Doped Semiconductors — Question 8

Question 8

Two-carrier Hall response can reverse before the majority population changes

In a low-field two-carrier model, $R_H=(p\mu_p^2-n\mu_n^2)/[e(p\mu_p+n\mu_n)^2]$. Take $\mu_n=3\mu_p$ and $p=4n$.

1. Determine the Hall sign even though holes are more numerous.

2. Find the population ratio for zero Hall voltage and explain why zero Hall voltage does not imply zero conductivity.

Question 8 - Solution

Two-carrier Hall response can reverse before the majority population changes

(a) Method and calculation

$$p\mu_p^2-n\mu_n^2=(4-9)n\mu_p^2=-5n\mu_p^2.$$$$\boxed{R_H=-5/(49en)<0}\quad\text{electron-like Hall sign}.$$

(b) Interpretation and further calculation

$$p\mu_p^2=n\mu_n^2\Rightarrow\boxed{p/n=(\mu_n/\mu_p)^2=9}.$$$$\boxed{\sigma=e(p\mu_p+n\mu_n)>0}\quad\text{even when }R_H=0.$$

Physical check

The Hall numerator weights mobilities quadratically, while ordinary conductivity weights them linearly. A single-carrier interpretation would misidentify the majority species.