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  • measuring output power

    Lets see if I have this right.....Using a VOX AC30VR for this test.....use the clean channel.....max out channel and master volume....max out tone controls on clean channel.....input a 1klz sine wave and increase generator level until the output sinewave starts to distort on scope....then measure the ac voltage across the 8 ohm dummy load......square it and then divide by 8.......so ac voltage of 16.4 volts squared is 268....divided ny 8 gives about 33.6 Watts.....This AC voltage can be measured with the AC section of a regular DMM??? Just want to clarify this.....The guy who had the amp claims it sounds"Wonky"...whatever that is........and that it doesn't seem to have the volume it once had......
    cheers,
    Bernie

  • #2
    You've got the right idea. Setting tone controls is not critical, but definitely tweak all controls a bit to see if you can juice any additional clean signal out of the output.

    Your formula is correct for AC RMS. If you have a voltmeter that measures true RMS, you're doing it right. If you use an analog scope, you'll likely measure AC voltage point-to-point (p-p), and you must multiply this value by 0.3535 in order to get the RMS value. Square the RMS value and divide by the load, as you said.

    In order to get the highest clean output, your load (speaker or dummy load) must be matched to the amp. If you've got a mismatched load, the output power will be less.
    --
    I build and repair guitar amps
    http://amps.monkeymatic.com

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    • #3
      Originally posted by xtian View Post
      You've got the right idea. Setting tone controls is not critical, but definitely tweak all controls a bit to see if you can juice any additional clean signal out of the output.

      Your formula is correct for AC RMS. If you have a voltmeter that measures true RMS, you're doing it right. If you use an analog scope, you'll likely measure AC voltage point-to-point (p-p), and you must multiply this value by 0.3535 in order to get the RMS value. Square the RMS value and divide by the load, as you said.

      In order to get the highest clean output, your load (speaker or dummy load) must be matched to the amp. If you've got a mismatched load, the output power will be less.
      Ok..I 've done it correct apparently...I used the AC function of my DMM....I don't think it is a true RMS meter.....but I figured it should be close enough..I also used the scope measurements as well...and they were 45Vp-p....when I did the calculations, they were only off by a watt or two...The amp actually sounds fine to me.....but....you know people.........I'll install the chassis back in the cab tomorrow and give it a run....Alsi I am using an 8 ohm dummy load...measuring exactly 8 ohms with the DMM....
      Cheers

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      • #4
        It can be difficult to troubleshoot "Wonky". You might ask the customer to be more specific.
        "I took a photo of my ohm meter... It didn't help." Enzo 8/20/22

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        • #5
          And the difference between 20 and 30 watts is hardly noticeable. Both loud and proud.

          In any combo, high volumes rattle the tubes and circuit aggressively. You may help the owner troubleshoot by using an external speaker cabinet. Good luck!
          --
          I build and repair guitar amps
          http://amps.monkeymatic.com

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          • #6
            And remember too that output power is only for the output section. people will demand you measure power at the onset of distortion, but when running the test signal through the preamp to get there, it can be difficult or impossible to drive the power amp fully without distortion. So if knowing exactly how much power is happening, one needs to bypass the preamp and input a signal directly into the power amp.
            Education is what you're left with after you have forgotten what you have learned.

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            • #7
              Originally posted by bsco View Post
              Ok..I 've done it correct apparently...I used the AC function of my DMM....I don't think it is a true RMS meter.....but I figured it should be close enough..
              I'm sure it's close enough (unless the sinewave is really distorted). If the meter is not true RMS it will be average sensing but callibrated RMS for a sinewave and will therefore read the same as a true RMS meter on a clean sinewave.

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              • #8
                Originally posted by xtian View Post
                If you've got a mismatched load, the output power will be less.
                Or more with this SS Vox if it's into a lower impedance. At least until the output IC dies or shuts down.

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                • #9
                  Off topic, slightly...

                  I've been building and tinkering with amps for some years now. Okay so there's one thing I never understood when I lived in ignorance, prior to my amp-days. And I still get confused when I finished a build and the first question people ask is how many watts?. It hurts my feelings actually... *puts a wrist over the forehead*
                  In this forum everyone is entitled to my opinion.

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                  • #10
                    "how many watts?"

                    Reply: Mains current draw or output power?

                    I like when amps have a current draw sticker over at the power cord.

                    'Hey, it says 150 watts'

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                    • #11
                      Originally posted by Jazz P Bass View Post
                      'Hey, it says 150 watts'
                      My reply to such statement is usually. Eh... yes... it does...
                      In this forum everyone is entitled to my opinion.

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