2
$\begingroup$

I'm having some trouble finding details on just how the Peltier effect (also known as the Peltier-Seebeck effect or the thermoelectric effect) works on a physical level. Am I correct in thinking that it has something to do with the small energy barriers that occur at ohmic contacts? I've searched extensively and the best information I can find is either lacking in details or hidden behind a paywall I can't afford.

What I mostly want to know here is: What causes the charge carriers on the cold side to pick up energy from the crystal lattice? What phonon-electron or phonon-hole interaction occurs?

The Seebeck effect (converting temperature to a potential difference) makes more intuitive sense to me, as diffusion-driven high-energy/high-density charge carriers move across the semiconductor (or metal), but this would seem to imply that it would be something that occurs in the bulk of the material, rather than at the junctions. And I don't know if just reversing that even makes sense to explain the Peltier effect (which is the reverse, current to temperature). Since the Peltier and Seebeck effects are considered to be the same thing operating in two different directions, a diffusion-based explanation would seem not to work.

|cite|improve this question
$\endgroup$

migrated from electronics.stackexchange.com Apr 24 '18 at 15:11

This question came from our site for electronics and electrical engineering professionals, students, and enthusiasts.

  • 1
    $\begingroup$ In semiclassical electron theory, the electronic thermal conductivity (not lattice - no phonons considered) of metals links the electrical current, thermal current, and entropy current together. The result can then be applied rather generally to the Seebeck, Peltier, and Thomson effects. Note that an explanation of the Seebeck effect based on diffusion (normally explained in the context of classical Drude free electron theory) is actually not correct, so your intuition is actually leading you astray. Again, I would suggest a good solid state physics text such as Ashcroft and Mermin. $\endgroup$ – Jon Custer Apr 24 '18 at 21:58
  • $\begingroup$ @JonCuster The Drude model based on diffusion of "classical" free electrons (or holes) is not that bad for semiconductors with low carrier concentrations. $\endgroup$ – Pieter Apr 25 '18 at 6:31
  • $\begingroup$ @Pieter what constitutes a "low" carrier concentration here? I imagine something on the order of 1e10 cm⁻³ like in (room-temperature) intrinsic crystalline Si would count? What about heavily doped Si, on the order of 1e20 cm⁻³? $\endgroup$ – Hearth Apr 25 '18 at 12:40
  • 1
    $\begingroup$ @Pieter - the Drude model is not that bad for metals some times either. Until it runs into problems, and then you look under the hood and realize where all the bodies are buried, and that they were papering over the various factors of 100 here and there that managed to mostly cancel each other out except when they didn't... $\endgroup$ – Jon Custer Apr 25 '18 at 13:31
  • $\begingroup$ @Felthry Any doping of silicon would be low enough that the charge carriers can be described with a Maxwell-Boltzmann velocity distribution. For the small-bandgap alloys that are used in thermovoltaics the situation may be a bit different, but M-B statistics is desirable for a large Seebeck coefficient. In metals (degenerate Fermi gas), the Seebeck coefficient is often too low and the thermal conductivity too high for a good figure of merit. $\endgroup$ – Pieter Apr 25 '18 at 14:14
1
$\begingroup$

Here is my understanding of the Peltier effect, obviously, grossly simplified. I'll be happy to be corrected.

When electrons cross a junction from a material with a lower electrochemical potential to a material with a higher electrochemical potential, they must lose some kinetic energy, which should lead to the cooling of the junction.

Similarly, electrons crossing from a material with a higher electrochamical potential to the material with a lower electrochemical potential must be gaining some kinetic energy and therefore cause local heating.

Although the Seebeck effect is also caused by the difference in electochemical potentials, its explanation does not directly follow from the Peltier's.

My understanding is that it is based on the difference in the dynamic equilibrium of the hot and cold junctions. Heating of a junction has a greater effect on the electrons in the material with the lower electrochemical potential and, correspondingly, causes a shift in the dynamic balance.

The resulting difference in the dynamic balance point between the two junctions forces re-distribution of electrons between the junctions, and since, this process continues as long as one of the junctions is being heated, there is a continuous re-distribution of the electrons, i.e., current.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I'm coming from an electrical engineering viewpoint here so just to make sure I'm following: By electrochemical potential, do you mean the work function? Or maybe the electron affinity? This still leaves me a bit unsure as to why this kinetic energy comes from lattice phonons... or are you saying the lattice energy is unaffected (directly anyway), and the only cooling done is of the thermal energy of the charge carriers? $\endgroup$ – Hearth Apr 25 '18 at 12:36
  • $\begingroup$ @Felthry Yes, your interpretation of my response, regarding work function/affinity and energy of electrons vs lattice, is correct. But, again, it is just an attempt to explain this phenomena in terms I am familiar with, rather than a claim that this is exactly what's happening. And, again, I'll be happy to hear a better explanation which could be understood by engineers. $\endgroup$ – V.F. Apr 25 '18 at 13:24
0
$\begingroup$

No, it does not depend on junctions. The elements work with $p$- and $n$-type materials arranged like this:

enter image description here

The electrical current goes back and forth between the hot plate and the cold plate. But the conductors are arranged in such a way that the charge carriers always go in the same direction, carrying heat. So the heat flow in this figure is upward, for all semiconducting connections.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Okay. But what makes them carry heat? Why can the cold plate be colder than ambient? That is, why do they absorb phonons instead of just getting their energy from the electric field? $\endgroup$ – Hearth Apr 24 '18 at 21:16
  • $\begingroup$ @Felthry The random velocities are larger at the hot end and lower at the cold end of the semiconductor. When there is no current, this gives the Seebeck voltage and an electric field that gives the charge carriers a drift velocity that is superimposed on their random thermal velocities. At open circuit, the diffusion and the drift current are in equilibrium. With a net current, the thermodynamics is easier than a microscopic view, but I would say that cooling of the cold end is accomplished by moving the fastest carriers in the M-B distribution to the warm end. $\endgroup$ – Pieter Apr 25 '18 at 14:28
  • $\begingroup$ Unfortunately this answer does not answer the questions risen by the asker. A bit sad that it solely focuses on a Peltier cooler/heater module instead of focusing on the Peltier effect itself. $\endgroup$ – thermomagnetic condensed boson Sep 20 '18 at 19:16
  • $\begingroup$ Also in the comment "At open circuit, the diffusion and the drift current are " do you mean a closed circuit? Because there is no current in an open circuit (unless you consider some thermal noise and things like that, I suppose). Also not sure why you focus on M-B distribution, shouldn't it be a Fermi Dirac one? For metals at least, M-B would not apply. For semiconductors it depends on the doping, etc. $\endgroup$ – thermomagnetic condensed boson Sep 20 '18 at 19:18
  • $\begingroup$ @coniferous_smellerULPBG-W8ZgjR Thermoelectric effects are small in metals because of Fermi-Dirac statistics. $\endgroup$ – Pieter Sep 20 '18 at 21:51

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.