# Electron and hole concentrations in semiconductors

This is a continuation from the previous tutorial - introduction to semiconductors.

The electron concentration in a semiconductor is the number of conduction electrons in the conduction bands per unit volume of the semiconductor, and the hole concentration is the number of holes in the valence bands per unit volume of the semiconductor.

The concentrations of electrons and holes in a semiconductor are determined by many factors, including the bandgap and band structure of the semiconductor, the types and concentrations of the impurities doped in the semiconductor, temperature, and any external disturbances to the semiconductor.

Density of States

Because electrons are subject to the Pauli exclusion principle, which requires that no more than one electron can occupy the same quantum-mechanical state, the number of electrons in a particular energy band is determined by both the number of available states in that band and the probability of occupancy for each state.

In a bulk semiconductor, the number of electron states in a given energy band is linearly proportional to the volume of the semiconductor. Therefore, a very useful concept is the density of states, which in a three-dimensional system like a bulk semiconductor is the number of states per unit material volume.

In a bulk semiconductor, the density of electron states within the energy range between $$E$$ and $$E+\text{d}E$$ for $$E\ge{E}_\text{c}$$ near the conduction-band edge is

$\tag{12-15}\rho_\text{c}(E)\text{d}E=\sum_\text{c}\frac{4\pi(2m_\text{c})^{3/2}}{h^3}(E-E_\text{c})^{1/2}\text{d}E=\frac{4\pi(2m_\text{e}^*)^{3/2}}{h^3}(E-E_\text{c})^{1/2}\text{d}E$

where $$m_\text{c}$$ is the density of states effective mass of an electron in a conduction band $$\text{c}$$ and $$m_\text{e}^*$$ is the density of states effective mass for electrons in all equivalent conduction bands.

For example, there are six equivalent conduction-band minima in Si and four in Ge, but only one in a direct-gap semiconductor like GaAs. Therefore, $$m_\text{e}^*=6^{2/3}m_\text{c}=1.08m_0$$ for Si, $$m_\text{e}^*=4^{2/3}m_\text{c}=0.55m_0$$ for Ge, and $$m_\text{e}^*=m_\text{c}=0.067m_0$$ for GaAs, where $$m_0$$ is the free electron mass.

Similarly, the density of states within the energy range between $$E$$ and $$E+\text{d}E$$ for $$E\le{E}_\text{v}$$ near the valence-band edge is

$\tag{12-16}\rho_\text{v}(E)\text{d}E=\sum_\text{v}\frac{4\pi(2m_\text{v})^{3/2}}{h^3}(E_\text{v}-E)^{1/2}\text{d}E=\frac{4\pi(2m_\text{h}^*)^{3/2}}{h^3}(E_\text{v}-E)^{1/2}\text{d}E$

where $$m_\text{v}$$ is the density of states effective mass of a hole in a valence band $$\text{v}$$ and $$m_\text{h}^*$$ is the density of states effective mass of holes in all valence bands that are degenerate at the valence-band edge.

In both direct-gap and indirect-gap semiconductors including GaAs, Si, and Ge, there are normally two hole bands, known as the heavy-hole band and the light-hole band, of different effective masses that are degenerate at the valence-band edge. Therefore,

$\tag{12-17}m_\text{h}^*=(m_\text{hh}^{3/2}+m_\text{lh}^{3/2})^{2/3}$

where $$m_\text{hh}$$ and $$m_\text{lh}$$ are the effective masses in the heavy-hole and light-hole bands, respectively. We have $$m_\text{h}^*=0.56m_0$$ for Si, $$m_\text{h}^*=0.31m_0$$ for Ge, and $$m_\text{h}^*=0.52m_0$$ for GaAs.

Carriers in Equilibrium

In thermal equilibrium, the probability of occupancy for a given electron state at an energy $$E$$ is described by the Fermi-Dirac function $$f(E)$$ given in (12-1) [refer to the introduction to semiconductors tutorial]. Therefore, the concentration of conduction electrons (negatively charged carriers) whose energies fall between $$E$$ and $$E+\text{d}E$$ is

$\tag{12-18}n_0(E)\text{d}E=f(E)\rho_\text{c}(E)\text{d}E\qquad(\text{m}^{-3})$

and the concentration of holes (positively charged carriers) whose energies fall between $$E$$ and $$E+\text{d}E$$ is

$\tag{12-19}p_0(E)\text{d}E=[1-f(E)]\rho_\text{v}(E)\text{d}E\qquad(\text{m}^{-3})$

Note that the probability of finding a hole at an energy $$E$$ is $$1-f(E)$$ because a hole is an unoccupied electron state. The total concentrations of electrons and holes in thermal equilibrium are, respectively,

$\tag{12-20}n_0=\displaystyle\int\limits_{E_\text{c}}^\infty{n_0}(E)\text{d}E=\int\limits_{E_\text{c}}^\infty{f(E)}\rho_\text{c}(E)\text{d}E=\int\limits_{E_\text{c}}^\infty\frac{\rho_\text{c}(E)\text{d}E}{\text{e}^{(E-E_\text{F})/k_\text{B}T}+1}$

and

$\tag{12-21}p_0=\displaystyle\int\limits_{-\infty}^{E_\text{v}}{p_0}(E)\text{d}E=\int\limits_{-\infty}^{E_\text{v}}{[1-f(E)]}\rho_\text{v}(E)\text{d}E=\int\limits_{-\infty}^{E_\text{v}}\frac{\rho_\text{v}(E)\text{d}E}{\text{e}^{(E_\text{F}-E)/k_\text{B}T}+1}$

Using (12-15) and (12-16), the electron and hole concentrations given by (12-20) and (12-21) can be expressed as

$\tag{12-22}n_0=N_\text{c}(T)F_{1/2}\left(\frac{E_\text{F}-E_\text{c}}{k_\text{B}T}\right)$

$\tag{12-23}p_0=N_\text{v}(T)F_{1/2}\left(\frac{E_\text{v}-E_\text{F}}{k_\text{B}T}\right)$

respectively, where $$N_\text{c}$$ and $$N_\text{v}$$ are the effective densities of states for conduction and valence bands, respectively, defined as

$\tag{12-24}N_\text{c}(T)=2\left(\frac{2\pi{m}_\text{e}^*k_\text{B}T}{h^2}\right)^{3/2},\qquad{N_\text{v}(T)}=2\left(\frac{2\pi{m}_\text{h}^*k_\text{B}T}{h^2}\right)^{3/2}$

and $$F_{1/2}(\xi)$$ is the Fermi-Dirac integral of order $$1/2$$ defined as

$\tag{12-25}F_{1/2}(\xi)=\frac{2}{\sqrt{\pi}}\displaystyle\int\limits_0^\infty\frac{x^{1/2}\text{d}x}{\text{e}^{(x-\xi)}+1}=\text{e}^\xi\frac{2}{\sqrt{\pi}}\int\limits_0^\infty\frac{x^{1/2}\text{d}x}{\text{e}^x+e^\xi}$

The relations given in (12-22) and (12-23) for the electron and hole concentrations are very general. They are valid for a semiconductor in thermal equilibrium with its environment no matter whether the semiconductor is doped with impurity or not. They are also valid for any value of $$E_\text{F}$$ with respect to $$E_\text{c}$$ and $$E_\text{v}$$.

It can be seen clearly from these relations that the values of $$n_0$$ and $$p_0$$ are strongly dependent on the value of the Fermi energy $$E_\text{F}$$. The impurities in a semiconductor affect the electron and hole concentrations through changing the value of the Fermi energy $$E_\text{F}$$.

From (12-22) and (12-23), it can be clearly seen that both $$n_0$$ and $$p_0$$ are determined by the Fermi-Dirac integral $$F_{1/2}(\xi)$$. For this reason, the characteristics of this integral as a function of its variable $$\xi$$ are plotted in Figure 12-3.

For $$\xi=0$$, we find that $$F_{1/2}(0)=0.76515$$. We see that $$F_{1/2}(\xi)\approx\text{e}^\xi$$ for large negative values of $$\xi$$. This approximation has an error of less than $$1\%$$ for $$\xi\le-3.6$$. Therefore, when the Fermi level is far away from both band edges so that $$(E_\text{c}-E_\text{F})/k_\text{B}T\ge3.6$$ and $$(E_\text{F}-E_\text{v})/k_\text{B}T\ge3.6$$, the electron and hole concentrations given in (12-22) and (12-23) can be approximated to an accuracy of better than $$99\%$$ by the following expressions:

$\tag{12-26}n_0=N_\text{c}(T)\text{e}^{-(E_\text{c}-E_\text{F})/k_\text{B}T}$

$\tag{12-27}p_0=N_\text{v}(T)\text{e}^{-(E_\text{F}-E_\text{v})/k_\text{B}T}$

In an intrinsic semiconductor, impurities contribute negligibly to the electron and hole concentrations. Practically all of the conduction electrons in an intrinsic semiconductor come from thermal excitation from the valance bands. Consequently, there are as many holes as electrons so that $$n_0=p_0$$.

For most semiconductors of interest, we have $$E_\text{g}\gg{k}_\text{B}T$$ at a temperature below the melting temperature of the semiconductor. Under these conditions, we can apply the relation $$n_0=p_0$$ to (12-26) and (12-27) to find that the Fermi level of an intrinsic semiconductor is given by

$\tag{12-28}E_\text{Fi}=\frac{E_\text{c}+E_\text{v}}{2}+\frac{k_\text{B}T}{2}\ln\frac{N_\text{v}}{N_\text{c}}=\frac{E_\text{c}+E_\text{v}}{2}+\frac{3k_\text{B}T}{4}\ln\frac{m_\text{h}^*}{m_\text{e}^*}$

which lies very close to the middle of the bandgap.

Therefore, we have

$\tag{12-29}n_0=p_0=n_\text{i}(T)\approx\sqrt{N_\text{c}N_\text{v}}\text{e}^{-E_\text{g}/2k_\text{B}T}=2\left(\frac{2\pi{k_\text{B}T}}{h^2}\right)^{3/2}(m_\text{e}^*m_\text{h}^*)^{3/4}\text{e}^{-E_\text{g}/2k_\text{B}T}$

for an intrinsic semiconductor, where the intrinsic carrier concentration, $$n_\text{i}$$, is a function of temperature.

Example 12-2

Calculate the values of the effective densities of states $$N_\text{c}$$ and $$N_\text{v}$$ for GaAs at $$300\text{ K}$$. Use the results to find the electron and hole concentrations and the Fermi level for intrinsic GaAs at $$300\text{ K}$$.

For GaAs, $$m_\text{e}^*=0.067m_0$$, and $$m_\text{h}^*=0.52m_0$$. Using the constants $$m_0=9.11\times10^{-31}\text{ kg}$$, $$k_\text{B}=1.38\times10^{-23}\text{ J K}^{-1}$$, and $$h=6.626\times10^{-34}\text{ J s}$$, the effective densities of states at $$T=300\text{ K}$$ can be calculated from (12-24) to be

\begin{align}N_\text{c}&=2\times\left[\frac{2\pi\times0.067\times9.11\times10^{-31}\times1.38\times10^{-23}\times300}{(6.626\times10^{-34})^2}\right]^{3/2}\text{ m}^{-3}\\&=4.35\times10^{23}\text{ m}^{-3}\end{align}

and

\begin{align}N_\text{v}&=2\times\left[\frac{2\pi\times0.52\times9.11\times10^{-31}\times1.38\times10^{-23}\times300}{(6.626\times10^{-34})^2}\right]^{3/2}\text{ m}^{-3}\\&=9.41\times10^{24}\text{ m}^{-3}\end{align}

At $$T=300\text{ K}$$, the bandgap of GaAs is $$E_\text{g}=1.424\text{ eV}$$, and $$k_\text{B}T=25.9\text{ meV}$$. Therefore,

$\frac{E_\text{g}}{k_\text{B}T}=\frac{1.424}{25.9\times10^{-3}}=54.98\qquad\text{and}\qquad\frac{E_\text{g}}{2k_\text{B}T}=27.49$

Because the Fermi level of an intrinsic semiconductor lies close to the center of the bandgap, $$(E_\text{c}-E_\text{Fi})/k_\text{B}T\approx(E_\text{Fi}-E_\text{v})/k_\text{B}T\approx{E}_\text{g}/(2k_\text{B}T)=27.49\gg3.6$$. Therefore, the approximations given in (12-26) and (12-27) and, consequently, those given in (12-28) and (12-29) are all valid in this situation. For this intrinsic GaAs, we then find from (12-29) that

\begin{align}n_0=p_0=n_\text{i}&=\sqrt{4.35\times10^{23}\times9.41\times10^{24}}\times\text{e}^{-27.49}\text{ m}^{-3}\\&=2.33\times10^{12}\text{ m}^{-3}\end{align}

and from (12-28) that

$E_\text{Fi}-\frac{E_\text{c}+E_\text{v}}{2}=\frac{25.9}{2}\times\ln\frac{9.41\times10^{24}}{4.35\times10^{23}}\text{ meV}=39.8\text{ meV}$

Because the center of the bandgap is at $$(E_\text{c}+E_\text{v})/2$$, this intrinsic Fermi level is $$39.8\text{ meV}$$ above the center of the bandgap. The reason for this shift of $$E_\text{F}$$ above the bandgap center is that $$m_\text{e}^*\lt{m}_\text{h}^*$$ for GaAs. Compared with the bandgap of $$1.424\text{ eV}$$, this shift away from the bandgap center is small, verifying the statement that the intrinsic Fermi level lies very close to the center of the bandgap.

In an extrinsic semiconductor, however, $$n_0$$ and $$p_0$$ are different from $$n_\text{i}$$ because of the contribution of carriers from the impurities in a semiconductor.

An impurity atom that can be positively ionized to contribute a conduction electron is a donor, and one that can be negatively ionized to contribute a hole to the valence bands is an acceptor.

In general, the requirement for charge neutrality in a semiconductor leads to the following relation:

$\tag{12-30}n_0+N_\text{a}^-=p_0+N_\text{d}^+$

where $$N_\text{a}^-$$ is the concentration of the immobile negatively ionized acceptors and $$N_\text{d}^+$$ is that of the immobile positively ionized donors.

When $$N_\text{d}^+\gt{N}_\text{a}^-$$, the semiconductor is an $$n$$-type semiconductor with $$n_0\gt{p}_0$$. In an $$n$$-type semiconductor, electrons are the majority carriers, and holes are the minority carriers.

When $$N_\text{a}^-\gt{N}_\text{d}^+$$, the semiconductor is a $$p$$-type semiconductor with $$p_0\gt{n}_0$$. In a $$p$$-type semiconductor, holes are the majority carriers, and electrons are the minority carriers.

The Fermi level of an intrinsic semiconductor lies very close to the middle of the bandgap and is only a weak function of temperature.

In contrast, the Fermi level of an extrinsic semiconductor is a function of the types and concentrations of the impurities. In an $$n$$-type semiconductor, it moves towards the conduction-band edge; in a $$p$$-type semiconductor, it moves towards the valence-band edge.

Up to a moderate doping concentration, the Fermi level remains in the bandgap. Such a semiconductor is called a nondegenerate semiconductor.

For a nondegenerate semiconductor, no matter whether it is intrinsic or extrinsic, (12-26) and (12-27) are valid for $$n_0$$ and $$p_0$$, respectively. Therefore, the carrier concentrations of a nondegenerate semiconductor in thermal equilibrium satisfy the following law of mass action:

$\tag{12-31}n_0p_0=n_\text{i}^2(T)$

The values of $$n_0$$ and $$p_0$$ in a nondegenerate semiconductor can be found by solving (12-30) and (12-31) simultaneously.

Then, from (12-26) and (12-27), the Fermi level of a nondegenerate semiconductor can be found:

$\tag{12-32}E_\text{F}=E_\text{c}-k_\text{B}T\ln\frac{N_\text{c}}{n_0}=E_\text{v}+k_\text{B}T\ln\frac{N_\text{v}}{p_0}$

This relation is valid for both intrinsic and extrinsic situations so long as the semiconductor is nondegenerate. In the intrinsic case, this relation is equivalent to (12-28), as $$E_\text{F}=E_\text{Fi}$$ for an intrinsic semiconductor. In the extrinsic case, it is valid for both $$n$$-type and $$p$$-type semiconductors.

In the case when $$N_\text{c}\gg{N}_\text{d}^+-N_\text{a}^-\gg{n}_\text{i}$$, the semiconductor is nondegenerate and has $$n_0\approx{N}_\text{d}^+-N_\text{a}^-\gg{p}_0\approx{n}_\text{i}^2/n_0$$. The Fermi level for such a nondegenerate $$n$$-type semiconductor shifts from $$E_\text{Fi}$$ toward the conduction-band edge; it can be approximated as

$\tag{12-33}E_\text{F}\approx{E}_\text{c}-k_\text{B}T\ln\frac{N_\text{c}}{N_\text{d}^+-N_\text{a}^-}$

In the case when $$N_\text{v}\gg{N}_\text{a}^--N_\text{d}^+\gg{n}_\text{i}$$, the semiconductor is nondegenerate and has $$p_0\approx{N}_\text{a}^--N_\text{d}^+\gg{n}_0\approx{n}_\text{i}^2/p_0$$. The Fermi level for such a nondegenerate $$p$$-type semiconductor shifts from $$E_\text{Fi}$$ toward the valence-band edge; it can be approximated as

$\tag{12-34}E_\text{F}\approx{E}_\text{v}+k_\text{B}T\ln\frac{N_\text{v}}{N_\text{a}^--N_\text{d}^+}$

These approximations fail when the net impurity concentration is sufficiently high to make the semiconductor degenerate or when it is too low so that $$n_0$$ and $$p_0$$ remain close to $$n_\text{i}$$.

Example 12-3

A piece of $$n$$-type GaAs is doped with a net impurity concentration of $$N_\text{d}^+-N_\text{a}^-=5\times10^{18}\text{ m}^{-3}$$. Is it degenerate or nondegenerate? Find its electron and hole concentrations and its Fermi level at $$T=300\text{ K}$$. How much is the shift of the Fermi level, measured form the intrinsic Fermi level, caused by the doping of the impurity?

From Example 12-2, we know that $$N_\text{c}=4.35\times10^{23}\text{ m}^{-3}$$ at $$T=300\text{ K}$$. We also find from Example 12-2 that $$n_\text{i}=2.33\times10^{12}\text{ m}^{-3}$$.

This $$n$$-type GaAs is nondegenerate because $$N_\text{d}^+-N_\text{a}^-\ll{N}_\text{c}$$ for the given impurity concentration.

The general procedure for finding $$n_0$$ and $$p_0$$ is to solve the simultaneous equations of $$n_0-p_0=N_\text{d}^+-N_\text{a}^-$$, from (12-30), and $$n_0p_0=n_\text{i}^2$$, from (12-31), with the known values of $$N_\text{d}^+-N_\text{a}^-$$ and $$n_\text{i}$$.

However, because $$N_\text{d}^+-N_\text{a}^-\gg{n}_\text{i}$$ for the given problem, we find that

$n_0\approx{N}_\text{d}^+-N_\text{a}^-=5\times10^{18}\text{ m}^{-3}$

and

$p_0=\frac{n_\text{i}^2}{n_0}=1.1\times10^6\text{ m}^{-3}$

The Fermi level for this nondegenerate $$n$$-type GaAs can be found by using (12-32):

$E_\text{F}=E_\text{c}-25.9\times\ln\frac{4.35\times10^{23}}{5\times10^{18}}\text{ meV}=E_\text{c}-294.6\text{ meV}$

Compared with the intrinsic Fermi level, $$E_\text{Fi}$$, found in Example 12-2, we find that

\begin{align}E_\text{F}-E_\text{Fi}&=\frac{E_\text{c}-E_\text{v}}{2}-294.6\text{ meV}-39.8\text{ meV}\\&=\frac{E_\text{g}}{2}-334.4\text{ meV}\\&=\frac{1424}{2}\text{ meV}-334.4\text{ meV}\\&=377.6\text{ meV}\end{align}

Therefore, the Fermi level of this $$n$$-type GaAs is shifted away from the intrinsic Fermi level by $$377.6\text{ meV}$$ toward the conduction-band edge.

In a heavily $$p$$-doped semiconductor, the Fermi-level can move into the valence band. Similarly, the Fermi level can move in to the conduction band in a heavily $$n$$-doped semiconductor. When the Fermi level lies within a valence band or within a conduction band, we have a degenerate semiconductor.

The electron and hole concentrations are still given by (12-22) and (12-23), respectively. Nevertheless, the law of mass action expressed in (12-31) and the position of the Fermi level given by (12-32) are not valid for a degenerate semiconductor because either (12-26) or (12-27) can have a significant error when the Fermi level lies above the conduction-band edge or below the valence-band edge.

Example 12-4

What is the impurity doping concentration required for $$n$$-type GaAs to become degenerate at $$300\text{ K}$$?

An $$n$$-type semiconductor becomes degenerate when its Fermi level lies at or above its conduction-band edge: $$E_\text{F}\ge{E}_\text{c}$$. From (12-22), we find that this condition requires that

$n_0\ge{N}_\text{c}(T)F_{1/2}(0)=0.76515N_\text{c}(T)$

at a given temperature $$T$$.

For GaAs at $$T=300\text{ K}$$, we find that $$n_0\ge3.33\times10^{23}\text{ m}^{-3}$$ from this relation because $$N_\text{c}=4.35\times10^{23}\text{ m}^{-3}$$, as found in Example 12-2.

Because $$n_0\gg{n}_\text{i}\gg{p}_0$$ in this situation, we find from (12-30) by neglecting $$p_0$$ in comparison to $$n_0$$ that the required impurity concentration for $$n$$-type GaAs to become degenerate at $$300\text{ K}$$ is simply

$N_\text{d}^+-N_\text{a}^-=n_0\ge3.33\times10^{23}\text{ m}^{-3}$

Carriers in Quasi-Equilibrium

Electrons and holes in excess of their respective thermal equilibrium concentrations can be generated in a semiconductor by current injection or optical excitation. When this situation occurs, the carriers will relax toward thermal equilibrium through both intraband and interband processes.

Intra-conduction-band relaxation allows electrons to reach thermal equilibrium among themselves through electron-electron collisions and electron-phonon interactions, while intra-valence-band relaxation allows holes to also reach thermal equilibrium among themselves through similar processes. The time constants of such intraband relaxation processes are generally in the range of 10 fs to 1 ps, depending on the concentration of the excess carriers.

Thermal equilibrium between electrons and holes is reached through electron-hole recombination processes, the time constants of which typically vary from the order of 100 ps to the order of 1 ms, depending on the properties of the specific semiconductor and the carrier concentration.

Consequently, thermal equilibrium in the conduction bands and that in the valence bands can be separately reached in less than 1 ps, but complete thermal equilibrium for the entire system would not usually be reached for at least a few hundred picoseconds.

If the external excitation persists, the semiconductor can reach a quasi-equilibrium state in which electrons and holes are not characterized by a common Fermi level but are characterized by two separate quasi-Fermi levels.

In such a quasi-equilibrium state, instead of a single Fermi-Dirac distribution function given in (12-1) [refer to the introduction to semiconductors tutorial] for both conduction and valence bands, the probability of occupancy in the conduction bands and that in the valence bands are described by two separate Fermi-Dirac distribution functions:

$\tag{12-35}f_\text{c}(E)=\frac{1}{\text{e}^{(E-E_\text{Fc})/k_\text{B}T}+1}$

for the conduction bands, and

$\tag{12-36}f_\text{v}(E)=\frac{1}{\text{e}^{(E-E_\text{Fv})/k_\text{B}T}+1}$

for the valence bands, where $$E_\text{Fc}$$ and $$E_\text{Fv}$$ are quasi-Fermi levels for the conduction and valence bands, respectively.

As an intrinsic property of the band structure, the densities of states, $$\rho_\text{c}(E)$$ and $$\rho_\text{v}(E)$$, given in (12-15) and (12-16) for the conduction and valence bands, respectively, are independent of equilibrium or nonequilibrium of the carriers. Therefore, in quasi-equilibrium, the electron concentration as a function of energy is

$\tag{12-37}n(E)\text{d}E=f_\text{c}(E)\rho_\text{c}(E)\text{d}E\qquad(\text{m}^{-3})$

and the hole concentration as a function of energy is

$\tag{12-38}p(E)\text{d}E=[1-f_\text{v}(E)]\rho_\text{v}(E)\text{d}E\qquad(\text{m}^{-3})$

The total concentrations of electrons and holes in quasi-equilibrium are, respectively,

$\tag{12-39}n=\displaystyle\int\limits_{E_\text{c}}^\infty{n}(E)\text{d}E=\int\limits_{E_\text{c}}^\infty{f}_\text{c}(E)\rho_\text{c}(E)\text{d}E=\int\limits_{E_\text{c}}^\infty\frac{\rho_\text{c}(E)\text{d}E}{\text{e}^{(E-E_\text{Fc})/k_\text{B}T}+1}$

and

$\tag{12-40}p=\displaystyle\int\limits_{-\infty}^{E_\text{v}}p(E)\text{d}E=\int\limits_{-\infty}^{E_\text{v}}[1-f_\text{v}(E)]\rho_\text{v}(E)\text{d}E=\int\limits_{-\infty}^{E_\text{v}}\frac{\rho_\text{v}(E)\text{d}E}{\text{e}^{(E_\text{Fv}-E)/k_\text{B}T}+1}$

Using (12-15) and (12-16), the electron and hole concentrations for a semiconductor in quasi-equilibrium can be expressed in a form similar to that of (12-22) and (12-23):

$\tag{12-41}n=N_\text{c}(T)F_{1/2}\left(\frac{E_\text{Fc}-E_\text{c}}{k_\text{B}T}\right)$

$\tag{12-42}p=N_\text{v}(T)F_{1/2}\left(\frac{E_\text{v}-E_\text{Fv}}{k_\text{B}T}\right)$

We find from (12-41) and (12-42) that the electron concentration, $$n$$, and the hole concentration, $$p$$, in a quasi-equilibrium state are completely quantified by the quasi-Fermi levels, $$E_\text{Fc}$$ and $$E_\text{Fv}$$, respectively.

We then find

$\tag{12-43}np=N_\text{c}(T)N_\text{v}(T)F_{1/2}\left(\frac{E_\text{c}-E_\text{Fc}}{k_\text{B}T}\right)F_{1/2}\left(\frac{E_\text{Fv}-E_\text{v}}{k_\text{B}T}\right)$

In the situation when $$(E_\text{c}-E_\text{Fc})/k_\text{B}T\ge3.6$$ and $$(E_\text{Fv}-E_\text{v})/k_\text{B}T\ge3.6$$, the quasi-equilibrium electron and hole concentrations given in (12-41) and (12-42) can be approximated to an accuracy of better than $$99\%$$ by

$\tag{12-44}n=N_\text{c}(T)\text{e}^{-(E_\text{c}-E_\text{Fc})/k_\text{B}T}$

$\tag{12-45}p=N_\text{v}(T)\text{e}^{-(E_\text{Fv}-E_\text{v})/k_\text{B}T}$

Then

$\tag{12-46}np=N_\text{c}(T)N_\text{v}(T)\text{e}^{-(E_\text{g}-\Delta{E}_\text{F})/k_\text{B}T}=n_\text{i}^2(T)\text{e}^{\Delta{E}_\text{F}/k_\text{B}T}=n_0p_0\text{e}^{\Delta{E}_\text{F}/k_\text{B}T}$

where

$\tag{12-47}\Delta{E}_\text{F}=E_\text{Fc}-E_\text{Fv}$

is the separation between the quasi-Fermi levels.

Because of the splitting of the quasi-Fermi levels in a quasi-equilibrium state, the law of mass action given in (12-31) is no longer valid but is replaced by (12-46).

Note that these approximations are not valid if the quasi-equilibrium electron and hole concentrations are high enough to push any one of the quasi-Fermi levels to the vicinity of any band edge or beyond. Such a situation can happen even in an intrinsic semiconductor under high electrical or optical excitation of carriers.

In a quasi-equilibrium state, high concentrations of electrons and holes that are in excess of equilibrium concentrations can be generated by electrical or optical excitation.

Compared with the equilibrium electron and hole concentrations, $$n_0$$ and $$p_0$$ given in (12-22) and (12-23), respectively, we find that $$n\gt{n}_0$$ if $$E_\text{Fc}\gt{E}_\text{F}$$ and $$p\gt{p}_0$$ if $$E_\text{F}\gt{E}_\text{Fv}$$.

Therefore, the existence of quasi-equilibrium electron and hole concentrations that are higher than the equilibrium concentrations is characterized by the splitting of quasi-Fermi levels with $$\Delta{E}_\text{F}\gt0$$. Quasi-equilibrium in a semiconductor is maintained when the carrier generation rate is equal to the carrier recombination rate.

Example 12-5

An equal number of excess electrons and holes of a concentration of $$\Delta{n}=\Delta{p}=5\times10^{18}\text{ m}^{-3}$$ is generated at $$T=300\text{ K}$$ in an intrinsic GaAs sample by optical excitation. Find the quasi-Fermi levels, $$E_\text{Fc}$$ and $$E_\text{Fv}$$. What is the separation between these quasi-Fermi levels?

Because $$n_\text{i}=2.33\times10^{12}\text{ m}^{-3}\ll\Delta{n}=\Delta{p}$$, we have $$n=p=n_\text{i}+\Delta{n}\approx\Delta{n}=5\times10^{18}\text{ m}^{-3}$$ in this situation.

From the values of $$N_\text{c}=4.35\times10^{23}\text{ m}^{-3}$$ and $$N_\text{v}=9.41\times10^{24}\text{ m}^{-3}$$ found in Example 12-2 for GaAs at $$300\text{ K}$$, we find that $$n\ll{N}_\text{c}$$ and $$p\ll{N}_\text{v}$$. Therefore, both $$E_\text{Fc}$$ and $$E_\text{Fv}$$ are still sufficiently far away from the band edges so that (12-44) and (12-45) are valid. We then find from (12-44) that

\begin{align}E_\text{Fc}&=E_\text{c}-k_\text{B}T\ln\frac{N_\text{c}}{n}\\&=E_\text{c}-25.9\times\ln\frac{4.35\times10^{23}}{5\times10^{18}}\text{ meV}\\&=E_\text{c}-294.6\text{ meV}\end{align}

and from (12-45) that

\begin{align}E_\text{Fv}&=E_\text{v}+k_\text{B}T\ln\frac{N_\text{v}}{p}\\&=E_\text{v}+25.9\times\ln\frac{9.41\times10^{24}}{5\times10^{18}}\text{ meV}\\&=E_\text{v}+374.2\text{ meV}\end{align}

Therefore, $$E_\text{Fc}$$ lives at 294.6 meV below the conduction-band edge, and $$E_\text{Fv}$$ lies at 374.2 meV above the valence-band edge. Because the bandgap of GaAs at $$300\text{ K}$$ is $$E_\text{g}=1.424\text{ eV}$$, the separation of these two quasi-Fermi levels is

\begin{align}\Delta{E}_\text{F}=E_\text{Fc}-E_\text{Fv}&=E_\text{c}-E_\text{v}-294.6\text{ meV}-374.2\text{ meV}\\&=E_\text{g}-668.8\text{ meV}\\&=755.2\text{ meV}\end{align}

The next tutorial covers the topic of carrier recombination.