I was wondering on how to measure the power of an amp
i was told that the amp should be connected to a dummy load
then the voltage is measure across the 8 ohm dummy load. The voltage measured is AC
then the voltage is divided by the 8 ohm to get current
then P=IV for to calculate the power for the amp
Is this possible or is there any other way?
i was told that the amp should be connected to a dummy load
then the voltage is measure across the 8 ohm dummy load. The voltage measured is AC
then the voltage is divided by the 8 ohm to get current
then P=IV for to calculate the power for the amp
Is this possible or is there any other way?
power measurements
Yup. That is the basic easy way. Besure that the device (DVM, whatever) is accurate at the frequency you are measureing. (Not all DVMs are accurate over the whole audio range) If you are using an oscilloscope note that you must divide the peak-to-peak voltage by 2.83 to get the RMS voltage. To save a step you can square the RMS voltage and divide that answer by 8 (ohms) to get power in watts. Then you can use different values of dummy loads to see if the amp puts out more power at different load resistances. Your source sould be a sine wave with little distortion. Ideally, your dummy load resister should have no or very little inductance. However, at low audio frequencies a wire wound resistor will probably not have enough inductance to mess up your measurements.
If you are testing a commercial amp be aware that it might not put out full power for very long. So get your set up set up, turn the amp up, take your measurement, turn the amp back down and then do your math. (This is the voice of BITTER experience!)
Yup. That is the basic easy way. Besure that the device (DVM, whatever) is accurate at the frequency you are measureing. (Not all DVMs are accurate over the whole audio range) If you are using an oscilloscope note that you must divide the peak-to-peak voltage by 2.83 to get the RMS voltage. To save a step you can square the RMS voltage and divide that answer by 8 (ohms) to get power in watts. Then you can use different values of dummy loads to see if the amp puts out more power at different load resistances. Your source sould be a sine wave with little distortion. Ideally, your dummy load resister should have no or very little inductance. However, at low audio frequencies a wire wound resistor will probably not have enough inductance to mess up your measurements.
If you are testing a commercial amp be aware that it might not put out full power for very long. So get your set up set up, turn the amp up, take your measurement, turn the amp back down and then do your math. (This is the voice of BITTER experience!)
Ohm my god I like the law...
Also for a lot of people the formula U = RI is a dear friend (all of them are variations of the basic Ohm's law). If then P=IU (current multipled by voltage) then why not use P = U(sqr)/R as I solved from U=RI is I=U/R giving P=UI moving on to P=UU/R. Since this is peak and the RMS it usually more interesting and the RMS value for sinusoidal is the peak divided by 2(sqrt) (which is why FETHead says 2.83 for peak-to-peak which is twice the peak and twice the root of 2 is 2.83) is is easy to calculate as the root of 2 squared is 2 again...
Hence for RMS values using peak voltage:
P=U(sqr)/2R
This is the quick one I think and even works in my head since it usually (or sometimes) is things like 20 Volts at 8 Ohms which is 20*20/2*8 = 25 W, nice and easy numbers...
Sorry for the lengthy reply/UrSv
Also for a lot of people the formula U = RI is a dear friend (all of them are variations of the basic Ohm's law). If then P=IU (current multipled by voltage) then why not use P = U(sqr)/R as I solved from U=RI is I=U/R giving P=UI moving on to P=UU/R. Since this is peak and the RMS it usually more interesting and the RMS value for sinusoidal is the peak divided by 2(sqrt) (which is why FETHead says 2.83 for peak-to-peak which is twice the peak and twice the root of 2 is 2.83) is is easy to calculate as the root of 2 squared is 2 again...
Hence for RMS values using peak voltage:
P=U(sqr)/2R
This is the quick one I think and even works in my head since it usually (or sometimes) is things like 20 Volts at 8 Ohms which is 20*20/2*8 = 25 W, nice and easy numbers...
Sorry for the lengthy reply/UrSv
None of the DMM I have are good much above 120Hz. Using a tone generator at 120Hz works fine of course. 
As FEThead alluded some amps may do nasty things (like power limit, or break) if operated at full power for any length of time. Commercial amps in particular often aren't designed for much above a 15% duty cycle....you'll note that the "5x100W" amp is rated for only 265W from the line.
As FEThead alluded some amps may do nasty things (like power limit, or break) if operated at full power for any length of time. Commercial amps in particular often aren't designed for much above a 15% duty cycle....you'll note that the "5x100W" amp is rated for only 265W from the line.
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