17.2. Semiconductors in Equilibrium#

17.2.1. Structure of a semi-semiconductor#

Symbol

Concentration of

\(p\)

Holes in the valence bands of \(Si\)-atom lattice

\(n\)

Free electrons in the conduction band

\(N_A\)

Acceptor atoms, usually class \(\text{III}\)

\(N_D\)

Donor atoms, usually class \(\text{V}\)

\(N_A^-\)

Ionized acceptor atoms, having collected \(e^-\) (usually from neighboring \(Si\) atoms)

\(N_D^+\)

Ionized donor atoms, having released \(e^-\)

Usually, \(N_A\), \(N_D\), \(N_A^-\), \(N_D^+\) only depends on the space coordinate, as a consequencen of doping. The concentration of free electrons and holes may be time dependent instead (…).

17.2.1.1. Intrinsic semiconductor#

Lattice of \(\text{Si}\) atoms. Some free electrons have left the valence bands of the atoms and entered the conduction band. For a new free electron, a new hole is left behind in the lattice. If the medium is electrically neutral the numbers of free electrons and holes are locally equal, and thus their concentrations

\[n_i(\mathbf{r}) = p_i(\mathbf{r}) \ .\]

17.2.1.2. Doped semiconductors#

17.2.1.2.1. \(p\)-type#

The semiconductor is doped with atoms of class \(\text{III}\), replacing some of the \(\text{Si}\) atoms in the lattice. These atoms introduce a hole in the lattice and are prone to collect \(e^-\) from neighboring \(\text{Si}\) atoms.

Under full-ionization condition, all the doping atoms are ionized, and thus \(N_A^- = N_A\). In a electrically netural region with no donor atoms, the concentration of the holes is equal to the sum of the concentration of the ionized acceptor atoms and the free electrons,

\[p = n + N_A^- \ .\]

In a standard \(p\)-type semiconductor, \(N_A >> n_i\), and thus \(p \sim N_A^-\).

17.2.1.2.2. \(n\)-type#

The semiconductor is doped with atoms of class \(\text{V}\), replacing some of the \(\text{Si}\) atoms in the lattice. These atoms introduce a “extra” loosely bound electron in the lattice and are prone to release it in the conduction band of the semiconductor.

Under full-ionization condition, all the doping atoms are ionized, and thus \(N_D^+ = N_D\). In a electrically netural region with no acceptor atoms, the concentration of the free electrons is equal to the sum of the concentration of the ionized donor atoms and holes,

\[n = p + N_D^+ \ .\]

In a standard \(n\)-type semiconductor, \(N_D >> p_i\), and thus \(n \sim N_D^+\).

17.2.1.2.3. Full ionization#

todo

17.2.2. Electric charge density#

(17.1)#\[\rho(\mathbf{r},t) = -q n(\mathbf{r},t) + q p(\mathbf{r},t) + q N_D^+(\mathbf{r},t) - q N_A^-(\mathbf{r},t) \ ,\]

being

  • \(q\) the elementary charge, the charge of the electron (here the opposite to get a positive numerical value of \(q\) in Coulomb)

  • \(n(\mathbf{r},t)\) the number volume density of the mobile electrons

  • \(p(\mathbf{r},t)\) the number volume density of the holes in the valence bands of the lattice

  • \(N_D^+(\mathbf{r},t)\) the number volume density of the positive donor ions

  • \(N_A^-(\mathbf{r},t)\) the number volume density of the negative acceptor ions

Usually, \(N_D^+(\mathbf{r})\), \(N_A^-(\mathbf{r})\), corresponding to the density of the donor and acceptor atoms in the lattice, that have fixed positions.

Remark. todo Discuss the values of \(N_D^+\) and \(N_D\) in terms of energy levels at different temperatures. Add this discussion to full-ionization section or in another section and then point to it?