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I am using this peltier device to ramp an aluminum plate down to -10C in a vacuum environment. I chose this particular device because it had the largest 'Qout' of any single-stage peltier I could find (400W!!), however I am having a hard time achieving my goal of -10C.

I am generally operating at about 20V and 14A with a large heat sink on the back (3"x3" Al water block feeding to a 10-gal ice water reservoir AKA fishtank). In this configuration, I can usually achieve between -5 and +2C, but never -10C.

Now I wonder if I would have better luck using a multi-stage peltier? Still becoming familiar with the technology, but from what I can tell, you tend to sacrifice Qout in exchange for increasing dT and vise versa.

If so, could I expect to see my cold-side temp drop if I switched to something like this?

Thanks!

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  • \$\begingroup\$ Something seems wrong. By the datasheet you should have I=22A at V=20V. Are you measuring the voltage at the tec? \$\endgroup\$ – τεκ Jan 18 '18 at 18:37
  • \$\begingroup\$ That was something that I noticed as well. No, I am simply reading the output from my master power supply. I will measure later today and update with that value. There is about 30 feet of wiring between my power supplies and the TEC @τεκ \$\endgroup\$ – Austin Prater Jan 18 '18 at 19:25
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You can see the problem of multi-stage if you look at your original device. Your device requires 700 W (28 V at 24.8 A), and if you are pulling heat out of your load at 400 W, your back side heat sink (or second device) must dissipate 700 + 400 W or 1100 W. So stacking will give you diminishing returns.

Your present approach should be able to get you to -10 if you can improve heat removal from your heat sink. You could try increasing the flow rate or getting a bigger block, or both. You will need to keep the hot side of your device down around 40 degrees C.

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  • \$\begingroup\$ Thanks @JohnBirckhead -- I'll try tossing a small heat sink on the back-side of the water block. I'm currently using ice water to feed into the block, but maybe I should look into using some sort of refrigerant, instead? \$\endgroup\$ – Austin Prater Jan 18 '18 at 20:43
  • \$\begingroup\$ Kind of defeats the purpose. Remember that you are absorbing 1100 W, (like a hair dryer on high) so you'd better be melting your ice in the fish tank pretty fast. You need to improve the thermal conductivity between the water and the device. I would try some thermal grease between the plate and the block first if you haven't already. How hot is the block? \$\endgroup\$ – John Birckhead Jan 18 '18 at 22:43
  • \$\begingroup\$ Hey John, just wanted to update this thread. This whole system is meant to perform inside a large vacuum chamber. I've been doing all of these tests at atmosphere -- and while I knew that being in vacuum would help minimize my losses, I didn't think it'd help as much as it did. Upon pumping down to 10E-3 Torr I've been able to achieve -10C easily! The healthy layer of silicone thermal grease helped a great deal, too. Thanks! \$\endgroup\$ – Austin Prater Jan 23 '18 at 17:55
  • \$\begingroup\$ Glad to hear it! \$\endgroup\$ – John Birckhead Jan 23 '18 at 19:14
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Yes. In fact, this is done quite often in ultra-low-noise image sensors used in bio-analysis and such (we are talking massively expensive sensors that are very, very sensitive).

Similar question: Can Peltier devices be cascaded?

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  • \$\begingroup\$ Thanks for the input Joren Vaes. I guess I'm not fully understanding how to interpret each particular peltier's 'performance parameters'. I want to achieve a low temperature, which means I need a large dT. However, in order to remove that heat from the system, I also need a large heat pumping capability. How can these parameters have an inverse relationship? It seems that high dT capability and high Qout capability should be synonymous? Do I want a peltier that is moderately rated in both of these areas and not biased to Qout or dT? Thanks for the help. \$\endgroup\$ – Austin Prater Jan 18 '18 at 18:45
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    \$\begingroup\$ The peltier element has to "pump" the heat Q against the temperature difference. This is easier the smaller the temperature difference is. \$\endgroup\$ – Janka Jan 18 '18 at 19:51

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