'273:
I did not mean that you had the theory wrong - you in fact agreed with the diode's increased voltage drop under load. I was referencing the OP's failure to understand the voltage drop.
You are arguing with words you put in the OP's mouth. I never said the voltage across the diode drops with increasing current output. In fact it increases, just not near as fast as it would for the wire shunt you suggest. I said that the (effective) resistance across the diodes decreases with current output, which is good. The ideal diode would start breaking down at 0.7 V and hold that voltage drop as the current increased further, acting like the perfect "pressure relief valve", but I challenge anyone to find such a component that would work at these high currents.
I already answered your question about referencing Vreg in the wrong place. I still have the factory-designed path thru the ammeter to the battery. I just added an additional current path that bypasses the ammeter when needed.
