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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 50W RMS into 8 ohms, driving an 8 ohm 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:

Power = V^2 / R

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

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1 Answer 1

up vote 5 down vote accepted

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

50 = V^2 / 8

leads to

V = sqrt( 400 )

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

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

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

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