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I want to cool some electronics with peltier elements in order to increase their potential working temperature compared to the surroundings. Unfortunately the surface where these elements are distributing heat is smaller than the peltier element surface. Now I was wondering if that influences the maximum power the peltier element can "heatsink". If the surface of the device is half the surface of the element, is the maximum power being able to transported with a given $\Delta T$ half the potential maximum power of the element? Or is it still the same?

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  • \$\begingroup\$ How did you come to conclusion that you needed a peltier element? You realize that you will still need to put some kind of cooler on the hot side of the peltier, right? Chances are, you can put the very same cooler directly on the components you want to cool down. \$\endgroup\$ – Dmitry Grigoryev Oct 22 '15 at 13:41
  • \$\begingroup\$ @DmitryGrigoryev: I want to extend the usable temperature range of my modules, i.e. if they can be used in a temperature range from 0°C-40°C, I want to use them now in temperatures from -40°C to 40°C \$\endgroup\$ – arc_lupus Oct 22 '15 at 14:07
  • \$\begingroup\$ You mean you want to heat up your electronics with the peltier element if needed? \$\endgroup\$ – Dmitry Grigoryev Oct 22 '15 at 14:14
  • \$\begingroup\$ Yes, that would be one application of the peltier elements I want to add. Something I am concerned about is that the current max. temperature will be lowered. Is that true? \$\endgroup\$ – arc_lupus Oct 22 '15 at 14:27
  • \$\begingroup\$ Not necessarily. The peltier element may be able to keep your electronics' temperature within limits, while being itself hotter than that on the other side. It may effectively help if your modules are rated at 40°C, and the ambient temperature is also near that value. As for the heating, I never heard of a peltier element being used for it, simple resistors prove to be much cheaper and quite efficient. \$\endgroup\$ – Dmitry Grigoryev Oct 22 '15 at 15:20
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A Peltier element consists of two thin ceramic plates with small "cubes" of different materials, which are separated by a small airgap:

enter image description here
Broken peltier element by Frank Andre (Own work) [Public domain], via Wikimedia Commons

Heat is transferred through the cubes from one plate to the other, but not laterally from one cube to its neighbor due to the air gap. Also the ceramic would need to have an exceptional thermal conductivity to spread heat from a small spot over the entire surface, which it does not have. (I guess even copper would not be sufficient, because the plates are so thin)

Also, each cube has the same "cooling power", so the power will not increase where your hot component touches the peltier element.

So, I'd advise you to use a heat spreader, like a (thicker) plate of copper between the peltier element and the hot component.


Please also keep in mind this:

A peltier element pumps a constant amount of heat energy per second (or has a constant "pumping power") from the cool to the warm side:

$$P_{cool}=const.$$

This is one parameter given in each datasheet.

On the other side, head is also transferred back by ordinary thermal conductivity, which depends on the temperature difference:

$$P_{thermal}(T)=C\cdot\Delta T$$

As result, the actual, total (usable) pumping power also depends on temperature difference:

$$P(\Delta T)=P_{cool}-C\cdot\Delta T$$

At a certain temperature difference, both processes cancel out each other, so the actual, total pumping power is zero. This temperature difference is also given in each datasheet.

As example: A 100W peltier element with a maximum temperature difference of 60°C can remove only 50W, when the temperature difference is 30°C.

(There are more effects, but this is the most simple thing one should keep in mind)


Another note: A peltier element can get pretty cold and could cause drops of dewed water near you component. This is another reason why you should use a heat spreader

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  • \$\begingroup\$ Wouldn't that mean that if I can operate a specific device up do 40°C, and now I am adding a peltier element to it, it will not be able to operate anymore up to this temperature range (because $\Delta T$ between radiator and device is now bigger)? \$\endgroup\$ – arc_lupus Oct 22 '15 at 14:10
  • \$\begingroup\$ @arc_lupus The unpowered device is a poor thermal conductor but when powered it works better than a solid block of silver because it generated the extra thermal gradient. You have to dump your components heat and the pump generated heat so your heat sink needs to be bigger. Depending on the ambient, heat output and peltier device you need to select the heatsink to dissipate the heat. \$\endgroup\$ – KalleMP Oct 22 '15 at 18:49
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To spread the heat across the peltier surface, you might add an aluminum plate to spread the heat. The following explains why you shouldnt bother.

A peltier element is, in this order:

1.) a heater

2.) a thermal shortcut

3.) a thermal transfer unit

The full wattage of the peltier element gets added to the thermal load of the air cooler, increasing its temperature. One side of the peltier is going to be slightly cooler than the other side, probably a few degrees C, certainly not more than say 20 degrees C.

If you have, say, 20W dissipating in your ICs, and introduce 12W with your peltier, your air cooler now has to content with 32 W total, increasing its delta-T by more than half. The peltier element has to give a delta-T of this difference, which it doesnt.

One can look at the data sheet of a peltier element, say CP85338. If you look at the graph at PERFORMANCE (Th=50°C), run at 8.5A @ 8.6V, meaning it introduces 72W of power into the system.

You see it can reach 70°C delta-T, but then it does not pump any heat, it only introduces 72W of heat into the system.

Now, lets look at a real life system: 22°C Air, forced air cooled heatsink of about 0,5 K/W. We have a 72W peltier CP85338 in between the chip and the heatsink. Let´s assume all heat gets transferred via heatsink.

Chip @ 40W:
With just the heatsink, the chip would have a delta-T of 20K. With heatsink and peltier pumping 40W @ 30K difference, the heatsink has to move 112W, so it runs at +56K, and the chip at +26K.

Chip @ 100W: With just the heatsink, the chip would have a delta-T of +50K. With heatsink and peltier, the heatsink has to move 172W, so it runs at +86K. At 70W pumped, our trusty peltier now has only a difference of 5K! So our chip would run at +81K instead of 50K.

TLDR: you cant buy it preconfigured because it makes no sense.

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  • \$\begingroup\$ What do you mean with "the peltier element has to give a delta-T of this difference, which it does not"? Why not? \$\endgroup\$ – arc_lupus Oct 22 '15 at 14:11
  • \$\begingroup\$ The higher the temperature differnce across the peltier, the less efficient it is, because it is a thermal shortcut and heat flows across the peltier from the warm side to the cold side. The settling point is the balance when the thermoelectric heat flow gets counterbalanced by the thermal heat flow. \$\endgroup\$ – Posipiet Oct 23 '15 at 6:41
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All nice answers. But the most important thing in sizing a peltier is the size of the heatsink on the other side. A peltier only transfers heat and you've got to get rid of it on the other side.. if you don't get rid of it the peltier just heats up more.

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  • \$\begingroup\$ The sizing of the heat sink on the other side is not the problem, it is able to handle the heat power. \$\endgroup\$ – arc_lupus Oct 22 '15 at 20:26

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