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The split load cathodyne phase inverter made easy

Discussion in 'Amp Tech Center' started by peteb, Apr 9, 2018.

  1. SSL9000J

    SSL9000J Tele-Meister

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    Outstanding, thank you sir!
     
  2. peteb

    peteb Friend of Leo's

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    this is tube operation made simple:


    AC:


    ground-cathode resistor-12ax7-plate resistor-ground





    the signal on the plate and cathode develop as equal AC current flows thru the resistors.


    if the resistors are equal the signals will be equal.


    by symmetry, one can see that the plate and cathode will develop opposite +/- voltages.







    this is meant to be accessible to people no matter their level in electronics.








    if it doesn't make sense, please ask questions.
     
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  3. SSL9000J

    SSL9000J Tele-Meister

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    I'm still confused about the concept of ground. I thought it had to do with the path of least resistance/equal potential sought by the charged electrons. But I'm totally misunderstanding the idea, because if that were the case, they wouldn't flow both out of and into the ground. Meaning there must exist a potential difference between the plate resistor ground and the cathode resistor ground. Gonna go scratch my head and read Electronics One-Seven some more.
    Thanks for your help!
     
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  4. RLee77

    RLee77 Friend of Leo's

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    Don’t worry, you’re not misunderstanding the idea, it’s the above idea itself that’s a misunderstanding.
     
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  5. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    I would say that it is probably not right for you to be confused about the concept of ground, it is just that a doctrinaire and confused concept has been present in many post in this thread.

    The concept that AC comes out of ground, and goes back there, the way it has been represented below in the quote, should confuse any rational thinker.
    That the current flows in a loop circuit is unargued, and that at one point in this circuit a connection to a common ground - that could be seen as a reference point - is also not a problem.
    So it would be fair to say it flows from this reference point, around the circuit, untill it reaches this point again, so - from & back to - but not - into & back out of.

    An AC source, one that may well be referenced to ground is missing from the

    circuit. And this is confusing.

    What the circuit might be better of showing is;

    (Ground reference point) - cathode resistor - (12AX7 cathode - AC modulations source at grid - 12AX plate) - plate resistor - (AC short circuit to ground reference point ) - DC source - (ground refence point.)

    But this is not simple, this is too simplified - not sufficient enough to be thorough, not really helpfull, and not an explination that I would choose to use other than maybe to help towards dispelling a missconception.

    But most of all - it has almost nothing to do with helpfully explaining how a cathodyne circut works!

    I hope this helps to appease your confusion.....though probably not;):twisted:.
     
    Last edited: Apr 27, 2018
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  6. SSL9000J

    SSL9000J Tele-Meister

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    "You never want to be the smartest person in the room." - unknown. I certainly qualify.
    From the good ol' Radiotron Designer's Handbook, 3rd edition, (because a picture is worth a thousand words,)
    We have AC modulations, which I'm assuming are our guitar signal, and DC bias both seeking a ground reference point. The coupling caps on the cathode & plate are blocking the DC, therefore the resistors become the path of least resistance to ground; somewhat ironically I might add. The AC gets filtered through the same coupling caps, and a portion of it passes through to the next stage, in this case the power tubes. The lower frequencies cannot pass through the caps, similarly to DC. Thus R1 & R2 become the path of least resistance, once again, leading to ground.
    Am I even close? Or is every word wrong?

    Thanks again!
     
  7. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    "If you think you have learned all there is to know, you haven't learnt enough." - even more unknown. I humbly agree.

    One would hope that looking at the circuit diagram should negates the need of trying to explain everything in a simple manner using words.

    I would say that you seem to be pretty close, using the loose framework of the unspecific schematic and the basic discussed concepts

    I find this following post sort of said everything that the thread title requires, and that the whole ground issue is somewhat irrelevant.
     
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  8. peteb

    peteb Friend of Leo's

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    RLee, I thought you got it.


    What are you not getting?



    Why don't you explain yourself by explaining your comment.


    That would be a common courtesy of a responsible poster.
     
  9. peteb

    peteb Friend of Leo's

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    This is a great question and deserves discussion. Thanks
     
  10. peteb

    peteb Friend of Leo's

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    Thanks Bendyha. That helps me to see where you are coming from. The path you describe above seems a lot like mine. Are we in disagreement? It does not look like we are in disagreement to me.
     
  11. peteb

    peteb Friend of Leo's

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    Consider the basic facts


    The signal voltage at the plate develops because signal current ( AC) flows thru the plate resistor. Ohms law applies here.

    Like wise


    The signal voltage develops at the cathode in the same exact way.


    The AC signal current flow in the cathode resistor is the same signal current flow in the plate resistor.

    The current law holds true here. All of the AC current thru the cathode resistor flows thru the tube and thru the plate resistor, there are no alternate paths or turn offs in this part of the circuit.


    Basically, there is equal AC current flow at both ends of both resistors. There is no other way about it. That is final and that is a fact. If anyone wants to disagree with that fact, that would be a good point to bring up, if anyone wanted to.
     
  12. peteb

    peteb Friend of Leo's

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    The cathode resistor connects between the cathode and ground, this is agreed.



    The plate resistor connects between the plate and a B+ node.



    One of the significant results of this thread is the understanding that the b+ nodes are an effective ground for AC.

    It is evident that the B+ nodes remain at or very close to zero volts AC at all times, thru out all playing conditions.





    If anyone is in disagreement that B+ nodes are ground for AC, then it would be good to bring that up.







    What is ground? Zero voltage potential and BOTH an endless source of electrons AND a place that will accept endless amounts of electrons.
     
  13. peteb

    peteb Friend of Leo's

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    All of the pieces are in place, AC current is flowing in and out of ground at the non tube end of BOTH the plate and cathode resistors.





    Then what remains is what is causing current to flow from ground, away from ground, and then back to ground. Electrons wouldn't do this on there own without a voltage potential to pull them or something, some reason for the current to flow.

    Why does the AC current flow thru the tube?






    Well, that's what a tube does. It is an AC current generator.
     
  14. peteb

    peteb Friend of Leo's

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    The tube generates the signal current(s) that flows thru the plate and cathode resistors.



    The signal voltages that develop on the plate and cathode result from signal flow outside the tube, that was generated inside the tube.



    That's tube ops!
     
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  15. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    AC current is flowing..out of ground
    AC current is flowing..out of ground
    AC current is flowing..out of ground

    No...don't see it. Don't you mean the magic smoke ?

    upload_2018-4-28_18-11-36.png
     
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  16. SSL9000J

    SSL9000J Tele-Meister

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    If I may use an analogy with a common metaphor for current flow in an electric circuit:
    Imagine a trough, with a bit of water in it and an X marked halfway from one end to the other. Tilting the trough one way causes the water to flow into one side of the "X", and out of the other. Tilting it the other way causes the water to flow in the opposite direction relative to the "X."
    The X represents the PI tube, the ends of the trough are ground, tilting it back & forth represents alternating current and the water is, of course, electrons. They're flowing back & forth through the tube with equal and opposite amounts of current. (More or less.) Thus a positive current through the plate resistor causes a negative current through the cathode resistor, 180 degrees out of phase, and vice versa. Couple those to the input grids of the power tubes and voila! Push-pull amplification.

    I think I'm starting to get it.
     
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  17. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    AC current is flowing in and out of ground....you say.

    So, do you also think it would be right to say;

    DC current is flowing in and out of ground....as well ?

    Would this not then be the same ground that the AC and DC is flowing into and out of ?

    How does that work? Would not the AC flowing out of ground sometimes get confused and flow up into the wrong resistor.....one that DC is meant to be flowing into?

    Then the DC would get all ...whats the word...ACIDIC...ACNED ?
     
    Last edited: Apr 28, 2018
  18. RLee77

    RLee77 Friend of Leo's

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    I tried, in several posts, #59 and #80 for example, and I didn’t feel the need to repeat myself.
    I respect your quest for knowledge, but I would offer that to be a “responsible poster” (your term), one would, after taking some measurements and devising a personal theory of operation, validate your theory with the abundant articles and principles available from proven texts, before posting your personal theories as fact.
    That’s the reason I posted again — to prevent confusion for others.

    The hands-on approach that you are doing is a great way to learn, however.
     
    Last edited: Apr 28, 2018
  19. RLee77

    RLee77 Friend of Leo's

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    No... a tube is not a generator or source. It just varies its conductivity when a grid signal is applied, allowing more or less of the b+ current to flow to ground. Remove the b+ and no current flows, even with signal on the grid. The source is b+, and current from a source node does not flow back to itself, which is what you've postulated.
     
    Last edited: Apr 28, 2018
  20. LudwigvonBirk

    LudwigvonBirk Tele-Holic

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    Word!
     
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