I just spent a lot of time writing comments/questions to your previous post and then I saw this!
Sorry. I was kicking myself too..
I can't see doing this without the micro but maybe it's possible.
Possibly some special cases (like the 70A->80A example) would be feasible, but I agree there's still too much complexity in the general case.
Why do you think the dummy load would stay on when the car disconnected? We're using a one-shot triggered by the falling edge of the pilot to turn on the dummy circuit. Wouldn't the dummy load get turned off at the end of our timing sequence the same as any other time?
It's only an issue with the really stripped-down (analogue) version - as you point out, the problem is actually worse than that since there need to be two different dummy loads for connected-but-not-charging and charging-active cases, and it needs to transition from one to the other (as well as the case of transitioning to none at all which is what I was worrying about).
The fact that the very short stretch 'just works' without any dummy load at all (because the power supply reservoir acts as a kind of sample-and-hold, remembering what the level was at the end of the high pulse) maybe suggests other approaches.
For a wacky approach, the micro can create dummy load by running the ADC at maximum rate and modulating the CPU clock speed to achieve the required voltage - as it happens, the CPU running at max speed consumes just over the amount of current we need to burn.
Could we solve the glitch problem by using silicon transistors instead of FETs to transition to and from the dummy load? Instead of switching on a new circuit of resistors, transistors could manage the current. As I think about this, it would be more complicated and introduce other problems. A filter approach might be better.
Another idea to fix the glitch (maybe half-baked, I've only just thought of it): the power supply reservoir in practice tracks the level of the high pilot, so we can provide another FET to drive the output from our power reservoir - turn this on momentarily to cause the transition on the output, then turn on the main FET once input and output are at approximately the same level.