## Laser water cooling, part 1: Peltier element theory

I just ordered a laser engraver/cutter from China. Â While I wait for the slow boat to come into port, I thought it would be a good idea to start preparing the infrastructure. Â These products don’t include any hardware for the required water-cooling, other than a submersible pump. Â Most people just say “just have a few gallons of water circulating, and you’ll be fine.” Â I say “if it’s worth doing, it’s worth overdoing.”

In the spirit of overdoing it, I remembered that I had a peltier element-cooled fridge in my lab for a while, and that the parts were still kicking around somewhere. Â Peltier coolers are really neat. Â They can use an electrical current to drive a difference in temperatures on the opposite plates. Â I found an interesting page with a simplified model for calculating their performance across a range of conditions. Â It’s interesting to note that you can’t just drive it harder to move more heat. Â Because the modules are consuming energy themselves, self-heating can (and often does) overwhelm the coolers ability to move heat. Â After spending a couple days understanding the models, I discovered an error in the page, so I’m going to re-post corrected equations.

This equation computes the temperature of the “cold side” of the cooler. Â The model depends on several variables:

I = Drive current (Amps)

Rp = Cooler resistance from the data sheet (Ohms)

Q1 = Thermal load that we’re trying to cool (Watts)

C1 = Thermal conductivityÂ from load to ambient (Watts/Â°C)

Cp = Thermal conductivityÂ through the peltier (Qmax/Î”Tmax) from the data sheet

Ch = Thermal conductivityÂ of the heatsink on the hot sideÂ (Watts/Â°C)

T1 = Temperature of the object being cooledÂ (Â°C)

T3 = Temperature of the ambient environment (Â°C)

I’m going to ignore P. Â For now, it suffices to say that it’s a constant that models the peltier junction’s performance. Â (Watts/Amp)

The article that I got this model from doesn’t really explain the terms at all, so I’m going to try to interpret the pieces of it. Â First of all, the T3 term references the entire model to ambient. Â If you used something else as a counterpoise, I’ll call it, to the peltier you could remove this term and model it.

The middle term models the heat flow out of the hot-side heatsink. Â The numerators of this term are the thermal load (Q1) and the self-heating from the peltier cooler’s current (see Ohm’s law for the inspiration of this). Â The entire term is divided by the thermal conductivityÂ of the heatsink, Ch. Â Thermal conductivityÂ is a very useful specification, as it tells us the Â°C across the device per Watt. Â In the case of a heatsink, that is referenced to ambient.

The first term models the heat flow from the load through the peltier. Â Again, there is the Q1 term, as we have to get the thermal load through the device. Â Second, we have half of the peltier’s self-heating. Â My assumption is that only half of the peltier’s self-heating has to travel all the way through the device. Â Finally, the -P*I term models the active cooling (the point of this whole thing). Â These are all divided by the combination of the thermal conductivityÂ of the load-peltier junction and the peltier’s internal thermal conductivity.

The upshot of this, is that we can model the performance of our system built around a given peltier junction given load, heatsink performance, and ambient temperature.

Now, let’s talk about P. Â We can derive P entirely from information commonly found in peltier junction datasheets. Â There really isn’t much more to say about this, just plug in the values…

Now, let’s make all of this a little less abstract… Â The image above is the module that I’ve salvaged. Â It’s easy enough to lookup the model number and get the data sheet.

Using the specification table, directly from the data sheet, we can calculate P to be 14.47 (for 25Â°C). Â For now, let’s also choose some values for the other parameters. Â Let’s say we want to cool a load that is producing 10 Watts, with a rather poor heat sink that has .2 Watt perÂ Â°C of conductivity (would be listed as 5 Â°C per Watt as resistance).

If we graph T1 versus drive current, I, we can see that the optimal current from the cooler is 0.8 Amps (red line). Â Unfortunately, if we compare it to Q1/Ch (which would happen if we just put a heatsink on the load) it’s almost 6Â Â°C hotter than without the peltier cooler… Â boo. Â The laser is estimated to produce about 200 Watts (20 times more than the 10 Watt example) of heat. Â The problem, ultimately, is that the hot-side heatsink matters. Â A lot.

Now, what happens if we find a much better heatsink? Â They’re expensive, but you can find .1Â Â°C per Watt heatsinks on Digi-key. Â This would be 10 Watts perÂ Â°C in terms of conductivity. Â With the better heatsink, the cold side is down to -17Â Â°C! Â The heat sink is more than able to shunt the heat from the load and the cooler.

All that is obviously super-awesome, but what if we want to calculate what the ideal drive current is for a peltier cooler? Â The Itec equation, above, will give us this information. Â We still need the P and Rp terms from the data sheet, and the thermal conductivity of the peltier and the heatsink. Â But, you’ll notice that thermal load and the ambient temperature are not factors in the equation. Â Therefore, the ideal drive current (and, therefore maximum temperature drop) are not affected by those factors.

I hope this has been at least somewhat interesting. Â Stay tuned for part two, where I investigate whether it’s really feasible (practically and economically) to move over 200 Watts with peltier elements. Â Also, I’m planning some research into how to characterize the efficiency of radiators in liquid cooling setups. Â If you have any insight, please leave it in the comments. Â Also, I started to write a mac application to model peltier cooler systems. Â I probably won’t finish it unless it seems like something people would want. Â Leave messages in the comments if you would pay a few dollars for something like that.