The conductivity of semiconductors can be increased, and in some cases, decreased by many orders of magnitude through the addition of impurities, or dopants. Single-component semiconductors are from column IV, with a net valence of four. There are two possibilities for a dopant:

• Donor (n-type): If an impurity from column V can be substituded into a lattice site, and the additional valence electron can be dissociated from its parent atom, the electron becomes free to move throughout the crystal lattice. This is donor activation, providing one additional negative carrier to the lattice. The donor activation reaction is: $$ D \leftrightarrow D^+ + e^-. $$ Such a dopant is known as a donor, or \( n \)−type(negative). Examples of \( n \)-type dopants are P and As in Si or Ge.
• Acceptor (p-type): If an impurity from column III is substituted onto a lattice site, and the missing valence electron is compensated by one from the electron sea, a hole becomes free to move throughout the crystal lattice. This acceptor activation provides one additional positive charge carrier (hole) to the lattice. The reaction here is $$ A \leftrightarrow A^- +h^+. $$ Such a dopant is known as an acceptor, or \( p \)−type (positive) Examples of \( p \)-type dopants are B, Al, Ga, In in Si and Ge.

The donor atom is introduced into the crystal (through e.g. addition into the Si melt) in order for it to give up its additional electron for conduction. At zero temperature, the conduction electron remains bound; it requires finite thermal energy to dissociate the electron from the donor. An order of magnitude for the activation enery can be estimated by assuming that the electron is bound as it is to the nucleus in a (singly charged) hydrogen atom. Recalling the bonding energy or an electron in a hydrogen atom (Rydberg constant), $$ \begin{align} E_n &= -\mathbb{R}/n^2, \\ \mathcal{R} &\equiv \bigg( \frac{ \hbar e^2 }{ 8\pi\varepsilon_0 } \bigg)^2 \frac{ 1 }{ 2m_e } \frac{ 1 }{ n^2 } \end{align} $$ Recognize here (and it is good to remember) that $$ e^2/(4\pi\epsilon_0)= 14.4 \text{eV} \cdot \text{A}, $$ and combined with \( \hbar^2 / (2 m_e) = 3.81 \text{eV}\cdot \text{A}^2 \), the Rydberg constant in vacuum is \( \mathcal{R}=13.6eV \). The effect of placing th eimputity inside the crystal host is to increae the dielectric constant by a factor ϵr, which weakens the Coulomb attraction by shielding, and to decrease the kinetic energy of the electron by the renormalization factor of the effective mass, \( m' \). Assuming \( n=1 \), the binding formula then becomes $$ \Delta E_D = \mathcal{R} \frac{ m'_e }{\epsilon^2_r}, $$ The same approach is relevant for \( p \)-type dopant activation, \( \Delta E_A \), where one would substitute the hole effective mass.

Charge Neutrality: Complete and Incomplete Ionization

By Gauss's law, a sample at zero electric field mush maintain charge nuetrality, with no net charge density. The charge neutrality condition for doped semiconductors $$ n + N^-_A = p + N^+_D $$

• Complete Ionization: The best case for doping is where dopants are available whith small activation energies compared with the thermal energy, \( \Delta E_{A,D}/(k_BT) \llless 1 \). This is the case for shallow donors such as Sb in Si at room temperature, or shallow acceptors such as B in Si. Here we can write $$ \begin{align} N^+_D &= N_D \quad (n-\text{type}), \\ N^-_A &= N_A \quad (p-\text{type}), \end{align} $$ where the concentrations \( N_A \) and/or \( N_D \) are controlled by the concentration of dopants added. This implies, respectively for \( n \)- and \( p \)-type material, $$ \begin{align} n &\simeq N_D \quad (n-\text{type}), \\ p &\simeq N_A \quad (p-\text{type}). \end{align} $$
• Incomplete Ionization: We can use the Fermi-Dirac distribution functions to account for incomplete ionization of dopants. Rememer that \( f(E) \) gives the probability of occupancy for a given state by an electron. For a given level \( E_i \), \( f(E_i) \rightarrow 0 \) if \( \mu \llless E_i \) and \( f(E_i) \rightarrow 1 \) if \( \mu \gggtr E_i \). Again we can consider two cases:
  • Donor: If the donor level is occupied by an electron, the electron has not dissociated from the parent atom, and the donor is not ionized. If the donor level is unoccupied, it is ionized. Ionization is then the complement of occupancy for the donor energy \( E_g - \Delta E_D \). $$ \begin{align} N^+_D &= N_D\big(1 - f(E_g - \Delta E_D)\big), \\ N^+_D &= N_D\bigg(1 - \frac{ 1 }{ \exp(\frac{E_g-\Delta E_D - \mu}{k_BT})+1 }\bigg). \end{align} $$
  • Acceptor: If the acceptor level is unoccupied by an electron, the hole has not dissociated form the parent atom, and the acceptor is not ionized. If the acceptor level is occupied, it is ionized: an electron jumps up out of the valence band to sit on the acceptor level, leaving a mobile hole behind. Ionization is equivalent to occupancy for the acceptor energy \( \Delta E_A \): $$ \begin{align} N^-_A &= N_A f(\Delta E_A), \\ N^-_A &= N_A \frac{ 1 }{ \exp(\frac{\Delta E_A - \mu}{k_BT})+1 }. \end{align} $$

Dominant Carrier

In most cases, it makes sense to assume that the material is dominant \( n \)- or \( p \)-type. This implies that one carrier type can be neglected, e.g. for \( p \gggtr n \), $$ \begin{align} p &\simeq N^-_A, \\ p &\simeq N_A \frac{ 1 }{ \exp(\frac{\Delta E_A - \mu}{k_BT})+1 }. \end{align} $$ Now, we can substitute in or the expression in terms of the chemical potential, $$ \begin{align} p &\simeq N_A \frac{ 1 }{ e^{\Delta E_A/{k_BT}}p/N_V+1 }, \\ p &\simeq \frac{ N_A N_Ve^{-\Delta E_A/k_BT} }{ p + N_Ve^{-\Delta E_A/k_BT} }. \end{align} $$ Defining \( q \equiv e^{-\Delta E_A/(k_BT)} \), $$ \begin{align} p^2 + q N_V p - q N_A N_V &= 0, \\ \frac{ qN_V }{ 2 } \bigg( -1 \pm \sqrt{1 + 4N_A/(qN_V)} \bigg) &= p. \end{align} $$ There are two relevant limits. For low doping, the second term in the radical is negligable, so we can Taylor expand $$ \begin{align} p &= \frac{ qN_V }{ 2 } \bigg(-1 \pm \sqrt{1 + 4N_A/(qN_V)} \bigg) \simeq \frac{ qN_V }{ 2 } (-1 + 1 + 2N_A/(qN_V)), \\ p &\simeq N_A \quad N_A \llless N_V e^{-\Delta E_A/(k_BT)}, \end{align} $$ which is the limit of complete dopant ionization. For high doping, the \( N_A \) term under the radical dominates, $$ \begin{align} p &= \frac{ qN_V }{ 2 } \sqrt{4N_A/(qN_V)} \simeq \sqrt{N_A N_V q}, \\ p &\simeq \sqrt{N_AN_V} e^{-\Delta E_A/(2k_BT)} \quad N_A \gggtr N_V e^{-\Delta E_A/(k_BT)} \end{align}. $$ This shows that the effectivenes of doping becomes poor when hole concentrations beyond \( p = N_V e^{E_A/(k_BT)} \) are attempted.