Split Load PI

[Mats]

Hi
One friend of mine and I had a look at a GA-40 amp schematic.
I noticed that there are some different versions of this amp.  In an early version GA-40 had a Split Load PI,
1952 and some years after that.
Im looking at the PI, I have not seen this kind of PI connection before.
The 1 Meg resistor, before PI, is grounded and not connected like the 1 Meg in an Fender 5E3 or
lets say the 10 Meg in a Magnatone 213.
Whats the pros and cons of this kind of connection of PI of the early GA-40 amp?
/Mats

[sluckey]

I believe you have discovered an error in that schematic. If the cathode resistor is 47K (as it should be) then the voltage on the cathode will be a big positive number. Since the grid is tied to ground (zero volts) the tube would be cut off and no signal could pass thru the tube.

If you believe that ≈.23 note next to the cathode resistor, that would mean that resistor would be about 47Ω (unbelievable). That would create a gross imbalance between the two signals that drive the output tubes.

The grid cannot be referenced to zero in a split load PI because the cathode voltage will be a big positive number. You must run the grid at a positive voltage that is a little less than the cathode voltage. You can do this by bootstrapping the grid to the cathode (like the 5E3), or you can use a fixed bias solution with a voltage divider between B+ and ground (like the cathode follower in the Fender 6G15 Reverb Unit, or the Paul C mod).

Here's the Paul C mod applied to a 5E3 PI...
     http://home.comcast.net/~seluckey/amps/5e3/5e3.pdf

[Mats]

Hi
Thanks for answer.
The schematic is a bit unclear but the Split Load PI is:
R24 47K resistor
R23 47K resistor
Tube is a 1/2 6SN7 and tube before PI is a 6SJ7.
So its just an error in the schematic ?
/Mats

[Jack_Hester]

I have three of these amps.  Two have been repaired and brought back to working condition.  The third is completely stock and non-working.  I left it that way, because it was my unmolested model to make my schematic drawing, as I couldn't read the original schematic(s) very well for this model.  It was something of a group effort, here, to correct it.  And, I reconciled it with all three amps.  Mostly with a meter on the resistors, as the colors are not what they were.  Nor are my eyes.

Jack

[sluckey]

I have three of these amps.

Jack, I would really like to know what the voltage readings are for V4 pins 1, 2, and 3. Would you mind doing that?

[Jack_Hester]

Jack, I would really like to know what the voltage readings are for V4 pins 1, 2, and 3. Would you mind doing that?

Sorry, can't get to them at the moment.  They are stored away, and I have been unable to do any shop work since the holidays.  I looked to see if I had recorded any readings, but can find none.  I'll try to get readings as soon as I can.

Jack

[sluckey]

OK, thanks. I'm asking because your schematic has the same obvious error as the original. Maybe it's not an error after all?

[VMS]

Here's the old thread where you can find some voltages:

http://el34world.com/Forum/index.php?topic=12452.25;

[Jack_Hester]

OK, thanks. I'm asking because your schematic has the same obvious error as the original. Maybe it's not an error after all?

I'm fairly sure that it's not an error.  I spent a lot of time hand-tracing to verify the components and their placement.  But, that's now been a few years, and memory does not serve me as well these days. 

Jack

[shooter]

Since the grid is tied to ground

I'm lost on the grid to ground, I see what is typical on most preamp gain stages at the grid, then if you split the typical 100k plate  and put half on the cathode, half on the plate, wouldn't that give you ballpark equal signals?

[sluckey]

Since the grid is tied to ground

I'm lost on the grid to ground, I see what is typical on most preamp gain stages at the grid, then if you split the typical 100k plate  and put half on the cathode, half on the plate, wouldn't that give you ballpark equal signals?

The grid is tied to ground thru a 1M resistor. IOW, there is zero volts on the grid. Yes, you would have equal signals on plate and cathode. Now suppose that 1mA (very believable) flows thru that tube to create those equal signal levels. 1mA thru a 47K resistor results in a 47VDC drop across the resistor. So, you have positive 47v on the cathode and zero volts on the grid. See my dilemma yet?

[Willabe]

So, you have positive 47v on the cathode and zero volts on the grid. See my dilemma yet?

In a common cathode preamp tube gain stage, to control the tubes output, the grid has to be less voltage then the cathode. Then the ACV signal from the guitars PUP's are like stepping on the gas pedal, increasing the amps output.   


               Brad   

[jazbo8]

Now suppose that 1mA (very believable) flows thru that tube to create those equal signal levels. 1mA thru a 47K resistor results in a 47VDC drop across the resistor. So, you have positive 47v on the cathode and zero volts on the grid. See my dilemma yet?

I think Jack is correct - there is no error in the schematic. V4 is a standard concertina/cathodyne stage, and the voltages for the amplifier are shown here, from which the current for V4 is ~0.245mA. But I think it may be a bit on the low side, since the measurements were taken with the filament voltage at 6V only.

[sluckey]

Looks like I may have to change my thinking on this circuit. I always thought the grid would need to be one or two volts negative of the cathode for best linearity. I've seen direct coupled cathodynes, fixed bias cathodynes, and bootstrapped cathodynes, but this is the only zero volt grid cathodyne I've ever seen. I have a bootstrapped cathodyne PI in my shop that will be easy to change to this circuit. I'll let you know what I find out.

[sluckey]

I played around with this cathodyne using the zero volt grid circuit shown in GA-40 schematic. And it works... sorta! The maximum clean signal I could get out of it was 4V. Maybe this is the reason this is the only cathodyne I've ever seen the grid run at zero volts.

Here's the test circuit I used. This is actually my L'il Maggie amp. I used a test switch just for convenience. The tube is different from the GA-40 tube but I believe the results would be similar for any suitable triode.



With the switch as shown (ie, typical bootstrapped cathodyne) the circuit will produce a clean 45V pp signal. That's plenty to drive most all small/medium output tubes. I didn't take any pics because it's just a boring symmetrical sinewave.

Switch to the TEST position and things change quickly. Now the grid is referenced to zero volts (1M grid leak resistor connected to ground). And the maximum clean signal is only 4V pp! At 6V pp, the bottom of the sinewave is noticably flattened at about zero volts. By the time you reach 10V pp, the flattening at the bottom is even more severe. Further increasing the signal just results in more flattening. The tube is definitely going into cutoff on the negative portion of the sinewave.

The following pics show what I described above. The scope probe was connected to the top side of the 100K cathode resistor. Plate waveforms were just a mirror image of the cathode waveforms.







I was using a D/L for a speaker load and never played guitar with this assymetrical signal distortion. The circuit did not behave exactly as I originally thought it would, but I'm convinced the zero volt grid cathodyne is pretty useless to drive most typical output tubes. Simply shifting the grid voltage up allows for a much greater clean output and all it costs is one resistor.

I'd love to hear some other discussion about this. I probably don't have it exactly right and I know I've left out a lot of details that could stand to be filled in.

[VMS]

Thanks sluckey for taking the time to demostrate the circuit.

Because GA40 doesn't use bootstrapping or other ways to center-bias the cathodyne it is probably biased near the cut-off.

I don't know how to draw loadlines but it is probably a bit similar to slo100 gain-stage with 39k cathode resistor:

http://ampbooks.com/mobile/classic-circuits/soldano-slo-preamp-3/

[shooter]

Thanks Sluckey, 4v wouldn't even push a el84.  if you had used the 47k vs your 100k would that push the asymmetry even more?  I would think the signal would be even less than the 4vac?

[sluckey]

You can get a lot more signal out. It just isn't pretty. I took it up to 45V because that was the max I could get in bootstrap mode. At about 45V the signal starts running into the B+ rail and becomes clipped at the top as well as the bottom. Maybe it's usable. But I bet it does not sound loud and clean. But back in '54 you only had to be a little louder than a D-28.

Maybe one of the breadboard guys will build this up and give it a real workout. But I believe the limitations of the circuit are why this one GA40 is the only circuit I have ever seen that uses a zero volt grid cathodyne. I wonder if anyone can find another example?

I may fire it up again and play a guitar thru it. The only reason I didn't do that yesterday is because I no longer trust my ears. But working with radar for 40 years makes me trust what I see on a scope. Unfortunately, I've seen some ugly stuff on a scope that still sounded OK to me!   

[shooter]

40 years makes me trust what I see on a scope

+1......til I started building tube amps 
The 1st time I hooked a scope to an amp with an issue I was poopin in my pant, thinking WTF did I get myself into

[jazbo8]

Maybe one of the breadboard guys will build this up and give it a real workout. But I believe the limitations of the circuit are why this one GA40 is the only circuit I have ever seen that uses a zero volt grid cathodyne. I wonder if anyone can find another example?

Perhaps it is not often used in guitar amps, but there are hundreds of hi-fi/PA amplifiers from Bogen, Dynaco, Fisher, Scott, etc. that used it, since the stage has no gain, it has to be paired wiith other gain stages, in the GA40, e.g., it is diven with a pentode. There is nothing wrong with the the plain ol' cathodyne, it's just how you use it...

[sluckey]

but there are hundreds of hi-fi/PA amplifiers from Bogen, Dynaco, Fisher, Scott, etc. that used it

Please show me one.

[PRR]

> I have not seen this kind of PI connection before.

If you do it like that, the large cathode resistor ensures the tube will carry low current. Then you will have very small drops on both 47K resistors. The cathode voltage approaches (B+)/Mu. Is that an 6SN7? Then Mu is 20. Voltage division goes as 1+20+1. Cathode voltage is 1/22 of B+. Assume B+ is 250. Cathode voltage is about 11V. With a 470K load, you can't get 10V peak on each side.

If that is really the way these are wired, they left a significant amount of output power on the table. Maybe the asymmetric action is so obnoxious that "it sounds loud".

If you change the tube to 6SL7 (Mu=60) then cathode sits at 4V and you can't get much over a couple volts peak.

> hundreds of hi-fi/PA amplifiers from Bogen, Dynaco, Fisher, Scott, etc. that used it

The split-load, yes.

But this plan with *grid at zero voltage* is not common.

A fair number of amps return the gridleak near the cathode (as with Sluckey's 2.7K+100K string). This levers the grid (and cathode) up to 50V-100V, allowing good swing.

An alternate trick beloved in hi-fi is to direct-couple the split-load's grid to the previous plate. 1st plate sits at a significant positive voltage, the SL needs a significant positive voltage, just wire them together! The trouble is that the 1st plate wants to sit at 1/2 to 2/3 of B+, while the SL grid wants to sit nearer 1/4 of B+. You can jimmy bias values, add bleeders or dividers, and make it fit. In hi-fi it avoids one R-C bass-cut, which makes it easier to apply huge NFB.

A third trick is to tie grid near ground and run the cathode resistor down to a negative supply line. +240V and -80V would work fine. But we don't need any -80V for anything else, and another power supply for just the one stage is silly. (I have seen this in instrumentation, where various supplies are needed for other chores, and the DC levels work out better.)

FWIW, Scott clung to the Floating Paraphase long after fashion favored the split-load cathodyne; but he did do the split for the 380. Direct-coupled too. This is perhaps easier to do with a Pentode as the 1st stage (you can push a Pentode's plate pretty low and still get strong current through it), but I have seen it done with a Triode at the 1st stage.

Note 1st plate sits at 60V and PI cathode sits at 50V. This does not add-up and may be a measurement mistake. We might expect 50V and 60V-- cathode sits positive of grid-plate node. Too tired to sort through more (or more-accurate) schematics.

[jazbo8]

> hundreds of hi-fi/PA amplifiers from Bogen, Dynaco, Fisher, Scott, etc. that used it

The split-load, yes.

But this plan with *grid at zero voltage* is not common.

A fair number of amps return the gridleak near the cathode (as with Sluckey's 2.7K+100K string). This levers the grid (and cathode) up to 50V-100V, allowing good swing.

An alternate trick beloved in hi-fi is to direct-couple the split-load's grid to the previous plate... [snip]

As usual, you are right, I neglected to mention that most hi-fi amps were in fact direct coupled and/or used in situations where very little PA drive voltage were required. But here is the Heathkit A-9 with the "grid at zero voltage" (well actually at the same potential as the cathode):

[Mats]

Hi
Thanks for replays and testing.
This is a good forum, for me as beginner, to get your answers.
Yes PI looks strange.
Like someone suggests, perhaps its up for a breadboard.
Anyway Gibson did keep this design for some years before changing
to another PI.
I have listen to some early GA-40 on Youtube and they sound good.
Thank you all from a Spring warm, Sweden.
/Mats