Measuring Output power using a scope

Discussion in 'Amps/Cabs Tech Corner: Amplifier, Cab & Speakers' started by RobRoyMcCoy, Jun 30, 2006.


  1. RobRoyMcCoy

    RobRoyMcCoy Member

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    Hi All,

    I have been trawling the internet to find out a consistent and agreed way to express output power w.r.t. guitar amps. I have done a search on this forum but I am still confused.

    I have measured the output waveform of my THD Flexi-50 using my scope at 27.5V peak to peak at the onset of clipping. I have put a 1kHz sine wave in the front end, terminated the amp into a 5ohm dummy load. I am running a pair of JJ KT88 in the power stage biased at approximately 65% of max plate dissipation at idle. The dummy load is a pair of 10 ohm wirewound power resistors connected in parallel. Each is rated at 50W.

    My confusion is this:

    If I take the peak voltage value I get 13.75V (27.5/2). Using V squared over R to calculate power I get 13.75*13.75/5 = 37.8W which I think is the peak power. For RMS power I use 13.75*0.707 or 9.72. So using the same formula I get 9.72*9.72/5 = 94.47/5 = 18.9W RMS.

    However, THD say this is a 50W head with EL34 and say up to 80W with KT88.

    Am I approaching this problem correctly or are my measurement lower than expected? Or do the amp manufacturers express power in a different way?

    My next step will be to run the load at 20 ohms, change the OP transformer tap and redo the measurement. I will also try it with EL34's.

    Any thoughts appreciated.

    Rob
     
  2. VacuumVoodoo

    VacuumVoodoo Member

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    Your calculations are correct. You didn't mention what tubes you used but 18W output looks about right for 6V6 or EL84 tubes. KT66 with cathode bias would give similar result too.
     
  3. donnyjaguar

    donnyjaguar Member

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    The 5 Ohm load sounds a bit wrong, but I'm not familiar with this particular amp. If the amplifier was engineered for 8 Ohms you really need to use a load of that value and re-calculate. I have 2-4-8 Ohm loads and its interesting to see what power output you get at each impedance.
     
  4. RobRoyMcCoy

    RobRoyMcCoy Member

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    Thanks guys. I am using KT88's so that is why I think the power output is low as I was expecting at least 50W peak. Do amp manufacturers typically express power output as peak or RMS when they say 50W (for example)?

    I agree that the 5 ohm is not ideal but it is a dummy load so I do not have to listen to a 50W (or whatever) sine wave. As a power resistor it is also relative frequency independent or non-inductive. I will repeat using the dummy load at 20 ohms connected to the 16ohm tap of the OT and then wait for my family to leave town and do the same test using my 8 ohm speaker in a 1 x 12!

    Thanks

    Rob
     
  5. RobRoyMcCoy

    RobRoyMcCoy Member

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    Thanks for that.

    So refering back to my measurements, does that mean that in order to express the power output of my amplifier I take the peak to peak voltage of 27.5 and use that in my formula?

    Therefore power output = (27.5 x 27.5)/5 = 151.25W (I don't believe this is right)

    or should I use the peak voltage = (13.75 x 13.75)/5 = 37.8W?

    Rob:messedup
     
  6. John Phillips

    John Phillips Member

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    Use the RMS voltage. For a pure sine wave that is the peak-peak voltage divided by the square root of 2 - approximately 1.41.

    So if your P-P is 27.5V, your RMS voltage is 19.4V, so the power output into 5 ohms is 75.6W.

    (You'll notice that this is exactly half the power calculated from the P-P voltage - which in theory you could reach if the amp was driven so hard the output became a pure square wave, but in practice it can never reach that due to power supply and OT limitations - because the 'square root' drops out of the equation when you square the voltage to calculate the power. This makes it very easy to get the RMS power from the P-P voltage.)
     
  7. ClinchFX

    ClinchFX Gold Supporting Member

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    Hi Rob,

    For a true sine wave, RMS voltage is peak voltage divided by the square root of 2, which equals peak voltage multiplied by .707 (the inverse of the square root of 2). There is an explanation here and here.

    RMS voltage will be 13.75 x .707 = 9.72V


    So the output power is (9.72 x 9.72)/5 = 18.89W. This seems too low. I remember doing the same measurement on a Marshall too many years ago to remember what model, and on my sons' Bassman 50 and 100 amplifiers, and got figures that came pretty close to the rated power.

    Did you try this measurement because you think the amp is low on power, or was it just curiosity? It sounds as if you have the amp biassed more for class A than class B, and this will cause a dramatic reduction in output power. THD describe the amp as class A/B.

    Peter
    ClinchFX
     
  8. John Phillips

    John Phillips Member

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    Peak-to-peak, not zero to peak.

    That's why you got one-quarter of the power :).
     
  9. ClinchFX

    ClinchFX Gold Supporting Member

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    Peak, not peak to peak. Check the links I posted.

    Peter.
    ClinchFX
     
  10. John Phillips

    John Phillips Member

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    Regardless, it's wrong.

    RMS voltage is calculated relative to the P-P voltage, not the zero-to-peak voltage.
     
  11. ClinchFX

    ClinchFX Gold Supporting Member

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    John, I can give you a heap more links to university texts that agree with those two links, but they go into some heavy maths, and I don't think anyone would appreciate a ten page post explaining it.

    Suffice to say that the textbooks are correct. Peak voltage is the unsigned measurement from zero to either peak, and peak-to-peak is the unsigned measurement from the positive peak to the negative peak. RMS voltage is .707 times peak. This is one of the first things you learn when studying AC theory.

    Peter.
    ClinchFX
     
  12. John Phillips

    John Phillips Member

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    I'm not going to argue theory with you either for the same reasons.

    (FWIW, refering to one of those links, mains voltage @240V RMS is 340V P-P, not 680V which it would be if the RMS refered to peak.)

    Just look at the figures. That amp is putting out 75W, not 18W.

    I did EE at university too BTW.

    That's all I'm going to say here.
     
  13. VacuumVoodoo

    VacuumVoodoo Member

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    Taking zero-peak value implies integration over 1/2 period of the sine waveform, peak-peak implies integration over whole period. Somebody is neglecting a factor of 2 somewhere in his thinking ;)
     
  14. ClinchFX

    ClinchFX Gold Supporting Member

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    OK,

    What will happen with power calculations if you full wave rectify the signal? (ignoring the fact that it would sound terrible.) Remember, a dummy load doesn't care what direction the current flows.

    Peter.
    ClinchFX
     
  15. VacuumVoodoo

    VacuumVoodoo Member

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    Sinewave with period T when rectified becomes a periodic waveform with period T/2. The integral formulae are HERE
     
  16. donnyjaguar

    donnyjaguar Member

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    RMS voltage measurements are taken using the 0 axis to peak value of one half of the sinewave x0.707. This may be easily proven by putting your AC voltmeter on the output of the amplifier. The whole purpose of RMS is so you can work an AC signal into Ohm's law and have the results make sense. To my knowlege this isn't conjecture.

    DJ
     
  17. glaswerks

    glaswerks Gold Supporting Member

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    FWIW, I came up with 18.87 watts.

    ((27.2 (p/p) / 2.8) ^2 ) / 5

    Gary
     
  18. RobRoyMcCoy

    RobRoyMcCoy Member

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    Ok guys, thanks for the great feedback. I am still unclear as to the correct expression of power for my amp and how other manufacturers express their power outputs, but as a result of this discussion I have some ideas to explore to further my understanding;

    1) Adjust the bias to find maximum power output. I had set at close to 70% based on Aiken's advice on his web page. I have had good results with this approach. However I can measure with my scope the impact on peak to peak voltage levels whilst adjusting bias. In line with this I could put some EL34's in, biased as per the THD spec and should expect to 50W so I can use that as a benchmark.
    2) Try a different load. I can switch my load to 20 ohms or use 1 of the 50W resistors at 10 ohms (for a short time).

    The reason I tried this was not because I had any doubt over the output of the amp. It is terrifyingly loud with EL34's and I haven't had the KT88's up at full volume through a speaker except to momentarily test that it would all hold together. I will post results here.

    Thanks

    Rob
     
  19. ClinchFX

    ClinchFX Gold Supporting Member

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    Yes Old Tele man, agree completely. You explained it in much more readable terms than I ever could.

    Sorry for disappearing from the discussion, it was nearly midnight here, and I had a big day at work today.

    Peter
    ClinchFX
     
  20. donnyjaguar

    donnyjaguar Member

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    I'm resisting the temptation to introduce power factor to obfuscate the discussion further. :)
     

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