So a silicon diode gives a constant 0.7V drop *independent of current*. Somehow that little factoid never registered before.
So with no load and 2 diodes in series, the inverter would see 1.3V less and would not shut down. Nifty.
Think I need two bridge rectifiers to get 1.3V drop as each one would act like a single diode? Although each rectifier has 4 diodes, there would be 2 series strings in parallel = 1 diode?
According to the "manual", the inverter should operate down to 11V before shutdown, or 12.3V at the alternator/battery with this device. That shouldn't be an issue as the alternator goes into "charge mode" at 13.9-15.5V as soon as it detects a battery V < 12.56.
I may try this with a smaller rectifier from another project first (if I can find it). The inverter's inline fuse is 8A for the cig. lighter plug&wire. (Inverter only goes up to 400W with the alligator clips and heavier wire)
And yes this is an el cheapo Canadian Tire (similar to Harbor Freight up here in the Frozen North) inverter. Have tested the inverter with a 13.8V power supply, and it does work fine as long as the input V is within range.
EDIT: just remembered that the other project diodes are Schottky because I wanted as little V drop as possible. So will prolly place an order for some of these (only 75A but much cheaper in Canada, slow boat from China though):
ebay.ca