Is a single thermocouple limited to generating a maximum potential difference of only ~6V at room temperature?

Question

It is well-known that the Seebeck effect, which is the fundamental operating principle of thermocouples, relates the temperature gradient across a thermocouple to the potential difference by:

$$ \Delta V = -S \Delta T $$

It is also well-known that the element with the highest melting point is tungsten, which has a melting point of $3,695 \text{ K}$. Based on my (rudimentary) understanding of the literature in the field, Seebeck coefficients of close to $1700 \mu \text{V/K}$ can be reached, but this is already considered extremely high (and perhaps close to the limits possible?), and in any case tungsten has a Seebeck coefficient of only around $2.5 \mu\text{V/K}$. Ignoring this and assuming that somehow we could get a high-seebeck-coefficient material to those temperatures, one finds that the practical maximum potential difference (via the aforementioned equation) is around $\text{6.12 V}$. This seems like an unusually low value?

Note that this is for a single thermocouple, of course multiple thermocouples together could generate a higher potential difference in combination, but for a single thermocouple this seems to be a hard limit? I can think of various ways I could be wrong:

• $\Delta V = -S \Delta T$ is an approximate equation that doesn't account for the full thermoelectric response of materials?
• NASA's MMRTG for the Perseverance rover appears to be able to reach 30 V, while the decay heat of plutonium appears to be around 700K, much less than the melting point of tungsten, so perhaps my assumptions are incorrect?
• The Seebeck coefficient is temperature-dependent, that is, $S = S(T)$ in general. Thus, while I ask about room temperature performance of materials, one may achieve higher values at higher temperatures?
• A thermocouple uses two dissimilar conductors, not a single conductor, so I should modify my result with $\Delta V = -(S_2 - S_1)\Delta T$ instead, which gives a larger (if still small, $\approx \text{10 V}$) maximum potential difference?

Any responses confirming or disproving my result are welcomed!

Other consulted SE questions:

• Is it possible to build a thermoelectric nuclear power plant?
• Absolute Seebeck coefficient of metals
• Why does Seebeck coeffcient increase with Temperature?
• Does the length of the wire in the Seebeck Effect impact the voltage generated?
• Coupling of two different metals

Answer 1

You are missing, amongst other things, that in general thermoelectric genetators are made of tens to hundreds of n-p pairs, each generating about 500 micro volts (bismuth telluride). They are not made of a single material.

Also, the fusion point of a material is unrelated to its Seebeck coefficient. Insulators have usually very high Seebeck coefficients, much higher than the one you linked to, but they are useless as thermoelectric materials because they have a high resistivity.

Edit to improve my answer without modifying my original one (above this edit):

• Yes, $V=S\Delta T$ is an approximation that holds true only in very simplified cases where the geometry is idealized and where the Seebeck coefficient does not depend on temperature. So this approximation starts to worsen as $\Delta T$ increases. Nevertheless, it's still "okayish" to get an intuition on what can be measured (the voltage).

• I do not know about this NASA TEG (thermoelectric generator) in particular, but I know it's powered by a radioactive source that acts as the hot spot, and radiation towards space acts as the cold spot, enabling this TEG to operate (thanks to the $\Delta T$). This TEG is certainly made of several n-p pairs electrically in series, thermally in parallel. The voltage generated by a pair (in the ballpark of 350 $\mu$ volts) has to be multiplied by the number of such pairs to get a more accurate estimate of how much voltage it can generate, given a $\Delta T$. Let's imagine that there are 500 n-p pairs, this generates 0.175 V per K of temperature difference. To reach 30 V, you would need a $\Delta T$ of about 170 K across the TEG. Hard, but not out of reach. I'm guessing the true numbers aren't that far from this estimate (i.e. same order of magnitude). By the way, it is not clear as to why you believe that the melting point has to do with the magnitude of the Seebeck coefficient. What makes you think there is a direct relationship, other than the fact that a greater $\Delta T$ might be applied while still in the solid state?

• Usually yes, at higher temperature the Seebeck coefficient increases. But you face other problems, such as the increase in the electrical resistivity, stability of all the materials involved. Also, the "best" commercial TE material at room temperature (bismuth telluride) is not the best at high temperatures.

• Right, with a thermocouple you indeed get a voltage reading proportional to the difference in the Seebeck coefficients of the 2 materials. Usually n and p semiconductors are chosen because they have opposite sign in their Seebeck coefficient, so that the voltage adds up instead of getting hindered, you get a higher voltage reading or a higher voltage in the case of a TEG usage.

Answer 2

Thermoelectric effect in metals is small, but there are also semiconductors available, and a Seebeck-effect refrigerator can be run in reverse (apply a thermal difference to the module and see a voltage on the terminals).

A module with multiple junctions like this example can be expected to output a few watts at 7V, given enough temperature difference on its plates. It's a multiple-in-series array of thermocouples, not a single.