Measure Power Output Using Clip Light

Measuring the voltage when it clips is the difficult part - even with a true RMS meter. When it clips you’re already putting out slightly more RMS voltage than it can unclipped. My solution is to measure peak voltage (with a diode and a cap before the multimeter, set on DC). Run it into hard clipping, then calculate using v^2/(2RL). You will not be overestimating output voltage that way, if anything it will be down a bit from the diode drop.
 
Measuring the voltage when it clips is the difficult part - even with a true RMS meter. When it clips you’re already putting out slightly more RMS voltage than it can unclipped. My solution is to measure peak voltage (with a diode and a cap before the multimeter, set on DC). Run it into hard clipping, then calculate using v^2/(2RL). You will not be overestimating output voltage that way, if anything it will be down a bit from the diode drop.

OK thanks. And what formula is that for? Haven't done much math since I passed my ham radio extra class license some years ago and my 1st class FCC license even farther back.
 
Average power of a sine wave, given the peak voltage and load resistance. The voltage won’t be quite sinusoidal when clipping, but the peak voltage will be the same as if it wasn’t clipping.

If you do the test with music or short bursts instead of a sine wave you will get inflated values because the power supply will be at the unloaded value. It is less under full load. Many manufacturers will rate their amps this way, however. The full load values can be quite disappointing.
 
I don't have an oscilloscope. Is it accurate to some degree to crank up the input level of a sine wave until the amplifier clip light comes on?

Yes you can do this ...

Put your ac voltmeter across your speaker leads, use a dummy load if possible but a speaker will work in a pinch,

Turn up the sine wave until the clip light comes on, then turn it down just a hair so the clip light is off...

You should see a relatively steady state voltage on your meter.

The reading will be RMS volts ... then use Watt's equations to calculate your power ... P = (E x E) / R ... So, if you are reading 20 volts cross an 8 ohm load... (20 X 20) / 8 == 400 / 8 == 50 watts RMS.

This won't be as accurate as using a scope to see the clipping. Most amps turn on the clipping light at like 95% power, but it will give you a pretty good ballpark figure of the amplifier's power.

BTW... be careful if you're using speaker ... Be sure to protect your hearing, a 1khz sine wave at 50 watts can cause permanent hearing damage.
 
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Measuring the voltage when it clips is the difficult part - even with a true RMS meter. When it clips you’re already putting out slightly more RMS voltage than it can unclipped. My solution is to measure peak voltage (with a diode and a cap before the multimeter, set on DC). Run it into hard clipping, then calculate using v^2/(2RL). You will not be overestimating output voltage that way, if anything it will be down a bit from the diode drop.
I used that way when I didn´t have a scope yet ... some 50 years ago.
Quite close, but "optimistic".
Reduce peak voltage found by 10% , reason being that when driven balls to the wall, the fully saturated transistor connects speaker to supply rail ... which at that doubled consumption (full squarewave) presents *significant* ripple, which is fake or useless power of course.
40V rails can easily show 4V peak ripple, sometimes twice as much, depending on supply capacitance.
Peak ripple voltage gets added to what would be "just clipping" peak value.