brulaz wrote:
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?
With a bridge rectifier wired up in the usual (obvious) way, with the "DC Output" corners going to the inverter and the "AC Input" side to the power supply, at any given time two of the diodes are forward biased and two are reverse biased. The forward biased ones each have a ca. 0.7V drop across them, whatever the forward voltage of the particular diode is, and are in series. The two reverse biased ones have basically the full supply voltage across them but of course are not conducting since they're reverse biased and can basically be ignored.
Parallel connections do not come into play here as they would for resistors; diodes are by design not ohmic. Two forward biased diodes in parallel still have the same 0.7V drop, but each one carries (ideally) half the current. It's the same idea as putting two batteries in parallel; the voltage stays the same.
8A implies a bit under 15W of power dissipation. A heat sink of some sort would seem to me to be a good idea.