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Well I guess we have all asked the question and stared at the 100W label on the box ??? I guess its kinda.... well not really sure.... but the tone is beautiful

Today I know how to answer that thanks to Gerald Weber, in a crude-ish way

Take your Digital Multi-Meter and connect it directly to your speaker and select AC - volts

Set amp volume to just before it distorts and play a few chords/Riffs/love songs and try to determine the average voltage

The formula is Volts2 / Ohm (speaker Imp) = W

eg. V = 15 and Ohm = 8

15x15 = 225 /8 = 28.12W

Of course these are RMS watts, not the Chinese equivalent!

Thanks Gerald (where ever you are)

    Thanks Attila.. I'll be using this method soon out of curiosity!
      Pleasure Norman

      For the sake of balance I was challenged on the topic with the following from Steerpike on the AV forum:

      Digital multimeters are spectacularly bad & inaccurate at measuring audio. They generally cannot measure above 1kHz, and have a low sampling frequency (3Hz or thereabouts).

      Additionally, very few meters read RMS - they measure peak AC with a scale correction for sine wave only. If they can measure real RMS it will be an advertised feature and they will be expensive products.

      You'd do better to put a small capacitor across the meter (5uF or so, and connect it via a diode to the speaker terminals. Now measure the *DC* voltage at a point where you can only JUST hear it starting to distort, now back off the volume atiny bit.
      You will be measuring the clipping threshold of the amplifier.

      Multiply this by 0.7 to get an RMS value. Square it, and divide by the speaker impedance.

      That is the peak average output power ( which some errneously call RMS power).


      The amplifier may not be able to maintain this power for sustained periods owing to it not being able to cool off well enough.




      Ok So I took up the challenge and here is the verdict on both tests using a Stratocaster on my Carolina Songbird

      1) Gerald Method:

      VAC = 8.6

      8.6 x 8.6 = 73.96 / 8 = 9.25 Watt Peak Average Power

      2) Steerpike Method

      VDC = 11

      11 x 0.7 = 7,7 x 7.7 = 59.29 / 8 = 7.41 Watt Peak Average Power


      In the Steerpike method I used a 1n4007 diode and 4.7uf cap


      Personally I like the 9.25 reading better.... sounds more kick ass than 7.41 but there is a 19% difference

      love to have a scope right now
        Don't need a scope, just get a sine wave generator and an analog multimeter ?
        Best alternative is something like this or this to turn your PC into a proper digital recording 'scope.
        Easiest to buy but a bit more expensive here.
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