how to measure amp power

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Topic: how to measure amp power

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manauta
Senior Member
Member # 1728

posted

Can I ask a real dumb question?????

How exactly are we to determine the output power of a given amplifier???

For voltage:
output AC voltage of the amps( speaker leads)?
or
input DC volatge (input battery power)?

For current:
output AC current (speaker leads)?
or
input DC current (with a clamp-on ammeter on input DC power lead)?

And now what formula do we use?
WATTS = VOLTS x AMPS

And after you get the Wattage do we have to multiply that number by 0.707?

To get the actual power?????

Please help me better understand this subject....

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Posts: 1059 | From: Crestview, Fl | Registered: May 2000 | IP: Logged |

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Doctorbass
Senior Member
Member # 1357

posted

I use a simple methode (fast) to calculate the max total RMS power at output of an amp.

Count the total ampere of fuse you have(ex: 2 green fuse of30 amp =60amp) and multiply that result by the voltage you have at input (DC power) ex: 13,8V you shoudl obtain 828W for this exemple. That is the total power you have at input of an amp. We know that class AB amp have an efficiency around 50-65%. To obtain the MAX RMS TOTAL power your amp give just multiply the total pouer od the input by the efficiency..

ex: 828W * 0.60 = 496W RMS

I toke 0,6 as the mean the AB class efficiency..

If it's a class D amp, the efficiency is around 65-80%..

another way (more accurate) is to use a scope and a voltmeter and to load the output of your amp just before you hit the limit of distortion(sinewave become distored) You will need to put a bass freq like 50Hz at input with test cd or signal generator and you multiply the RMS voltage and the RMS amp at oupput and you get the result..

Doc [Big Grin]

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Posts: 643 | From: Quebec city | Registered: Mar 2000 | IP: Logged |

http://www.eatel.net/~amptech/elecdisc/caraudio.htm

thrtyhz
New Member
Member # 8737

posted

If you would like more info on this, here is a great web site:
http://www.eatel.net/~amptech/elecdisc/caraudio.htm
I have done it this way : amp output voltage x amp output voltage divided by speaker resistance = power.

But first you have to get the speaker resistance which is: amp output voltage divided by the amperage = resistance.

For this test you should play a constant frequency so the results would only be close for that freq. For more accurate results please see the above web site.

Posts: 35 | From: Al | Registered: Feb 2002 | IP: Logged |

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