12BH7 Long Tailed Pair Clipping Early

Discussion in 'Shock Brother's DIY Amps' started by andrewRneumann, Nov 20, 2021.

  1. andrewRneumann

    andrewRneumann Tele-Afflicted

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    Hello All,

    I've been working on a LTP using a 12BH7 driven by a 12AX7 cathode follower. Here's the current design:
    12BH7-LTP.png
    The load line (for the left triode with Ra = 27K) should look like this:
    12BH7 LTP Load Line.png
    The bias is sitting at Vgk = -9.2V with -10.8V of negative signal swing available before the tube should cut off at Vgk = -20V. As this is a LTP, the headroom is actually doubled, so the signal should be able to swing -21.6V negative before cut off is reached. The total input headroom should be around 40Vp-p if this were perfectly center biased. Under those conditions the unloaded output signal swing should be about 200Vp-p.

    So I am confused by why I am getting clipping on the negative signal swings like this:
    12BH7 traces.jpg

    The yellow trace is the input signal which is 36Vp-p. You can see that the output signal (green) of the left triode is really clipping it's positive peaks and sloped heavily. This is loaded by a 1M volume control turned all the way down, so no grid current from the following stage should be charging the 1n coupling cap. I can't for the life of me figure out why this is happening. Why is it clipping so early and sloping like that? Am I missing something obvious?
     
    Last edited: Nov 20, 2021
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  2. peteb

    peteb Friend of Leo's

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    Are you saying it is the positive or negative peaks?

    I read negative and then positive.



    In my own clean power testing on a scope, on a single ended power tube, I expect to see a slope on one end and a horizontal clip on the other end.

    not totally sure to say which end is which until I scoped class AB to compare.

    class AB cuts horizontal on the top and bottom, which makes the flat top saturation, and thus the sloped cut is cutoff.

    my uncertainty is that in scoping an AB signal, why do you see the saturation and not see the cutoff?



    The bottom line is I think you are looking at cut off and cut off will slope like that.



    why isn’t it saturating?
     
  3. andrewRneumann

    andrewRneumann Tele-Afflicted

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    Yeah sorry I wrote that in a confusing way. The negative input signal is clean. The output signal positive peaks (which correspond to the input signal negative peaks) are clipped.

    This is driving a 1M volume control turned all the way down. So the following tube never receives a signal and never draws grid current. There should be no clipping of the positive signal due to "saturation" of the following tube.

    To my eye, the clipping is cut-off of the LTP... but why? Shouldn't I get more headroom out of this thing? Clearly, I don't fully understand what is going on.
     
  4. peteb

    peteb Friend of Leo's

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    Hi Andrew,

    I enjoy grappling with tube concepts like cut off, saturation and bias. I tend to look for mathematical relationships and not so much load diagrams even though I like a good graph.


    Let’s try to make some sense of this.

    I see

    bias - 9.2
    Cut off -20
    Where does the cutoff come from? Load line diagram?
    Negative Headroom equals -20 minus -9.8 = -10.2

    why is the headroom doubled in an LTP?

    imput signal 36 volts peak to peak

    why use peak to peak? Isn’t RMS more standard?


    I wouldn’t be concerned with clipping in the next tube. What about saturation in this tube? If it is center biased cut off and saturation will onset at the same point.


    It appears to not be center biased.

    my analysis:

    36 volts peak to peak would be .717*18= 12 VAC RMS.

    You predicted your negative headroom to be 10.2.


    12 VAC RMS > 10.2 leads to cutoff


    My questions

    why isn’t it center biased? It looks like it is biased too cold, too close to cut off.

    and why would an TPI have double the headroom?
     
  5. andrewRneumann

    andrewRneumann Tele-Afflicted

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    There is local negative feedback from the shared cathode. This means that Vgk does not match the input signal, but is a smaller version of it. For instance, a 10Vpp input signal may only cause Vgk to swing 5Vpp. I’m not sure the input headroom is exactly doubled, but it should around there somewhere.

    I agree that the tube appears to be biased close to cut-off based on the output waveform. But with the bias measuring -9.2V it seems to be more central. Something else is going on that I don’t have a grasp on.

    I don’t mix RMS measurements with readings from tube characteristics charts. The load line analysis is accurate—Vgk should be able to swing 20Vpp and the input signal swing should be about double that (40Vpp).
     
  6. Tom Kamphuys

    Tom Kamphuys Tele-Holic

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    In what circuit is it in? E.g. is the right input used for feedback?
     
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  7. 2L man

    2L man Tele-Holic

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    You should use DC setting on oscilloscope and it would be easier to understand what it show! 50V DC setting and zero line on bottom of the screen should show true Anode output signal.

    Oscilloscope on anode and probe set to 10M show reliable craphs. On control grid, 10M sometimes it can mess the circuit operating point.

    Obviously green output peak voltage is almost 250V and on resistive loadline the output voltage can not come higher than the operative voltage is.

    If green is after the capacitor then slope show the discharge.

    There is something strange on green when numerically it reads Vpp 14,4? If input is set to 50V that clipped signal is already 140Vpp and without clipping it would be about 200Vpp?
     
    Last edited: Nov 21, 2021
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  8. sds1

    sds1 Tele-Afflicted

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    Would you mind elaborating on this? I have an analog scope and I've only ever used it with AC coupling enabled.
     
  9. 2L man

    2L man Tele-Holic

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    When DC-mode is selected the reference stay at zero so what we see is true voltage sweep. When AC-mode is selected the voltage graph comes both side of the reference.

    For example if amp B+1 is 400V you can use 50V setting on DC and if you set reference to the bottom of the screen you can look all triode anodes and see the DC and the signal which sweep both side of the DC.

    When a Single Ended power tube Anode is measured there comes fascinating effect when voltage sweep can come almost double what the plate idle voltage is and it is when bias current stores energy to Output Transformer which then gets released when tube is driven.

    To set the reference, when an old analog oscilloscope is used, the mode selection should be set to GND and the sweep line comes visible and then its position is set usually to the bottom of the screen when amplifier uses only positive voltage.

    This reference can be set down below the screen and now when adjusting a point in DC-signal, usually the peak is watched and counted how many "squares" it is lowered below the screen. Doing this it is possible to use better resolution.

    Modern digital scopes usually show somewhere where the reference is.
     
    Last edited: Nov 21, 2021
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  10. sds1

    sds1 Tele-Afflicted

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    Thanks @2L man , are there probe voltage rating ground or other safety issues with turning off AC coupling, enabling GND mode that we should be aware of?

    Coming from someone dumb who just shorted out a floating DC supply last week using his ground (EARTH) probe on the DC-... :(
     
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  11. Ten Over

    Ten Over Tele-Holic

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    Local negative feedback, a.k.a. cathode degeneration, a.k.a. cathode current feedback is very minimal because the long tail (8k1 resistor) approximates a constant current source. Since there is very little change in current across the shared cathode resistor, there is very little change in the voltage that would create negative feedback.

    Here is the real reason for twice the input voltage:
    LTP Figure 1 Schem PNG.png
    LTP Figure 1 Dialog PNG.png
     
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  12. Ten Over

    Ten Over Tele-Holic

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    The 'scope shot looks a lot like bias shift from grid current. Try measuring the grid-to-cathode voltage of the inverting stage with no signal and then measure it with that 18Vp input signal that you used before.
     
  13. 2L man

    2L man Tele-Holic

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    I think it does not have any effect to probe or oscilloscope voltage rating and GND-mode just "disconnects" input inside scope. In my experience older CRT oscilloscopes are very strong against abuse.

    Tube amps which have less than 1000V can not damage them but it is good to always use probe at 10× because it stress circuit less.

    I recall oscilloscope input rating is at least 1500V and probe at 10× make it 10 times more.
     
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  14. andrewRneumann

    andrewRneumann Tele-Afflicted

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    Thanks everyone for thoughtful posts.

    Noted. I will try the measurement again with no probe on the grid and see if the output changes.

    What capacitor are we talking about? The coupling cap? There is a 1M resistance to ground following the coupling cap... would that not prevent any bias shifting in the cap? I am going to completely disconnect the following circuit to take it out of the equation.

    It is reading Vpp=144V.

    That's a spot on analysis. For the sake of not driving into the details, I simply considered the affect of the shared cathode as a type of local negative feedback. It is not the same as an unbypassed cathode resistor, as you rightly point out, but it does have a similar effect does it not? And doesn't the cathode voltage have to change by 50% of what the input signal is? I don't consider that "very little change." Maybe that's not what you meant.

    I'm going to try some DC shots with long time domains to try and capture any shifts. I'm not certain there is a shift going on--as I said the coupling caps are not letting much current through with 1M to ground after them. Is it possibly a bias shift in the power supply?

    I'm also probably misunderstanding coupling cap bias shift--thought it was more of an affect when the following stage drew grid current through the coupling cap. Back to the books...
     
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  15. andrewRneumann

    andrewRneumann Tele-Afflicted

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    It is being driven by a cathode follower and each output cap is loaded by 1M to ground. There is no negative feedback applied to the right triode.
     
  16. Ten Over

    Ten Over Tele-Holic

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    Yeah, the cathode voltage has to change by something around 50%, but most of that comes from the voltage change across the 8.1k tail resistor. Let's say that the cathode voltage changed by 0.5V. The voltage change across the 1.6k cathode resistor is only .082V while the voltage change across the 8.1k tail resistor is 0.418V.

    To test how much cathode degeneration is coming from the shared cathode resistor, bypass it. You will see very little change in the gain when you bypass the cathode resistor, leading one to suspect that very little cathode degeneration is taking place.
     
  17. Ten Over

    Ten Over Tele-Holic

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    I'm talking about the 22nF input capacitor. The grid side of that capacitor charges negative when grid current occurs which shifts the bias to more negative and closer to cut-off. It's real easy to detect with a DMM measuring from the grid to the cathode.
     
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  18. andrewRneumann

    andrewRneumann Tele-Afflicted

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    I don’t understand it the exactly same way as you I guess. I don’t see AC signal current flowing in the bias or tail resistors at all. The ideal current source in your analysis is steady at the DC bias current even when AC signal is applied. So I don’t see AC signal voltages being developed across either of those resistors.

    The 50% signal voltage appearing on the cathode is caused by other triode. I should have said “shared cathode”, not “shared cathode resistor”. The resistors really just set up the proper bias and ensure relative equality between the two triodes.

    At least that’s how I see it…

    EDIT: I think I’m confusing the ideal with the real. The ideal is a constant current source attached to the shared cathode. The ideal current source has infinite resistance. The real is a finite resistance which does not have constant current. So some AC signal will appear across the bias resistor and tail resistor. This AC signal across the bias resistor is why the input resistance is boot-strapped to a much higher value than the grid leak resistor alone.
     
    Last edited: Nov 21, 2021
  19. andrewRneumann

    andrewRneumann Tele-Afflicted

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    I will try to detect whether a bias shift is happening, but I have my doubts. The yellow trace is the input signal probed directly on the grid and it looks pretty symmetrical. There’s no clipping of the input signal so there shouldn’t be any appreciable bias shift. There might be grid current flowing, but I’m driving this with a cathode follower so I think the CF can supply the necessary current without disturbing the average charge of the input cap.

    I found that I could not measure the voltage on the grid with a DMM without getting a spurious result. I could only get what seemed to be reasonable result by measuring the voltage at the base of the bias resistor. They are in blue on the schematic. But… apparently the bias isn’t what I think it is if it cutting off so quickly.

    Still, I’m open to all possibilities and admit I’m learning a bit of this as I’m going. This is my first LTP from scratch. There’s also a high likelihood of operator error on the scope work.

    I’m going to post some more test results tonight.
     
  20. Ten Over

    Ten Over Tele-Holic

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    I have noticed that we just don't seem to be on the same wavelength. but I'm sure that it is my fault.
    LTP Figure 3A Schem PNG.png
    LTP Figure 3A Dialog PNG.png
     
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