2
$\begingroup$

I was wondering if there is a mathematical relation between the Seebeck coefficient of a material and its electrical resistivity (or conductance). For me it makes sense intuitively since (as far as I understand) the Seebeck coefficient and the resistivity are explained by relatively similar quantum mechanical processes.

$\endgroup$
1
$\begingroup$

It turns out that the Seebeck coefficient can't quite be written in terms of the resistivity because while the resistivity depends on the basic transport parameters (effective mass, scattering, charge density), the Seebeck coefficient depends on the relative rates of transport for electrons/holes above versus below the Fermi level. This means that many more details (or at least more assumptions) of the material are needed to compute a Seebeck coefficient. The fact that a Seebeck coefficient can be either positive or negative for materials with similar resistivities is a clue to how the complex microscopic details are critical. For example, a semiconductor could have the same resistivity whether it is n-doped vs p-doped, but these would often lead to opposite signs of Seebeck coefficient.

$\endgroup$
0
$\begingroup$

There is definitely a relation between the resistivity and the Seebeck coefficient. The fact that it may be insanely complicated does not impede its existence.

There is a well known formula called the Mott formula that relates the Seebeck coefficient to the conductivity (the inverse of the resistivity). It states that $S \propto \left ( \frac{\partial \ln (\sigma(\epsilon))}{\partial \epsilon} \right )_{\epsilon=E_\text{f}}$ for metals. There is a version of it for semiconductors and generalizations in some circumstances. It is widely used and appears in solid state/condensed matter textbooks.

$\endgroup$

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.