30 May 2025

Fermi-Dirac Distribution and Carrier Equilibrium

Fermi occupation, equilibrium electron and hole concentrations, and the mass-action law.

bsc semester-iv mj-7 semiconductor-devices fermi-level carrier-concentration

The Fermi level $E_F$ is the electron chemical potential. For a single-particle state of energy $E$, thermal equilibrium gives the Fermi-Dirac occupation probability

\[\boxed{f(E)=\frac{1}{1+\exp[(E-E_F)/(k_BT)]}}.\]

At $E=E_F$, $f=1/2$ at every nonzero temperature. If $\Delta=E-E_F$, then

\[f(E_F+\Delta)+f(E_F-\Delta)=1,\]

and the slope at the midpoint is

\[\left.\frac{df}{dE}\right\rvert_{E_F}=-\frac{1}{4k_BT}.\]

Thus increasing temperature broadens the transition about $E_F$ without changing its midpoint. In the limit $T\to0$, $f(E)$ becomes a step: states below $E_F$ are occupied and states above it are empty. A hole requires an empty valence-band state, so its probability is

\[\boxed{1-f(E)=\frac{1}{1+\exp[(E_F-E)/(k_BT)]}}.\]

Electron concentration

The electron concentration is the number of available conduction-band states weighted by their occupation:

\[n=\int_{E_c}^{\infty}g_c(E)f(E)\,dE.\]

For a non-degenerate semiconductor with $E_c-E_F\gtrsim3k_BT$, the exponential in the denominator is large and

\[f(E)\simeq\exp\!\left[-\frac{E-E_F}{k_BT}\right].\]

For a three-dimensional parabolic band,

\[g_c(E)=\frac{1}{2\pi^2}\left(\frac{2m_n^*}{\hbar^2}\right)^{3/2}\sqrt{E-E_c}.\]

Set $\epsilon=E-E_c$. Separating the band-edge factor gives

\[n=\frac{1}{2\pi^2}\left(\frac{2m_n^*}{\hbar^2}\right)^{3/2} e^{-(E_c-E_F)/(k_BT)} \int_0^\infty\epsilon^{1/2}e^{-\epsilon/(k_BT)}\,d\epsilon.\]

With $u=\epsilon/(k_BT)$,

\[\int_0^\infty\epsilon^{1/2}e^{-\epsilon/(k_BT)}\,d\epsilon =(k_BT)^{3/2}\Gamma\!\left(\frac32\right) =(k_BT)^{3/2}\frac{\sqrt\pi}{2}.\]

Using $h=2\pi\hbar$ then yields

\[n=2\left(\frac{2\pi m_n^*k_BT}{h^2}\right)^{3/2} e^{-(E_c-E_F)/(k_BT)}.\]

Defining the effective conduction-band density of states

\[N_c=2\left(\frac{2\pi m_n^*k_BT}{h^2}\right)^{3/2},\]

we obtain

\[\boxed{n=N_c e^{-(E_c-E_F)/(k_BT)}}.\]

Hole concentration

For the valence band, available hole states are weighted by $1-f(E)$. With $\epsilon=E_v-E\ge0$,

\[p=\int_{-\infty}^{E_v}g_v(E)[1-f(E)]\,dE =N_v e^{-(E_F-E_v)/(k_BT)},\]

where

\[\boxed{N_v=2\left(\frac{2\pi m_p^*k_BT}{h^2}\right)^{3/2}}, \qquad \boxed{p=N_v e^{-(E_F-E_v)/(k_BT)}}.\]

Mass action and the intrinsic Fermi level

Multiplying the non-degenerate equilibrium concentrations cancels $E_F$:

\[\begin{aligned} np &=N_cN_v\exp\!\left[-\frac{(E_c-E_F)+(E_F-E_v)}{k_BT}\right]\\ &=N_cN_v e^{-E_g/(k_BT)} =n_i^2. \end{aligned}\]

Hence

\[\boxed{n_i=\sqrt{N_cN_v}\,e^{-E_g/(2k_BT)}}.\]

For intrinsic material, $n=p$ and $E_F=E_i$. Equating the electron and hole expressions and taking logarithms gives

\[\boxed{E_i=\frac{E_c+E_v}{2}+\frac{k_BT}{2}\ln\!\left(\frac{N_v}{N_c}\right)}.\]

Because $N_c\propto(m_n^)^{3/2}$ and $N_v\propto(m_p^)^{3/2}$,

\[\boxed{E_i=\frac{E_c+E_v}{2}+\frac{3k_BT}{4}\ln\!\left(\frac{m_p^*}{m_n^*}\right)}.\]

The intrinsic level is exactly at midgap only when the effective densities of states are equal. The relations $n=N_c e^{-(E_c-E_F)/(k_BT)}$, $p=N_v e^{-(E_F-E_v)/(k_BT)}$, and $np=n_i^2$ use the non-degenerate approximation; degenerate material requires Fermi-Dirac integrals.

Solved Problems

1. Occupation on opposite sides of the Fermi level

At $T=300\ \mathrm K$, find the occupation probabilities at $E=E_F\pm0.100\ \mathrm{eV}$. Take $k_BT=0.0258520\ \mathrm{eV}$.

For the level above $E_F$,

\[f(E_F+0.100)=\frac{1}{1+e^{0.100/0.0258520}} =0.02047.\]

For the level below $E_F$,

\[f(E_F-0.100)=\frac{1}{1+e^{-0.100/0.0258520}} =0.97953.\]

Both probabilities are dimensionless and lie between $0$ and $1$. Their sum is $1.00000$, which checks the exact symmetry $f(E_F+\Delta)+f(E_F-\Delta)=1$; the positive energy offset lowers occupation.

2. Intrinsic level and equilibrium carriers

At $300\ \mathrm K$, a semiconductor has $N_c=2.80\times10^{25}\ \mathrm{m^{-3}}$, $N_v=1.04\times10^{25}\ \mathrm{m^{-3}}$, and $E_g=1.120\ \mathrm{eV}$. Find $n_i$ and the intrinsic-level displacement from midgap. Then find $n$ and $p$ if $E_F-E_i=+0.250\ \mathrm{eV}$.

First,

\[n_i=\sqrt{N_cN_v}\,e^{-E_g/(2k_BT)} =6.676\times10^{15}\ \mathrm{m^{-3}}.\]

The intrinsic-level displacement is

\[E_i-\frac{E_c+E_v}{2} =\frac{k_BT}{2}\ln\!\left(\frac{N_v}{N_c}\right) =-0.01280\ \mathrm{eV}.\]

The negative sign places $E_i$ slightly below midgap because $N_v<N_c$. For $E_F-E_i=+0.250\ \mathrm{eV}$,

\[\begin{aligned} n&=n_i e^{(E_F-E_i)/(k_BT)} =1.058\times10^{20}\ \mathrm{m^{-3}},\\ p&=n_i e^{-(E_F-E_i)/(k_BT)} =4.214\times10^{11}\ \mathrm{m^{-3}}. \end{aligned}\]

The positive Fermi-level shift increases electrons and suppresses holes. The product check gives $np=4.457\times10^{31}\ \mathrm{m^{-6}}=n_i^2$.

Descriptive Questions

State the Fermi-Dirac distribution and derive its value and slope at $E=E_F$.
Derive the non-degenerate conduction-electron concentration from the density-of-states integral.
Obtain the hole concentration and the equilibrium mass-action law.
Derive the intrinsic Fermi level and explain why it need not lie exactly at midgap.

Numerical Problems

Spectroscopic occupation data give the Fermi-function slope $df/dE=-8.00\ \mathrm{eV^{-1}}$ at $E=E_F$. Use $[df/dE]_{E_F}=-1/(4k_BT)$ and $k_B=8.61733\times10^{-5}\ \mathrm{eV\,K^{-1}}$ to infer the carrier temperature.

Final answer: $T=-[4k_B(df/dE)_{E_F}]^{-1}=362.6\ \mathrm K$.
At $300\ \mathrm K$, locate a state whose occupation probability is $0.900$ relative to $E_F$.

Final answer: $E-E_F=-0.05680\ \mathrm{eV}$; the state lies $56.80\ \mathrm{meV}$ below $E_F$.
The Fermi-edge sensitivity can be written as $-df/dE=[4k_BT]^{-1}\operatorname{sech}^2[(E-E_F)/(2k_BT)]$. At $300\ \mathrm K$, find its full width at half maximum in energy.

Final answer: Half maximum occurs when $\cosh[(E-E_F)/(2k_BT)]=\sqrt2$, so the full width is $4k_BT\operatorname{arcosh}(\sqrt2)=4k_BT\ln(1+\sqrt2)=0.09114\ \mathrm{eV}=91.14\ \mathrm{meV}$.
Calculate $N_c$ at $300\ \mathrm K$ for $m_n^*=0.26m_0$.

Final answer: $N_c=3.327\times10^{24}\ \mathrm{m^{-3}}$.
In a non-degenerate parabolic conduction band, the equilibrium electron population per energy interval is proportional to $w(\epsilon)=\epsilon^{1/2}\exp[-\epsilon/(k_BT)]$, where $\epsilon=E-E_c$. At $300\ \mathrm K$, find the most probable excess energy and the mean excess energy.

Final answer: Maximizing $w$ gives $\epsilon_{\rm mp}=k_BT/2=0.01293\ \mathrm{eV}$. The mean is $\langle\epsilon\rangle=k_BT\,\Gamma(5/2)/\Gamma(3/2)=3k_BT/2=0.03878\ \mathrm{eV}$.
At $300\ \mathrm K$, the intrinsic level is measured to lie $10.0\ \mathrm{meV}$ above midgap. Assuming parabolic bands, infer the effective-mass ratio $m_p^/m_n^$ from $E_i-E_{mid}=(3/4)k_BT\ln(m_p^/m_n^)$.

Final answer: $m_p^/m_n^=\exp[4(0.0100)/(3k_BT)]=1.675$.

All distribution identities, carrier-equilibrium reductions, and numerical values are reproducible in the accessible Maxima worksheet for Fermi-Dirac carrier equilibrium.