To go from PP to SE, ground one power tube's grid ...
That only works if you have cathode-biased output tubes.
What will always work is to have a volume control for the one side you want to "turn off" then have a coupling cap from the wiper to the grid and grid reference resistor (often a 220k in cathode biased amps, or maybe a 100k in a fixed bias amp, which then runs to the bias voltage).
If you use a volume control after the existing coupling cap, then have a new coupling cap from the volume's wiper to a new coupling cap (which is 10 times the original coupling cap value), then you can have push-pull, single-ended or anything in between.
First, you don't want to use a PP OT as a SE OT because of the way it's made the DC saturates it. BUT... if you have a PP OT and and one of the output tubes [is] idling each half of the primary still has the same "equal but opposite" DC and that's not a problem?
As I've edited it, this is correct.
If you turn off the drive signal to one side, it continues to idle and that gives equal and opposite current through both halves of the primary at idle. Power supply hum cancels and the core doesn't tend to saturate.
Regardless, forget the "ground the grid" part. This is basically right when thinking about a.c., but implies the wrong thing with regards to bias. If you had a fixed bias amp and did this, the tube with the grounded grid has 0v bias and tries to melt itself.
Second, If the amp is AB then when you simply ground the grid of one power tube then the ON tube isn't reproducing the whole signal, Basically it isn't center biased? It's just doing it's half of AB?
You don't have to change the bias ...
necessarily.
There is technically a difference between an amp that is designed from the ground up to be class A and a similar amp designed to be class AB. But
for a small enough input signal, the driven tube does not cut off. Since class A is defined as a tube being on for the whole input signal cycle, you might call this class A operation.
What a designer does to make a class AB amp is generally raise the supply voltage and/or lower the OT impedance, and raise the amount of bias voltage. The tube is idled closer to cutoff, and a relatively small signal stays class A. For a bigger input signal, one side is driven to cutoff while the other side is turned on harder. The phase inverter is designed to be able to deliver this bigger signal voltage.
So when you turn off the drive to one side, you can't drive the remaining side but so far until you get sever distortion from driving the remaining tube hard into cutoff. But you know what? You
wanted less output by running single-ended, so this isn't a problem.
Third, If the OT is a 8K CT, does this mean one tube "sees" 4K to CT and the other "sees" 4K to CT?
Yes, but only when each tube is conducting. When the tubes have a big enough drive signal that one side is driven to cutoff, the remaining side sees plate-to-plate impedance divided by 4, or 2kΩ for your example.
If you want one tube to do all the signal you'd need a different primary load.
If you wanted
maximum output power, then you'd use a different load. But you don't want single-ended operation for maximum output power, cause that's what your push-pull setup was giving you.
So your best plan is leave the speaker load alone. Best load and biasing for maximum single-ended output power is probably irrelevant, because your supply voltage is likely too high. That would have been raised to give a boost to the push-pull output power.
Again, since the SE setting is probably for lower volume and a different sound, it seems pointless to try to optimize it to get another watt or 3.
