determinning wattage

Thread Starter

Mr.oaktree

Joined Dec 10, 2014
68
can anyone tell me how i can determine the wattage output of audio amplifier ?? all i possess is a basic multimeter and a old analog multimeter . any advice most welcome .
 

studiot

Joined Nov 9, 2007
4,998
Every item of electrical equipment should have a 'rating plate' somewhere on it.
It is a legal requirement in the UK for mains driven equipment.

Take 50% of this figure and divide by the number of channels.

So a two channel amp that draws 200 watts from the mains works out at {0.5*200}/2 = 50 watts per channel.
 

Thread Starter

Mr.oaktree

Joined Dec 10, 2014
68
ah yes that would be conveinent wouldnt it .... the problem is that i installed two new power transistors in one channel and the other channel is complete different set of transistors . i tried to match as closely as possible but seems one channel (the new channel) has signifigantly more power than the other .
 

studiot

Joined Nov 9, 2007
4,998
Do I remember a thread about this here?

No offence but have you adjusted the balance?

Maybe there is a problem there as well.

Maybe the second channel was also sick?

Post details for help.
 

Thread Starter

Mr.oaktree

Joined Dec 10, 2014
68
Do I remember a thread about this here?

No offence but have you adjusted the balance?

Maybe there is a problem there as well.

Maybe the second channel was also sick?

Post details for help.
yes thread is n projects , {repairing amplifier} no balance is not adjusted. my thought also that other channel may be sick as well . if i could read max output each channel id have a direction to head i think
 

studiot

Joined Nov 9, 2007
4,998
1khz may be too high.


Many analog voltmeters have a db scale, or they have an AC volts scale.

So take the same input signal, as low a frequency as you can perhaps around 100hz, and connect it to each channel (or both) and measure the voltage across the same load (speaker) at reasonable volume so that you are measuring a good few ac volts on your meter ac volts scale and post the results.

Whilst you are at it, with the speaker connected but no signal measure the DC volts across each speaker.
This should be zero but a loss of bias setting could show up here in the weak channel.
 

studiot

Joined Nov 9, 2007
4,998
Surely you have what my grandmother used to call a hurdy gurdy?

(any audio source will do, CD, record player, radio, guitar.................)
 

studiot

Joined Nov 9, 2007
4,998
Yes, at some level that you can compare on your voltmeter.

It will not show low levels of audio signal, only moderate to high.

But also rememeber that the frequency response of your meter will probably not be more than 200hz so your signal must be this low.
 

studiot

Joined Nov 9, 2007
4,998
Use the 'output' socket if you have one. (it is used for measuring output, it doe not mean the meter outputs anything)

This has a DC blocking capacitor in series with the socket so the only responds to AC.
 

cmartinez

Joined Jan 17, 2007
8,257
In a perfect world... I'd use a sensor that would measure current flowing to the speaker connected to one channel of an oscilloscope, and I'd use the other channel to measure the voltage reaching the speaker. Then I'd run a sinusoidal audio wave of an average frequency (for the human ear) through the amp and turn the volume all the way up.... do some simple calculations and obtain rms wattage... I'd even have the luxury of varying the frequency source just to obtain a response curve too... in a perfect world
 

MagicMatt

Joined Sep 30, 2013
117
Surely wattage is fairly irrelevant? I thought that was just how much power it could handle before it went "pop!". What determines output volume isn't the wattage, it's the gain! A 20W per channel amp can sometimes be "louder" than a 50W per channel amp when fed with the same input signal, but the 50W per channel amp will handle more before you start hitting the limit of the components.

I would say your new channel doesn't have more power, it probably has more gain, in which case you may be able to just bias that with a resistor or two and match the channels.
 
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