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How to calculate output power for a push pull AB1 guitar amp

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  • #16
    Originally posted by Old Tele man View Post
    ...the "effective" load impedance (Zo') is the combined affects of OT reflected impedance (Zoo/4) and the combinational loading of the tube(s) itself (also called 'load factor'):

    Zo = Zoo/4 <---purely an OT and speaker function

    % = (rp/(rp+Zo))^2 <---'load factor', a tube & OT function

    Zo' = Zo * %

    ...thus the "full" original equation is:

    Po = (gm*Vg)^2 * (rp/(rp+Zo))^2 * Zo

    ...which simplifies to:

    Po = Zo' * (gm*Vg)^2

    ...by letting Zo' = Zo * (rp/(rp+Zo))^2

    ...and, remember that the Vg signal is assumed to be an RMS-value, not a peak value!
    Google isn't giving me much help in finding information on combinational loading. Anyone care to explain what it is? Wild guess: Is it the load that the oppsite tube puts on the working one?

    Is this the reason that if you take a 50W P-P pair amp and simply parrallel two tubes to make a quad, it will still be a 50W amp? If you ignore the combinational loading part of the equation: double the tubes, double the transconductance, double the power.

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    • #17
      ...go visit PENTODE PRESS and look at the 'Vacuum Tube Archeology' postings...there's an article explaining what "loading factor" is and which old college textbook you can find the source information in.
      ...and the Devil said: "...yes, but it's a DRY heat!"

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      • #18
        Originally posted by Ptron View Post
        Is this the reason that if you take a 50W P-P pair amp and simply parrallel two tubes to make a quad, it will still be a 50W amp? If you ignore the combinational loading part of the equation: double the tubes, double the transconductance, double the power.
        Well, I'm trying to do the numbers but I must be screwing something up. I'm getting ~69W with two EL34s, Vg=36V-pk. Then when I add a pair of tubes ( double gm, halve Rp) I get 244W!

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