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Old 12th June 2007, 01:20 AM   #1
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Default calculation of output power?..

hi
anyone who know how to measure the exact watts of a certain amplifier?.. do it has a difference in rms power and normal power?..

regards..
ryan..
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Old 12th June 2007, 06:02 AM   #2
jnb is offline jnb  Australia
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If you bring the output up on a scope and have, for example a 1kHz sinewave at 2Vpeak at the input then you can turn up the volume until you can see the distortion, but back up a little and keep in mind that distortion can be harder to see on a low feedback amp. You should have a dummy load connected when doing this or your results will be higher than they should be.

Centre the output on the scope and measure the peak voltage from the zero volt line, then multiply this by 0.707 and square the result then divide by the impedance of the dummy load to get the power into that load.
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Old 13th June 2007, 04:10 AM   #3
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thanks.. this formula maybe help..correct me if i'm wrong..
(VLOAD )squared / 2(RLOAD )
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Old 13th June 2007, 04:58 AM   #4
jnb is offline jnb  Australia
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Power equals Volts squared divided by resistance.

In other words, Power equals voltage times current ie: W=VxI . But I=V / R . Therefore W=(V)squared/R.

if you use the RMS voltage, ie: Vpeak x 0.707, you'll get real watts. Simply use Vpeak to get the peak power output.

By the way, peak power output and PMPO are not the same. Anyway, I don't think PMPO is a useful measurement for HiFi.
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Old 13th June 2007, 06:32 AM   #5
bernhix is offline bernhix  Philippines
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hi,

I have a question about the AC voltage measured by a DVM (Digital Voltmeter), Is this an RMS value?
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Old 13th June 2007, 08:09 AM   #6
jnb is offline jnb  Australia
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As far as I know, DMMs try to give an RMS figure but some are better than others. Unless one actually claims to do RMS, it is more likely to give an average figure, which is close to RMS.

Ordinary DMMs are also likely to be best at mains frequency sinewaves. Be cautious measuring other frequencies and AC which is not a sinewave,
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Old 13th June 2007, 08:19 PM   #7
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Default hi everyone

this is very simple and best way to calculate class AB power amplifiers output powers

The Formula (for 8Ù RMS Power) = VAC² * 0,0975

VAC is Alternative power supply voltage!! (power transformer ½ secondary voltage)
----------------------------------------------------------

Ex= the amp work with (+) and (-) 100VDC =
(VDC to VAC convert formula= 100/1,414 = 70VAC)

Power transformer secondary voltages =
0VAC - 70VAC - 140VAC (usually, 70V - 0V - 70V)


The output power @ 8 ohms load=
70² * 0,0975 =
4900 * 0,0975 = 477W max power @ 8 ohms load
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Old 14th June 2007, 02:08 AM   #8
Tony is offline Tony  Philippines
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to test the power output of your amp, you got to have a dummy load, say 8ohms, this resistor should at least be 50watts rated.

then, you have to have a signal source with sine wave output with at least 2volts output capability.

a scope will come very handy...

so with a dummy load connected, you can inject input signal, gradually increasing input voltage while monitoring the output waveform.

read the ac peak voltage before onset of clipping, that is when tops of the sinewaves flatten, in which case back down on the input until you get the maximum unclipped sinewave.

take care that your amp will run hot...

good luck and cheers..
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Old 14th June 2007, 08:52 AM   #9
AndrewT is online now AndrewT  Scotland
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Default Re: hi everyone

Quote:
Originally posted by serhatkay
this is very simple and best way to calculate class AB power amplifiers output powers

The Formula (for 8Ù RMS Power) = VAC² * 0,0975

VAC is Alternative power supply voltage!! (power transformer ½ secondary voltage)
----------------------------------------------------------

Ex= the amp work with (+) and (-) 100VDC =
(VDC to VAC convert formula= 100/1,414 = 70VAC)

Power transformer secondary voltages =
0VAC - 70VAC - 140VAC (usually, 70V - 0V - 70V)


The output power @ 8 ohms load=
70² * 0,0975 =
4900 * 0,0975 = 477W max power @ 8 ohms load
Hi,
I cannot follow why your formulae should give the output power. Could you explain further?
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Old 14th June 2007, 08:56 AM   #10
AndrewT is online now AndrewT  Scotland
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Quote:
Originally posted by jnb
Power equals Volts squared divided by resistance.
.................... I don't think PMPO is a useful measurement for HiFi.
PMPO gives no useful information.
In my view PMPO is of no use for anything, much less comparing the power output of magical speakers that claim to deliver hundreds of times more power than they consume.
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