Quote:
Originally Posted by MPaulHolmes
I'm almost done with a power section layout for the SR controller. I'm sending it to Advanced Circuits, since they do stuff with 4 ounce copper now. The plan is to drill holes and mill the sides and top flush of an approximately 10"x1"x2" aluminum block, send it in to be type 3 anodized, and use that as the heat spreader. The company that does it can do 1 to 2 mil thick anodizing, and they say it's 2000-3000V/mil of isolation, and as hard as saphire! The SR control boards should be here either tomorrow or Thursday.
Oh I just did some figurin' about how much energy the aluminum block can absorb. That will give me a loose idea as to how much energy can be dumped into it in short bursts:
Specific Heat Capacity of Aluminum is 900 J/(m*degC). My Aluminum block will have a mass of 0.889 kg. If I consider a change of 30 degC for the aluminum block to be acceptable for short periods of time, the amount of energy the block can absorb is:
900*0.889*30 J = 24000 J.
Now, at 1000amps, if it's wasting 2000 watts of power (oh please don't be that bad!), it will take the heat spreader 12 seconds before the temperature has raised by 30 degC. At that point, the software can throttle back a bit, and allow the energy to slowly seep away.
I'm not sure if my reasoning is right, but it makes sense to me! Please tell me if I'm not understanding something! ya!
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Great news about the anodizing! Get the thinnest layer they offer.
I'm worried about using energy calculations for cooling design. But, as mentioned earlier, I'm sleep deprived and may be missing something. The thermal resistance of the heat spreader will cause the temperature near the MOSFETs to be much higher than that at the other end of the spreader. Don't the energy calculations assume that the heat is spread through the entire header immediately?
If true, then the aluminum spreader can take 24KJ but the temperature at the MOSFETs might be 300C and the other end of the spreader might be 30C...until it all reaches thermal equilibrium, that is.