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  • #16
    Originally posted by Chuck H View Post
    It does occur to me that I will need to add another zener to seperate the two -PI swings for accuracy. Otherwise the two halves pass through what amounts to a full wave rectifier prior to clamping making the voltage seen by the zener 1.4X the desired clamping voltage. The -PI swings would clamp before the power tubes were in cutoff.
    You don't need the extra zener. The zener won't charge up to x 1.4 like a capacitor. I thought the tubes wouldn't cut off ernough if the zener voltage was too low but if you also have a zener across the cathode resistor to limit the bias shift and have the second zener (a volt or two higher than the cathode zener) connected to ground rather than the cathode it should work fine.

    Check the plots - 18W stock.pdf, Cathode zener.pdf, Chuck zener.pdf, Cathode + Chuck zeners.pdf
    Last edited by Dave H; 06-30-2011, 10:24 PM. Reason: Plots added

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    • #17
      I'm sorry Chuck H, I think you've lost me with 'using the above #2 circuit, to achieve bias shift clamping AND PI negative swing clamping with a low impedance discharge path during cutoff' and 'Any positive voltage above the zener voltage on the cathode will be present on the grids.'

      Are you just describing the negative going grid voltage excursion on each output valve being clamped via the #2 circuit?

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      • #18
        Dave, thank you for responding. I know where your going but I'm past that. My current circuit is the Paul Ruby clamp for the PI negative swing at a ZV just above bias voltage. The bias voltage is clamped by another zener across the power tube cathode circuit. The two "mods" work very well together. The zener across the cathode circuit aloow me to choose where I want to clamp the bias (which is right at the onset of clipping) and the PR circuit is then set according to that bias voltage so the tubes are sure to be in cutoff when the PR zeners open. I'm not unhappy with the results but I'd like to use my own ideas to get them if I can. Paul Ruby was very gracious to put his circuit into the public domain since it's probably, at this time, the most eloquent way to avoid grid conduction that I know of. Still, there's something gratifying about getting the results you want with your own idea.

        If you look at the circuit you'll see that there is a 0V reference with two out of phase voltages going through a pair of diodes. It's a full wave rectifier circuit. The DC at the end of the two standard diodes will be (roughly) 1.4X of the voltage from either side to the 0V reference. Since these two standard diodes meet at the top of a single zener, the zener will see it's breakover voltage BEFORE I want it too... Unless I add another zener to isolate either side all the way to the top of the power tube cathode circuit.

        trobbins, you have been a great help and support in disecting and evaluating this. Thank you.

        The zener works both ways... Any positive voltage on the power tube cathode that breaks over the zener will be present on the power tube grids so the bias will be fixed at that zeners voltage. Likewise, any negative swing from the PI that breaks over that zener will have a much lower impedance to ground and clamp the swing. Reducing the amount of charging on the coupling caps and allowing them time to discharge. As noted, I will need another zener in the circuit for the reasons stated above. The foreward voltage drop of the standard diodes guarantees the swing voltage will be above the bias voltage before clamping, so the tubes should be in cutoff before any clamping. It should work.
        "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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        • #19
          Hi Chuck H,

          Just to clarify your comments - "My current circuit is the Paul Ruby clamp for the PI negative swing at a ZV just above bias voltage." - I think you mean the zener in your circuit #2 will have a rating of 2ZV (where ZV is the zener clamped voltage level across the common cathode resistor).

          And I think "so the tubes are sure to be in cutoff when the PR zeners open", would be 'when the PR zener conducts'.

          I also still don't appreciate your comments regarding the PR zener conducting in the 'diode direction'. The 'diode' in the PR zener in circuit #2 is back-to-back with the diodes that connect to the grids - so there will be no current ever flowing in that manner (apart from leakage).

          Ciao, Tim

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          • #20
            Originally posted by Chuck H View Post
            If you look at the circuit you'll see that there is a 0V reference with two out of phase voltages going through a pair of diodes. It's a full wave rectifier circuit. The DC at the end of the two standard diodes will be (roughly) 1.4X of the voltage from either side to the 0V reference. Since these two standard diodes meet at the top of a single zener, the zener will see it's breakover voltage BEFORE I want it too... Unless I add another zener to isolate either side all the way to the top of the power tube cathode circuit.
            We are nearly on the same page. I should have posted the circuit I used for the simulation. Here it is - Jaffa20 Chuck.pdf. It's just a slightly different implementation of the Paul Ruby mod. If you disconnect the + end of my VZ2 'zener' and connect it to the power tube cathodes then it is your circuit. It will work fine like that if the voltage of VZ2 is doubled. I've checked by changing the circuit and re-running the simulation with VZ2 set to 35V. The waveform is the same. If an 18V zener is used in my circuit or a 35V one in your circuit then clipping at the PI output will be symmetrical as VZ2 will clamp the -ve peaks at the same voltage as the valves clamp the positive peaks by grid conduction. For your actual circuit you may have to adjust the zener voltages to match your cathode voltage if it is much different from mine.

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            • #21
              Thanks Dave. Like I said, I'm past that. Your circuit is functionally the same as what I'm using. The physical differences are that I'm not using a standard diode in series with the cathode zener and I'm using a standard diode and a zener off each PI output to 0V with the ZV just above the cathode ZV. But functionally it's pretty much the same. Don't it work great. I'm gonna go plug in and crank it for a bit right now since no one's home
              "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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