Donor dopants and σ = neμe

Donor dopants in n-type silicon

An n-type semiconductor is made by adding a small amount of a pentavalent (group V) impurity such as P, As or Sb to Si or Ge.

1. The donor atom replaces a Si atom in the lattice. Four of its five valence electrons form covalent bonds with the four neighbouring Si atoms.
2. The fifth electron has no bond to join. It stays only weakly bound to the positive donor ion, like the electron in a hydrogen atom but in a medium of high permittivity ($\varepsilon_r = 11.9$) and with a small effective mass.
3. Its binding (ionisation) energy is therefore tiny, about 0.045 eV for P in Si (hydrogen model: $E_b = 13.6\,\frac{m_e^*/m_e}{\varepsilon_r^2}$ eV).
4. In the band diagram this electron occupies a donor level $E_d$ just below $E_c$ (about 0.05 eV).
5. At room temperature $kT = 0.026$ eV is enough to ionise almost every donor, so each donor gives one electron to the conduction band and becomes a fixed positive ion $D^+$. Hence $n \approx N_d$.
6. By the mass action law $np = n_i^2$, the hole concentration falls far below $n_i$, so electrons are the majority carriers.

            Ec ----------------------
                 e-  e-  e-   (free)
            Ed  - + - + - + -  ~0.05 eV
                (ionised donors)

            EF  .......... (near Ec)

            Ev ----------------------

     Si -- Si -- Si
     |     |     |
     Si -- P+ -- Si    e- (5th electron,
     |     |     |         loosely bound)
     Si -- Si -- Si

Proof of $\sigma = ne\mu_e$

Consider a bar of n-type material of cross-section $A$, with electron concentration $n$, in an electric field $E$.

1. The field accelerates electrons, but collisions with the lattice limit them to an average drift velocity $v_d$ opposite to $E$. The drift mobility is defined as

$$\mu_e = \frac{v_d}{E} \;\Rightarrow\; v_d = \mu_e E$$

(From the Drude model, $v_d = \frac{e\tau}{m_e^*}E$, so $\mu_e = \frac{e\tau}{m_e^*}$, where $\tau$ is the mean free time.)

2. In time $\Delta t$, electrons travel $v_d\Delta t$. All electrons in the volume $A v_d \Delta t$ cross a given section. Charge crossing:

$$\Delta Q = e\,n\,A\,v_d\,\Delta t$$

3. Current density:

$$J = \frac{\Delta Q}{A\,\Delta t} = n e v_d = n e \mu_e E$$

4. Ohm's law in point form is $J = \sigma E$. Comparing:

$$\sigma = n e \mu_e$$

If holes are also present, their contribution adds: $\sigma = e(n\mu_e + p\mu_h)$. In n-type material $n \gg p$, so $\sigma \approx ne\mu_e \approx N_d e\mu_e$.

Example: Si with $N_d = 10^{16}$ cm$^{-3}$, $\mu_e = 1350$ cm$^2$/Vs gives $\sigma = 10^{16}\times1.602\times10^{-19}\times1350 = 2.16$ S/cm.