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We performed an undergrad experiment where we looked at the resistance $\rho$ and Hall constant $R_\text H$ of a doped InAs semiconductor with the van der Pauw method. Then we cooled it down to around 40 K and did temperature-dependent measurements up to around 270 K. We were asked to create the following three plots from our measurements and interpret them.

This is conductivity $\sigma = 1 / \rho$ versus the inverse temperature $T^{-1}$. I see that increasing the temperature (to the left) decreases the conductivity. I do understand that higher temperatures do that since the electrons (or holes) have more resistance due to phonon scattering. However, since higher temperatures mean a higher amount of free electrons, I would think that $\sigma$ should go up, not down.

http://chaos.stw-bonn.de/users/mu/uploads/2013-12-07/plot1.png

The density of holes $p = 1/(e R_\text H)$ does increase with the temperature, that is what I would expect:

http://chaos.stw-bonn.de/users/mu/uploads/2013-12-07/plot2.png

And the electron mobility $\mu = \sigma R_\text H$ decreases with the temperature as well:

http://chaos.stw-bonn.de/users/mu/uploads/2013-12-07/plot3.png

Now, I am little surprised that even though $p$ goes up with $T$, $\mu$ and $\sigma$ go down with $T$. Are the effects of phonon scattering and other things that increase the resistance that strong?

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  • $\begingroup$ Maybe the "more free electron" part is somewhere near 20K? Maybe it is a single-crystal so the direction of flow is important? Maybe 4-point-method or whatever method you use for I/V characteristic measurement has some flaws such that contact points generate reverse-effect(s)? $\endgroup$ Commented Dec 7, 2013 at 20:17
  • $\begingroup$ I assume that we used the 4-point-method, it was called van der Pauw in the manual. I added it in the text. The semi-conductor was InAs. We did 8 measurements with various directions of current and voltage for the resistance and 10 for the Hall constant, with changed direction of the magnetic field and with no field at all. $\endgroup$ Commented Dec 7, 2013 at 20:36
  • $\begingroup$ Oh, and if the “more free electron” part is around 20 K, why does $p$ go up when $T$ goes over 100 K? $\endgroup$ Commented Dec 7, 2013 at 20:37
  • $\begingroup$ Yes, Im wrong about 20K part as it cannot be that low.Im sorry. What about irregularities between metal and semiconductor caused by doping? $\endgroup$ Commented Dec 7, 2013 at 20:57
  • $\begingroup$ I would understand that $p$ does not rise above a given value. However, it does not fall below a certain value. What do you mean with those irregularities? $\endgroup$ Commented Dec 7, 2013 at 22:33

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Phonon scattering goes up a lot as temperature increases -- faster than electron numbers increase in the conduction band.

Keep in mind that phonons obey the Bose-Einstein distribution, so their numbers scale like

$$N_{BE}=\frac{1}{e^{\frac{\hbar\omega}{k_b T}}-1}$$

In the large $T$ limit, this becomes

$$\frac{k_b T}{\hbar\omega}$$

So their numbers roughly scale linearly with temperature at "high temperature". For phonons, "high temperature" means above the Debye temperature, but that's only ~650K for silicon; you're a good chunk of the way there at room temperature.

However, electrons follow a Fermi-Dirac distribution, so you'd expect their numbers to scale like.

$$N_{FD}=\frac{1}{e^{\frac{\epsilon}{k_b T}}+1}$$

In the large T limit, this goes to $\frac{1}{2}$.

There's also a chemical potential for the electrons that limits their numbers. Phonons have no such restriction; given the energy, you can have as many phonons as you want.

Even if you're not talking about high temperatures, note that the $N_{BE}>N_{FD}$ is always true.

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