THE HEATSINK GUIDE: Peltier Guide, part 3

Variables that affect TEC performance

Another thing that is important to realize is that TECs are affected by voltage and temperature. I refer to the Melcor site as a place to read up on how to calculate TEC performance based upon voltage, absolute temperature of the TEC, the number of thermocouples in the TEC, and something described as the geometric factor. I won't bother to rewrite all of those relationships - here is the link if you are interested.

You might be saying to yourself, "Dude - the manufacturer gives you that information - why bother?". Well, the fact is that TEC performance is very sensitive to temperatur; the curve you get from the manufacturer might be for constant hot side temp of 50C - your hot side temperature might be a lot different, your voltage might be different - and Q_Max and DT_Max are all tied up in that. The plot shown to the right is for a TEC with Q_Max = 72 watts @ 20C hotside. It shows the solved cold side temperature difference versus ambient using a 33 watt load (CPU heat), and a heatsink resistance of R_Heatsink of 0.5 C/W (see the heatsink information section for details about heatsink performance measurements).

The only variable is ambient temperature (ranging from 20°C to 60°C). This shows that your TEC works better when it is hotter, but moreover the total system performance changes by 8°C relative to ambient over the 40°C span! The temperature is important because it affects the Seebeck coefficient electrical resistivity of the thermocouples as well as the thermal conductivity of the substrate.

The voltage obviously is important because it affects the enforced temperature difference.

Two parameters we haven't looked at until now is the maximum allowed electrical current I_max through the device (exceeding the current will damage the TEC), and the geometry factor G.

The number of thermocouples and the geometry factor help to describe the size of the device - more thermocouples means more pathways to pump heat - the geometry factor is not explained by Melcor. They offer the factor (G) for their devices - but that doesn't help when trying to calculate performance for another manufacturer's TEC. One thing I did observe about G is that it is related to the density of thermocouples per square area and it is also related to the thickness of the TEC. After looking at the Melcor data I finally discovered that G = I_max/50. It is a perfect match for every Melcor TEC. When I went to make the above plot for the Tellurex TEC (using the Melcor relations), I had to play with G a little to get the right curve - 3.9/50 = 0.078 , but I found that G = 0.084 was about right to match the Tellurex chart.

This "empirical" determination of the geometry factor G is clearly a a hack - but it is all I have - if anyone knows the calculation of G more specifically please email me.

Melcor has a downloadable program called Aztec that can handle all this for you automatically, but I didn't like the choices of independent versus dependent variables they used. As it turns out I am trying to calculate hot side temp and cold side temp - not continually guess at what they might be - but hey - Aztec works and it's free. The other obvious problem is that it is only for Melcor TECs (but other TEC manufacturers, such as KryoTherm, offer similar software for download as well).

Thus I went to the Melcor information page and spent some looking over their equations trying to come up with a few quick rules of thumb. I finally realized that the sensible thing to do was to implement their equations into a few custom Excel VBA functions. These functions are the basis for all of the plots shown in this article - details follow later. One final note - I used the Melcor supplied values for the Seebeck coefficient, resistivity and thermal conductivity - all of this applies to Bismuth-Telluride TECs only!

For another TEC flavour we would need to adapt those values - but Bismuth-Telluride is the only material commonly used for TECs that are suitable for temperature ranges common in electronics cooling.

System Integration of TEC with CPU and Heatsink

We should know the following: CPU (or graphics chip) power output, heatsink thermal resistance, TEC parameters, ambient temperature. That is all we need.

If we don't know the CPU or GPU power output - then we look on the web, manufactures publish it. A very good page where you can find processor electrical specifications of all common CPUs is Chris Hare's Processor Electrical Specification page.

Getting your heatsink's thermal resistance could be tricky - some manufacturers specify the thermal resistance of their heatsinks, but the values are often not very precise, or "optimized" for marketing purposes. Check out the heatsink information page for more information on thermal resistance, and how to calculate it. You must know your TEC parameters, at the very least Q_Max and hopefully more.

Lets suppose we have the Tellurex TEC curve from above. Also suppose our CPU is at 15 watts (keep in mind that the power usage of current CPUs is much higher!), our heatsink has R_Heatsink = 0.5 °C/W and ambient temperature is 25°C.

Here is the most simple method:
Interpolate along the TEC curve to the CPU output (15 watts) and find that DT = 45C. Look at the total power output and see that it is about 43 watts to the heatsink. 43 watts*0.5C/W = 21.5°C. Thus the heatsink will be 25 + 21.5  = 46.5°C. The TEC is enforcing a 45°C difference and thus the cold-side temp of the TEC will be 46.5 - 45 = 1.5°C. That's pretty cold - good stuff for an overclocker; but you might encounter problems with condensation. See the last part of the Peltier Guide for details about condensation problems.

Let us now look at a less favourable example: Suppose your have a poor quality heatsink; suppose R_Heatsink is 1.5 °C/W - then your heatsink will be 65C over ambient 65+25  = 90°C and then your coldside temp would be 45°C. What if you didn't use the TEC? Then your CPU would be 15watts * 1.5°C/W = 22.5°C over ambient or at 47.5°C. In that case it is probably is not worth using the TEC because you are dumping 30 extra watts into your case and drawing 3 amps off of your power supply.

In the next part, we analyze in more detail in which situations a peltier element will help cooling, and in which situations it won't.

Go to part 4 of the Peltier Guide (or return to part 2)

Note: This article was originally written by "Bo", a visitor of The Heatsink Guide, who wishes to remain anonymous, and only slightly modified by me. Thank you Bo for sharing this article with us!