_geek
    Commented Jun 11, 2023 at 18:10
  • \$\begingroup\$ I suppose since you're unlikely to be able to get that near the theoretical limit anyway, just giving the rule of thumb figure as 75% is good enough. Any practical amplifier is going to lose more efficiency than that just from the additional diodes and resistors to counter crossover distortion. \$\endgroup\$
    – Hearth
    Commented Jun 11, 2023 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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