The charge carriers in semiconductors are electrons ($n$ from "negative" with charge $-e$) and holes ($p$ from "positive" with charge $e$), whose chaotic thermal motion is slightly distorted by external fields. The elementary charge $e=1.60 \cdot 10^{-19}~\text{C}$. To describe the transfer processes, the carrier mobility $\mu$ is introduced, which determines the dependence of its average velocity $v$ on the applied external electric field $\mathcal{E}$:
\[ |v| = \mu |\mathcal{E}|.\]The mobility depends on the temperature and the concentration of defects in the crystal lattice. Thus the volume concentration of electrons $n$ (dimension $1/\text{m}^3$) and the volume concentration of holes $p$ (dimension $1/\text{m}^3$) in a semiconductor with impurities, depend on the temperature and the change in Fermi energy $\Delta E_F$, which characterizes the influence of impurities, as follows:
\[ n = n_i \sqrt{\frac{T}{T_0}} e^{\frac{\Delta E_F}{k_\text{B}T}}, \quad p = n_i \sqrt{\frac{T}{T_0}} e^{-\frac{\Delta E_F}{k_\text{B}T}}.\]For silicon, $n_i = 1.1 \cdot 10^{10}~\text{cm}^{-3}$ and $T_0 = 300~\text{K}$.
Using Ohm's law $R = U/I = \rho l/S$, express the resistivity $\rho$ of a semiconductor through $\mu_n$, $\mu_p$, $n$, and $p$.
Semiconductors are often doped, i.e. impurities ($|\Delta E_F| \gg k_B T$) are added to them so that some carriers can be neglected relative to others. A doped semiconductor is called by the letter of the dominant carrier: $p-\mathrm{Si}$ for silicon which has significantly more holes than electrons and $n-\mathrm{Si}$ for silicon which has significantly more electrons than holes. Impurities that shift the Fermi level so that there are more electrons than holes are called donors and their concentration is denoted by $N_D$. Conversely, impurities that increase the concentration of holes are called acceptors and are denoted by $N_A$. In this case, the condition of electroneutrality is satisfied
\[ n - N_D = p - N_A, \]which, if the condition $e^{|\Delta E_F| / k_B T} \gg 1$ leads to the expression $p=N_A - N_D$ in the $p$ doped semiconductor and to the expression $n=N_D - N_A$ in the $n$ doped semiconductor.
Thus, in $p-\mathrm{Si}$ and $n-\mathrm{Si}$ the sign of the charge of the charge carriers is different, hence the Hall voltage $V_H$ will have a different sign. Consider a rectangular sample of a doped semiconductor with thickness $W=0.50~\text{mm}$ and cross-sectional area $A=2.5\cdot 10^{-3}~\text{cm}^2$ be in a magnetic field $B=1.0\cdot 10^{-4}~\text{T}$ and a current $I=1.0~\text{mA}$ flowing through it.
