I'm looking to add a peltier element to one of my projects. However, I want to sometimes heat and sometimes cool, depending on the ambient temperatures on both sides of the element.

Image from Wikimedia commons.

Anyway, my first thought was to use a H-Bridge and then reverse the power for reversing the flow of heat. However, looking at the diagram above, it looks like it may be polarized. (I have no idea, though.)

Conclusion one: since you'd be flipping the power, the P-type would act like a N-type and vice versa. This doesn't seem logical, but I've never taken an electronics class, so it may very well be true. P and N probably are treated differently (chemically).

Conclusion two: it is polarized because the P-type always has to be attached to the positive side (and vice versa) so you can't flip it.

Is either one right? Can I use an H-Bridge for this?

  • \$\begingroup\$ That image from Wikipedia is wrong. There isn't a 'hot' or 'cold' side - it depends how you hook it up. You can easily see that if you mirror the element shown around it center, it's exactly the same element but with 'hot' and 'cold' flipped. \$\endgroup\$ – RJR Aug 6 '14 at 1:58

A TEC is polarized in the sense that how it is connected matters. If you want to be able to heat and cool an element then a full Hbridge will work. This will allow you to pump current both directions across a TEC's terminals.

If you apply a positive voltage to a TEC in one polarization then side A will get warm and side B will cool. If you then reverse polarity then side A will cool and side B will get warm. So if you just need heat pumped in one direction then a half bridge or even direct connection will be fine.

Edit: Notice if you apply a positive voltage from one electrical connection to the other you are starting at "P". If you reverse connections you are still starting from "P" but now hot and cold will be flipped.

  • \$\begingroup\$ I want the functionality both ways, so a full H bridge would be needed. \$\endgroup\$ – Anonymous Penguin Aug 6 '14 at 2:19
  • \$\begingroup\$ @ACD, Are there linear Hbridges? I've only run TEC's from a linear signal. I've heard that PWM is not so good for TEC's, because besides heat transfer they also have some internal heating. (which you'd like to be small.) The heat transfer is linear in the current but the internal heating goes as I^2. (So PWM generates more heat for a given heat transfer.) I've never really looked into it, have you run a TEC from an Hbridge? Maybe it's not that bad? \$\endgroup\$ – George Herold Aug 6 '14 at 2:20
  • \$\begingroup\$ @George Herold if this is an issue, you could filter the output of the H-bridge and you would get a continuous voltage from 0 to Vcc. With ripples probably. But this is better than a square signal if the TEC behaves as you said. \$\endgroup\$ – Blup1980 Aug 6 '14 at 5:32
  • \$\begingroup\$ This seems to be wrong. A Peltier element is basically a chain of diodes. Breakdown reverse voltage of a diode is usually several times higher than forward opening voltage. Amount of heat absorbed is $W=PIt$, with $P$ being Peltier coefficient, $I$ the current. If you just reverse the voltage, you'll most likely not have any noticeable current, and thus your TEC won't operate at all (you'll have $W\approx0$). Also see this image, notice "Hot Side". \$\endgroup\$ – Ruslan Aug 6 '14 at 8:24
  • \$\begingroup\$ @Ruslan this is simply not true. I have designed plenty of active regulation circuits for laser diodes that both cool or heat the diode by reversing current. Generally they say "hot side" so an orientation is known. They are telling you if you apply a positive voltage from Red to Black, that side will get hot. \$\endgroup\$ – ACD Aug 6 '14 at 12:29

They are polarized, but the only "bad" thing about reversing the polarity is, the hot side gets cold and vice-versa. So you can use one side of a peltier as a heater or cooler by simply reversing the polarity. So if you were using it to cool a CPU, and have the polarity wrong, you will be heating the CPU side instead of cooling it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.