Theoretically speaking, the maximum voltage swing across the load will give you the power output, assuming the output stage and power supply is not the limiting factor.
P = I2 * R
P = I2 * R
Simple....
Connect a tone-genrator at the input, and an 8 Ohm resistor and an osciloscope at the output of the amp.
Cranck the output open 'till the sinus looks not yet distorted (1% to 3% distortion in the real world), measure the Vout, preferably with a RMS voltmeter, and do the math using Jon Snell's formula...
Connect a tone-genrator at the input, and an 8 Ohm resistor and an osciloscope at the output of the amp.
Cranck the output open 'till the sinus looks not yet distorted (1% to 3% distortion in the real world), measure the Vout, preferably with a RMS voltmeter, and do the math using Jon Snell's formula...
No need or point in an RMS voltmeter, or a voltmeter at all - simply read the p-p value off the scope screen - then divide by 2.828 to get the RMS value, square that, and then divide by 8 to get the RMS power to 8 ohms.
For example:
60V p-p = 21V RMS (near enough - just over)
21 x 21 = 441.
Divide by 8 gives about 55W.
If all you're stuck with is a cheap voltmeter that doesn't do RMS accurately, use it to measure PEAK voltage and convert to equivalent sine wave RMS. A peak detector can be made from a diode (lower the Vf the better), a 10-100k resistor and a good quality 0.1 to 1uF cap. Let it pick off the peak value on a DC scale when you are already clipping, and calculate P= V^2/(2*R). The value of the fundamental, when unclipped, cannot exceed this so it doesn't matter how far into clipping it actually is. If you overdrive when reading AC even on an accurate RMS reading meter you'll get an inflated power number. Without a scope or distortion analyzer it's difficult to tell when clipping begins and the peak detector method removes all the guesswork. So what about the diode? Let's see if you can get within half a volt of the right number with anything less than the 'proper' equipment - won't be any worse than any other cheapie method, and at least it's on the conservative side.
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No great accuracy is needed 😀
You don't spec. an amp as 51.236 watts - you would spec. it as 50W.
If you did want accuracy?, you'd also have to measure the distortion in order to get any kind of accuracy - but the simple scope method is all you ever really need.
You also wouldn't need a true RMS meter anyway as it's a sine wave you're measuring, and a normal meter would be fine with that (depending on it's frequency response - and that applies to a true RMS meter as well).
partially....
An accuracy of 0.25 Watts for a 50 Watt audio amp is almost ridiculous... But knowing whether it puts out 48 or 52 watts might be... in that respect a scope is not accurate enough. Remember the accuracy of a scope is about 5% at best.
A note of caution.
Many amplifiers will not withstand sustained testing under sine conditions so be very careful if you attempt this.
I can think of at least one way to allow a pretty good measurement result, and all without loading the amp for more than a couple of seconds at a time. You would also monitor the output via a speaker and series resistor to determine when the amplifier distorts or clips.
Many amplifiers will not withstand sustained testing under sine conditions so be very careful if you attempt this.
I can think of at least one way to allow a pretty good measurement result, and all without loading the amp for more than a couple of seconds at a time. You would also monitor the output via a speaker and series resistor to determine when the amplifier distorts or clips.
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