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Finding the proper load for a pentode push-pull output stage

Discussion in 'Tube Audio' started by kward, Jan 26, 2016.

  1. kward

    kward AK Subscriber Subscriber

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    Let's say I have a fixed biased EL34 push pull output stage with 380V on the plates, and 250V on the screens. What is the procedure to find the optimal plate-to-plate load needed for the output transformer?

    When I run through what I think is the proper procedure I get significantly different results than supposedly what the data sheet says is optimal. So either the data sheet is wrong or I am calculating it wrong. So I thought I would pop this question up to the forum gurus for some help.

    Can someone walk me through this, with a bit more precision than just guessing or merely interpolating the various columns on the data sheet? Yeah, I think you can get pretty accurate results through interpolating the on the data sheet, but I would like to understand the "proper" way to do this if all you have is a characteristic curve plot at the given screen voltage you want.

    How does the procedure change if I have a cathode biased stage with some known resistor size in the cathode circuit?

    Thanks.
     
  2. thorpej

    thorpej AK Subscriber Subscriber

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    Well, I'm far from an expert, but the way I understand it, it goes a little something like this.
    1. Draw the max dissipation curve on the plate characteristics graph.
    2. Draw a load line for your class-B operation that, for your plate voltage, passes just below the apex in the max dissipation curve, (i.e. parallel to the tangent of the apex, but merely very close instead of infinitely close :) ). Then Ohm's law tells you what this load is where the load line hits zero plate volts.
    3. Draw a load line for 2x the class-B load, which is your class-A load line, and slide it up in the usual manner until you reach your desired bias point (~70% max dissipation).
    4. Your plate-to-plate load is then 2x your class-A load.
    Adjust and repeat as necessary until you find a load whose resulting curve looks about right. Seems to me that one key thing would be ensuring that the selected bias point is actually far enough away from the cut-off point so that the other tube in the pair doesn't see the class-B load "too early" (which seems like it would send the tube over the max dissipation line). You might not be able to find a suitable bias point for the "ideal" class-B load, at which point you have to compromise. Another compromise might be if you can't draw a class-B load line that parallel to the tangent of max dissipation given your plate voltage.

    It seems to me that the ideal load basically "hugs" the max dissipation line at the transition from class-A load to class-B load, like the Enterprise sling-shotting around the Sun.

    I find it easier to start with the class-B line because it's the one that you don't have to slide around... you can pretty easily visualize where the line has to go from your plate voltage past the curve.

    Sorry, I'd have to burn a little more mental bandwidth on how the changing operating point that comes along with cathode bias would change things, although intuitively it seems like it would pull the load line further away from the max dissipation curve as the other tube cuts off and you see the class-B load.
     
    Last edited: Jan 26, 2016
  3. kward

    kward AK Subscriber Subscriber

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    How is the plate load 2x the class A load? I think that may be the flaw in my thinking since I was ass/u/ming the full plate load would be 4x of the class A load. So, yeah, that's sort of what I did, except for that little factor to convert class A load to plate-to-plate load.

    Here's my exact procedure. To start with, I want a fixed biased EL34 class AB1 output stage. I want to maximize the output power given the predetermined plate voltage and screen voltage of 380V and 250V respectively. So I eyeball the plate curves and then fix a point on the graph at the bias point. In this case, I've chosen that to be at 380V plate and 50 mA plate current.

    Next I eyeball where a straight line (the load line) will intersect the knee of the characteristic curve at the 0V grid line. I chose it this way since that will maximize the voltage swing allowed and will hopefully also maximize power output. (Yes I know the load line doesn't actually follow a straight line--it's more an oval shape probably, but I am understanding a straight line approximation is good enough for this procedure...well unless that's the flaw.) Then I draw vertical lines from the bias point down to the X axis, and also from the 0V grid line intersection point down to the X axis. Also draw horizontal lines from the bias point over to the Y axis and from the 0V grid intersection point to the Y axis. So the graph now looks like this:

    upload_2016-1-26_20-22-3.png

    The load that the pentode sees (for one tube) is then the difference in plate voltage between the red vertical lines divided by the difference in plate current between the blue horizontal lines (i.e., Ohm's law), which is:
    (380V - 58V)/(210mA - 50 mA) = 322V/161mA = 2K

    But this is the load for a single tube. Since this is really a push-pull pair, the actual plate-to-plate load is 4x of that or 8K (impedance of the full winding as seen from plate to plate is square of turns ratio--or is that the flaw in my logic?).

    8K seems high for an EL34 output stage, at least according to the data sheet table, which says it should be "about" 3.5K under very similar conditions.

    upload_2016-1-26_20-20-4.png


    So I surmise I must be doing something wrong. :dunno:
     
  4. jazbo8

    jazbo8 Well-Known Member

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    The load line you drew and the multiplication factor you used were incorrect - they're a mishmash of both class A and class B loads and wrong for both cases. For class B, the load line must intersect the x-axis (Va) at Ia = 0mA, to learn how to draw it properly and calculate the Ra-a, please refer to this page.
     
  5. prelius

    prelius AK Subscriber Subscriber

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    Kevin,
    For for PP amp, the Ra-a is 2x, not 4x of Ra. Otherwise, your calculations are pretty close with the recommended values from the datasheets.
    What you built is an one tube load line(Ra), and it represents 50% of the work cycle of a PP amp, therefore, you need to double the Ra, not quadruple it.
    Cheers, Paul.
     
    Last edited: Jan 26, 2016
  6. thorpej

    thorpej AK Subscriber Subscriber

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    ¯\_(ツ)_/¯ It's what I was taught :) Sorry, I'm a software guy! But it kind of logically makes sense... if the impedance of the entire winding is X while both tubes are conducting, then the impedance of half the winding while both tubes are conducting is X/2. When the other tube stops conducting, the impedance of the entire winding thus becomes X/4 because the impedance ratio is the square of the turns ratio.

    The flaw in your logic is that you drew an A load line through the knee of the 0V grid curve... notice how the right end of your load line doesn't end at your desired plate voltage... You want to draw the B load line from your desired plate voltage though the knee of the 0V grid curve. If I draw a line between 350V through the knee of the 0V grid curve, I end up cross 0 plate volts at 245mA, which is 1428 ohms. So the class A load (i.e. both tubes conducting) would be 2856 ohms, so a line drawn between 350V and 122mA gives us the slope of the class A load line. Sliding that up to meet your desired 50mA idle current, looks like a bias voltage of -17.5V or so, and the A and B load lines intersect at the 200V mark right around 105mA. Since that is the class A load for just one tube, I double the load to get the P-P load, so 5.7K. So it looks to me like a 5K P-P transformer would work pretty well (and would tilt the scale in favor of less distortion).

    I suspect that if you plotted the operating point from the Mullard data sheet, the B load line would be quite far above the knee of the 0V grid curve (maximize linear operation at the expense of power efficiency).
    Scan.jpg
     
    Last edited: Jan 26, 2016
  7. kward

    kward AK Subscriber Subscriber

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    I see my logic flaw now. It is operating in either class A or B depending on output power. When in class A, plate impedance one tube sees is one half of total primary (because both sides of the primary are conducting). But when in class B, a single tube sees one quarter of the total plate to plate primary impedance since half of the windings are no longer conducting. So there are really two load lines at play for full power output, that together form a composite load line

    Excellent. I appreciate the quick responses. Thanks everyone.
     
  8. dcgillespie

    dcgillespie Fisher SA-100 Clone Subscriber

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    In addition to thorpej's information, it is best to determine the optimum quiescent bias current by use of a transfer curve generated from the Class B load line chosen, rather than just arbitrarily picking a value, which may not be the low distortion operating point. A transfer curve is generated from a graph where Ia is shown along the vertical axis, and Eg1 is shown along the horizontal axis, with both values representing 0 at the point of axis intersection. From the point of maximum Ia where Eg1=0, a tangential line is draw to the curve produced by the load line plotted on the graph. Where the line intersects the Eg1 axis is the optimum bias voltage, which from the family of plate curves can then be translated back into the optimum quiescent current for the most linear operation of the stage using element voltages and load chosen.

    Dave
     
  9. kward

    kward AK Subscriber Subscriber

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    Ultimately Dave, this is an attempt of mine to see if I can determine/derive the best values for an Enhanced Fixed Bias controlled output stage while at the same time building my first son an amp that he requested after seeing the Fisher 400 output stage based amp I built my second son (that amp used your derived EFB values from your Fisher 400 thread). Now suddenly my first son wants an amp too! So right now I wasn't too worried about getting the exact optimal bias point because I was just going to determine that empirically on the bench with my distortion analyzer anyway. But now that you provided this information, I will analyze it this way too--as that might give me a better starting point.
     
  10. thorpej

    thorpej AK Subscriber Subscriber

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    Ah, if you're using EFB, then I think there isn't anything special you need to do to account for being "cathode biased" ... any bias shift caused by the addition of the 5 or 10 ohm resistors used to measure current should be negligible. It would be a different story if the output stage were cathode biased in the traditional sense, because as the current increased, the operating point shifts. Dave wrote an article about this and how some output stages are designed around it ... I think he compared the Dynaco ST-35 vs the Heathkit AA-151 ... the latter having load and operating point that works very well with the shift that comes with increased cathode current and thus would not really get a performance benefit from EFB.
     
  11. kward

    kward AK Subscriber Subscriber

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    Yeah, I am just curious if there is some additional steps for cathode bias, or is the procedure identical? Certainly the load line will change a bit with cathode bias since there is more resistance in the circuit. For example in the data sheets, the optimal plate to plate load differs by 2K sometimes between cathode bias and fixed bias.
     
  12. jazbo8

    jazbo8 Well-Known Member

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    But it is not really cathode bias in the traditional sense, after all it is called EFB for a reason.:)
     
  13. kward

    kward AK Subscriber Subscriber

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    Right but if I were not to do EFB but rather only straight up cathode bias? Just looking for understanding on how the calculations are done.
     
  14. thorpej

    thorpej AK Subscriber Subscriber

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    I can't imagine the procedure is fundamentally any different, but you just have to consider the left-ward shift in the operating point as the current increases. For example, the RCA 6L6GB data sheet does show a difference like you say... 6.6K primary for 360V/270V fixed bias vs 9K for 360V/270V cathode bias. I think this is to account for the pivoting of the class B load line as the cathode current increases (and thus the bias voltage gets dragged more negative), although I'd defer to @dcgillespie on that one...
     
  15. thorpej

    thorpej AK Subscriber Subscriber

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    Put another way, I don't think it's really the added resistance at the cathode that changes anything ... after all, it's typically minuscule compared to the actual plate load (though it perhaps behaves a bit differently being a resistive rather than reactive load?) The real issue is how the bias shifts once the current really starts to flow when you enter class B operating conditions.
     
  16. kward

    kward AK Subscriber Subscriber

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    Makes sense. The max current the pair pulls at max power in true cathode bias is also, I don't know, maybe 20% less than same parameters under fixed bias. So it does seem that the load line must be at least a little different.

    Anyway good discussion.
     
  17. gadget73

    gadget73 AK Subscriber Subscriber

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    I'm not going to pretend to understand load lines and such. Its always been something that just goes over my head, but the old RCa 6L6 datasheet does give info for push-pull tubes in both self bias and fixed bias. They suggest the same load for both modes with the same voltages. Based on that, it seems like the bias method doesn't grossly affect the load that the tube needs to work into.

    http://www.r-type.org/pdfs/6l6.pdf
     
  18. thorpej

    thorpej AK Subscriber Subscriber

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    The section that lists the same load for cathode and fixed bias is the push-pull Class A1 section. Push-pull Class AB1 shows different loads for fixed and cathode biased. This makes sense because you don't enter class B operating conditions in a push-pull Class A1 configuration (note that zero signal vs max signal plate and screen current is very similar, just like single-ended Class A operation).
     
  19. gadget73

    gadget73 AK Subscriber Subscriber

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    If you roll down a little further, it lists it for AB1 push-pull mode as well. I see 6600 or 8500 ohms, depending on the screen volts. Same for both modes.
     
  20. thorpej

    thorpej AK Subscriber Subscriber

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    Oh hey, you're right, my bad. Huh, it's really interesting how they generate almost the same max signal power output for a lower max plate and screen current compared to the fixed bias configuration. (It's also kind of funny how the frumpy old metal can 6L6 has higher design max ratings than the 6L6G / 6L6GB.)
     

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