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Trying to make a more accurate bias chart

Discussion in 'Amp Tech Center' started by peteb, Jul 17, 2017.

  1. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    One of many moot points one could have chosen to clarify, but as I have pointed out, he is taking a rectified view of AC amplification. But never having come across it before in the 125 years of electronic development that I have read up upon, I have decided to opt out. I think with this halfwave, positively oriented amplification concept that bias might be .75V, but now that things are being refered to DC and in RMS (new buzz word) I couldn't say what that might be.




    Of course the major factor not yet mentioned is the efficiency of phase inverters to do the top half descending of the swing with more POWER than the bottom half trying to ascend.




    But being a Fender, where the tubes are mounted hanging upside down, then of course the lower half of the wave is now the top, and so the lower (now higher) wave has the greater


    inetia of Conduction Angle on the falling slope.




    This is then not a positive and negative halfwave thing, but a shift of phase to the ascending/descending halves of the wave form.




    True balance of phase inverter load is then naturally only obtainable with the tube in a horizontal orientation, being sure to keep the two sides of the triode also horizontal -this


    more of a long-tailed-pear anomally, as the cathodyne splits on one side of a double triode only - this one would then naturally have to be lower than the other side, that would be driving it.:twisted:;)
     
    Last edited: Jul 26, 2017
  2. peteb

    peteb Friend of Leo's

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    Thanks Bendyha, I lot of your answer goes over my head, I try to keep things simple.
     
  3. peteb

    peteb Friend of Leo's

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    Thanks Rob, I can't argue that, but what about the 6G2? How much of that 56.5 cathode voltage is sneaking around the side of the tube up to the grid? See that 55 making its way to the grid?


    Reading and knowing the AC on the grids is really telling about tube operation during playing conditions. None of the main amp sites have this covered at all Rob, you could lead the way and be the first.




    Back to the 6G2's PI and reading the DC off of the schematic and using it to infer the AC audio signal on the grids.




    I have only measured how far the power tubes are opening, but what I consistently see is the max signal voltage reaching half+ of the bias voltage, meaning that the power tubes at full volume, full signal are opening up 50% or a little more. Applying this to preamp tubes, the typical preamp tube bias is 1.5-2.1 volts, so it would be reasonable to assume that the max signal voltage is around 1 volt, a very convenient number , especially when one wants to consider gain. These typical preamp tubes have +2 volts on the cathode and zero volts on the grid for an overall bias of -2v DC. This is what Rob, Bendyha, and Clint are all agreeing is typical.



    Now look at the 6G2 schematic. And at the PI tube in particular. The cathode voltage is not 2 volts, it is 56.5 volts, quite different. Look at the grid, there is no voltage called out. If this were a typical cathode biased stage, the grid should have no DC volts, which would make the bias for this Cath PI stage -56.5 volts, quite different than the 1.5 volts you three say is typical for Cath PI. If we are discussing the 6G2 cathodyne phase inverter, the observation the three of you are making appears non relevant.


    Back to the 6G2 and its cathodyne PI. I've measured the DC volts on the grid of a 6G2, it measures 23 volts. I wasn't surprised. LTPI have more elevated voltages on the cathode and the grid than this, so what's the big deal?
    The 55 volts on the node near the cathode, partially gets on the grid. I also measured the cathode voltage on my 6G2 to be 74 VDC (+31% of schem). The bias is then 23 VDC - 74 VDC = -51 VDC.

    Taking it to the next step, the AC signal voltages are not on the schematic, but if this PI stage has a 51 volt bias, my standard line of reasoning, is that a good target for a max signal voltage would be half of this, 51/2= 25 VAC.
     
  4. peteb

    peteb Friend of Leo's

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    Rob, or Bendyha, or Clint, can any of you show me one cathodyne PI with a bias voltage of 1.5 v?

    The cathodyne PI used on 5E3,6G2,BF Princeton, BF Princeton reverb all have elevated cathode voltages in the range of 50 volts.


    https://robrobinette.com/How_The_5E3_Deluxe_Works.htm
     
  5. clintj

    clintj Friend of Leo's

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    The 6G2 PI grid will have 55V on it at idle. If you meter the grid directly, your meter provides a leak path to ground which pulls down the voltage to that 23V number you see. That's why the specified test point is at the junction of the 1M and 56K resistors.

    A roughly center biased triode should show virtually no grid current, therefore there should be no voltage drop across the 1M grid leak resistor normally.

    The bias voltage of a tube is the difference between grid and cathode, not the cathode voltage only. There's old amps with grid biased preamp stages that show 0V on the cathode, and it's also possible, but uncommon, to run a power tube as mixed bias (a combination of lower cathode voltage from a smaller bias resistor and a fixed negative voltage supplied to the grids).
     
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  6. robrob

    robrob Poster Extraordinaire Ad Free Member

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    The reason I mentioned the typical bias voltage was to make the point if you don't understand the difference between cathode and bias voltage maybe you shouldn't be trying to develop new tube amplifier theory.
     
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  7. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    You can easily obtain a 1/2 watt, 1 gigaohm resistor to make a high impedance voltmeter probe, and then measure in millivolts.(Assuming you have a 1M input imedance now). This will enable you to make fairly close (0.4% off) readings across a 1M resistor without loading it down, and your false 23V will then probably read as 55mV ( x 1000 = 55V)
     
    Last edited: Jul 27, 2017
  8. RLee77

    RLee77 Friend of Leo's

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    That's 23 Heisenvolts. :twisted:
     
    Last edited: Jul 26, 2017
  9. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    Not to be confused with parallel leek.........................................................or a meter leek.
    upload_2017-7-27_10-39-28.png upload_2017-7-27_10-44-32.png
     
  10. peteb

    peteb Friend of Leo's

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    And once again, I've said I'm not making any new theories, I'm just taking what's out there and refining it, by looking at things with more detail than is usually done.
     
  11. peteb

    peteb Friend of Leo's

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    Good info there Clint, thanks.


    I still have a little problem with it however, a question.


    If 6g2 cathodyne PI has a bias of 1.5 volts, limiting the input audio signal volatage to 1.5 VAC or less, and the PI sends 20 VAC of audio signal to the power tubes. How does the PI induce a loss as cathodyne PI are known to do?



    The workings of the cathodyne PHase inverter is not the emphasis of this thead. The purpose of this thread is to try to quantify how much a tube "opens up" in operation. That goal of the thread was successfully accomplished by measuring the audio signal on the grid and dividing it by the bias voltage, which appears to be the most accurate way to quantify how much a tube "opens up".


    How did the cathodyne PI become involved. I was pointing out that bias voltages on tubes give information as to what to expect for the magnitude of the AC signal entering the tube. I was just using the cathodyne phase inverter as a place to apply this method, which was interesting because using this method could possibly explain the unusual behavior of the cathodyne PI, which is the only tube stage known to induce a loss.



    Maybe I'm wrong, the cathodyne PI does have a small bias like the other preamp tubes? Maybe the cathodyne phase inverter does NOT induce a loss as a lot of people seem to believe.


    I think either, the cathodyne PI has a significantly larger bias than other preamp tubes, or the cathodyne PI actually does NOT induce a loss. I think one or the other has to be true. The cathodyne PI can't have a small 1.5 bias AND induce a loss, it appears that could not possibly happen.
     
  12. peteb

    peteb Friend of Leo's

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    Looking at the 6G2 schematic, I agree the 55 minus the 56.5 gives a 1.5 volt bias, even though it's hard to measure. No current in the grid resistor, is a requirement for this, fine.



    Like I said in the last post, I don't see how the cathodyne PI induces a loss, it actually grows the voltage just like a good pre amp tube should.
     
  13. clintj

    clintj Friend of Leo's

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    Well, it shouldn't be too hard to measure the AC voltage into and out of the cathodyne and compare them.
     
  14. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    Fact : Cathodyne phase inverters will ALWAYS have less input than output !

    Fact : The bias voltage on the 6G2 phase inverter is 1.5V.

    Fact : The diference in voltage between the control-grid and the cathode is the bias voltage.

    Fact : To get 100V swing out of a phase inverter, you have to put 100V swing into the control grid, plus the bias 1.5V. So an input of 101.5V. Thus a loss of 1.5V.

    Fact : You still don't understand the relationship between input, output & bias.
     
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  15. robrob

    robrob Poster Extraordinaire Ad Free Member

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    Pete, this thread is like your "Best 2 Prong Cord" thread where you tried to convince us a two prong power cord plus a death cap is safe.

    I know this sounds harsh, but I want readers of this thread to know I disagree with about 95% of what the op has posted in this thread.
     
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  16. peteb

    peteb Friend of Leo's

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    Thank you Clint, excellent suggestion. Actually, I just did that.


    But thank you


    That is a constructive comment, I would like to hear more of those.
     
  17. peteb

    peteb Friend of Leo's

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    Thank you Bendyha, but there is something wrong in there, and I'm about to figure it out.



    Your wrong
     
  18. peteb

    peteb Friend of Leo's

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    Congratulations Rob, you just expressed an opinion.



    You are not backing up your words at all, nada!


    You haven't even tried to assess my analysis of the openness of the tube by studying the relationship between grid signal AC voltage and cathode DC voltage during playing conditions. I think it is pretty solid and I think you will have a hard time disproving even a little part of it.



    I see you give out helpful advice to other tdprier's, but when I post, you do your best to try to work against me as best as you can. I'm fine with that, because when you try to disprove anything I say, it's actually very helpful, so I thank you!




    My results about the openness of the tube is 100 percent based on measurements taken under playing conditions, on four different amps, there was 100 distinct data points in all.
     
    Last edited: Jul 27, 2017
  19. clintj

    clintj Friend of Leo's

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    Ok then. You've built a theory upon several wrong assumptions regarding tube operation, rejected several attempts to provide you with reading material and widely acknowledged sources, and greeted any contrary hard evidence with condescension. Now you're arguing with two respected members, one of which has freely provided a wealth of info and knowledge that has helped with several members' builds and designs because they were trying to help you by pointing out problems with your interpretations and you don't seem to like your preconceptions challenged.

    I'm out.
     
  20. peteb

    peteb Friend of Leo's

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    the cathodyne phase inverter is enigmatic, and POSSIBLY misunderstood by even the experts.


    It starts with the DC bias voltage on the grid, it is noticeably absent from the schematic.


    As Clint points out, what you measure is not what you get because taking the measurement affects the result, but that does not keep people such as Weber, showing that voltage on a voltage chart.




    I measured nearly equal audio signals, VAC, on the anode, the cathode, the grid, and even the node inbetween the cathode and the grid. They all seemed to max out around 9 VAC, with the volume of the amp halfway up. This would correspond to a max signal voltage, being fed onto the grid, of around 18 VAC. I predicted above, in post 103 that this would be around 25 VAC.

    What I learned?

    Just like when I was studying the power tubes, measuring the audio signal during playing conditions was extremely enlightening, if you want to see what is happening inside the amp, Or the tube, examine the signal voltage and the bias voltage during playing conditions, you will see the tube doing its tubing thing.



    Does the cathodyne phase inverter insert a loss, a gain, or neither into the overall amplifier system?

    Judging from the signal voltages entering and exiting the PI tube, the gain appears to be close to unity, which is a gain of one, which means no gain.

    This result is consistent with the expectation that the cathodyne PI is not a source of gain.




    What does this say about the idle bias of the cathodyne phase inverter? If the audio signal appearing on the grid, which affects the operating bias of the tube, is greater than the idle bias, then the portion of the signal that is larger than the idle bias, will not open the tube any further, meaning that this part of the signal does not get amplified.


    I order to amplify the full signal, the max signal Needs to be less than the idle bias V.



    Applying this to my measurements, the idle bias voltage needs to be more than around 18 volts or the amp ain't gonna work.




    It would be really interesting to see voltages from other people so we can compare.


    If you don't have voltages to share, please don't try to be overly critical of my results



    Thank you
     
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