All - I'm struggling to make sense of the load that a parafeed output tube "sees". Here's how I *understand* things:
An amplified AC signal appears at the plate of the output tube. The AC signal is referenced to signal ground (e.g., the chassis) which we treat as 0V. Accordingly, there is a voltage potential difference between plate and signal ground -- and if these two points (plate and ground) are connected with a conductor, current will flow. In a choke-loaded parafeed setup, there are 2 paths to signal ground: (1) plate -> parafeed cap -> output transformer (OT) -> cathode bypass cap -> ground; and (2) plate -> plate choke (PC) -> last PSU cap -> ground.
To make things simpler, we can assume that all caps in the above two paths allow AC to pass freely, and so we can ignore them. That leaves two simplified paths to ground: (1) through the OT; and (2) through the PC. Both the OT and the PC have impedance, which for the sake of this thought experiment we can think of simply as resistance. So we have 2 parallel resistors to ground.
Let's suppose the PC has fixed 100 "resistance units" (the actual units don't matter for this thought experiment - all that matters is the ratio of units of one path to the other). The OT primary's "resistance units" (impedance) are variable and depend on the load put on the OT secondary. Let's suppose that the OT is loaded so that its primary also has 100 "resistance units". So we then have two parallel paths to ground, each of 100 units. This means that the tube "sees" an effective resistance of 50 units to ground.
This suggests that, unlike in a non-parafeed setup where there is only one path to ground via the non-parafeed OT, the presence of the PC in parallel with the OT in a parafeed setup limits the plate loading effect (the load that the tube "sees") of changes to the OT primary impedance. Example: OT primary "resistance units" are increased from 100 to 10K. In a non-parafeed setup, the tube now "sees" a 10K load. But in a parafeed setup with a PC having 100 "resistance units", the tube "sees" 100 units in parallel with 10K units, which works out to about 99 units -- i.e., at really high primary impedances, plate load = PC impedance.
Is this correct? Or am I way out to left field (again)?
MTIA, Derek