# A Hall resistance coefficient vs Seebeck coefficient signs for n-p type

#### fluidistic

Gold Member
I am totally confused about whether it even makes sense to speak of n and p-types materials, when refering to whether the majority of charge carriers are electrons or holes.

We can read all over the place that the Hall effect can tell us whether a material is p or n-type, just by checking the sign of the Hall resistance $R_H$. We hear the same story regarding the Seebeck effect and the Seebeck coefficient sign. If it's positive then electrons are the majority, while if it's positive, holes are the majority.

However, what these sources do not say, is that in reality there is no reason whatsoever for $R_H$ and $S$ signs to match. Even in the simplest metal one can think of: lithium. Li displays a positive $S$ indicating a p-type material while the Hall resistance is negative, indicating an n-type material, for the same temperature. What gives? What's going on?

Some highly cited sources claim that the Seebeck coefficient yields the correct answer while the Hall resistance coefficient doesn't:
Highly cited paper 1 said:
In contrast to the Hall effect, the sign of the Seebeck effect unambiguously corresponds to the prevailing type of charge carriers, irrespective of the mechanism of their transport
("Reliable measurement of Seebeck coefficient in semiconductors").

But then, other highly cited papers claim the very opposite:
Highly cited paper 2 said:
Lithium is one of the simplest metals, with negative charge carriers and a close reproduction of free electron dispersion. Experimentally, however, Li is one of a handful of elemental solids (along with Cu, Ag, Au etc.) where the sign of the Seebeck coefficient (S) is opposite to that of the carrier.
(https://arxiv.org/abs/1311.6805)

I do not understand anything. I do understand that $R_H$ and $S$ can be calculated from first principles and yield different signs, but I do not understand how this relates to the charge carriers majority. Something is very fishy here.

On another book (Fundamentals of thermoelectricity by Behnia), they mention that the Hall effect and the Seebeck effects probe different electrons, i.e. electrons that aren't in the same places in the Fermi surface. Maybe that's a hint towards reaching some understanding of what's going on. But I do not understand the picture of that chapter, the y-axis is not labelled and there is no description in the text to fill in the conceptual gap.

So what is going on? How on Earth can we fix this contradiction? Who is right, if anyone? Are the claim both wrong? Incomplete (as I suspect)? Does it even make sense to speak of an n and p-type material?

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#### Fred Wright

In a semiconductor at high enough temperature electrons will have high enough energy to excite across the band gap. When that happens there will be both n-type carriers in the conduction band and p-type carriers in the valence band such that the resultant thermopower (absolute value of Seebeck coefficient) will be compensated (reduced) because the two contributions subtract. In a heavily doped semiconductor, where the dopants produce many majority carriers (could be either n-type or p-type) the thermopower will be reduced at high temperature due to the excitation of minority carriers of opposite sign. Although there are fewer minority carriers than majority carriers, they have a larger thermopower.

#### fluidistic

Gold Member
In a semiconductor at high enough temperature electrons will have high enough energy to excite across the band gap. When that happens there will be both n-type carriers in the conduction band and p-type carriers in the valence band such that the resultant thermopower (absolute value of Seebeck coefficient) will be compensated (reduced) because the two contributions subtract. In a heavily doped semiconductor, where the dopants produce many majority carriers (could be either n-type or p-type) the thermopower will be reduced at high temperature due to the excitation of minority carriers of opposite sign. Although there are fewer minority carriers than majority carriers, they have a larger thermopower.
I appreciate your insight, but I fail to see how this helps to answer my questions in any way.
In short, you are saying that in semiconductors, S(T) goes to 0 when T increases.
While in heavily-doped semiconductors, S(T) will decrease (or increase) when T increases, depending on the minority carriers (either holes or electrons).

But I feel this completely misses the point of my questions. It seems that you are assuming a direct link between the Seebeck coefficient sign and the charge carrier type (like Datta and Lundstrom do) exists. Many authors claim that the sign of the Hall coefficient of resistance is the one that indicates whether the material is of n or p-type, etc. I do not want to repeat what I wrote in my first post.

#### fluidistic

Gold Member
I have seen that, I am actually the asker of that question on that website! I thought the question was answered until I found out that many authors had a different take on it, as I detail here on PF (and as I mention to Jon Custer who took a look at the references I suppose, but I have had no feedback).

"Hall resistance coefficient vs Seebeck coefficient signs for n-p type"

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