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Calculating gain of class A and class B region of class AB output stage

Discussion in 'Tube Audio' started by Alan0354, Dec 29, 2017.

  1. Alan0354

    Alan0354 Super Member

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    Hi

    I have been working on the gain calculation of a tube class AB output stage and I cannot proof that the gain is equal when the stage is in class A region and class B region. This is my work:

    Gain class A and B.jpg
    You can see the gain of the two region is different.

    For class A region, I use Fig.33b in p437 of this article:https://pearl-hifi.com/06_Lit_Archive/07_Misc_Downloads/Push_Pull_Theory_MIT.pdf

    For class B, I just use the fact the one tube is off, half of the primary winding is out of the picture and the primary impedance see by the single valve is Rpp/4 and calculate the gain as shown.

    Please comment on the derivations.

    Thanks
     

     

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  2. kward

    kward AK Subscriber Subscriber

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    Are the gains supposed to be equal? My intuition says they would not be equal because the slope of the load lines in class A and class B portions are not the same. For the gains to be the same, wouldn't the slopes of the load lines need to be the same?
     
    Last edited: Dec 30, 2017
  3. Alan0354

    Alan0354 Super Member

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    Hi Kavin

    Yes, the slope of class B is steeper as shown in the Valve Wizard article, but in Class A, both tube is working, that's why you see in Fig.33b in the article I linked shows the Thev voltage source is 2ueg.

    Valve Wizard1.JPG



    I am not sure the gains are supposed to be equal, BUT a lot of old harmonics will be created if the two gains are not the same. I cannot imagine they are not equal. But for the life of me, I cannot get the two conditions to produce the same result.

    Thanks
     
    Last edited: Dec 30, 2017
  4. thorpej

    thorpej Super Member

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    Are you talking about voltage gain? It's clear from the a load line drawn on composite characteristics curves that the Class A region has more voltage gain than the Class B region. But I always understood that the reason for using AB operation was to allow you to scoot the quiescent operating point to the right and allow you to have the resulting total peak-to-peak voltage swing without melting the tube. But because nothing is free, it comes at a cost -- higher distortion (which is all odd harmonic because the even harmonics disappear in the transformer) that you have to clean up with some amount of negative feedback (pick your favorite flavor).
     
  5. PakProtector

    PakProtector AK member

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  6. thorpej

    thorpej Super Member

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    Yes, we know what they mean. But when discussing AB, it's handy to refer to "the A portion" and "the B portion" because it's a lot shorter to write than "the portion of the cycle where both halves of the circuit are conducting" and "the portion of the cycle where only one half of the circuit is conducting" because, for all intents and purposes, those portions of the cycle behave exactly as the corresponding portions of the cycle in Class A or Class B operation.

    No need to get your knickers all in a twist, geez.
     
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  7. Alan0354

    Alan0354 Super Member

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    Hi Jason
    Yes, I am talking about voltage gain. That is what I see in the equations that gain is always higher in class A region than in class B region.

    Can you show me how you read the load line drawn on composite characteristics curves that class A region has more voltage gain than class B?

    Thanks

    Alan
     
  8. Celt

    Celt Peanut Head Staff Member Super Mod

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    Alan and Douglas....please quit the quarreling and following one another's threads. This has GOT TO STOP!
     
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  9. thorpej

    thorpej Super Member

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    The way I’ve always done it: because the A lines represent where both halves are conducting, you can consider those to be overlapping... so snip the lower half AB line using your favorite Photoshop-alike, and scoot it up vertically only so they overlap... and you then have the full “squashed lightning bolt” shape of the full 360 degree cycle.

    Is that what you mean?
     
  10. thorpej

    thorpej Super Member

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    Alan, as far as voltage gain is concerned, I would expect the voltage gain to be higher in the A region because of the higher impedance, just like in any other garden-variety voltage amplifier.
     
  11. thorpej

    thorpej Super Member

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    That said, depending on what you’re trying to do, just drawing the upper 180 degrees might be sufficient. For example, if you draw just a class B line, you will know the peak-to-peak voltage swing, and armed with that and the turns ratio of the output transformer, you can make a fairly good estimate of max power output (though it almost always ends up being a little lower because we don’t have frictionless pulleys).
     

     

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  12. kward

    kward AK Subscriber Subscriber

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    Here's my reasoning. Clearly when the amp is at full power, the load the "on" tube sees reflected through the transformer is 1/2 of what both tubes see at lower power where both are conducting. So it seems clearly black and white to me that because the reflected load changes every cycle at max power output, that the gain of the stage changes every cycle as the output stage crosses over from both to one tube conducting.
     
  13. maxhifi

    maxhifi AK Subscriber Subscriber

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    Hi Alan,

    I had a look at your calculations, and following your logic, the only reasonable conclusion is that as the amplifier transitions from 360 degrees of conduction to 180 degrees, the gain should drop by about half, given the high internal resistance of a beam power tube. (i.e., the step change from A= gm * Rpp/2 to A= gm * Rpp/4)

    If this actually happened in practice, the performance curves which are shown on tube data sheets would make no sense, and an amplifier without massive negative feedback would work poorly.

    One thing the equations don't consider is that gm is not constant, and another that the equation you're using for gain of a pentode is always given as being valid for small signals, but never as a large signal model. Another thing to consider is there is no sharp transition from A to B, but rather a gradual one. I have been really busy first with work, then with Christmas, but I plan on having a closer look at this when I have the chance. I feel like something is being left out of this analysis, although I can't put my finger on what it is.

    -Max
     
  14. Alan0354

    Alan0354 Super Member

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    Hi Jason

    I went back and look at the composite load line on the KT88. I take the class A load line of the top and bottom graph ans "SUM" them together. That is the Itop - Ibottom

    You can see I drew the current summing line, it looks like a straight continuous line connecting the top and bottom composite ( class B ) line to become a straight line. After summing, there is no "lightning bolt" shape.

    But the straight composite load line really does not imply no distortion. That has to be calculated from the load line intersecting the plate curves.

    Composite class A line summing.jpg
     
  15. Alan0354

    Alan0354 Super Member

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    Hi Guys,

    Thanks for putting up with this. What I am addressing is a separate thing from distortion of the tubes, it is on top of the tube distortion. Changing gain at crossover point will actually change the wave form and it is big. From calculation, the two gains are not close. I'll post with some numbers after I worked it out.

    This drawing is what I am concern with. In the drawing, the gain in class A region is 2, gain in class B region is 1. I drew a triangle wave as input to make it easier to see. The green is the output. You can see it's big distortion even if you assume the tube has no distortion:

    Tube cross over.jpg

    It WILL look like this even though the tubes has no distortion if the gain is different.

    But if you look at the output waveform of PP tube amp, the sine wave looks reasonably good. So something is not adding up.
     
  16. PakProtector

    PakProtector AK member

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    Maybe if you can answer the following, something will 'click' and/or make sense.
    1. Why have you justified the gain is 1 or 2?
    2. What is required to deliver gain?
    3. Plot gm from idle to full power
    For a single tube
    For the sum of both tubes

    IOW, since it is clearly possible to build an open-loop pentode stage that runs in class AB that is relatively low in distortion, the theory that states its distortion must be high, as shown in that graph, is clearly in need of replacement with one that matches delivered performance.

    There is a simple 'Theory' proverb that states exactly what should be done with such theories...LOL

    It is further proof that there is good reason that Class A is worth the trouble...and not the 'marketing department' Class A that is actually just low power AB.

    cheers,
    Douglas
     
    Last edited: Dec 30, 2017

     

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  17. Alan0354

    Alan0354 Super Member

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    Hi Max

    With all these, I was just thinking about SE last night. 3 6L6GC in parallel, option of using parafeed to get rid of the DC current constrain. 300V to 350V +B, 200V screen, Local NFB to lower the output impedance of pentode mode ( no UL), might consider GNFB to raise damping factor more..........I might get like 20W of class A. Too bad I don't see any reasonable priced OPT that has primary resistance below 1K, or else, I can think of 4 tubes in parallel.

    Maybe I should build two, one PP and one SE. I just hope with LNFB and maybe GNFB, damping factor can be low enough to tame my speakers. I really think 25W is enough. The crappy kid amp just barely no enough. I don't listen that loud.
     
    Last edited: Dec 30, 2017
  18. trobbins

    trobbins Active Member

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    Also note that this is mid-band gain. It will get more interesting out of band as more output transformer parameters become influential.
     
  19. Alan0354

    Alan0354 Super Member

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    Ok, I finally did the calculation of driving the output transformer push pull with
    1) Output impedance ( rp) = very high by using current source of gm X input voltage.
    2) Output impedance is ZERO using a voltage amp with output impedance = ZERO.
    3) Output impedance of each side = Rpp/2.

    Here is the result:

    Gain Class A vs B.jpg

    You can see ONLY the case output impedance = 0 that gain of class A = class B.

    Looks like the LOWER the output impedance, the better matching the gain between class A region and class B region.

    Let me know what you think.
     
  20. PakProtector

    PakProtector AK member

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    Wow...what a nice set of equations. Unfortunately those are not the numbers you are( should be ) looking for Alan. Why has your model switched from pentode to triode-style sources? That is not a good idea either...and I will also remind you that there plenty of pentodes with plate resistance below that of triodes( as in less than 5kOhm ). That plate resistance is not something that will drop out harmlessly. You are unfortunately barking up the wrong tree...:)

    Why don't you go and have a real look at the stuff Bill has put up. I have seen notices via the Joe List that what you seek is in his archive.
    cheers,
    Douglas
     
    Last edited: Dec 31, 2017

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