Fermi-Dirac distribution

The Fermi-Dirac distribution gives the probability that a quantum state of energy $E$ is occupied by an electron, or by any other fermion, in thermal equilibrium at temperature $T$:

$$f(E)=\frac{1}{1+\exp [(E-E_F)\mathrm{/}{k}_{B}T]},$$

where $E_F$ is the Fermi level and $k_{B}T$ the thermal energy, 25.85 meV at 300 K. Because no two electrons can occupy the same state, states far below $E_F$ are filled and states far above it are empty, and temperature blurs the boundary between them over a few $k_{B}T$. Multiplied by the density of states, the distribution gives the number of electrons in a semiconductor's conduction band and of holes in its valence band. Applied separately to electrons and holes under injection, through quasi-Fermi levels, it decides where a laser or LED has optical gain and how wide its emission spectrum is.

Values and shape

The occupation is exactly one half at $E_F$ at any temperature above absolute zero, and the distribution is symmetric about that point: a state an energy $\mathrm{\Delta }E$ below $E_F$ is empty with the same probability that a state $\mathrm{\Delta }E$ above it is filled, $1-f(E_F-\mathrm{\Delta }E)=f(E_F+\mathrm{\Delta }E)$. At 300 K:

The occupation falls from 0.9 to 0.1 over $2\ln 9k_{B}T=4.39k_{B}T$, 114 meV at 300 K. Its derivative $-df\mathrm{/}dE$ peaks at $E_F$ with the value $1\mathrm{/}(4k_{B}T)$ and has a full width at half maximum of $3.53k_{B}T$, 91 meV. At absolute zero the distribution becomes a step, with every state below $E_F$ filled and every state above it empty. Temperature changes the tail most: a state 0.1 eV above $E_F$ is occupied with probability $2.9\times {10}^{-7}$ at 77 K, 0.020 at 300 K and 0.052 at 400 K.

Where it comes from

The distribution follows from the exclusion principle. A single-particle state of energy $E$ in contact with a reservoir at temperature $T$ and chemical potential $\mu$ holds either no fermion or one, so its grand partition function has only two terms, $1+e^{-(E-\mu )\mathrm{/}{k}_{B}T}$, and its mean occupation is

$$⟨n⟩=\frac{e^{-(E-\mu )\mathrm{/}{k}_{B}T}}{1+e^{-(E-\mu )\mathrm{/}{k}_{B}T}}=\frac{1}{e^{(E-\mu )\mathrm{/}{k}_{B}T}+1}\mathrm{.}$$

In semiconductor physics the chemical potential of the electrons is called the Fermi level, $E_F=\mu$. For bosons, any number of which can occupy one state, the sum runs over all occupation numbers and gives the Bose-Einstein distribution $1\mathrm{/}(e^{(E-\mu )\mathrm{/}{k}_{B}T}-1)$; photons in thermal equilibrium have $\mu =0$, which leads to Planck's law of blackbody radiation. Far above $\mu$ both reduce to the Maxwell-Boltzmann factor $e^{-(E-\mu )\mathrm{/}{k}_{B}T}$:

A Fermi-Dirac occupation never exceeds 1, while the Bose-Einstein occupation grows without limit as $E$ approaches $\mu$.

The Boltzmann approximation

When $E-E_F$ is several $k_{B}T$, the 1 in the denominator is negligible and $f(E)\approx e^{-(E-E_F)\mathrm{/}{k}_{B}T}$. The approximation overestimates the occupation by the factor $1+e^{-(E-E_F)\mathrm{/}{k}_{B}T}$, by 5.0% at $3k_{B}T$ and 0.67% at $5k_{B}T$. In a non-degenerate semiconductor the Fermi level lies in the gap, several $k_{B}T$ from both band edges, so every conduction-band state is in this tail, and integrating over the density of states gives

$$\begin{array}{rl}{\displaystyle n}& {\displaystyle =N_c{e}^{-(E_c-E_F)\mathrm{/}{k}_{B}T},}\\ {\displaystyle p}& {\displaystyle =N_v{e}^{-(E_F-E_v)\mathrm{/}{k}_{B}T},}\\ {\displaystyle np}& {\displaystyle =N_c{N}_v{e}^{-E_g\mathrm{/}{k}_{B}T},}\end{array}$$

with the effective densities of states $N_c$ and $N_v$ tabulated in the density of states entry. The product $np=n_i^2$ does not depend on the Fermi level, and its exponential dependence on the bandgap makes thermally generated currents very sensitive to temperature: the diffusion part of an InGaAs photodiode's dark current, proportional to $n_i^2$, doubles about every 6.5 K near room temperature, and the generation part, proportional to $n_i$, about every 13 K.

Carrier density and degeneracy

Without the approximation, the electron density in a parabolic band is

$$\begin{array}{c}{\displaystyle n=N_c{\mathcal{F}}_{1\mathrm{/}2}(\eta ),\eta =\frac{E_F-E_c}{k_{B}T},}\\ {\displaystyle {\mathcal{F}}_{1\mathrm{/}2}(\eta )=\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{x^{1\mathrm{/}2}dx}{1+e^{x-\eta }},}\end{array}$$

where ${\mathcal{F}}_{1\mathrm{/}2}$ is the Fermi-Dirac integral of order one half, normalized so that ${\mathcal{F}}_{1\mathrm{/}2}(\eta )\to e^{\eta }$ for $\eta \ll -1$. The Boltzmann formula overestimates the density at a given Fermi level by 1.7% at $\eta =-3$, 12% at $\eta =-1$ and 31% at $\eta =0$, where $n=0.765N_c$; it stays within 5% while $n$ is below $0.14N_c$. Inverting the relation gives the Fermi level for a given density:

Once $n$ exceeds $0.77N_c$ the Fermi level lies inside the band and the semiconductor is degenerate. In GaAs, whose light electron effective mass gives $N_c=4.7\times {10}^{17}$ cm⁻³, the electrons become degenerate at densities reached in lasers and heavily doped contacts. The Joyce-Dixon approximation,

$$\begin{array}{rl}{\displaystyle \eta }& {\displaystyle \approx \ln r+\frac{r}{\sqrt{8}}-4.950\times {10}^{-3}{r}^2}\\ {\displaystyle }& {\displaystyle +1.484\times {10}^{-4}{r}^3-4.426\times {10}^{-6}{r}^4,}\\ {\displaystyle r}& {\displaystyle =\frac{n}{N_c},}\end{array}$$

reproduces the exact values to within 0.001 up to $r=5$ and 0.01 at $r=10$.

Deep in the degenerate regime, ${\mathcal{F}}_{1\mathrm{/}2}(\eta )\to \frac{4}{3\sqrt{\pi }}{\eta }^{3\mathrm{/}2}$, and the Fermi level measured from the band edge becomes the Fermi energy of a degenerate electron gas, $E_F-E_c=({\mathrm{\hslash }}^2\mathrm{/}2m^{\ast })(3{\pi }^{2}n{)}^{2\mathrm{/}3}$, independent of temperature. A metal is the extreme case. The conduction electrons of copper, $8.47\times {10}^{22}$ cm⁻³ with a mass close to that of a free electron, have a Fermi energy of 7.0 eV, equivalent to a temperature of about 82,000 K, so at room temperature only the electrons within a few $k_{B}T$ of $E_F$, a fraction of about $k_{B}T\mathrm{/}{E}_F=0.4\mathrm{\%}$, can change state. This makes the electronic heat capacity of a metal proportional to temperature and far smaller than the classical $\frac{3}{2}{k}_B$ per electron.

Holes, donors and charge neutrality

The probability that a valence-band state is empty, and so holds a hole, is

$$1-f(E)=\frac{1}{1+\exp [(E_F-E)\mathrm{/}{k}_{B}T]},$$

the same function mirrored about $E_F$, and the hole density is $p=N_v{\mathcal{F}}_{1\mathrm{/}2}((E_v-E_F)\mathrm{/}{k}_{B}T)$. A donor level is not quite a Fermi-Dirac state, because it can bind an electron of either spin but not two; the fraction of donors left ionized is $1\mathrm{/}(1+2e^{(E_F-E_D)\mathrm{/}{k}_{B}T})$, the factor 2 coming from the spin degeneracy. The Fermi level of a doped semiconductor is fixed by charge neutrality,

$$p+N_D^{+}=n+N_A^{-},$$

the positive charge of holes and ionized donors balancing the negative charge of electrons and ionized acceptors. For silicon doped with ${10}^{16}$ cm⁻³ of phosphorus, whose donor level lies 45 meV below $E_c$, the solution at 300 K ($N_c=2.8\times {10}^{19}$ cm⁻³) puts $E_F$ 205 meV below the conduction band and 160 meV below the donor level, so 99.6% of the donors are ionized and $n=1.0\times {10}^{16}$ cm⁻³. Device simulators solve the same equation self-consistently with the electrostatic potential.

Quasi-Fermi levels in lasers and LEDs

When current or light injects carriers, the electrons in the conduction band and the holes in the valence band each reach thermal equilibrium among themselves within a picosecond or less, long before they recombine, so each population follows a Fermi-Dirac distribution with its own quasi-Fermi level, $E_{Fn}$ for electrons and $E_{Fp}$ for holes. A photon of energy $h\nu$ is then amplified rather than absorbed if the upper state of the transition is more likely to be full than the lower one, which holds for $E_g<h\nu <E_{Fn}-E_{Fp}$ (Bernard and Duraffourg). For bulk GaAs with parabolic bands ($N_c=4.7\times {10}^{17}$ cm⁻³, $N_v=7.0\times {10}^{18}$ cm⁻³, $E_g=1.424$ eV) at 300 K:

Because $N_c$ is fifteen times smaller than $N_v$, the electron quasi-Fermi level moves into the conduction band while the hole level is still in the gap. In this estimate the band edge becomes transparent at $1.2\times {10}^{18}$ cm⁻³, which is the carrier density behind a laser's transparency current; quantum wells, band nonparabolicity and the shrinking of the gap at high carrier density shift the numbers in real devices without changing the balance of the two distributions.

At the low densities of an LED the carriers are non-degenerate, and their Boltzmann tails multiplied by the joint density of states give a spontaneous emission spectrum proportional to $\sqrt{h\nu -E_g}{e}^{-(h\nu -E_g)\mathrm{/}{k}_{B}T}$. Its full width at half maximum, $1.8k_{B}T$, is 46.5 meV at 300 K, or 27 nm at 850 nm.

Common questions

What is the Fermi-Dirac distribution?

The function that gives the probability that a quantum state of energy $E$ is occupied by a fermion, such as an electron, in thermal equilibrium: $f(E)=1\mathrm{/}(1+\exp [(E-E_F)\mathrm{/}{k}_{B}T])$. It is close to 1 well below the Fermi level, close to 0 well above it, and one half at $E_F$.

What is the value of the Fermi-Dirac distribution at the Fermi level?

Exactly one half, at any temperature above absolute zero. In semiconductor physics this is the usual definition of the Fermi level.

When can the Boltzmann approximation be used?

When the states of interest lie more than about $3k_{B}T$ above the Fermi level, where it overestimates the occupation by 5% or less. For a whole band, the Boltzmann formula for the electron density is within 5% while $n$ is below about $0.14N_c$: $6.5\times {10}^{16}$ cm⁻³ in GaAs and $3.9\times {10}^{18}$ cm⁻³ in silicon.

What is the difference between Fermi-Dirac and Bose-Einstein statistics?

Fermions such as electrons obey the exclusion principle, so a state holds at most one of them and the occupation $1\mathrm{/}(e^{(E-\mu )\mathrm{/}{k}_{B}T}+1)$ never exceeds 1. Bosons such as photons and phonons can share a state without limit, and their occupation $1\mathrm{/}(e^{(E-\mu )\mathrm{/}{k}_{B}T}-1)$ diverges as $E$ approaches $\mu$. Both approach the Maxwell-Boltzmann factor when $E-\mu$ is many $k_{B}T$.

How does temperature change the Fermi-Dirac distribution?

It sets the width of the step. At absolute zero the occupation drops from 1 to 0 at $E_F$; at 300 K it falls from 0.9 to 0.1 over 114 meV. The occupation of states far above $E_F$ rises steeply with temperature: 0.1 eV above $E_F$ it is $2.9\times {10}^{-7}$ at 77 K and 0.020 at 300 K.

References: N. W. Ashcroft and N. D. Mermin, Solid State Physics (Saunders, 1976). S. M. Sze and K. K. Ng, Physics of Semiconductor Devices, 3rd ed. (Wiley, 2007). B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 3rd ed. (Wiley, 2019). L. A. Coldren, S. W. Corzine and M. L. Mašanović, Diode Lasers and Photonic Integrated Circuits, 2nd ed. (Wiley, 2012). M. G. A. Bernard and G. Duraffourg, "Laser conditions in semiconductors," Physica Status Solidi 1, 699 (1961). W. B. Joyce and R. W. Dixon, "Analytic approximations for the Fermi energy of an ideal Fermi gas," Applied Physics Letters 31, 354 (1977). J. S. Blakemore, "Approximations for Fermi-Dirac integrals, especially the function F1/2(η) used to describe electron density in a semiconductor," Solid-State Electronics 25, 1067 (1982).