LTP bias Q

[nickb]

If I missed something in my scan of the thread, forgive me. There is a problem lurking here that's fundamental to the idea, I think.

Biasing the signal for clipping one side or another at the PI necessarily is separated from the power tube grids by the coupling caps. The coupling caps wash out any DC level changes, so what goes to the grids is always volt-second centered and pure AC. Grid blocking is an effect caused by the grid acting like a diode in that it conducts DC pulses out of the incoming coupling caps on positive peaks and shifting the DC offset on the coupling caps. Feeding the coupling caps an asymmetrically clipped signal to counteract this grid conduction doesn't help the grid-blocking DC shift because all the DC is removed. All it can do is to shift the overall size of the signal fed to the grids, and that will cause some reduction in grid blocking just because the signal's smaller.

So (just speculating without simulating it) I think that you can't do anything on the PI side of the caps that couple to the output grids that help any more than making the peak to peak signal smaller. The caps' DC removal washes out any advantages you thought you were getting.

I think...

I hate to go tete a tete, especially since my neurons are still entangled by all the quantum physics malarkey the other day. But..here's the rub, I did simulate it before I spoke and found that it does seem to work:



The average DC level on both sides of he cap is the same aside from a constant offset so there is no net charging of the capacitor.


I think...

[The Dude]

......Just bouncing my "what if"s off the higher minds.

I shall then kindly remove myself from the conversation.

[uneumann]

I hate to go tete a tete, especially since my neurons are still entangled by all the quantum physics malarkey the other day. But..here's the rub, I did simulate it before I spoke and found that it does seem to work:

[ATTACH=CONFIG]34644[/ATTACH]

The average DC level on both sides of he cap is the same aside from a constant offset so there is no net charging of the capacitor.


I think...

Not sure what you mean by "work". Maybe I'm missing something, but change of bias in a resistor-tail LTPI will change the PI grid voltages, which also leads to change of headroom. I think your simulation shows a shift in clipping (hard to see, so I speculate a bit). The shifted clipping charges the output cap and the output voltage shift (your blue line) is visible at the tail of the burst.

Basically I think RG and teemuk have this right... you can shift PI clipping from center, to top, or bottom of the waveform by altering PI bias. But you can't control the output tube grid drive baseline as long as your AC coupled to the output tubes. No matter what waveform the PI creates, 50% of the waveform area will be above the static grid level and 50% will be below. A cap can't source or drain continuous current. Now, if you have DC coupling, it's a different story.

[R.G.]

It was that suspicion that I didn't fully grasp all of it that led me to be so wishy-washy... I'll go have a look

Did you simulate the grid conduction with a fake 0V-threshold diode?

The reason this grabbed me is that I ran into a real-life version of this just recently, although not on tube drive. I did a PCB to replicate the entire preamp circuits of the Vox Beatle, and this includes the repeat percussion, which has a JFET through a cap at one input. If the JFET is not biased properly, negative-going parts of the input signal DC clamp the incoming signal to a negative voltage, making a thing similar to grid blocking, for the same reasons.

It took a long time to figure that out.

I'll go do some sim.

[nickb]

It was that suspicion that I didn't fully grasp all of it that led me to be so wishy-washy... I'll go have a look

Did you simulate the grid conduction with a fake 0V-threshold diode?

The reason this grabbed me is that I ran into a real-life version of this just recently, although not on tube drive. I did a PCB to replicate the entire preamp circuits of the Vox Beatle, and this includes the repeat percussion, which has a JFET through a cap at one input. If the JFET is not biased properly, negative-going parts of the input signal DC clamp the incoming signal to a negative voltage, making a thing similar to grid blocking, for the same reasons.

It took a long time to figure that out.

I'll go do some sim.

Yes, I did use a diode. And in the interests of full disclosure, here is the simulation schematic:-
-

[nickb]

Not sure what you mean by "work". Maybe I'm missing something, but change of bias in a resistor-tail LTPI will change the PI grid voltages, which also leads to change of headroom. I think your simulation shows a shift in clipping (hard to see, so I speculate a bit). The shifted clipping charges the output cap and the output voltage shift (your blue line) is visible at the tail of the burst.

Basically I think RG and teemuk have this right... you can shift PI clipping from center, to top, or bottom of the waveform by altering PI bias. But you can't control the output tube grid drive baseline as long as your AC coupled to the output tubes. No matter what waveform the PI creates, 50% of the waveform area will be above the static grid level and 50% will be below. A cap can't source or drain continuous current. Now, if you have DC coupling, it's a different story.

The blue line is the unmodified system and shows the cap is charged at the end of the burst. The green is Chuck's idea. Clearly the cap is not charged for that case.

In all fairness, I was simulating just the effect of driving with a asymmetrically clipped waveform, which was the core of the idea as I saw it.

Whether it's possible to get the PI to behave like this is a different question in my mind. I haven't tried it but if the current on one side of the PI goes to zero then the other side but must be at max since it has constant current tail. So you would end up with a symmetric clipped waveform, but I still think it will work.

It would be hard to make it perfect, but it would be better than nothing.

[Chuck H]

I hate to go tete a tete, especially since my neurons are still entangled by all the quantum physics malarkey the other day. But..here's the rub, I did simulate it before I spoke and found that it does seem to work:

[ATTACH=CONFIG]34644[/ATTACH]

The average DC level on both sides of he cap is the same aside from a constant offset so there is no net charging of the capacitor.


I think...

I was just thinking this at work today. With a reduced differential I would think that would also reduce charging. But the simulations absolutely don't take everything into account and I think in a real world situation it's R.G.s interpretation. If you think about what's actually happening, how would one even begin to account for power tube grid starvation on the opposite side of the coupling cap without confounding PI operation so much as to be detrimental? I'll still test this next time I have an amp open on the bench.

[nickb]

The real problem is the the PPPI can't be modified to behave in the desired fashion. Either the currents are very small lowering gain and increasing output impedance or the plate resistors get small reducing the gain unacceptably.

[R.G.]

I'm tied up like crazy with the day job at the moment. I'll try to do a more detailed sim that I did before and see if I can understand the issue better.

[Chuck H]

What I didn't take into account in my initial consideration is that the PR mod is gating on the power tube grid side of the coupling cap. So it's not just gating the cap, it's gating the power tube grid. I hoped it was possible to induce the same clip with circuit values to prevent the coupling cap from charging. But it's not the PI action charging the cap, it's the grid trying to draw current that charges the cap. I think. Maybe. So the PR mod is doing more than just clipping to provide "off time" for the cap to discharge. It prevents the power tube grid from trying to draw current from the PI output above the zener voltage. Which you can't do by just clipping the PI on the opposite side of the coupling cap. Did I get that right?

[uneumann]

What I didn't take into account in my initial consideration is that the PR mod is gating on the power tube grid side of the coupling cap. So it's not just gating the cap, it's gating the power tube grid. I hoped it was possible to induce the same clip with circuit values to prevent the coupling cap from charging. But it's not the PI action charging the cap, it's the grid trying to draw current that charges the cap. I think. Maybe. So the PR mod is doing more than just clipping to provide "off time" for the cap to discharge. It prevents the power tube grid from trying to draw current from the PI output above the zener voltage. Which you can't do by just clipping the PI on the opposite side of the coupling cap. Did I get that right?

Chuck - I think you have it, but I'd word it differently. The problem is clearer to me thinking about signal currents through the caps. There are two caps - the coupling cap and the cathode bypass cap.

First consider the coupling cap. Regardless of what the PI signal looks like (due to bias or whatever), positive swings eventually cause output tube grid current. Once that happens, the coupling cap charges asymmetrically and causes a voltage shift (negative) on the grid. The grid shift increases output tube bias and pushes the tube toward cutoff. I don't think the PR mod does much to stop that, in fact it may make it worse since it limits how cold the bias gets, therefore aids grid current flows and ensuing grid shift.

The second cap - the bypass cap is another story. It charges since asymmetric currents flow thought it due to the tube operating in class AB. In class AB, more signal current flows on positive grid swings than negative grid swings. That current asymmetry causes the cathode cap voltage to shift (increase). (Caps can't source or sink net signal current.) This also pushes the tube toward cutoff. The PR mod does address this. It limits the cap shift (or bias increase) that can occur.

I think the point is that that PR mod impacts the bypass cap, but the grid shift problem is not impacted much and caused by a different mechanism. It's still early, but I think I got this right.

[Chuck H]

The second cap - the bypass cap is another story. It charges since asymmetric currents flow thought it due to the tube operating in class AB. In class AB, more signal current flows on positive grid swings than negative grid swings. That current asymmetry causes the cathode cap voltage to shift (increase). (Caps can't source or sink net signal current.) This also pushes the tube toward cutoff. The PR mod does address this. It limits the cap shift (or bias increase) that can occur.

I think the point is that that PR mod impacts the bypass cap, but the grid shift problem is not impacted much and caused by a different mechanism. It's still early, but I think I got this right.

Just to be sure we're talking about the same circuit, the PR mod is a zener in series with a standard diode, cathodes facing each other. One of these circuits is parallel to each power tube grid load. A zener voltage is chosen a volt or two greater than the bias voltage so the zeners don't conduct until the power tube is in cutoff.

I think you might be confusing this with the zener across the cathode resistor to put a ceiling on the rising voltage. Some call this the "Chuck H" mod, but that's a misnomer since I certainly didn't invent it and actually got the idea from Ken Gilbert (kg). I've also seen something similar on the old version of the nyquistplot site. What I did do that was unprecedented on the interwebs was to combine these two ideas by valuing the cathode resistor zener just at the point of clipping and then value the PR mod zeners a volt higher to achieve predictable operation with a guarantee of no audible zener clipping. Still pretty obvious and no big deal but it seems to have helped a few people with their buzzy 18 waters so I'm glad for that.

Apologies if you are indeed talking about the PR mod and I simply still don't get it

[uneumann]

Just to be sure we're talking about the same circuit, the PR mod is a zener in series with a standard diode, cathodes facing each other. One of these circuits is parallel to each power tube grid load. A zener voltage is chosen a volt or two greater than the bias voltage so the zeners don't conduct until the power tube is in cutoff.

I think you might be confusing this with the zener across the cathode resistor to put a ceiling on the rising voltage. Some call this the "Chuck H" mod, but that's a misnomer since I certainly didn't invent it and actually got the idea from Ken Gilbert (kg). I've also seen something similar on the old version of the nyquistplot site. What I did do that was unprecedented on the interwebs was to combine these two ideas by valuing the cathode resistor zener just at the point of clipping and then value the PR mod zeners a volt higher to achieve predictable operation with a guarantee of no audible zener clipping. Still pretty obvious and no big deal but it seems to have helped a few people with their buzzy 18 waters so I'm glad for that.

Apologies if you are indeed talking about the PR mod and I simply still don't get it

My bad - my mistake - I was indeed talking about a zener across the cathode cap.
Now for that second cup of coffee...

[Dave H]

First consider the coupling cap. Regardless of what the PI signal looks like (due to bias or whatever), positive swings eventually cause output tube grid current. Once that happens, the coupling cap charges asymmetrically and causes a voltage shift (negative) on the grid. The grid shift increases output tube bias and pushes the tube toward cutoff. I don't think the PR mod does much to stop that, in fact it may make it worse since it limits how cold the bias gets, therefore aids grid current flows and ensuing grid shift.

The PR mod prevents grid bias voltage shift by ensuring that the output tube's clipped grid signal is symmetrical about ground (for cathode bias) as it is when the grid signal isn't clipped so there is no bias shift. Without the PR mod the bias shift can be enough to push the output tubes into Class C, i.e. they are both turned off for a small part of the cycle resulting in horrible crossover distortion (fizz?).

It doesn't have to be the PR mod, any other method which forces the grid signal to stay symmetrical under overdrive should work but I can see too many drawbacks with fudging the PI bias or output level. A post PI MV or crossline MV could work to remove the fizz as long as it's only turned down just enough to keep the grid drive roughly symmetrical but not enough to clip the PI before the output tubes.

Edit:
I'll have to learn to type faster. Chuck beat me to it

[R.G.]

Got a few moments to run the simulator. First, I verified that bias shift happens. It does.
Then I put in the PR anti-shifter on the "grid" circuit, and verified that it prevents bias shift. It does.
Then I clipped the waveform on the PI side of the coupling cap and checked for bias shift - which now happens again.

As I thought, the bias shift is a variant of DC clamping with a diode similar to the old TV signals' DC restoration circuits. It happens with any waveform, clamping one peak of the AC waveform to the clamping voltage - in this case, the cathode voltage on the output tube. This always produces a DC offset as the clamping happens.

So the answer to the OP is: no, you can't do anything to the signal going to the power tube grid on the outside of the coupling cap to prevent bias shift as the tube grid starts conducting on positive peaks. Well, nothing except clipping the entire waveform so it never gets big enough to cause grid conduction in the first place.

Anything you do has to be done on the grid side to prevent grid conduction from pumping charge from the coupling cap.