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excessive plate dissipation causes redplating - does this power dissipation from AC or DC?

Discussion in 'Amp Tech Center' started by peteb, Jul 20, 2018.

  1. robrob

    robrob Poster Extraordinaire Ad Free Member

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    pete, I think the root of your confusion on this issue is your thinking in terms of AC and DC when in reality a power tube is putting out a varying DC signal. The output transformer turns the varying part of the DC signal into AC--but that comes later.

    When discussing power tubes just think in terms of varying DC and it's much simpler to understand what's going on. The electrons slamming into the plates to cause them to glow red are ALL DC, so red plating is caused by DC current, not AC + DC.

    Every electron that hits the plate causes a temperature rise which ends up as wasted heat. It doesn't matter if the electrons passed through the control grid at idle or with a guitar signal on it.

    BAM!

    Next peteb thread please :D
     
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  2. LudwigvonBirk

    LudwigvonBirk Tele-Holic

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    beware the PSYOP.
     
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  3. Old Tele man

    Old Tele man Friend of Leo's

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    An excellent, easy-to-understand, technically correct analogy: delta(▲)-DC. Sometimes we ( I? ) get too pedantic...
     
    Last edited: Jul 24, 2018
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  4. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    upload_2018-7-24_22-49-50.png
     
  5. Mr BC

    Mr BC Tele-Meister Gold Supporter

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  6. Old Tele man

    Old Tele man Friend of Leo's

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    Yeah, pedagogues do have a tendency to be pedantic.
     
    Last edited: Jul 25, 2018
  7. TeleTucson

    TeleTucson Tele-Afflicted

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    I thought we'd get back to this at some point. :) It's of course still possible (and often extremely useful) to decompose signals into AC and DC components regardless of offsets, etc. And for power delivered to a simple passive load it is often useful to decompose the power into DC power and RMS AC power, etc., and for communications engineering and noise analysis this is pretty essential. But to get at power dissipation in a tube, where the voltage that the current drops across in the tube is *not* given by a passive impedance multiplied by current, this decomposition is not the most graceful way to figure out power dissipation. I think this is where Pete is getting hung up.

    You can't really figure out the plate dissipation without load lines, signal power being delivered, etc.

    Pete, here is a nice (constructive) description which might be helpful, -

    http://www.unclespot.com/moreBIAS.html

    But
     
    Last edited: Jul 24, 2018
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  8. peteb

    peteb Friend of Leo's

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    Thanks Rob, I agree.


    I don’t think you are disagreeing with me.



    I said redplate energy is DC energy.


    I think you agree with that. No?



    You said, the AC is later.


    That is 100% what I have been saying:



    The pt apples DC to the tube, the tube then apples AC to the load, the OT.






    How about that Rob? we are in agreement.



    Elpico also said it is DC. All three of us agree. We could probably say that we have reached a consensus.


    It wasn’t that hard of a question. Look up AC plate dissipation in audio tubes.


    It doesn’t exist.
     
  9. peteb

    peteb Friend of Leo's

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    Thanks for this.

    This is what I have been saying.



    Your post confirms what I was leaning toward, DC current flow causes plate dissipation and redplating, AC current flow does not.



    Thanks for posting and I am glad that we are in agreement.
     
    Last edited: Jul 24, 2018
  10. peteb

    peteb Friend of Leo's

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    This is good.
     
  11. TeleTucson

    TeleTucson Tele-Afflicted

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    But with a load on the tube, on the average the electrons won't hit the plate as hard (and thus less heat) when there's signal present, noticeable mostly as you approach clipping ... :)
     
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  12. peteb

    peteb Friend of Leo's

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    I like this because we can use this description to explain why AC current doesn’t add to plate dissipation and DC current does. And it’s irregardless of whether the energy is put on the tube or comes from it.




    AC current component flows from ground, thru the cathode, the grid, plate and beyond, the same exact current path as the DC current flows thru the tube but yet the DC contributes to heat dissipation and redplate condition of the plate and AC does not. Why?


    Using Rob’s electron collision model description:


    At idle the DC electrons hit the plate at 60 mph, at signal pulse 70 mph, at an anti pulse 50 mph.



    IF the signal is symmetrical between positive and negative, class A, the speeding up and slowing down will have no effect on current flow. But if the signal is non symmetrical (class AB) due to cut off, the current increases and then more DC energy gets transferred. That’s the connection I see between AC and plate dissipation. AC in class AB causes more DC pLate dissipation.



    I’m not sure.
     
    Last edited: Jul 24, 2018
  13. LudwigvonBirk

    LudwigvonBirk Tele-Holic

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    Good #152 post.
     
    Last edited: Jul 24, 2018
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  14. TeleTucson

    TeleTucson Tele-Afflicted

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    OK – this is usually described with some simple trig and integrating over a cycle (averaging), but I’ll go the route of “no math”.

    The dialog immediately above would suggest that AC “averages out” for heating and makes no difference. I tried to comment on this earlier, and it’s not quite the case. Here’s why (no math, no load lines, no RMS, no integrating sin^2 over a cycle, etc., etc. -- just words):

    The cyclical variation of tube current when a signal is present means the average tube current is about the same as the DC. When the average tube current is unchanged, the voltage drop across a load from the power supply is also the same on the average, so the tube voltage is the same – on the average – as the DC value. This is what Pete measures. And it’s not because he has a bogus “non-RMS” voltmeter, or is not using an oscilloscope.

    However, here’s the thing. When the there is a significant AC signal and tube current is at a maximum in an AC cycle, because of the increased "IR" voltage drop across the load at that part of the cylcle, the tube voltage drops during that part of the cycle to its low value. Yes - it will still continue up and down and be the same as the DC on the average, etc., etc., but the issue is how many electrons see lower voltage, and how many see higher voltage. Similarly, when the tube current is at a minimum in the AC cycle, the voltage drop across the load is a minimum and this means the voltage on the tube is at a maximum. Remember what current is - it is the number of electrons in a time interval (i.e., a portion of a cycle, for example).

    So if you think about this, this means that – on the average – when there is a signal that results in an AC cycle of the voltage (on top of DC, so yes, Robrob, it is still net flow in one direction, OK, OK, ….), more electrons traverse the tube when the voltage on the tube is at a minimum of the cycle than traverse the tube when it is at a maximum. Because the voltage electrons see when they traverse the tube translates into their energy on impact at the plate, and the heat they produce on impact at the plate, this means that when the signal is present, even though the AC current variation averages to zero, there is less heat generated on the plate. Oh, and guess what - the delta in power is driving the load rather than heating the tube. Energy is conserved and thermodynamics is satisfied, amazing ....

    See - no math. But this means that the more signal, the less redplating. (And yes, this discussion of load voltage drops ignores phase lags due to complex impedance, and was simplified to the case of a dummy resistive load R for the purists out there, and also ignored inherent tube nonlinearities, etc., etc.).

    Go ahead & have at me. Tell me I'm all washed up, I can take it! :)
     
    Last edited: Jul 25, 2018
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  15. peteb

    peteb Friend of Leo's

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    This sounds good, but why exactly is there less heat generated?





    “the delta in power is driving the load rather than heating the tube.”




    You aren’t saying dissipation is less because more input energy is transformed to audio power. That has been floated around by a few people.



    THe DC energy at the plate is more important to me than the AC energy at the plate.


    The DC energy is the input energy to the tube. The AC energy is not the input energy to the tube it is the output energy of the tube.


    The DC input energy is used in efficiency calculations, bias calculations and redplate calculations.



    AC energy on the plate is not used for bias calculations, not used for redplate calculations, and will only work as output energy in an efficiency calculation, it is not the input energy.
     
    Last edited: Jul 25, 2018
  16. TeleTucson

    TeleTucson Tele-Afflicted

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    Just take your time, read it again, more carefully. When the signal is present, there are more electrons per unit time (averaged over a cycle of the signal) that have low energy, and fewer that have high energy. It's their energy when they slam into the plate (which is given by the time-varying tube plate to cathode voltage at the moment they accelerate in their flight through the vacuum) that makes the heat. So signal means less heat. I think this is what this entire thread has been trying to get at.
     
    Last edited: Jul 25, 2018
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  17. peteb

    peteb Friend of Leo's

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    We were already trying concepts like this, but I don’t think you are talking about this.



    I will read slower.
     
  18. peteb

    peteb Friend of Leo's

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    I reread the first passage and second.


    I can’t follow the reason for less heat with more signal. Other than converting more energy to audio power and less to heat. This really helps to balance energy equation but otherwise I am skeptical.

    Or just based on kinetic energy. Lower speed, less heat. What causes the lower speed, the signal pulling down the plate voltage?
     
  19. TeleTucson

    TeleTucson Tele-Afflicted

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    Yes, dynamically during a cycle. But you don't see the change in voltage in the DC because it averages out. However, when the AC signal is present, more electrons made their journey from the cathode to the plate at low voltage than at high voltage. So less kinetic energy banging into the plate on the average when the signal is present.
     
  20. peteb

    peteb Friend of Leo's

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    This is confusing.


    More electron flow when what is at low voltage?

    More electrons flow with a positive grid, which is when the plate is at it lowest, but the screen does the drawing when the plate goes low.
     
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