calculation of output power?..

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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.
 
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.
 
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,
 
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
 
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..:D
 
Re: hi everyone

serhatkay said:
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?
 
jnb said:
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.
 
boylestad states that "the square of voltage across load divided by two times of the resistance of load is the output power.. so it means (Vload )squared / 2(Rload )...
we even get its efficiency regarding to its power output dissipated
over power delivered by transformer to the circuit times 100%..

we all know that all class A,B,AB amplifier only have an efficiency higher and below or close to 75%..therefore power of a certain amplifier will not be exceed on transformers power..

but i was wondering about PMPO's.. why do they give incorrect descriptions with regards to wattage?.. an exaggerated power..
 
(Vload )squared / 2(Rload ), where Vload is the peak signal voltage, happens to give the same result as (Vloadpeakx0.707)squared/Rload. I don't know why one would go about it that way but I guess if the shoe fits, wear it. (BTW, I don't agree with the apparently unnecessary confusion).

With regards to PMPO, I think it's just a way for manufacturers to use big numbers. After all, if your filter bank has the energy to put out a zillion watts for a zillionth of a second, I think some would be tempted to say so.
 
gaetan8888 said:
How do you calculate the RMS power output of a power amplifier from the DC rail voltage of a schematic ?
You could only guess.

If you start by assuming that the peak AC voltage touches the rail, then you could put that into the above formula.

It will not actually be that large though. Assuming the amp was meant to try to swing to the rails, signal voltage can be dropped across the output impedance of the amp, rail voltage can be lost across Vbe, and the rails may sag due to a finite source impedance.
 
There's no accepted way of measurering PMPO. I think it's just some advertising people very liquid saturday nights "think of a number"-game taken a few steps to far but supposedly most companies "measure" (estimate is more likely) it by adding the total 1ms peak output of all output channels added together.
 
gaetan8888 said:
How do you calculate the RMS power output of a power amplifier from the DC rail voltage of a schematic ?
build a good mathematical model of the proposed output stage. Then use this to predict a number of output conditions.

A good start is a BJT version of the modified Bensen spreadsheet, but there are probably a few others available out there.
 
gaetan8888 said:
How do you calculate the RMS power output of a power amplifier from the DC rail voltage of a schematic ?

The DC rail voltage on a schematic is usually quoted or measured under quiescent conditions and a nominal mains voltage. It can easily vary +/-5% due to mains voltage variations, depending on where you live.

When the amplifier is working to deliver maxiumum output power just below clipping, the average DC rail voltage will decrease (due to a finite source impedance of the transformer, rectifier and wiring) and there will be a significant ripple on the rail voltage, depending on the size of the reservoir capacitors.

There will also be a minimum voltage drop across the output devices when they saturate, depending on the (variable) load impedance, number and type of output devices.

A rough approximation for a typical amplifier is to subtract a few volts from the DC rail voltage, say 4 to 8 to allow for these losses, or sometimes even more. Square it, and divide by twice the load impedance. That will give you RMS power. ie.

Prms = ( Vpsnom - Vloss )^2 / ( 2 Rload )

The '2' factor is because we are providing a Peak voltage but we want to calculate the RMS power.

HTH
 
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