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I was looking in particular at Constantan with platinum when I was wondering how could something have a negative Seebeck coefficient.

I presumed that it must be linked to hole movement throughout the metal as opposed to electron movement but now I am wondering whether, with any pair of different metals, one would always have a positive Seebeck coefficient with regards to the second whilst the second would have a negative Seebeck coefficient with regards to the first. Would this be the case?

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    $\begingroup$ "Seebeck coefficient is X" is meaningless. Measurement always involves a pair of different metals. $\endgroup$ – Pieter Feb 17 '18 at 17:47
  • $\begingroup$ @Pieter Oh yes sorry, I meant relative to platinum for the example with constantan. However, I am struggling to see why it would ever be negative, regardless of any pairings of metals. $\endgroup$ – Terminus Est Feb 17 '18 at 19:13
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    $\begingroup$ For any pair of metals, if there is an effect for A relative to B, the Seebeck coefficient is negative to B with respect to A. So if you understand the positive effect, you understand the negative sign too. $\endgroup$ – Pieter Feb 17 '18 at 19:18
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    $\begingroup$ A simple explanation works when the charge carrier density is so low that the material can be regarded as a classical gas of electrons or holes. This is the case in relatively lightly doped semiconductors. The Seebeck coefficient is large then. For a Fermi gas, the effect is orders of magnitude smaller, requires different theory. $\endgroup$ – Pieter Feb 17 '18 at 20:31
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    $\begingroup$ could you update your question to reflect anything you've learned in this discussion? $\endgroup$ – pentane Feb 18 '18 at 1:04

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