


Silicon heat capture could allow cheap refrigeration and energy saving










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RE: Not that universal
By MrTeal on 1/22/2008 1:25:47 AM , Rating: 3
Let's suppose we have a CPU that is cooled by water, so a high wattage of heat dissipated by the CPU equals a low temperature increase in the water. But instead of some thermal compound, let's stick a piece of cardboard between the CPU and the waterblock. Our CPU when turned off is at ambient. We turn it on, and within a second the CPU temp rises by some wattage, which equals 5 degrees. Now there is a differential of 5 degrees between it and the water. A differential of 5 degrees means 1 degree's worth of heat wattage can leak past the cardboard and into the water, which for the purposes of this remains the same. The CPU has been coolde by 1 degree so it is 4 degrees above ambient. In the next second that the CPU is on it again rises by 5 degrees worth of heat wattage. This time the differential is 9, and therefore 1.8 degrees can leak over. It's at 7.2, we add 5, and a 12.2 differential means we dissipate 2.44 degrees. And so on until the CPU is 25 degrees hotter than the ambient water temperature and now the CPU dissipates as much heat into the water as it generates.

Ok, I think here's where you're getting it confused a little. In the simple cardboard example you have, the cardboard will have a thermal conductivity, k (kappa), given in W/(m*K). The thermal resistance (theta) is = length/(k*Area). So, in this case, say the cardboard is 1mm thick (l) and has a 4cm^2 area. The whole thing has a thermal resistance of 10K/W, so for every watt of power it dissipates, the temperature will rise 10 Kelvin (or degrees C).

There will be some delay involved due to the (small) thermal mass of the CPU die, but assuming that the CPU die has perfect contact with the top of the package, and the waterblock stays at ambient the whole time (ie, it has no thermal resistance), the temperature of the CPU die will be...
Tcpu = 20C + 10(K/W)*P, where P = power dissipated and T is in celcius. So, if your CPU is dissipating 40 watts, with a crappy piece of cardboard in between it and the waterblock it'll eventually reach an equilibrium, at 420C. Ouch.

For reference here, when you're insulating a transistor in a power amplifier or some such thing, you often have to electrically isolate the back of the transistor from the heatsink, since the transistor might be hooked to a voltage and you don't want to zap someone. One of the best materials was beryllium oxide, until they stopped using it on account of the cancer and all. It had a thermal conductivity of about 0.25K/W. With just one of those crappy elastomer pads you see on heatsinks, it's about 0.1K/W. Hence why people buy the fancy thermal grease.

Let's do the same thing with a regular silicone nanowire TEC with poor heat conductivity. The CPU turns on and it raises 5 degrees. 1 degree of heat on the CPU side leaks out so it's again at +4. Now here comes the tricky part - where it leaks out to - because I don't understand the seebeck effect. One of 2 things could happen.

If you think of it as a heat engine, you have power going in from the CPU, say 100W. You have electrical power out, and being exceedingly generous I'll say 10%, so 10W. The other 90W are heat that goes to the water block. You would still be dissipating 90W, and even if the element had a resistance of 1K/W, you'd still be looking at a 100C cpu once it stabilized. All that for 10 lousy watts of power.

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