Quote:
Originally Posted by MPaulHolmes
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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Sounds like a good estimate. It's a little conservative since you're not considering the heat transfer of the cooling system (fins or water cooling or whatever) that is taking heat out of the heat spreader. Also, if the turn on/off times are less, switching losses could be a lot less. So, that 2000W could be a lot lower. Of course, you're also not considering all the thermal resistances of the mosfet to the heat spreader and it's effect dynamically on the junction temperature. It should be lower with the anodizing scheme than the thermal pad. Very interesting...