Question is about bootstrapping the mosfet CF.
You won't be bootstrapping the MOSFET. You're
bootstrapping the previous stage's plate load.
It dawned on me how this circuit works: the same as the large apparent input resistance of a bootstrapped CF (with a grid reference returned to the junction of cathode bias and cathode load resistors).
Bootstrapping ExplainedRefer to the cathode follower diagram below. The output of the cathode follower is at the top of the 1kΩ resistor, and the 1MΩ input resistor is returned to the top of the 56kΩ cathode load.
Assume the cathode follower is a 12AX7 with a gain after the CF's feedback of 0.98. If a 1v input signal is applied to the grid, 0.98v appears at the cathode. The junction of the 1kΩ and 56kΩ will also rise about 0.98v * [56k / (56k + 1k)] = 0.96v.
The current drawn from the source though the 1MΩ will be that due to the difference between the input signal and resulting output, because they are of the same phase. 1v - 0.96v = 0.04v -> 0.04v / 1MΩ = 0.04
μA (micro-amp).
Pretend you are the previous stage, and the circuit is a black box. You have applied a 1v signal and have a resulting current draw of 0.02μA, which means the input resistance of the driven stage looks like 1v/0.04μA = 25MΩ. This
apparent resistance due to the circuit connection is the
bootstrapping effect. The presence of a portion of the in-phase follower output makes the circuit act as though a much larger resistance is present at the bootstrapped point.
Now realistically, gain may be slightly lower than 0.98 and largely depends on how much gain the device is throwing away in feedback to make the follower (the more gain, the closer to a final gain of 1, and the higher the bootstrapped resistance).
The Subject CircuitInstead of a grid reference resistor, a portion of the plate load resistor is being used as the bootstrapped resistance.
It seems the bootstrapped point (the part of the plate load connected to the follower's output) cannot be the B+ end of the plate load, because that's at AC Ground so output signal would be Zero.
The bootstrapped point cannot be at the plate of the previous stage, because there would be no difference in voltage between the input and output of the follower, so it would do nothing.
So it looks to me like tapping the plate load at the midpoint allows this circuit to function properly, and makes the portion of the plate load between the previous stage's plate and the follower output (R2 in Merlin's diagram) causes the bootstrapped resistance to both look like a very large plate load for the 1st stage and like a very large grid reference resistor for the follower.
Having talked through all this, the pentode's gain may get very large, but will be limited by its output voltage capability (does it run into cutoff or saturation). When you try it, it may be profitable to use a 50kΩ pot in place of the plate load, and adjust to find a suitable tapping point... I'm thinking you may run it towards less-than-full-effect.
This looks promising, but maybe even more so in an amp with a minimal preamp of just a pentode (and the follower) before the phase inverter.