Quote Originally Posted by qurious63ss View Post
So you measure temp at the base of heatsink? So that means the better your heatskink is at dissipating the heat, the lower the temp at the base of the heatsink should be, correct? So how does heat translate into the PIE formula? Power is still a function of current and voltage so how would lowering the temp of the heatsink affect power draw?
A CPU is not a resistive load, and actual draw varies from one microsecond to the next. There is also no way to determine 'I' for a CPU unless you instrument the VRM. So it's not practical for most people to use P=I*E since they can only guess at 'I'. We do know that a CPU does no mechanical work, so all of the input power is turned into heat. Basically the heatsink method gives that power directly.

Since we know (or can determine) the thermal resistance of the heatsink, we can determine the power dissipated by applying the thermal resistance Tr (in C/W) where C is the difference between the base and ambient temperature. The formula is Crise / Tr = Wload. The actual temperature at the base, along with ambient, is only used to determine the rise.

Lowering the temperature at the base will have some effect on total power, but not very much unless the change is large. It's always best to use a big cooler, but the goal is to get rid of the heat, not reduce power.

If for some reason you want to calculate amps input, and you know power (from the heat load) and E (from the Vcore) you could use I=P/E to find amps... but unless you are designing a VRM, I'm not sure why you would do that.