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Fender blues jr - convert to cathode bias?

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  • #16
    Originally posted by frus View Post
    well it's kind of circular definition, because if it WAS class A, then the plate/cathode current in full tilt wouldn't increase and voltage on the cathode wouldn't rise... OK, it would a bit because of the screen current.... maybe even considerably...?
    I know, right? I always figured it was just something I didn't get. Because I've cathode biased el84's at 12W each and then watched the cathode voltage rise with signal applied to the grid. Double or more. I don't remember the shift in Vp so I can't comment on current. But bias isn't current. It's voltage that controls current. A tube can have the same current across it in AB1 as it can in A. So a shift in bias doesn't need to reflect a significant change in current.
    "Take two placebos, works twice as well." Enzo

    "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

    "If you're not interested in opinions and the experience of others, why even start a thread?
    You can't just expect consent." Helmholtz

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    • #17
      Originally posted by frus View Post
      Most near-class A EL84 push-pull amps have no NFB, apparently beefier OT attributes to improvement of bass response; blues junior has the NFB
      Good point.
      "Take two placebos, works twice as well." Enzo

      "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

      "If you're not interested in opinions and the experience of others, why even start a thread?
      You can't just expect consent." Helmholtz

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      • #18
        Originally posted by Chuck H View Post
        a shift in bias doesn't need to reflect a significant change in current.
        It needs to, it shifts solely because the current through the cathode resistor rises; according to the Ohm's law, the voltage across the resistor has to rise too.
        Look at his picture:



        let's say bias voltage is around -9V, idle plate voltage (in fact the difference between plate and cathode) is 300V. Each tube draws around 32mA. So the total current draw is 64mA; cathode resistor must be 9/0.064 = 140 ohms. We apply the signal to the grids. If the signal's amplitue is lower than 6Vpp, the amp runs in class A (blue loadline); as one tube's current increases, the other's decreases. When the other's tube's current reaches zero, this tube's current is 64mA. But if we apply bigger signal, the current through this tube continues to rise (the other tube's current is still zero), it goes into class B (red loadline), and the current rises to some 130mA max, when it goes into overdrive and cannot supply more current. But as the average current increases, the voltage through cathode resistor must rise too. How much, depends on the slope of the loadline (I drew the usual 8k OT) and idle bias (if it's biased hotter, the blue loadline will go "up", and the difference between idle current and max current will be lower)

        just how the max current depends on the bias type (because it does, I measured more current and output with fixed bias than cathode,), I hope someone will chime in, because I don't know?

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        • #19
          The load line doesn't take into account that the plate voltage is falling at the same time.

          Edit: Your right to call me on the above statement. It should read that increased current through the cathode resistor doesn't need to mean a significant increase in power. As in watts.
          Last edited by Chuck H; 12-19-2012, 12:51 AM.
          "Take two placebos, works twice as well." Enzo

          "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

          "If you're not interested in opinions and the experience of others, why even start a thread?
          You can't just expect consent." Helmholtz

          Comment

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