Vref = 0.5v, phase current = 4A therefore sense resistor is 0.5/4 = 0.125ohm and dissipates 0.5 * 4 W = 2W so need to be 5W rated.
If it only = 2W then why do the res need to be 5W? I thought vref was up to 1v as well? if so then 1/4 = 0.25 hence the two 0.5r, but this would mean max current is 4W? probably got it wrong again :whistling:

Theres not much available in 2 or 5W resistors in that range as they cant be wirewound.......Why is it possible to get a 2W smd res 30p? dosnt make sense.

For the heat sinks I was going to reuse the originals, their about 50mm square and 140mm long with fan at each end. The 6203's mount on both sides with the fins top and bottom.

Does the shunt regulator have to be on each board or could I just put one on the PSU distribution board?