6.23 Semiconductors

Semiconductors are at the core of modern technology. This section discusses some basic properties of semiconductors that will be needed to explain how the various semiconductor applications work. The main semiconductor manipulation that must be described in this section is doping, adding a small amount of impurity atoms.

Figure 6.31: Vicinity of the band gap in the spectra of intrinsic and doped semiconductors. The amounts of conduction band electrons and valence band holes have been vastly exaggerated to make them visible.
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If semiconductors did not conduct electricity, they would not be very useful. Consider first the pure, or “intrinsic,” semiconductor. The vicinity of the band gap in its spectrum is shown to the left in figure 6.31. The vertical coordinate shows the energy ${\vphantom' E}^{\rm p}$ of the single-electron quantum states. The horizontal coordinate shows the density of states ${\cal D}$, the number of quantum states per unit energy range. Recall that there are no quantum states in the band gap. States occupied by electrons are shown in red. At room temperature there are some thermally excited electrons in the conduction band. They left behind some holes in the valence band. Both the electrons and the holes can provide electrical conduction.

Time for a reality check. The number of such electrons and holes is very much smaller than the figure indicates. The number $\iota_{\rm {e}}$ of electrons per quantum state is given by the Fermi-Dirac distribution (6.19). In the conduction band, that may be simplified to the Maxwell-Boltzmann one (6.21) because the number of electrons in the conduction band is small. The average number of electrons per state in the conduction band is then:

\begin{displaymath}
\fbox{$\displaystyle
\iota_{\rm e} = e^{-({\vphantom' E}^{\rm p}- \mu)/k_{\rm B}T}
$} %
\end{displaymath} (6.33)

Here $T$ is the absolute temperature, $k_{\rm B}$ is the Boltzmann constant, and $\mu$ is the chemical potential, also known as the Fermi level. The Fermi level is shown by a red tick mark in figure 6.31.

For an intrinsic semiconductor, the Fermi level is about in the middle of the band gap. Therefore the average number of electrons per quantum state at the bottom of the conduction band is

\begin{displaymath}
\mbox{Bottom of the conduction band:}\quad
\iota_{\rm e} = e^{-{\vphantom' E}^{\rm p}_{\rm gap}/2k_{\rm B}T}
\end{displaymath}

At room temperature, ${k_{\rm B}}T$ is about 0.025 eV while for silicon, the band gap energy is about 1.12 eV. That makes $\iota_{\rm {e}}$ about 2 10$\POW9,{-10}$. In other words, only about 1 in 5 billion quantum states in the lower part of the conduction band has an electron in it. And it is even less higher up in the band. A figure cannot show a fraction that small; there are just not enough atoms on a page.

So it is not surprising that pure silicon conducts electricity poorly. It has a resistivity of several thousand ohm-m where good metals have on the order of 10$\POW9,{-8}$. Pure germanium, with a smaller band gap of 0.66 eV, has a much larger $\iota_{\rm {e}}$ of about 3 10$\POW9,{-6}$ at the bottom of the conduction band. Its resistivity is correspondingly lower at about half an ohm-m. That is still many orders of magnitude larger than for a metal.

And the number of conduction electrons becomes much smaller still at cryogenic temperatures. If the temperature is a frigid 150 K instead of a 300 K room temperature, the number of electrons per state in silicon drops by another factor of a billion. That illustrates one important rule:

You cannot just forget about temperature to understand semiconductors.
Usually, you like to analyze the ground state at absolute zero temperature of your system, because it is easier. But that simply does not work for semiconductors.

The number of holes per state in the valence band may be written in a form similar to that for the electrons in the conduction band:

\begin{displaymath}
\fbox{$\displaystyle
\iota_{\rm h} = e^{-(\mu - {\vphantom' E}^{\rm p})/k_{\rm B}T}
$} %
\end{displaymath} (6.34)

Note that in the valence band the energy is less than the Fermi level $\mu$, so that the exponential is again very small. The expression above may be checked by noting that whatever states are not filled with electrons are holes, so $\iota_{\rm {h}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1-\iota_{\rm {e}}$. If you plug the Fermi-Dirac distribution into that, you get the expression for $\iota_{\rm {h}}$ above as long as the number of holes per state is small.

From a comparison of the expressions for the number of particles per state $\iota_{\rm {e}}$ and $\iota_{\rm {h}}$ it may already be understood why the Fermi level $\mu$ is approximately in the middle of the band gap. If the Fermi level is exactly in the middle of the band gap, $\iota_{\rm {e}}$ at the bottom of the conduction band is the same as $\iota_{\rm {h}}$ at the top of the valence band. Then there is the same number of electrons per state at the bottom of the conduction band as holes per state at the top of the valence band. That is about as it should be, since the total number of electrons in the conduction band must equal the total number of holes in the valence band. The holes in the valence band is where the electrons came from.

Note that figure 6.31 is misleading in the sense that it depicts the same density of states ${\cal D}$ in the conduction band as in the valence band. In reality, the number of states per unit energy range in the conduction band could easily be twice that at the corresponding location in the valence band. It seems that this should invalidate the above argument that the Fermi level $\mu$ must be in the middle of the band gap. But it does not. To change the ratio between $\iota_{\rm {e}}$ and $\iota_{\rm {h}}$ by a factor 2 requires a shift in $\mu$ of about 0.01 eV at room temperature. That is very small compared to the band gap. And the shift would be much smaller still closer to absolute zero temperature. At absolute zero temperature, the Fermi level must move to the exact middle of the gap.

That illustrates another important rule of thumb for semiconductors:

Keep your eyes on the thermal exponentials. Usually, their variations dwarf everything else.
If ${\vphantom' E}^{\rm p}$ or $\mu$ changes just a little bit, $e^{-({\vphantom' E}^{\rm p}-\mu)/{k_{\rm B}}T}$ changes dramatically.

(For gallium arsenide, the difference between the densities of states for holes and electrons is much larger than for silicon or germanium. That makes the shift in Fermi level at room temperature more substantial.)

The Fermi level may be directly computed. Expressions for the total number of conduction electrons per unit volume and the total number of holes per unit volume are, {D.30}:

\begin{displaymath}
\fbox{$\displaystyle
i_{\rm e} =
2 \left(\frac{m_{\rm ...
...e^{-(\mu - {\vphantom' E}^{\rm p}_{\rm v})/k_{\rm B}T}
$} %
\end{displaymath} (6.35)

Here ${\vphantom' E}^{\rm p}_{\rm {c}}$ and ${\vphantom' E}^{\rm p}_{\rm {v}}$ are the energies at the bottom of the conduction band, respectively the top of the valence band. The appropriate effective masses for electrons and holes to use in these expressions are comparable to the true electron masses for silicon and germanium. Setting the two expressions above equal allows $\mu$ to be computed.

The first exponential in (6.35) is the value of the number of electrons per state $\iota_{\rm {e}}$ at the bottom of the conduction band, and the second exponential is the number of holes per state $\iota_{\rm {h}}$ at the top of the valence band. The bottom line remains that semiconductors have much too few current carriers to have good conductivity.

That can be greatly improved by what is called doping the material. Suppose you have a semiconductor like germanium, that has 4 valence electrons per atom. If you replace a germanium atom in the crystal by a stray atom of a different element that has 5 valence electrons, then that additional electron is mismatched in the crystal structure. It can easily become dislocated and start roving through the conduction band. That allows additional conduction to occur. Even at very small concentrations, such impurity atoms can make a big difference. For example, you can increase the conductivity of germanium by a factor of a thousand by replacing 1 in a million germanium atoms by an arsenic one.

Because such valence-5 impurity atoms add electrons to the conduction band, they are called “donors.” Because electrical conduction occurs by the negatively charged additional electrons provided by the doping, the doped semiconductor is called “n-type.”

Alternatively, you can replace germanium atoms by impurity atoms that have only 3 valence electrons. That creates holes that can accept valence band electrons with a bit of thermal energy. Therefore such impurity atoms are called “acceptors.” The holes in the valence band from which the electrons were taken allow electrical conduction to occur. Because the holes act like positively charged particles, the doped semiconductor is called “p-type.”

Silicon has 4 valence band electrons just like germanium. It can be doped similarly.

Now consider an n-type semiconductor in more detail. As the center of figure 6.31 indicates, the effect of the donor atoms is to add a spike of energy states just below the conduction band. At absolute zero temperature, these states are filled with electrons and the conduction band is empty. And at absolute zero, the Fermi level is always in between filled and empty states. So the Fermi level is now in the narrow gap between the spike and the conduction band. It illustrates that the Fermi level of a semiconductor can jump around wildly at absolute zero.

But what happens at absolute zero is irrelevant to a room temperature semiconductor anyway. At room temperature the Fermi level is typically as shown by the tick mark in figure 6.31. The Fermi level has moved up a lot compared to the intrinsic semiconductor, but it still stays well below the donor states.

If the Fermi level would not move up, then the total number of electrons in the conduction band would not change. And there would be extremely few electrons in the donor states for that Fermi level. That is not possible, because all the other donor electrons cannot just disappear. In fact, the amount of electrons contributed by the donor states is dramatic; that is because there are so extremely few conduction electrons in the intrinsic case. The Fermi level has to move up significantly to explain the increase. An increase in Fermi level $\mu$ increases the number of electrons per quantum state (6.33) in the conduction band. The Fermi level has to go up far enough that the combined number of electrons in the conduction band and the donor states is just a bit more than the number of donor electrons.

But the Fermi level cannot move too close to the donor states either. For assume the contrary, that the Fermi level is really close to the donor states. Then the donor states will be largely filled with electrons. But at room temperature the gap between the donor states and the conduction band is comparable to ${k_{\rm B}}T$. Therefore, if the donor states are largely filled with electrons, then the states at the bottom of the conduction band contain significant numbers of electrons too. Since there are so many of these states compared to a typical number of donors, the number of electrons in them would dwarf the number of electrons missing from the donor states. And that is not possible since the total number of electrons cannot exceed the number of donor states by any noticeable amount. So the Fermi level has to stay low enough that the number of electrons $\iota_{\rm {e}}$ stays small in both the donor states and conduction band. That is as sketched in figure 6.31.

If more donors are added, the Fermi level will move up more. Light doping may be on the order of 1 impurity atom in a 100 million, heavy doping 1 in 10,000. If the donor atoms get too close together, their electrons start to interact. If that happens the spike of donor states broadens into a band, and you end up with a metallic “degenerate” semiconductor. For example, low temperature measurements show that phosphor donors turn silicon metallic at about 1 phosphor atom per 15,000 silicon ones. It may seem strange that impurity electrons at such a small concentration could interact at all. But note that 1 impurity in 15,000 atoms means that each $25\times25\times25$ cube of silicon atoms has one phosphor atom. On average the phosphor atoms are only about 25 atom spacings apart. In addition, the orbit of the very loosely bound donor electron is really far from the positively charged donor atom compared to the crystal spacing.

The upward shift in the Fermi level in n-type material has another effect. It decimates the already miserably small number of holes in the valence band, (6.34). Therefore the number of conduction band electrons provided by the valence band becomes many times smaller still than it was already for the intrinsic semiconductor. Essentially all conduction band electrons are provided by the donors. Also, almost all electrical conduction will now be performed by electrons, not holes. The electrons in n-type material are therefore called the “majority carriers” and the holes the “minority carriers.”

Figure 6.32: Relationship between conduction electron density and hole density. Intrinsic semiconductors have neither much conduction electrons nor holes.
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The fact that raising the amount of conduction band electrons lowers the amount of valence band holes may be verified mathematically from (6.35). That equation implies that the product of the electron and hole densities is constant at a given temperature:

\begin{displaymath}
\fbox{$\displaystyle
i_{\rm e} i_{\rm h} =
4 \left(\fr...
...)^3
e^{-{\vphantom' E}^{\rm p}_{\rm gap}/k_{\rm B}T}
$} %
\end{displaymath} (6.36)

This relationship is called the “law of mass action” since nonexperts would be able to make sense out of electron-hole density relation. And if you come to think of it, what is wrong with the name? Doesn’t pretty much everything in physics come down to masses performing actions? That includes semiconductors too!

The relationship is plotted in figure 6.32. It shows that a high number of conduction electrons implies a very low number of holes. Similarly a p-type material with a high number of holes will have very few conduction electrons. The p-type material is analyzed pretty much the same as n-type, with holes taking the place of electrons and acceptors the place of donors.

The law of mass action can also be understood from more classical arguments. That is useful since band theory has its limits. In thermal equilibrium, the semiconductor is bathed in blackbody radiation. A very small but nonzero fraction of the photons of the radiation have energies above the band gap. These will move valence band electrons to the conduction band, thus creating electron-hole pairs. In equilibrium, this creation of electron-hole pairs must be balanced by the removal of an identical amount of electron-hole pairs. The removal of a pair occurs through “recombination,” in which an conduction band electron drops back into a valence band hole, eliminating both. The rate of recombinations will be proportional to the product of the densities of electrons and holes. Indeed, for a given number of holes, the more electrons there are, the more will be able to find holes under suitable conditions for recombination. And vice-versa for holes. Equating a creation rate $A$ of electron-hole pairs by photons to a removal rate of the form $Bi_{\rm {e}}i_{\rm {h}}$ shows that the product $i_{\rm {e}}i_{\rm {h}}$ is constant. The constant $A$$\raisebox{.5pt}{$/$}$$B$ will depend primarily on the Maxwell-Boltzmann factor $e^{-{\vphantom' E}^{\rm p}_{\rm {gap}}/{k_{\rm B}}T}$ that limits the number of photons that have sufficient energy to create pairs.

This picture also provides an intuitive explanation why adding both donors and acceptors to a semiconductor does not double the amount of current carriers over just one type of doping alone. Quite the opposite. As figure 6.32 shows, if the number of holes becomes comparable to the number of electrons, there are not many of either one. The semiconductor behaves again like an intrinsic one. The reason is that adding, say, some acceptors to an n-type material has the primary effect of making it much easier for the conduction band electrons to find valence band holes to recombine with. It is said that the added acceptors compensate for the donors.


Key Points
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Doping a semiconductor with donor atoms greatly increases the number of electrons in the conduction band. It produces an n-type semiconductor.

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Doping a semiconductor with acceptor atoms greatly increases the number of holes in the valence band. It produces an p-type semiconductor.

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The minority carrier gets decimated.

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The Fermi level is in the band gap, and towards the side of the majority carrier.

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There is compensation in doping. In particular, if there are about the same numbers of electrons and holes, then there are not many of either.