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Next: Diffusion and drift of Up: From Semi-conductivity to Micro-electronics Previous: Band structure, mobility, effective

Fermi statistics, charge carrier concentrations, dopants

The electrons in a semi conductor are fermions, and so they follow Fermi-Dirac statistics. The probability of a particular energy state $\epsilon$ being occupied is therefore
\begin{displaymath}
f_{FD}(\epsilon) = \frac{1}{e^{(\, \epsilon-\epsilon_F) /kT}+1}
\end{displaymath} (49)

This distribution function takes into account that no two fermions may have the same wave function (occupy the same state). In a system consisting of $N$ electrons at zero temperature, all available states are occupied up to the Fermi energy level, $\epsilon_F$. Figure 15 displays the Fermi-Dirac distribution for different temperatures.

Figure 15: The distribution function for fermions at three different temperatures.
\includegraphics[width=0.99\textwidth]{FD-dist.eps}

It is a property of the Fermi-Dirac distribution that at all finite temperatures

\begin{displaymath}
f_{FD}(\epsilon_F) = \mbox{$\frac{1}{2}$}, \qquad \forall \quad T>0
\end{displaymath} (50)

so that the Fermi level $\epsilon_F$ may also be thought of as that level at finite temperature where half of the available states are filled.

Exercise 6.6
Verify this.

Figure 16 illustrates this for pure (intrinsic) insulators and semi conductors.

Figure 16: The position of the Fermi level is in the middle of the band gap for pure (intrinsic) insulators (a) and semi conductors (b).
\includegraphics[width=0.35\textwidth]{band_ef.eps}

If the multiplicity of energy states from $\epsilon$ to $\epsilon + d\epsilon$ is given by $g(\epsilon)d\epsilon$, then the number of electrons $n(\epsilon)d\epsilon$ with energies from $\epsilon$ to $\epsilon + d\epsilon$ is

\begin{displaymath}
n(\epsilon)d\epsilon = g(\epsilon) f_{FD}(\epsilon)d\epsilon...
...epsilon)\frac{1}{e^{(\,\epsilon-\epsilon_F) /kT}-1} d\epsilon.
\end{displaymath} (51)

Remember that in the case of free electrons in a three dimensional box (a Fermi gas, for example, a metal) we found that

\begin{displaymath}
g(\epsilon)d\epsilon = \frac{8\sqrt{2}\pi L^3m^{3/2}}{h^3}\sqrt{\epsilon}d\epsilon.
\end{displaymath} (52)

The number of electrons in a Fermi gas that have energies from $\epsilon$ to $\epsilon + d\epsilon$ is
$\displaystyle n(\epsilon)d\epsilon$ $\textstyle =$ $\displaystyle g(\epsilon) f_{FD}(\epsilon)d\epsilon$ (53)
  $\textstyle =$ $\displaystyle \frac{8\sqrt{2}\pi L^3m^{3/2}}{h^3}
\frac{\sqrt{\epsilon}d\epsilon}{e^{(\, \epsilon-\epsilon_F) /kT}+1}$  

This distribution is reproduced in figure 17.

Figure 17: The distribution of electron energies in a Fermi gas at various temperatures.
\includegraphics[width=0.7\textwidth]{electron-e.eps}

The Fermi energy, $\epsilon_F$, could be found by allocating the $N$ electrons sequentially to the available states at $T=0$ from energies $\epsilon = 0$ to $\epsilon = \epsilon_F$.

$\displaystyle N$ $\textstyle =$ $\displaystyle \int^{\epsilon_F}_{0}g(\epsilon)d\epsilon$ (54)
  $\textstyle =$ $\displaystyle \frac{8\sqrt{2}\pi Vm^{3/2}}{h^3} \int^{\epsilon_F}_{0}
\sqrt{\epsilon}d\epsilon$  
  $\textstyle =$ $\displaystyle \frac{16\sqrt{2}\pi Vm^{3/2}}{3h^3}\epsilon_F^{3/2}$  

The Fermi energy is therefore
\begin{displaymath}
\epsilon_F = \frac{h^2}{2m}\left( \frac{3N}{8\pi V}\right)^{2/3}.
\end{displaymath} (55)

The quantity $N/V$ is the density of free electrons.

What happens in a semi-conductor ?

Intrinsic Carrier Concentration
Semi-conductor behaviour is defined by the conductivity due to the electrons crossing the (narrow) band gap due to thermal excitations.

Our aim is to adapt the expression for $n(\epsilon)d\epsilon$ in a free electron system to be appropriate for electrons (holes) in a semiconductor.

We can then find the number of electrons in the conduction band and holes in the valence band, due to thermal excitation of electrons from the valence band to the conduction band. These "charge carriers" account for the conductivity of the material.

We know that in a semi-conductor, the electrons are nearly free, but not actually free. There is a periodic potential due to the lattice cores, as in figure 3. We have seen that the electrons acquire an effective mass in the semi conducting material $m^*$, which can be more or less than their free mass at rest $m_e$. Part of the effect of the periodic lattice potential can therefore be modelled by simply replacing the mass of the electron with its effective mass in the equation 52 for the multiplicity of states $g(\epsilon)d\epsilon$,

\begin{displaymath}
g(\epsilon)d\epsilon = \frac{8\sqrt{2}\pi L^3{m^*}^{3/2}}{h^3}\sqrt{\epsilon}d\epsilon.
\end{displaymath} (56)

and the Fermi energy is
\begin{displaymath}
\epsilon_F = \frac{h^2}{2m^*}\left( \frac{3N}{8\pi V}\right)^{3/2}.
\end{displaymath} (57)

Figure 18 shows how Fermi distribution $f_{FD}(\epsilon )$ maps to the band gap.

Why place the Fermi energy in the middle of the gap ?

We will prove shortly this actually where it is $\epsilon_F = \epsilon_g / 2$, in the intrinsic case.

However, for now, consider that the position of the Fermi energy is defined by equation 50, $f_{FD}(\epsilon_F) = \mbox{$\frac{1}{2}$}$. Then as $\frac{1}{2}$ is inbetween the filled valence band states and empty conduction band states (at $T=0$K), we note this is an intuitive place for the Fermi level.

Figure 18: The position of the Fermi level and the Fermi distribution $f_{FD}(\epsilon )$ in the band gap.
\includegraphics[width=0.6\textwidth]{band-dos.eps}

When we look at electrons in the conduction band, we note that here electrons are free and we can use the multiplicity of states as for a free electron system, provided we measure the density of states from the conduction band edge upwards.

For electrons in the conduction band

\begin{displaymath}
g(\epsilon) \propto \sqrt{\epsilon - \epsilon_g}
\end{displaymath} (58)

When we look at holes in the valence band, we note that here holes are free and we can use the multiplicity of states as for a free electron system, provided we measure the density of states from the valence band edge downwards.

For holes in the valence band

\begin{displaymath}
g(\epsilon) \propto \sqrt{-\epsilon }
\end{displaymath} (59)

In the conduction band, $(\epsilon - \epsilon_F) \gg kT$, so we may approximate

\begin{displaymath}
f_{FD}(\epsilon) \approx e^{-(\epsilon - \epsilon_F)/kT}
\end{displaymath} (60)

The number of electrons in energy interval $d\epsilon$ is now

\begin{displaymath}
n(\epsilon)d\epsilon = \frac{8\sqrt{2}\pi L^3{m^*}^{3/2}}{h^...
...ilon - \epsilon_g}\,e^{-(\epsilon-\epsilon_F) /kT}\, d\epsilon
\end{displaymath} (61)

The number of electrons is therefore

\begin{displaymath}
n_e = \int_{\epsilon_g}^{\infty} \frac{8\sqrt{2}\pi V {m^*}^...
...lon -\epsilon_g}\,
e^{-(\epsilon - \epsilon_F)/kT}\,d\epsilon
\end{displaymath} (62)

To solve this equation, we make the substitution
$\displaystyle x$ $\textstyle =$ $\displaystyle (\epsilon -\epsilon_g)/kT$ (63)
$\displaystyle \epsilon$ $\textstyle =$ $\displaystyle xkT - \epsilon_g$  
$\displaystyle dx$ $\textstyle =$ $\displaystyle \frac{d\epsilon}{kt}$  
$\displaystyle d\epsilon$ $\textstyle =$ $\displaystyle kTdx$  

Then
$\displaystyle n_e$ $\textstyle =$ $\displaystyle \int_{0}^{\infty} \frac{8\sqrt{2}\pi V {m^*}^{3/2}}{h^3}
\sqrt{xkT}\,
e^{-(xkT+\epsilon_g-\epsilon_F)/kT}\, kT\,dx$ (64)
  $\textstyle =$ $\displaystyle e^{(\epsilon_F-\epsilon_g)/kT} (kT)^{3/2} \frac{8\sqrt{2}\pi V {m^*}^{3/2}}{h^3}
\int_{0}^{\infty} x^{1/2}e^{-x}\, dx$  

Now, the integral $\int_{0}^{\infty} x^{1/2}e^{-x}\, dx = \Gamma(\mbox{$\frac{3}{2}$})
=\frac{\sqrt{\pi}}{2}$, therefore, we finally arrive at the number of electrons in the conduction band at a given temperature $T$.
\begin{displaymath}
n_e = 2V \left( \frac{m^*kT}{2\pi\hbar^2}\right)^{3/2}e^{(\epsilon_F-\epsilon_g)/kT}.
\end{displaymath} (65)

This is the number of electrons available to carry a current.

In an intrinsic semi-conductor, the probability of a hole in the valence band is the complimentary probability for an electron in the valence band, at finite temperature.

\begin{displaymath}
f_h (\epsilon) = 1 - f_{FD} (\epsilon) = \frac{1}{e^{-(\epsilon-\epsilon_F)/kT} + 1}
\end{displaymath} (66)

Exercise 6.7
Verify this.

As before, $-(\epsilon - \epsilon_F) \gg kT$, so

\begin{displaymath}
f_h (\epsilon) \approx e^{-(\epsilon_F-\epsilon)/kT}
\end{displaymath} (67)

The situation is depicted in figure 19.

Figure 19: The distribution of holes and electrons in a semi conductor. Note how the energy axis is relabelled so that the top of the valance band corresponds to zero energy.
\includegraphics[width=0.6\textwidth]{npnh.eps}

The number of holes in the valence band will be given by the multiplicity of states for holes in the valence band, multiplied by the probability distribution of a hole in the valence band.

\begin{displaymath}
n(\epsilon)d\epsilon = \frac{8\sqrt{2}\pi V{m^*}^{3/2}}{h^3} \sqrt{-\epsilon}\,
e^{-(\epsilon_F - \epsilon)/kT}\,d\epsilon
\end{displaymath} (68)

Note how the valance band corresponds to negative energy states in the expression above. The total number of holes requires an integration of the negative energy states.
\begin{displaymath}
n_h = \int_{0}^{-\infty} \frac{8\sqrt{2}\pi V{m^*}^{3/2}}{h^3} \sqrt{-\epsilon}\,
e^{(\epsilon - \epsilon_F)/kT}\,d\epsilon
\end{displaymath} (69)

The integration may be performed by making the substitution $x = \frac{-\epsilon}{kt}$
\begin{displaymath}
n_h = -V \frac{\sqrt{kT}}{2\pi^2}\left( \frac{2m_h^*}{\hbar^...
...F/kT} kT \int_{0}^{-\infty} x^{\mbox{$\frac{1}{2}$}}x^{-x}\,dx
\end{displaymath} (70)

As before, the integral is a gamma function $\int_{0}^{\infty} x^{1/2}e^{-x}\, dx = \Gamma(\mbox{$\frac{3}{2}$})
=\frac{\sqrt{\pi}}{2}$, therefore, we finally arrive at the number of holes in the valence band at a given temperature $T$.
\begin{displaymath}
n_h = 2V \left( \frac{m_h^*kT}{2\pi\hbar^2}\right)^{3/2}e^{-\epsilon_F/kT}.
\end{displaymath} (71)

This is the number of holes available to carry a current.

Exercise 6.8
Compare the equations 65 and 71, for the number of electrons and holes respectively. Are there forms symmetrical, as expected ?

Consider the product of the number of holes and electrons.

\begin{displaymath}
n_e n_h = 4V^2 \left( \frac{kT}{2\pi\hbar^2}\right)^3 (m^*_e m^*_h)^{3/2} e^{-\epsilon_g/kT}.
\end{displaymath} (72)

It is very important that the product of $n_e$ and $n_h$ is independent of the Fermi level. It drops exponentially as the band gap increases.

Exercise 6.9
What are the consequences of this ?

From now on, we will write

$n_e = n$ negative charge carriers
$n_h = p$ positive charge carriers

Intrinsic semiconductors
In an intrinsic semiconductor, all the electrons in the conduction band are thermally excited from the valence band.

Clearly, $n_e = n_h$, that is, $n=p$. From equations 65 and 71,

\begin{displaymath}
{m_e^*}^{3/2} e^{(\epsilon_F-\epsilon_g)/kT} = {m_h^*}^{3/2} e^{-\epsilon_F/kT}
\end{displaymath} (73)

Taking logs of both sides :
\begin{displaymath}
\frac{\epsilon_F-\epsilon_g}{kT} = \frac{-\epsilon_F}{kT} + \mbox{$\frac{3}{2}$}
\ln\left(\frac{m_h^*}{m_e^*}\right)
\end{displaymath} (74)

therefore
\begin{displaymath}
\epsilon_F = \frac{\epsilon_g}{2} + \mbox{$\frac{3}{4}$}kT\ln\left(\frac{m_h^*}{m_e^*}\right)
\end{displaymath} (75)

At $T=0$K, $\epsilon_F = \epsilon_F/2$



Example : Gallium Arsenide

$m_e^*$ = 0.072 $m_e$
$m_h^*$ = 0.4 $m_e$
$\epsilon_g$ = 1.0 eV
Therefore

\begin{displaymath}
\epsilon_F = \frac{\epsilon_g}{2} + 1.3kT.
\end{displaymath}

At room temperature, $kT = 0.027$ eV,

\begin{eqnarray*}
\epsilon_F &=& 0.5 + 0.034 \\
&=& 0.534 \\
&\approx& \frac{\epsilon_g}{2} \qquad \mbox{within 7\%}.
\end{eqnarray*}



In the intrinsic case

\begin{displaymath}
n_e = n_h = n_i = V \left( \frac{kT}{2\pi\hbar^2}\right)^{3/2} (m^*_e m^*_h)^{3/4}
e^{-\epsilon_g/2kT}
\end{displaymath} (76)

where $n_i$ is the intrinsic carrier density.
\begin{displaymath}
\ln n_i = \mbox{const} + \mbox{$\frac{3}{2}$} \ln kT -\frac{\epsilon_g}{2kT}
\end{displaymath} (77)

As the logarithm varies slowly with temperature, we find in the intrinsic region,
\begin{displaymath}
\ln n_i \propto \frac{1}{T}.
\end{displaymath} (78)

A plot of $\ln n_i$ versus $\frac{1}{T}$ gives a straight line with slope $-\frac{\epsilon_g}{2k}$, as in figure 20.

Figure: A plot of $\ln n_i$ versus $\frac{1}{T}$ for 0.5 $\Omega $m n-type silicon sample.
\includegraphics[width=0.6\textwidth]{nvsT.eps}



Example : Silicon and Germanium

$n_i(T)$ cm$^{-3}$ = $1.76 \times 10^{16} T^{3/2}e^{-4550/T}$ for Ge
$n_i(T)$ cm$^{-3}$ = $3.88 \times 10^{16} T^{3/2}e^{-7000/T}$ for Si



Extrinsic semiconductors - Substitutional dopants
Intrinsic semiconductors contain only the elemental material of their pure substance. Extrinsic semiconductors have been specially modified so that some of the atoms of the pure substance are substituted by carefully chosen foreign atoms called dopants. The concentration of dopants ranges from $10^{13}$/cm$^3$ to $10^{17}$/cm$^3$. Since the number of pure atoms is usually about $10^{23}$/cm$^3$, it is clear that the dopant concentrations are about 1 ppm to 0.1 ppb. This is exceedingly low. Clearly, the concentrations of unwanted impurities must be even lower than this. It is therefore understandable that microchips have to be fabricated from the purest material under carefully controlled clean conditions.

The presence of substitutional impurities changes the electronic structure of the semi-conductor locally around the impurity. Usually, the doping is intended to produce a hole state just above the valence band, or an electron state just under the conduction band.

Figure 21: A group V dopant (Phosphorous) in a group IV semiconductor (Si) showing the hydrogenic wave function of the donor impurity's electron.
\includegraphics[width=0.7\textwidth]{dopants3.eps}

For example, in a group IV semiconductor, each atom is tetrahedrally located with respect to its neighbours. Its valence electrons participate in covalent bonds with these neighbours. A substitutional donor impurity would have one extra valence electron. This additional electron would not be able to participate in the covalent bonding to the neighbouring atoms. To a first approximation, it occupies a hydrogenic orbital around its core, but with an extremely large dimension. The wave function of this hydrogenic orbital is typically some thousand angstroms (Figure 21). Usually, the doping is not of a sufficiently high concentration that the wavefunctions of the extra substitutional donor impurity electrons atoms overlap each other. These dopant electron states are therefore localised. As the dopant electron is well screened from its core, it is easily ionised to the conduction band. The position of the dopant state in the band gap is therefore just under the conduction band. There are some devices where the material is deliberately overdoped, so that the substitutional donor impurity electron wavefunctions overlap and form a band. Figure 22 illustrates the foregoing text.

Figure 22: Substitution of dopant atoms for crystal atoms a) Group IV semiconductors (eg. Silicon) and b) Group III-V semiconductors (eg. Gallium Arsenide).
\includegraphics[width=0.9\textwidth]{dopants.eps}

In extrinsic materials at moderate temperatures, the conduction is dominated by the electrons or holes originating from impurity or defect states in the lattice. Even when the temperature is too low for intrinsic electron-hole pair generation, these localised states within the band gap can be thermally ionised to generate free electrons or holes as shown in figure 23. This is the extrinsic regime of the semiconductor. Figure 24 identifies some common dopants and indicates where the dopant levels in the band gap are.

Figure 23: Free carriers created by : a) ionisation of donor states just under the conduction band to form electron carriers or c) ionisation of acceptor states just above the valence band to form hole carriers compared to b) the intrinsic semiconductor. Note the movement of the Fermi level as appropriate.
\includegraphics[width=0.8\textwidth]{dopants4.eps}

Figure 24: Some common dopants and their position in the band gap.
\includegraphics[width=0.9\textwidth]{dopants2.eps}



Ionisation energy of dopant states
We model a donor state as a positive core with one electron in the hydrogenic analogue scenario.

\begin{displaymath}
\frac{1}{R}\frac{d}{dr}\left( {r^2}\frac{dR}{dr}\right) +
\...
... \epsilon_0 r} +
E - \frac{\hbar^2 l(l+1)}{2mr^2} \right) = 0
\end{displaymath} (79)


\begin{displaymath}
E_n = -\frac{mZ^2e^4}{32\pi^2(\epsilon_0\epsilon_r)^2\hbar^2...
... \frac{1}{n^2} \right) = \frac{E_1}{n^2}, \ \ \ n=1,2,3,\cdots
\end{displaymath} (80)

The solutions are the radial wave functions of the hydrogen atom and their corresponding quantised energy states. These solutions are corrected for :
  1. $m_e \Longrightarrow m^*_e$ the effective mass of the electron.
  2. $Z = 1$ the effective charge of the donor impurity.
  3. $\epsilon_0 \Longrightarrow \epsilon_0\epsilon_r$ the screening of the core by the material. Thus we adjust the permitivity of free space to the permitivity of the material.
We need only consider the lowest energy Rydberg $s$-state where $n=1$ and $l=0$.
\begin{displaymath}
\frac{1}{R}\frac{d}{dr}\left( {r^2}\frac{dR}{dr}\right) +
\...
...\left(\frac{e^2}{4\pi \epsilon_0 \epsilon_r r} + E \right) = 0
\end{displaymath} (81)


\begin{displaymath}
E_1 = -\frac{m^*e^4}{32\pi^2(\epsilon_0\epsilon_r)^2\hbar^2}
\end{displaymath} (82)



Example : The Phosphorus donor impurity in Silicon
Relevant data :

$\epsilon_r$ = 11.7 for Si
$m^*_e/m_e$ = 0.19 near the bottom of the conduction band in Si
Therefore, we find
$\displaystyle E_1$ $\textstyle =$ $\displaystyle -\frac{me^4}{32\pi^2\epsilon_0^2\hbar^2}
\left(\frac{m^*_e}{m_e} \right)\left(\frac{1}{\epsilon_r} \right)^2$ (83)
  $\textstyle =$ $\displaystyle -13.6 \mbox{eV} \times \left(\frac{m^*_e}{m_e} \right)\left(\frac{1}{\epsilon_r}\right)^2$ (84)
  $\textstyle =$ $\displaystyle -0.019 eV$ (85)

So, according to this model, the donor impurity electron requires 0.019 eV to become ionised. The position in the band gap of the donor impurity level is therefore just 0.019 eV below the conduction band edge.



Majority carriers
In general, one impurity type dominates in an extrinsic semiconductor. The associated carrier is known as the majority carrier. For an $n$-type material, the majority carrier is an electron. For an $p$-type material, the majority carrier is a hole.



Position of the Fermi level in extrinsic semiconductors
We consider first the case of $n$-type doping with a substitutional donor impurity. Just as we did before in the case of the intrinsic semiconductor, we can calculate the number of electrons in the conduction band as a result of thermal excitation of electrons from the donor levels (ie. ionisation of the donors). As before, the Fermi-Dirac distribution gives the probability of populating the conduction band. We use the free electron multiplicity of states, which can describe the multiplicity of states in the case of a semiconductor with a band gap provided we replace the free electron mass with the effective mass $m^*$. Since the electrons are coming from the donor level, not the valence band, we replace the energy of the the valence band with that of the donor level, $\epsilon_d$. This situation is represented in figure 25. Compare it to the situation of the intrinsic semiconductor in figure 18

Figure 25: The position of the Fermi-Dirac distribution for the calculation of the number of electrons in the conduction band as a result of thermal excitation of electrons from the donor levels.
\includegraphics[width=0.45\textwidth]{band-dos2.eps}

Using similar arguments and mathematics as in the intrinsic case, we find the Fermi level in the case of $n$-type material :

\begin{displaymath}
\epsilon_F = \frac{\epsilon_d +\epsilon_g }{2} + \frac{kT}{2}\ln\left[ \frac{N_d}{2N_c}\right]
\end{displaymath} (86)

where $N_d$ is the density of donor levels and $N_c$ is the density of electron states in the conduction band which was previously calculated in equation 65.

Similarly we find the Fermi level in the case of $p$-type material :

\begin{displaymath}
\epsilon_F = \frac{\epsilon_a + 0 }{2} - \frac{kT}{2}\ln\left[ \frac{N_a}{2N_v}\right]
\end{displaymath} (87)

where $N_a$ is the density of acceptor levels and $N_v$ is the density of hole states in the valence band which was previously calculated in equation 71.

It turns out that the second terms in the last two equations are so small that they can be neglected. For an $n$-type semiconductor

\begin{displaymath}
\epsilon_F = \frac{\epsilon_d +\epsilon_g }{2}
\end{displaymath} (88)

and for a $p$-type semiconductor
\begin{displaymath}
\epsilon_F = \frac{\epsilon_a}{2}.
\end{displaymath} (89)

We will write the intrinsic material Fermi level as
\begin{displaymath}
\epsilon_i = \frac{\epsilon_g }{2}
\end{displaymath} (90)

as before.

Note
If an intrinsic semiconductor is doped with $n$-type impurities, not only does the number of electrons increase, but the number of holes decreases below the levels in the intrinsic conductor.
This can be understood by comparing figures 18 and 25. It turns out that in such an $n$-type semiconductor
\begin{displaymath}
\left.\begin{array}{l}
\mbox{density of} \\
\mbox{negativ...
...n_n = n_i \exp\left(\frac{\epsilon_F - \epsilon_i}{kT} \right)
\end{displaymath} (91)

and
\begin{displaymath}
\left.\begin{array}{l}
\mbox{density of} \\
\mbox{positiv...
...p_n = n_i \exp\left(\frac{\epsilon_i -\epsilon_F}{kT} \right)
\end{displaymath} (92)

where $n_i$ is the number of carriers in the intrinsic material. Note that $\epsilon_F - \epsilon_i$ will be positive and of the order $\frac{\epsilon_d}{2}$. Therefore
\begin{displaymath}
n_n \gg n_i
\end{displaymath} (93)

and as $\epsilon_i -\epsilon_F$ will be negative,
\begin{displaymath}
p_n \ll n_i
\end{displaymath} (94)

so that
\begin{displaymath}
n_n \gg n_i \gg p_n.
\end{displaymath} (95)



Law of Mass-Action for semiconductors

\begin{displaymath}
n_n p_n = n_i^2.
\end{displaymath} (96)

Exercise 6.9
Verify this.

Similarly, for a $p$-type semiconductor :

\begin{displaymath}
p_p n_p = n_i^2
\end{displaymath} (97)

and
\begin{displaymath}
p_p \gg n_i \gg n_p.
\end{displaymath} (98)

The temperature dependance of the charge carrier concentration, for example, for electrons in n-type material,

\begin{displaymath}
n_n = n_i \exp\left(\frac{\epsilon_F - \epsilon_i}{kT}\right)
\end{displaymath} (99)

may be made explicit by substituting in the form (equation 76) for $n_i$
\begin{displaymath}
n_n = V \left( \frac{kT}{2\pi\hbar^2}\right)^{3/2} (m^*_e m^*_h)^{3/4}
e^{(\epsilon_g-\epsilon_F/kT}.
\end{displaymath} (100)

Since from equation 86
\begin{displaymath}
\epsilon_F \approx \frac{\epsilon_d +\epsilon_g }{2},
\end{displaymath} (101)

the temperature dependence of $n_n$ is that of half the ionisation energy of the donor level.

Consider again the diagram (figure 20) showing the plot of $n_i$ versus $\frac{1}{T}$.

  1. The slope at low temperature gives half the ionisation energy of the donor level Thus, at temperatures below -100$^{\circ}$C, carriers are being excited from the donor impurity level just below the conduction band.
  2. The plateau (called the exhaustion region), is where all the donor levels are ionised. This level gives the doping density $N_d$.
  3. Above $\approx 300^{\circ}$C the electrons begin to be excited from the valence band across the band gap, just as in the intrinsic semi-conductor case. Here the material behaves like an intrinsic semiconductor with slope $\approx \frac{\epsilon_g}{2k}$ as expected.



Majority and Minority carriers
Semiconductors often have both donors and acceptors present. Usually one type dominates. An $n$-type semiconductor is one dominated by negative charge carriers.

\begin{displaymath}
n_n = N_d -N_a
\end{displaymath} (102)

where $N_d$ is the number of donor levels (concentration of donor impurities) and $N_a$ is the number of acceptor levels (concentration of acceptor impurities). Clearly
\begin{displaymath}
p_n = \frac{n_i^2}{N_d -N_a}.
\end{displaymath} (103)

Similarly for a p-type semiconductor
\begin{displaymath}
p_p = N_a -N_d
\end{displaymath} (104)

and
\begin{displaymath}
n_p = \frac{n_i^2}{N_a -N_d}.
\end{displaymath} (105)


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Next: Diffusion and drift of Up: From Semi-conductivity to Micro-electronics Previous: Band structure, mobility, effective
Simon Connell 2004-10-04