So I've been working on a project for many months, or at least the concept of it with some bench experiments in between when needed.
I've pretty much gotten it all figured out thanks largely to the people on this forum.
However there is 1 small caveat that I haven't figured out yet. It's not essential to the project but it would be seriously useful if I found a solution for it.
The basic problem is this

So let's say I have 2 vacuum tubes of the same type in push pull configuration like above.
The line between them is the output and it needs to remain at 0VDC.
So what I have here is the bottom triode being biased at a specific point on the load line, let's say V1 is -75 and V3 is -100.
This means that the grid of the bottom triode is 25v more negative than the cathode and it is operating at 75v. This is a reasonably linear area on the load line for a 6080 tube.
Tubes are not like solid state devices, they vary from one to another and drift over time. Therefore I need to use the opamp integrator "servo" on the left to bias the grid of the top triode.
The integrator samples its non-inverting input and tries to make its inverting input the same by adjusting its output which is feeding the grid of the top triode.
So effectively it will adjust the grid of the top triode until the output between the two triodes is at ground level or 0vDC.
The only time there is 0vDC between them is generally when both tubes are at an equal impedance, kind of like 2 resistors on a resistor divider.
However, what I want to do here is adjust the supply voltages to both tubes but still maintain that equal impedance.
The problem here is that while the top triode can be automatically adjusted via the integrator, the bottom triodes grid is fixed.
If I use an integrator on the bottom grid as well as the top one, the bottom one will go to 0vDC and the top one will do whatever it needs to get the output to be 0vDC, which in this scenario is to cut off conduction of the bottom tube or something equally wonky.
I tried simulating a situation where the bottom triode's grid is stabilized first before letting the bottom integrator take control but it still drops down to zero while the top integrator takes control off the offset.
I do not know why this happens exactly, I would need them both to take equal control for it to work.
I tried referencing the non-inverting input of the bottom integrator to the output of a current sense IC attached to a current source and then referencing the inverting input to another current sense IC on the cathode of the bottom triode, that way the offset mechanism wouldn't be functioning off of the output voltage but the current conduction and the servo would try to make the bottom triode conduct the current expected at the spot on the loadline that I chose for both tubes, but for some reason it didn't work either. It just cuts off the tube and for some reason both the opamp inputs were at the same potential even though the currents were not the same. Perhaps a simulated current sense IC failure?
So the only other method I can think of to have an automatic bias on both tubes is to reference the voltage difference between the top grid and the output and mirror the potential difference at the bottom triodes cathode and grid.
So for example lets say the top grid was -25v and the output was zero.
I would use some sort of mechanism that would measure a 25v difference and then keep the bottom grid at a 25v difference as well.
This way I can have the integrator keep track of the grid bias requirements at different points along the load line as I adjust the input voltages and then mirror them to the bottom.

I assume this method would reach an equalibrium without wonky results but the question is what sort of equalibrium would it reach.
Assuming it did what it needed to to keep 0V at the output and kept both triodes at an equal bias setting it could easy be at any point of current along the vertical axis.
If I had the supply voltage at 100v, would it settle at a -30v grid bias or a -40v grid bias? That's a 50ma difference in current conduction.
In either case I assume I could just adjust the reference of the non-inverting input of the integrator and adjust the idle current that way.
But that still leaves me with the question of what kind of circuit I can use to sample the voltage differential of the top triodes grid and the output and then mirror that differential on the bottom triode.
Any ideas?
I've pretty much gotten it all figured out thanks largely to the people on this forum.
However there is 1 small caveat that I haven't figured out yet. It's not essential to the project but it would be seriously useful if I found a solution for it.
The basic problem is this

So let's say I have 2 vacuum tubes of the same type in push pull configuration like above.
The line between them is the output and it needs to remain at 0VDC.
So what I have here is the bottom triode being biased at a specific point on the load line, let's say V1 is -75 and V3 is -100.
This means that the grid of the bottom triode is 25v more negative than the cathode and it is operating at 75v. This is a reasonably linear area on the load line for a 6080 tube.
Tubes are not like solid state devices, they vary from one to another and drift over time. Therefore I need to use the opamp integrator "servo" on the left to bias the grid of the top triode.
The integrator samples its non-inverting input and tries to make its inverting input the same by adjusting its output which is feeding the grid of the top triode.
So effectively it will adjust the grid of the top triode until the output between the two triodes is at ground level or 0vDC.
The only time there is 0vDC between them is generally when both tubes are at an equal impedance, kind of like 2 resistors on a resistor divider.
However, what I want to do here is adjust the supply voltages to both tubes but still maintain that equal impedance.
The problem here is that while the top triode can be automatically adjusted via the integrator, the bottom triodes grid is fixed.
If I use an integrator on the bottom grid as well as the top one, the bottom one will go to 0vDC and the top one will do whatever it needs to get the output to be 0vDC, which in this scenario is to cut off conduction of the bottom tube or something equally wonky.
I tried simulating a situation where the bottom triode's grid is stabilized first before letting the bottom integrator take control but it still drops down to zero while the top integrator takes control off the offset.
I do not know why this happens exactly, I would need them both to take equal control for it to work.
I tried referencing the non-inverting input of the bottom integrator to the output of a current sense IC attached to a current source and then referencing the inverting input to another current sense IC on the cathode of the bottom triode, that way the offset mechanism wouldn't be functioning off of the output voltage but the current conduction and the servo would try to make the bottom triode conduct the current expected at the spot on the loadline that I chose for both tubes, but for some reason it didn't work either. It just cuts off the tube and for some reason both the opamp inputs were at the same potential even though the currents were not the same. Perhaps a simulated current sense IC failure?
So the only other method I can think of to have an automatic bias on both tubes is to reference the voltage difference between the top grid and the output and mirror the potential difference at the bottom triodes cathode and grid.
So for example lets say the top grid was -25v and the output was zero.
I would use some sort of mechanism that would measure a 25v difference and then keep the bottom grid at a 25v difference as well.
This way I can have the integrator keep track of the grid bias requirements at different points along the load line as I adjust the input voltages and then mirror them to the bottom.

I assume this method would reach an equalibrium without wonky results but the question is what sort of equalibrium would it reach.
Assuming it did what it needed to to keep 0V at the output and kept both triodes at an equal bias setting it could easy be at any point of current along the vertical axis.
If I had the supply voltage at 100v, would it settle at a -30v grid bias or a -40v grid bias? That's a 50ma difference in current conduction.
In either case I assume I could just adjust the reference of the non-inverting input of the integrator and adjust the idle current that way.
But that still leaves me with the question of what kind of circuit I can use to sample the voltage differential of the top triodes grid and the output and then mirror that differential on the bottom triode.
Any ideas?