# Clarification on the Seebeck Effect

Alright, I've been interested in the Seebeck effect lately, so I've been trying to learn it. From what I understand, this is measured with the Seebeck Coefficient, which gives you the $\mu\textrm{V}$ (Millionth of a volt) per $\textrm{K}$ (Kelvin). For example (according to this), if I take Molybdenum and Nickel, with 1 Kelvin of difference, I will produce 25 $\mu\textrm{V}$.

This is where I need clarification, is this per contact (of any size)?

I'd assume that size DOES matter, at which point I'd ask, what unit of surface area is this in? (ex: $\mu\textrm{V}/\textrm{K}/\text{cm}^2$)

The only reason why I'd think that it is per contact, is that I can't find any unit of surface area.

• – Qmechanic Jun 27 '13 at 14:43

The thermoelectric effect is the direct conversion of temperature differences to electric potential differences (Seebeck effect) and vice-versa (Peltier effect). When considering the electrical currents and heat fluxes involved, there is a size dependency, but such is not the case for the temperature differences and the electric potential differences involved. The proportionality factor between the temperature difference and the electric potential difference is a material dependent constant.

• Alright, so if I double the number of contact points, regardless of if I change the total surface area, I'll double the output? – codesmith Jun 26 '13 at 3:28
• Putting two thermoelectric devices (both identical and subject to the same temperature difference) in series will double the voltage generated. No difference to regular batteries really. – Johannes Jun 26 '13 at 4:45
• @John - I agree with Johannes. Just to be really really clear: If you have a devices-in-parallel configuration, the voltage does not change; if you have a devices-in-series configuration the voltage does change. – Steve Byrnes Jun 27 '13 at 14:28

Your first paragraph is correct. The clarification I offer is that indeed, contact size does not matter for the voltage output. Not even the length of the material(s). Thus, if one stacks a lot of small thermoelectric material chunks in a small area, one should be able to produce a relatively high voltage with a given $$\Delta T$$. This is the principle of thermoelectric generators (TEGs). In that case, each material chunk is called a leg, and they are electrically connected in series, alternating between n and p-doped semiconductors, so that the total Seebeck coefficient of the TEG is the sum of the absolute values of the Seebeck coefficient of each leg. This is how such modules can offer voltage output of about 0.1 V/K, which is way higher than $$\approx 10^1$$ or $$10^2 \mathrm{\mu V/K}$$.

There is a drawback however in using small cross-sectional sized legs. While the voltage does not depend on the size, the electric (and thermal) resistance does depend on size. So that when the TEG is part of a circuit and used a power source, the output power will be affected by the size of the thermoelectric legs making up the TEG. Maximizing power output is thus an optimization problem, where several parameters compete with or against each other.

Lastly, regarding the already accepted answer by Johannes, one has to be careful in that the Peltier effect is not the opposite of the Seebeck effect. Indeed, stictly speaking, the Peltier effect needs a current to take place, while the Seebeck effect involves a difference of electric potential. The latter does not imply an electrical current.

As an example, take an open circuit consisting of a single material. If you apply a temperature difference between each end, the Seebeck effect is going to take place, creating a non zero $$\Delta V$$ but, in the steady state, no current will flow whatsoever. The Peltier effect on the other hand requires a current, two dissimilar materials (else it's called the Thomson effect), and will manifest itself as a heating or cooling at the junction(s) between the materials. But if the temperature of these junctions are also controlled externally, it is possible to have a uniform temperature distribution across the full circuit (so $$\Delta T=0K$$) while there's a current passing through it. Clearly, the Peltier effect cannot be the opposite of the Seebeck effect, because the Seebeck effect would not take place if the whole circuit was kept at a same temperature, i.e. no voltage would (nor current) arise in that case.

Therefore, be cautious when you read sentences like the first sentence in Johannes' answer. They are widespread but not quite accurate.