Understanding "grid drive" to drive, 2, 4, 6 power tubes

[nsubulysses]

Ok, so say you have this very common type of LTP design for your phase inverter for a 4 power tube amp. So now you want to make a 6 power tube version of the same amp. What could be changed to make the PI work better for 6 power tubes vs 4?

- increasing R1 and R2 to 150K or 220K ?

- increasing R3 and R4 82K/100K to 100K/120K ?

-some sort of change in the R5-R8 region to bias different for bigger maximum peak to peak signal before PI clipping?

-altogether changing 12AX7 to 12AU7 or 12AT7 and change R5-R8 to accomodate the different tube?

I don't really understand grid drive in a current sense, i guess really only in a voltage sense. As in, if you have a big enough pk-pk signal the power tubes can acheive full power.

ALSO, would it also be common to change the 160K NFB resistor to a higher value, say 180K or 200K or somethign, because with a 6 power tube amp the pk-pk signal being fed back will be larger? It seems like this would make sense and it seems like when you look at 50W vs 100W Marshall JMP schematics the 50W has 47K NFB resistor and 100W has 100K.

Anyway, I'm just guessing. Any insights or explanations are very much appreciated.

[Justin Thomas]

One thing you could check is the Fender Super Twin, which uses an LTPI and drivers to boost the current...

Justin

[mozz]

That's a amp.My back hurts just looking at it.

[Jazz P Bass]

Here is the Super Twin schematic: Super Twin Reverb Schematic 015870.pdf

[Dave H]

I don't really understand grid drive in a current sense, i guess really only in a voltage sense. As in, if you have a big enough pk-pk signal the power tubes can acheive full power.

You don't need any more grid voltage to drive 6 tubes to full power than you do for 4 tubes and they don't need any more grid current drive either. What is needed is for the PI to be able to produce the same peak voltage without clipping across the lower value grid leak resistors needed for 6 power tubes. Notice on the Super Twin Reverb schematic Jazz posted that the power tube grid leak resistors are 33k each side. The max grid circuit resistance for one 6L6 in fixed bias is 100k so the grid leak resistor should be reduced to 47k for two tubes and 33k for 3 tubes. The Super TR schematic says the bias voltage is -61V so the PI has to be designed to produce 61V peak without clipping into a 33k resistor for full power output.

[Helmholtz]

Another reason for increased drive current capability would be class AB2 operation, where some real grid current is employed to push output.

[Mike Sulzer]

Depends on what you want to do with the amp. For clean playing, I do not think you need to change the drive much, if at all. But if you want to get the same sound on hard over drive, you need a proportional increase in drive current so that you can charge the coupling caps with the same time constant and maintain the same voltage across the output tube grid to ground resistors.

[Helmholtz]

Depends on what you want to do with the amp. For clean playing, I do not think you need to change the drive much, if at all. But if you want to get the same sound on hard over drive, you need a proportional increase in drive current so that you can charge the coupling caps with the same time constant and maintain the same voltage across the output tube grid to ground resistors.

Class AB2 operation is likely to produce blocking distortion with coupling caps. For this reason DC coupled drivers or transformer drive should be used.

[pdf64]

...The max grid circuit resistance for one 6L6 in fixed bias is 100k so the grid leak resistor should be reduced to 47k for two tubes and 33k for 3 tubes...

And the designer needs to bear in mind that the grid circuit resistance comprises the effective grid leak resistor + the effective bias supply output resistance. The latter looks to be much lower than is usual for a tube guitar amp.
I think that the harder the circuit conditions push the tube dissipation, the more of a consideration the grid circuit resistance limit becomes.

[J M Fahey]

Since 1972 I have been successfully applying: "need more power than 4 bottles can provide? ... go Solid State".
Just sayin'

[R.G.]

This is one of those situations where all that modelling stuff from circuits 101 comes in handy.

Looking at only one output tube, it's grid looks like a near-infinite resistance to ground being fed a microscopically small positive current from the rest of the innards of the tube, and one or more capacitors, effectively to ground. If the tube is set up with a near open circuit to its bias supply, the leakage makes the grid drift positive. In triodes with a cathode resistor, this can be useful as a "grid leak bias". For power tubes, we want better control of the conditions, so we introduce the grid leak resistor to he bias voltage.

The grid leak resistor on an output tube goes from the grid to the bias supply, and is sized so that the leakage current can be sucked out through the resistor to the bias supply with negligible change to the actual bias voltage, which happens on the grid wires themselves. The value of the grid leak resistor is determined by how much grid leak current exists, and that's in turn determined by how much the tube's internal leakage is. For higher leakage tubes, you use lower resistance grid leak resistors.

If you parallel output tubes, you parallel up the grid leakages, so you have to parallel up grid leak resistors. The external resistance driven by the driver, whatever it is, gets lower, and hence needs more drive current. Likewise the grid capacitances are paralleled up, so your driver needs to drive more current into the increased capacitances to get the grid voltage to move. For capacitors, I = C dv/dt, so to keep the same frequency response ("dv/dt") you have to provide correspondingly more current; not doing this makes the high frequency cutoff point of the grid circuit drop proportionately to the capacitance - the number of additional paralleled tubes.

So adding more tubes in parallel adds more DC loading because the grid leak resistance drops proportionate to the number of tubes (or should!) and the AC loading increases because there are now multiple paralleled capacitances, one set per tube added to the paralleling.

Back in the driver side, a triode as a PI output has some output impedance. If I'm remembering correctly, the plate resistance looks like it's in series with the actual plate, and the external plate resistor appears in parallel with the external plate pin to the supply, which is the same as ground for AC signals. There will generally be a series capacitor from the PI drive to the output tube grid to let the output tube have negative bias independent of the driver's plate voltage.

The PI may be differential, where both sides of the output tube are driven from plates, or it may be concertina, with one output tube plate driven from a plate and one from a cathode. There are other variations. For good modelling, you have to take into account the driving point output impedance for each situation.

With that as background, you can start making some guesses/estimates of the effect of paralleling additional tubes. Conservative design would parallel a new grid leak resistor with each additional ouput tube, to keep the added tubes from going into runaway. The additional grid leak loads and capacitances load down the drive impedance, meaning that the driver either has to supply more current or the output tubes can't be driven as hard as one tube could. It may be that you had enough reserve current drive in the original driver setup to ignore the new drive needs, in which case you don't have to do anything. It might be that you can accept the lower drive per tube, giving less-hard drive and lower high frequency response.

It may be that you can raise the voltage on the driving tube plate and try to drive the PI output to higher voltage to get more current through to the newly paralleled grids. It may be that you have to change PI tubes to variants that have lower rp and a lower Rp to get the increased current, and increase the signal drive voltage to this changed PI tube to get that much current out.

Which one of these you do depends on how much reserve current drive the original PI had, how much loss of HF response you can stand (including "can stand with the existing feedback stability situation"). There are enough "it could be" statements there that you have to sit down and start applying values to the various rp, Rp, grid leaks, and capacitances to get a good answer. Or you could do what most amp tweakers do - parallel the tubes and then easter-egg the PI values, and find some set of things that sound good enough.

As a caveat, the rp per driver tube, the leakage in the output tubes, the capacitances, and so on are all likely to be mildly non-linear. In a push-pull AB amp, you're certain that the grid loading and capactances change as you move over the crossover zone. But knowing the lumped, assumed-linear values gets you into the ball park.

And I apologize for this not being an answer. It's more of what to think about and how I at least would attack the issues.

[nickb]

To understand better you need to draw the load lines for the LTPI. I've drawn them quickly and very roughly below. Have a go yourself and you can take the time to do it more accurately. Also I see I didn't choose the DC conditions especially well but it'll serve to illustrate the point, I hope.

Using a 12AX7 the red line is the AC load estimated at 23.5k ohms ( 82K in parallel with 33k gird leak) and the blue line is the DC load line using 100K. For a max grid swing of 0v to -4V you get a plate voltage swing of 125V to 195V or 70Vpp when you need 122Vpp. Note that you have a 0.8V grid voltage change in one direction and 2.7V in the other. In other words it's badly distorted.
12AX7:


Here it is using a 12AT7 and a 30K plate resistor and AC load of 16K. You can now get 90V to 260V or 170Vpp for an input grid to cathode swing of 8Vpp. The distortion is much less.

12AT7:


In both cases the stage gain is low and that is why the Super Reverb* buffers the 12AX7 LTPI using cathode followers.

The choice of negative feedback resistor is just that, a choice. You are balancing distortion, power supply hum rejection, gain and sensitivity to component tolerances / aging.

Note: *I meant Super Twin Reverb

[Justin Thomas]

Since 1972 I have been successfully applying: "need more power than 4 bottles can provide? ... go Solid State".
Just sayin'

I think what you MEANT to say was, "more efficient speakers!"

Just don't sell me out for digital, okay?

Jusrin

[Old Tele man]

UNTIL the Class-AB1 operation becomes Class-AB2...all the power tube grids are VOLTAGE-driven into HIGH impedance loads.

But, when the PI pushes into AB2 operation, all the power tube grids become LOW-impedance loads, requiring CURRENT from the PI (usually provided by Cathode-follower circuits).

[Helmholtz]

Likewise the grid capacitances are paralleled up, so your driver needs to drive more current into the increased capacitances to get the grid voltage to move. For capacitors, I = C dv/dt, so to keep the same frequency response ("dv/dt") you have to provide correspondingly more current; not doing this makes the high frequency cutoff point of the grid circuit drop proportionately to the capacitance - the number of additional paralleled tubes.

The ~15pF grid capacitances of power tubes certainly don't require extra low impedance drive in guitar amps. MosFets would be a different matter.