I'm only trying to determine inefficiency due to the resistors, not including the buck's own inefficiency. Are you saying the buck's own inefficiency will affect inefficiency due to the resistors? Academic?
The load is 6A max, can be less. I assume losses are greater with higher load (correct?), so let's say 6A.
Since there's no electrical way to sync these OTS (off the shelf) modules, I can only manually sync them, and hope they don't drift.
LM2596 datasheet gives 4% accuracy. MP1584 doesn't say. So, let's say 4%. At 5V out, ~0.25V max. If one buck drifts up, and the other drifts down, that's 0.5V max difference.
You said to use (I^2)*R. Using the 1A figure from my ohm's law calc above, with a 0.25 ohm R:
watts loss per volt:
(1A^2) * 0.25R
= 0.25W
At 0.5V difference:
0.25W * 0.5V = ~ 1/8W max loss
with two x LM2596 at 5V.
Correct?
Why can't I use a smaller R?
Where do I plug in the 6A?
Thx