In a college laboratory I have been looking at the Hall effect in an undoped Germanium crystal. The Hall voltage induced on the crystal is known to be $$V_H = \frac{R_HBI}{t},$$ where B is the perpendicular magnetic field strength, I is the current across the crystal, $t$ is the hall element thickness and $R_H$ is the Hall coefficient given by

$$R_H = \frac{1}{pe} - \frac{1}{ne},$$ with $e$ being the electron charge and $p$ and $n$ being the carrier density for the positive and negative charge carriers respectively. I keep finding that my Hall voltage seems to be invariant to the change in magnetic field. It remains at some constant negative voltage for all values of $B$ (from 0 to 0.07 Tesla). Does anyone know why this happens? It seems contradictory to the formula I have been given. I hypothesise it is because increasing the magnetic field causes the 'holes' and the electrons to increase their positive or negative contribution to the Hall voltage respectively, so they effectively cancel each other out. Then the only reason we obtain a constant negative hall voltage is due to the generally higher mobility of the holes. However, I am unable to rectify this answer with the formula.

A hypothesis above would also adequately explain the invariance of the Hall voltage to changes in temperature, which is what was also found in the experiment.

  • $\begingroup$ Are you scanning $B$ or $I$ or both. Also, did you cross check your hypothesis with Ge properties? $\endgroup$ Commented Sep 7, 2016 at 12:07
  • $\begingroup$ I'm just scanning B. Additionally, how would I go about cross checking my hypothesis with these charts, none seem to be relating the applied magnetic field and carrier density or hall voltage. $\endgroup$
    – Cococabana
    Commented Sep 7, 2016 at 12:14
  • $\begingroup$ Well, I had in mind: more current -> higher temperature, so mobility vs temp,, e.g. could have been of interest (as before the edit, you wrote current) Now that it is the field. I don't think that in this range the field would change neither $p$ nor $n$ $\endgroup$ Commented Sep 7, 2016 at 12:35
  • $\begingroup$ How high is your constant voltage (at $B=0$ as well?) compared expected values putting your geometry and literature values for $p$ and $n$. Is it possible that you have $p=n$ and the constant voltage is zero within error bars? $\endgroup$ Commented Sep 7, 2016 at 12:39

1 Answer 1


In semiconductors with both electrons and holes contributing to the Hall effect, the above formula for the Hall coefficient is false. Even for equal electron and hole concentrations, n and p, the effects of the positive and negative charge carriers on the Hall coefficient will, in general, not cancel because the electron and hole mobilities enter the correct formula for the Hall coefficient. This can be seen in the correct formula for the Hall coefficient including the simultaneous effects of electrons and holes (s. e.g. Wikipedia Hall effect). Germanium has a much larger electron mobility than hole mobility. Therefore a negative Hall coefficient can be expected, even for n = p explaining the experiments as suspected.


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