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

  1. #1
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    How to calculate output power for a push pull AB1 guitar amp

    So, I've gotten to the stage where I want to know more intimately how amps work, and now I'm trying to figure out the calculated output power of tubes in a push-pull AB1 quartet. I've been reading Norman Crowhurst's thoughts on output tubes and power, but haven't found anything yet that makes me feel lucid as far as calculating output power is concerned.

    If I understand the push pull principles correctly and how the curves work on the load lines found on tube data charts, I think the output power for a PAIR could be calculated by taking the change in plate voltage that occurs from the quiescent bias point to the point where bias is 0 volts, and multiplying it by the change in plate current that occurs from the quiescent bias point to the point where bias is 0 volts. Then, multiply it by .707.... .....and that should be the output power for a pair of tubes in AB1. So put another way, it would be (chage in plate voltage) x (change in plate current) x .707.

    Is that correct? Or is there something that I'm missing?

    And then, if there is a quartet instead of a pair, is the output power then just double that of the pair of tubes?

    Thanks for your input,

    Anson

  2. #2
    Senior Member Old Tele man's Avatar
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    ...if you know some of the signal voltages in your amp, try this:

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

    where:
    Po = Power output, Watts(avg)
    Zo' = Effective reflected load impedance, Ohms
    Vg = PI drive signal to each power tube, V(rms)
    gm = Power tube avg transconductance, Ap-per-Vg

    Zo' = Zo*loading factor; Zo = Zpp/4
    ...and the Devil said: "...yes, but it's a DRY heat!"

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    Supporting Member tubeswell's Avatar
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    Heya Old Tele Man

    Forgive my ignorance of US convention in writing algebraic formulae on the net, but what does the "^" symbol mean?

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    Old Timer defaced's Avatar
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    ^ = exponent
    -Mike

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    Thanks for that. I've been searching for the transconductance value for an 3l34 and a 6550 though and have only found values for class A applications, and I'm looking for class AB1 applications. Any suggestions? I think I'll continue looking a little more.

    Anson

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    Supporting Member tubeswell's Avatar
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    EL34 and KT88/6550 datasheets

    Try these:

    http://www.r-type.org/pdfs/el34.pdf

    http://www.r-type.org/pdfs/kt88.pdf

    FWIW found at The National Valve Museum site here:

    http://www.valve-museum.org

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    Thanks for the links. So in trying to calculate output power from a pair of el34's, I did this:

    P output= (p-p effective load)*((g1 volts input*.707)*(transconductance))^2



    P output = 1000*((36*.707)(.011))^2 .......which equals about 91 watts..



    That's a huge value! Larger than I expected acutally. Is that right?

    How does the equation know that I'm interested in a PAIR of el34's and not just one? I'm assuming because if I just had one tube I would use a higher load (say 2000 ohms instead of 1000). Any ideas?


    Anson

  8. #8
    Supporting Member txstrat's Avatar
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    Usually I go to this website http://www.webervst.com/tubes/calcbias.htm fill in one of the forms or look in the tables below for the expected output power.
    No exact values for an amp with ratings in between, though.

  9. #9
    Noodle of Reality Steve Conner's Avatar
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    Quote Originally Posted by anson View Post
    That's a huge value! Larger than I expected acutally. Is that right?
    Anson
    No, because a pair of EL34s can't drive a 1000 ohm plate-plate load. They'll run out of current and the assumption of linearity underlying OTM's equation will break down. Try again with 4k or 6.6k.
    "Enzo, I see that you replied parasitic oscillations. Is that a hypothesis? Or is that your amazing metal band I should check out?"

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    Senior Member hasserl's Avatar
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    Quote Originally Posted by txstrat View Post
    Usually I go to this website http://www.webervst.com/tubes/calcbias.htm fill in one of the forms or look in the tables below for the expected output power.
    No exact values for an amp with ratings in between, though.
    That is for calculating the DC power dissipated thru the tube plates, not the power output, two different things.

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    Steve,

    As far as I know, the value in the equation that old tele man is using, "Zo' = Effective reflected load impedance, Ohms" is referring to the effective load, which is 1/4 of the p-p reflected impedance for a push-pulll pair, so thats why I used 1000 ohms instead of 4000 ohms. Old Tele man, you there to varify this?
    Anson

  12. #12
    Senior Member Old Tele man's Avatar
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    ...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!
    Last edited by Old Tele man; 08-15-2008 at 12:51 AM.
    ...and the Devil said: "...yes, but it's a DRY heat!"

  13. #13
    Senior Member booj's Avatar
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    Measure the signal voltage at the output jack into a resistive load at somewhere between 100 and 500 hz. (I like about 330) and watch an o- scopes picture just before clipping.

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    Senior Member Enzo's Avatar
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    Booj, I believe he is interested in learning how to calculate what to expect coming out of an amp under design. As opposed to just measuring it after the fact.

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    It seems that tele man gave him the info that he needed. However this can only really be used for didactic purposes, or if an amp is intended for operation well within design parameters.

    Between line voltage changes, tube inconsistencies, variations in transformer winds, etc., and the best you'll ever get is a rough approximation of what you'll actually wind up with in the real world. Damn things wind up all over the place really, which is why...i believe...most major mfg's are a little slow to publish hard output numbers. 2x6L6/EL-34's fixed bias = 40 watts, 4x6L6/EL-34's = 80 watts. As long as it's not seriously out of whack the just call it good enough!

    -Carl Z

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    Quote 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.

  17. #17
    Senior Member Old Tele man's Avatar
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    ...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!"

  18. #18
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    Quote 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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