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I have a question relating to this diagram: enter image description here

It shows the relationship between Temperature and Hall-Voltage of a P-doped Germanium Plate. The electron density n is constant from 10°C to 50°C, that's why the voltage is constant.

But Why exactly does the Voltage(Hall) decrease with higher temperatures and why is it getting negative (below 0) and increasing again. My understanding of it is somehow vague and would appreciate a detailed response.

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  • $\begingroup$ Hint: consider the hole and the electron density. (And the electron density is not constant below 50 C). $\endgroup$
    – user137289
    Commented Feb 12, 2018 at 15:41

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When you have this p-type doped Ge, you have practically only holes around room temperature so that the Hall coefficient ($R_H \propto V_H)$ is is approximately $$R_H=\frac {1}{ep} \tag 1$$ where $e$ is the electron charge, $$p=N_A$$ is the hole concentration, which is approximately equal to the acceptor doping. The electron concentration is negligible at room temperature. When you increase the temperature, you get more and more electrons in addition to holes so that also the electron conduction becomes significant. Then the simple relation (1) doesn't hold any longer. You have to replace it by $$ R_H=\frac {p-nb^2}{e(p+nb)^2} \tag 2$$ where $b=\mu_e/\mu_h$ is the ratio of the electron and hole mobility.

This explains the decrease in Hall voltage and change of sign at higher temperatures.

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  • $\begingroup$ Note that, in general, changes to the Hall coefficient should be expected as a function of temperature. For a semiconductor, changes in $n$ and $p$ are clear drivers. However, metals can also show changes in the Hall coefficient, mainly because at higher temperatures the charge carriers get to explore more of the band structure near the Fermi energy. As noted in Ashcroft and Mermin, aluminum has a region of Hall coefficient vs temperature where the primary charge carrier is positive. $\endgroup$
    – Jon Custer
    Commented Feb 12, 2018 at 16:00
  • $\begingroup$ @JonCuster - - Yes, this is very intriguing that even some metals seem also to have hole conduction. $\endgroup$
    – freecharly
    Commented Feb 12, 2018 at 16:37
  • $\begingroup$ Some level of hole conduction is common in many metals - it all depends on the shape of the Fermi surface and the band structure there. Lots of words like 'pockets of holes' can be found in papers (usually from the 1950s or 60s when people were into metallic band structure measurements). Then the details get simplified in textbooks and everybody forgets about holes in metals... $\endgroup$
    – Jon Custer
    Commented Feb 12, 2018 at 16:43
  • $\begingroup$ The formula doesn't explain the flip of the voltage sign or pretty much anything, it's just a mathematical description. Would it be possible to include the missing information in the answer? $\endgroup$
    – Nic Szerman
    Commented Dec 7, 2021 at 0:14
  • $\begingroup$ @NicSzerman - The formula explains it perfectly. In Ge the electron mobility is much larger than the hole mobility, thus b is much larger than 1. When you increase the temperature in p-doped Ge, the electron concentration n increases finally approaching the hole concentration p. Thus the value of the negative second term in the numerator of the formula becomes larger than the first term and the sign of the Hall coefficient changes from positive to negative. $\endgroup$
    – freecharly
    Commented Dec 8, 2021 at 20:27

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