4.2 Statistics of Electrons in Solids

Now we turn to the electron system in thermodynamic equilibrium. As we have seen, the electron state for a given energy band is characterized by the quantum number k, the average number of electrons in this state being

$f_0(\varepsilon _{{\bf k}})=\frac 1{\exp \left( \frac{\varepsilon_{\bf k}-\zeta }{k_BT}\right) +1} .$ (7)

The function (7) is called the Fermi-Dirac function. The chemical potential $\zeta$ is determined by the normalization condition

$N=\sum_{{\bf k}}\frac 1{\exp \left( \frac{\varepsilon _{\bf k}-\zeta }{k_BT}\right) +1} .$ (8)

The summation here should be done also over spin indices. The equation (8) defines the dependence of the chemical potential $\zeta$ on the electron density $n=N/{\cal V}.$ It is also convenient to introduce the density of electron states with the formula similar to Eq. (4)

$g(\varepsilon )=2\sum_{{\bf k}}\delta \left( \varepsilon -\varepsilon _{{\bf k}}\right)$ (9)

where we have taken into account spin degeneracy. We get

$n=\int_0^\infty g(\varepsilon )f_0(\varepsilon ) d\varepsilon .$ (10)

For the quadratic spectrum with the effective mass m we have (see Problem 4.2)

$g(\varepsilon )=\frac{\sqrt{2}}{\pi ^2}\frac{m^{3/2}}{\hbar ^3}\sqrt{\varepsilon} .$ (11)

Inserting this formula to the normalization condition (10) and introducing the dimensionless chemical potential $\zeta ^{*}=\zeta /k_BT$ we get the following equation for $\zeta ^{*}$
$n=\frac{\sqrt{2}}{\pi ^2}\frac{(mk_BT)^{3/2}}{\hbar ^3}{\cal F}_{1/2}(\zeta ^{*})$

where ${\cal F}_{1/2}(z)$ is a particular case of the Fermi integrals

${\cal F}_n(z)=\int_0^\infty \frac{x^n dx}{e^{x-z}+1}.$

Degenerated Electron Gas

Now we discuss the important limiting cases. The first one is the case for good metals or highly doped semiconductors in which the density of conduction electrons is large. The large density of conduction electrons means that $\zeta ^{*}\gg 1$ . That leads to the following approximate presentation for the Fermi function
$f_0(\varepsilon )=\Theta (\zeta -\varepsilon )$

where

$\Theta (x)=\left| \begin{array}{ll} 1, & \mathrm{if} x>0, \\ 0, & \mathrm{if} x<0 \end{array} \right.$

is the Heaviside unit step function. In this approximation we easily get
$\zeta _0=\frac{\hbar ^2k_F^2}{2m}=\frac{\hbar ^2}{2m}(3\pi ^2n)^{2/3}.$

This quantity is often called the Fermi level because it is just the border between the full and empty states. We will also use this word and denote it as $\epsilon _F.$ To get temperature dependent corrections one should calculate the integral in Eq. (10) more carefully. One obtains (see Problem 4.3)
$\zeta =\epsilon _F\left[ 1-\frac{\pi ^2}{12}\left( \frac{k_BT}{\epsilon_F}\right)^2\right] .$

Now we can check when our approximation is really good. To do it let us request that the second item in the brackets to be small. We get

$\frac{\epsilon _F}{k_BT}=\frac{\hbar ^2(3\pi ^2n)^{2/3}}{2mk_BT}\gg 1 .$ (12)

So we see that the gas is degenerate at big enough electron density and small effective mass. Note that the criterion has the exponential character and inequality (12) can be not too strong (usually 5-7 is enough). In a typical metal $n\approx 10^{22}$ cm^–3, $m\approx 10^{-27}$ g, and at room temperature $\frac{\epsilon _F}{k_BT}\approx 10^2.$

Non-Degenerate Electron Gas

Now we discuss the situation when the electron density in the conduction band is not very high and the electrons are non-degenerate. It means that $\zeta <0, \exp (-\zeta ^{*})\gg 1.$ In this case the Fermi distribution tends to the Maxwell-Boltzmann distribution
$f_0(\varepsilon )\approx e^{\zeta ^{*}}e^{-\varepsilon /k_BT}$

where

$\mu^*$ = $\ln \left[ \frac{4\pi^3 \hbar^3 n}{(2\pi m k_BT)^{3/2}} \right],$
$e^{\zeta^*}$ = $\frac{4\pi^3\hbar^3n}{(2\pi m k_BT)^{3/2}} .$ (13)

We see that the chemical potential of non degenerate electron gas is strongly temperature dependent. The degeneracy for room temperature is intermediate at $n\approx 10^{19}$ cm^–3. These formulas are not very interesting in typical semiconductors because electrons are taken from impurities which can be partly ionized. So the dependence n vs. T remains unknown. To get insight into the problem let us consider the band scheme of a typical semiconductor with one donor level $\varepsilon _D$ and one accepter one $\varepsilon _A$ , Fig. 3

Fig 3. Band scheme of a typical semiconductor.

(the origin of energies is the bottom of the conduction band). The most important feature is that in such situation we have both electrons (in the conduction band) and holes (in the valence band). The occupation factor for holes is
$f_0^{\prime }(\varepsilon )=1-f_0(\varepsilon )=\frac 1{e^{\frac{\zeta -\varepsilon }{k_BT}}+1}.$

It is natural to call the function $f^{\prime }(\varepsilon )$ as the Fermi function of holes. According to our prescription for energies, the electron energy in the conduction band is $\varepsilon =\hbar ^2k^2/2m_n$ , at the donor level $\varepsilon =-\varepsilon _D$ , at the accepter level $\varepsilon =-\varepsilon _A$ , while in the valence band $\varepsilon =-\varepsilon _G-\varepsilon ^{\prime }$ where $\varepsilon _G$ is the width of the forbidden gap while $\varepsilon ^{\prime }=\hbar ^2k^{^{\prime }2}/2m_p.$ If the also introduce the hole chemical potential, $\zeta ^{\prime }=-\varepsilon _G-\zeta$ we come to exactly the same form for the hole Fermi function as for electrons with the replacement $\varepsilon \rightarrow \varepsilon ^{\prime }, \zeta \rightarrow \zeta ^{\prime }$ . To get the position of the chemical potential now we should apply the neutrality condition
number of electrons = number of holes

which reads
$\int_{c.b.}g_n(\varepsilon )f_0(\varepsilon ) d\varepsilon +\sum_A\frac 1{e^{\frac{-\varepsilon _A-\zeta }{k_BT}+1}}=\int_{v.b.}g_p(\varepsilon ^{\prime })f_0(\varepsilon ^{\prime }) d\varepsilon ^{\prime }+\sum_D\frac 1{e^{\frac{\varepsilon _D+\zeta }{k_BT}+1}} .$

This equation is strongly simplified if both electrons and holes obey the Boltzmann statistics. Denoting $A=\exp (\zeta ^{*})\ll 1$ and assuming that $\exp (\zeta ^{*}+\varepsilon _G/k_BT)\gg 1$ we get

$\nu_nA+\frac{n_A}{\frac 1A\exp \left( -\frac{\varepsilon _A}{k_BT}\right) +1}=\nu _pA^{-1}e^{-\frac{\varepsilon _G}{k_BT}}+\frac{n_D}{A \exp \left( +\frac{\varepsilon _D}{k_BT}\right) +1}$ (14)

where we have introduced
$\nu _{n,p}=\frac{(2\pi m_{n,p}k_BT)^{3/2}}{4\pi ^3\hbar ^3}.$

Even now we have a rather complicated situation which depends on the relation between the energies and the temperature. In the following we analyze few most important cases.

Intrinsic semiconductor

It is the simplest case where there is no both donors and accepters. From the Eq. (14) we have

A = $\left(\frac{m_p}{m_n} \right)^{3/4} \exp \left( - \frac{\varepsilon_G}{2k_BT } \right) ,$
$\zeta$ = $-\frac{\varepsilon_G}{2} + \frac{3}{4}k_BT \ln \left( \frac{m_p}{m_n} \right) .$ (15)

We see that the chemical potential in an intrinsic semiconductor is close to the middle of the forbidden gap. Concentrations of the electrons and holes are the same
$n_i=n_T\exp \left( -\frac{\varepsilon _G}{2k_BT}\right) , n_T=\sqrt{\nu _n\nu _p.}$

The concentration n_T for the room temperature and the free electron mass is 2.44· 10¹⁹ cm^–3, it scales as (m_nm_p)^3/4T^3/2.

Extrinsic semiconductor

Let us assume that only donors are present and $\varepsilon _G\gg \varepsilon _D.$ In this case we get from Eq. (14)
$\nu _nA\left( Ae^{\frac{\varepsilon _D}{k_BT}}+1\right) =n_D.$

At very low temperatures the first term is the most important one, and
$A=\sqrt{\frac{n_D}{\nu _n}}e^{-\frac{\varepsilon _D}{2k_BT}}.$

We see that the chemical potential is near the middle of the distance between the donor level and the bottom of the conduction band, the concentration in the conduction band being
$n_n\approx \frac{n_D}Ae^{-\frac{\varepsilon _D}{k_BT}}=\sqrt{\nu _nn_D}e^{-\frac{\varepsilon _D}{2k_BT}}.$

At high temperatures we get
$A=\frac{n_D}{\nu _n}, n_n\approx n_D.$

The result is clear enough: at high temperature all the donors are ionized while at low temperatures electrons ''freeze-out'' into the donor states. The situation in accepter semiconductor is just the mirror one.

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$\mu^*$	=	$\ln \left[ \frac{4\pi^3 \hbar^3 n}{(2\pi m k_BT)^{3/2}} \right],$
$e^{\zeta^*}$	=	$\frac{4\pi^3\hbar^3n}{(2\pi m k_BT)^{3/2}} .$	(13)

A	=	$\left(\frac{m_p}{m_n} \right)^{3/4} \exp \left( - \frac{\varepsilon_G}{2k_BT } \right) ,$
$\zeta$	=	$-\frac{\varepsilon_G}{2} + \frac{3}{4}k_BT \ln \left( \frac{m_p}{m_n} \right) .$	(15)