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New Music Man 2475-130 bias question

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  • New Music Man 2475-130 bias question

    I have a Music Man 212-HD 130 (chassis 2475-130) with a tube PI. I'm installing a new quad of JJ EL34s.

    I understand that the schematic calls for 0.5vDC at the power tubes' cathodes. But I don't want to blindly set the bias, without checking the current in the tubes.

    I'm wondering how to evaluate the plate voltage, in order to choose a bias point, in amps such as this where the screens are radically lower than the plates. The plates have ~702v, and the screens have ~350v. So, in the bias calculator, do I enter 700v or 350v plates?

    BTW, I have confirmed 702v on plates (schematic confirms this), and 348v on screen grids.

    Schematic: https://www.thetubestore.com/lib/the...-Schematic.pdf


    In practice, biasing with the assumption of 350v plates proved too low--lots of crossover distortion. Biasing for 700v plates and a target of 50% (or about 18mA each) gives me a clean 120 watts output with no crossover distortion.
    --
    I build and repair guitar amps
    http://amps.monkeymatic.com

  • #2
    If you are getting no x-over at 50%, I would go with that.
    The .5V at the cathodes is not 'going blindly' as you know the value of the cathode resistor (10 ohms). So it is actually sensing the cathode current, and asking you to set for 50mA per pair, or 25mA per tube. When you are given a voltage spec. at the grids, that is when you are going blindly and current can be all over the place.
    In this case, at 700V plate, their spec works out to 70%. I prefer your 50%, and I think the tubes will too.
    Originally posted by Enzo
    I have a sign in my shop that says, "Never think up reasons not to check something."


    Comment


    • #3
      By "blindly", I meant relying on what the manufacturer stipulated back in the day, rather than actually understanding what's going on. Thanks, G1.
      --
      I build and repair guitar amps
      http://amps.monkeymatic.com

      Comment


      • #4
        Ok, I see what you mean then.
        So with your 18mA, and assuming the tubes are well matched, you would see about .36V at each of the cathode resistors.
        Any time you have a cathode resistor you have a convenient place to check the current via I = E/R (again assuming tubes are well matched if cathode resistor is shared).
        Originally posted by Enzo
        I have a sign in my shop that says, "Never think up reasons not to check something."


        Comment

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