# Calculating the peak output voltage from RMS power of an audio amplifier

I want to calculate the peak voltage output of an audio amplifier when driving at its maximum rated power, a loudspeaker.

For example, an amplifier rated at 50 W RMS into 8 Ω, driving an 8 Ω loudspeaker.

This wikipedia article tries to explain what RMS power actually means, but it's not clear. And I'm wondering if it's actually wrong where it says:

[RMS Power] is proportional to the RMS voltage

This seems wrong, since I know that the equation for power/voltage/resistance is:

$$P = \frac{V^2}{R}$$

So assuming that the amplifier does not distort, does not exceed 50 W RMS and so on, what is the peak AC voltage it would output?

• "RMS power" is an incorrect term. RMS voltage is proportional to the average of the power waveform, not the RMS of the power waveform. – endolith Jan 22 '16 at 20:51

With some assumptions (pure sine wave, resistive load, etc.) you can simply apply your formula.

$$50 = \frac{V^2}{8}$$

$$V = \sqrt{400}$$

hence $V = 20$. This is the DC voltage that puts 50 W in an 8 Ω load.

For an AC sine wave the peak (with respect to the average) is $\sqrt{2}$ times this value, or approximately 28 V. These peaks appear both in positive and negative direction, so the peak-to-peak voltage is twice this value, approximately 56 V.

This is the minimal DC supply a capacitor-coupled push-pull amplifier stage would need. A direct-coupled push-pull amplifier would need one positive and one negative 28 V supplies, a bridge amplifier would need one 28 V supply. (In each case, more is needed due to losses, margins, etc.)

So, you should calculate the output power of each channel amplifier as the formula below:

$$P=\frac{V^2}{2R}$$

V - peak output voltage,

R - speaker impedance.