Hello World ☆ 
This article is designed for beginners who want to learn about semiconductors.
I majored in physics during my university years and focused on theoretical physics.
However, during my master’s program, I unexpectedly delved into research on high-energy physics or particle physics experiments involving semiconductor detection devices.
This article compiles the knowledge I acquired about semiconductors during that time.
Given the potentially extensive content, feel free to explore based on your specific areas of interest. And most importantly, enjoy the read!
The higher the number of free electrons in the conduction band, the more electricity flows, and the higher the number of holes in the valence band, the more electricity flows.
As temperature increases, the shape of the Fermi-Dirac distribution changes, and electrons move more easily from the valence band to the conduction band, and holes move more easily from the conduction band to the valence band as well.
As these numbers increase, the flow of electricity becomes easier.
Therefore, the values of carrier density (free electrons and holes) in the conduction and valence bands are one of the key quantities that determine whether electricity flows easily. We will see how to obtain this.
Let’s look at some concrete examples to get a feel for it.
Let’s consider the case of a true semiconductor (simple and easy to handle, so this is enough to visualize a concrete example).
The Fermi energy \(E_F\) is located almost in the middle of the forbidden band, and the large difference in band gap energy is about 1 eV.
In other words, the \(E-E_F\) between the bottom of the conduction band and the Fermi energy is about 0.5 eV (500 meV). The \(k_BT\) has an energy of 25.9 meV at room temperature 27°C (300 K).
In other words, \(E-E_F >>k_BT \), so \(\frac{E-E_F}{k_BT} >>1 \).
Hence \(e^{\frac{E-E_F}{k_BT}} >>e^1=e≈2.71828>1 \).
$$f(E)=\frac{1}{1+e^{\frac{E-E_F}{k_BT}}}≈e^{-\frac{E-E_F}{k_BT}}・・・⑤$$
This form is that of the Maxwell-Boltzmann distribution.
This approximation makes the calculation easier.
Similarly, for a hole of energy E near the top of the valence band, since \(E-E_F >>k_BT \),
$$1-f(E)=1-\frac{1}{1+e^{\frac{E-E_F}{k_BT}}}≈e^{-\frac{E_F-E}{k_BT}}$$
can be approximated as
There is a reason why this Fermi-Dirac distribution is approximated to the Maxwell-Boltzmann distribution, and it is very difficult to do integration if the Fermi-Dirac distribution is used as it is.
In most cases, the approximation of the Maxwell-Boltzmann distribution is physically sufficient. Therefore, general reference books on semiconductor engineering at the university level rarely use the Fermi-Dirac distribution as it is.
Substituting equations (3) and (5) into equation (4) and using \(\hbar=h/2\),
$$n=\int_{E_C}^{∞}f(E)D_e(E)dE$$
$$=\frac{8\sqrt{2}π}{h^3}m^\frac{3}{2}e^{\frac{E_F}{k_BT}}\int_{E_C}^{∞}\sqrt{E-E_C}e^{-\frac{E}{k_BT}}dE$$
(*Substituting E-Ec for De(E) because it was necessary to consider E-Ec and the energy from the bottom of the conduction band).
Near the top of the conduction band, the probability of the Maxwell-Boltzmann distribution is almost zero, so there is no particular problem in setting the upper bound to ∞.
If we do this variable transformation \(x=\frac{E-E_C}{k_BT}\),
$$\frac{dx}{dE}=\frac{1}{k_BT}$$
$$∴dE=k_BTdx$$
Hence the integral is,
$$\int_{E_C}^{∞}\sqrt{E-E_C}e^{-\frac{E}{k_BT}}dE$$
$$(k_BT)^{\frac{3}{2}}e^{-\frac{E_C}{k_BT}}\int_{0}^{∞}\sqrt{x}e^{-x}dx$$
which is then.
$$\int_{0}^{∞}\sqrt{x}e^{-x}dx=\frac{\sqrt{π}}{2}$$
using the integral formula (if you have time, try to come up with this formula yourself),
The electron density is \(n=2(\frac{2πmk_BT}{h^2})^\frac{3}{2}e^{-\frac{E_C-E_V}{k_BT}}\). Summing up the previous coefficients,
$$N_C≡2(\frac{2πmk_BT}{h^2})^\frac{3}{2}$$
and defining it as,
$$n=N_Ce^{-\frac{E_C-E_F}{k_BT}}・・・⑥$$
This leads to the following equation.
The interpretation of this equation is that if we assume that there is an effective concentration of NC electronic states at the bottom of the conduction band E=EC, then we can regard it as being subject to the Maxwell-Boltzmann distribution.
Since we are considering the energy of the free electrons in the conduction band, we are considering E-EF>0.
Therefore, if we assume that NC electronic states are concentrated at the bottom of the conduction band and multiply them by a temperature-dependent probability, we obtain the electron density we want to find.
Therefore, we call NC the effective density of states of the conduction band. The graphical representation is as follows.
Notice : translate from Japanese to English
エネルギー : energy
実効状態密度 : Effective state density
ボルツマン分布 : Boltzmann distribution
存在確率 : Existence probability
電子密度 : Electron density
ホール密度 : Hole density
A similar procedure is used to obtain the density p of holes in the valence band.
$$p=\int_{-∞}^{E_V}(1-f(E))D_h(E)dE$$
$$≡N_Ve^{-\frac{E_F-E_V}{k_BT}}・・・⑦$$
Diagrammatically, it looks like this
Notice : translate from Japanese to English
エネルギー : energy
実効状態密度 : Effective state density
ボルツマン分布 : Boltzmann distribution
存在確率 : Existence probability
電子密度 : Electron density
ホール密度 : Hole density
Now, you may be wondering what this effective density of states is for.
The reason is that it simplifies the model compared to the following two figures shown previously.
Notice : translate from Japanese to English
エネルギー : energy
状態密度 : state density
存在確率 : Existence probability
電子密度 : Electron density
ホール密度 : Hole density
Notice : translate from Japanese to English
エネルギー : energy
状態密度 : state density
存在確率 : Existence probability
電子密度 : Electron density
ホール密度 : Hole density
With the original density of states (Fermi-Dirac distribution), the carrier density cannot be obtained without integration, but using the approximation (Maxwell-Boltzmann distribution), the integration could be done in a simple form.
As a result, the density of carriers can be easily obtained by using equations (6) and (7), since the integration is already done when expressed in terms of the effective density of states.
Incidentally, \(k_BT\) is 34.5 meV at 127°C (400 K), so this approximation can be used over a rather wide range.
In a true semiconductor, free electrons and holes are formed when electrons in the valence band move up to the conduction band.
At this point, the electron density n in the conduction band and the hole density p in the valence band are equal. Calling this density the intrinsic density ni, the following relationship holds.
$$n=p=n_i・・・⑧$$
From this we have \(n_i^2=np\).
This relationship, which holds in thermal equilibrium (stable state with no thermal exchange), is called the mass action law.
If we take the square root and substitute equations (6) and (7) into this equation,
$$n_i=\sqrt{N_CN_V}e^{-\frac{E_g}{2k_BT}}・・・⑨$$
is obtained. \(E_g=E_C-E_V\) is the band gap energy.
In other words, the true density is a constant determined by the 3 properties of the material (band gap energy, effective density of states in the conduction band, and effective density of states in the valence band) and temperature.
It turns out that the electron density and hole density can be easily determined with an understanding of the mass action law.
When measuring band gap energy in an experiment, free electrons in the conduction band emit light energy, go to the valence band, and recombine with holes.
When this happens, they emit light (semiconductors like silicon do not emit light). By examining the wavelength of this emitted light, we can determine the band gap energy.
fig.z Relationship between bands (zones) and energy
Notice : translate from Japanese to English
電子エネルギー : electron energy
伝導帯 : conduction band
自由に動ける : Free to move
バンドギャップ(禁止帯) : Band gap (forbidden band)
存在できない : Cannot exist
光 : Light
発光 : Emission
バンドギャップエネルギー : Band gap energy
価電子帯 : Valence band
ほとんど自由に動けない : Hardly free to move
バンド図 : Band diagram
In student experiments, etc., the natural logarithm is taken on both sides of equation (9) and transformed as follows.
$$\rm{ln} n=\rm{ln} \sqrt{N_CN_V}-\frac{E_g}{2k_B}\frac{1}{T}$$
The experimentally determined values in the above equation are the carrier density, effective density of states, and temperature.
Boltzmann’s constant is known from the beginning because it is a constant.
We plot this equation with the vertical axis \(\rm{ln} n\), the horizontal axis \(\frac{1}{T}\) and the intercept \(\rm{ln} \sqrt{N_CN_V}\), taking some experimental data.
Then the slope \(-\frac{E_g}{2k_B}\) can be obtained and the band gap energy can be obtained.
According to the law of mass action, the relation \(n=p\) holds, which means that the number of free electrons in the conduction band is equal to the number of holes in the valence band.
This was confirmed in the previous section.
Using this, we can also find the Fermi energy \(E_F\). Finding the Fermi energy leads to finding the number of carriers (electron density and hole density) in the conduction band and valence band.
Therefore, it is important to determine the Fermi energy.
To find the Fermi energy, simply substitute equations (6) and (7) into \(n=p\), take the natural logarithm on both sides, and solve for \(E_F\).
At this time, also substitute the effective density of states.
The masses of the holes and electrons are put as \(m_e^, m_h^*\)
as effective masses.
Consider, for example, a free electron (here an electron that has kinetic energy \(p^2/2m\) and is free to move around in a vacuum) as the effective mass.
Normally, we might think of a free electron in a vacuum, but we are now thinking of an electron in a semiconductor (bulk) or in a crystal. This is not a vacuum state, and there are atoms around the free electrons.
Notice : translate from Japanese to English
余った電子 : Surplus electrons
In other words, electrons in a crystal move freely as free electrons, but are affected by potentials due to atomic nuclei and other factors.
By incorporating the effect of this complex potential into the effective mass \(m_e^,m_h^\)
, it can be described mathematically as a free electron.
So, if you do the math, it is easy to get an idea of this effective mass.
Back to the story, if we take \(E_V=0\) and the top of the valence band as 0 and the origin, the Fermi energy we want to find is,
$$E_F=\frac{E_g}{2}+\frac{3}{4}k_BT\frac{m_h^}{m_e^}$$
is obtained.
Looking at the energy, the effective mass \(m_h^*\) of a hole is 5~10 times larger than the effective mass \(m_h^*\) of an electron, and \(k_BT\) has an energy of 25.9 meV at 300K.
On the other hand, the band gap energy is about 1eV=1000meV, so the first term is so large that the second term is negligible when compared to the first term.
Therefore, the Fermi energy of a true semiconductor is located in the center of the conduction band and the valence band.
At the end of Chapter 2, we introduced n-type and p-type semiconductors.
Here we will deepen our understanding, including a review.
So far, we have focused on true semiconductors.
The reason for this is that they have a simple structure and are easy to explain.
In a true semiconductor, free electrons exist in the conduction band (holes in the valence band) at room temperature.
However, the number of electrons is far fewer than in metals, which are all around us.
The low number of electrons means that electricity does not flow very well.
This is because of its high electrical resistance.
Therefore, this is not suitable for use as a device that conducts electricity.
Therefore, artificial modifications are made to the semiconductor to make it easier for electricity to flow.
Semiconductors with this artificial work are called n-type semiconductors or p-type semiconductors.
These semiconductors are also called compound semiconductors and are composed of two or more elements (a true semiconductor is composed of one element).
Let’s start with n-type semiconductors and use Si, a group IV element, as an example; let’s look at a group V element mixed in with a Si element.
Then there will be one extra electron, as shown in the figure below.
Group IV elements have four outermost electrons (four outermost electrons), while group V elements have five outermost electrons (five outermost atoms).
Since an element is stable if it has eight outermost-shell atoms, group-IV Si can behave as if it has eight outermost-shell electrons by covalent bonding among Si, and thus is stable.
The above figure shows a group-V atom, which has only one extra outermost-shell electron because it has five.
Normally, this electron is pulled around by the Coulomb force (the force that binds this electron) of the atomic nucleus.
If this one extra electron were free to move, we could create a semiconductor that can easily conduct electricity.
Since this electron is bound by the Coulomb force, it is in an energy level slightly below the bottom of the conduction band. This energy is called the donor level.
Since the Coulomb energy is about several tens of meV, at room temperature, the electron receives thermal energy and moves up to the conduction band, where it is free to move around as a free electron.
Therefore, the more doping (amount of mixing), the more easily electricity flows through the semiconductor.
This semiconductor is called an n-type semiconductor because it has many negatively charged (negative) electrons.
And the group V atoms mixed to make an n-type semiconductor are called donors.
In the medical world, there is something called transplantation.
This is a medical procedure in which a donor (donor) transfers organs and other items to a recipient (recipient) and plants them. In the case of semiconductors, group V atoms are called donors because they provide electrons.
A p-type semiconductor is the opposite.
Group III elements (3 outermost shell electrons) are doped into group IV silicon atoms (4 outermost shell electrons).
It is stable with 8 outermost-shell electrons, but in this case it has 7 outermost-shell electrons and is missing one electron.
This is what we call a hole. In fact, we are not talking about the existence of a positively charged particle, but rather the blank space created by the lack of one negative charge, and we are treating the hole as a positively charged particle.
Notice : translate from Japanese to English
13族元素 : group of 13 elements
Comparing the nuclei of Si and those of group III (number of protons that are positive charges), group III elements have one less positive charge.
This means that at absolute zero, holes are attracted by the Coulomb force of group III elements and are bound around group III elements as positive charges.
This energy (orbital) is called the acceptor level.
As the temperature rises, the holes shake off this Coulomb force and rise into the valence band.
As the number of holes in the valence band increases, electricity flows more easily.
The group III atoms that make this p-type semiconductor are called acceptors. It is called an acceptor because it accepts (accepts) electrons from other atoms.
The energy positions of the donor and acceptor levels are illustrated below.
At absolute zero, electrons and holes exist at these positions, and at room temperature they receive thermal energy and rise to the conduction band and valence band.
Since the energy difference between these levels is several tens of meV, the electron or hole receives that amount of thermal energy and moves up to the conduction band or valence band.
Notice : translate from Japanese to English
電子のエネルギー : Electron energy
伝導帯 : conduction band
禁止帯 : Inhibition band
価電子帯 : valence band
ドナー準位 : Donor level
アクセプター準位 : acceptor level
ドナー原子は+に帯電 : Donor atom is + charged
アクセプター原子は-に帯電 : Acceptor atom is – charged
It is important to note that the donor atom becomes positively charged after the electron is released
(similarly, when an electron in the valence band moves up to the acceptor level, a hole is created in the valence band, and the acceptor atom becomes negatively charged).
This is called ionization. The energy of several tens of meV required for this ionization is called ionization energy.
Semiconductors mixed with impurities such as donors and acceptors are called impurity or compound semiconductors.
So far we have considered the carrier density for genuine semiconductors.
Next, we will consider the carrier density of impurity semiconductors.
First, we will consider the case where only donors are doped (the same applies to the case where only acceptors are doped).
The energy level of the donor is close to the conduction band, and the electrons can move around as free electrons when they rise to the conduction band, receiving this small energy difference of a few tens of meV (ionization energy).
Statistical mechanics can be used to determine the number of electrons that rise from the donor level to the conduction band, but an approximation is used here.
As an approximation, we consider that the carrier density n in the conduction band of an n-type semiconductor can be expressed by equation (4) as in the case of a true semiconductor.
$$④=n=\int_{E_C}^{∞}f(E)D_e(E)dE$$
$$\int_{E_C}^{∞}\frac{1}{1+e^{\frac{E-E_F}{k_BT}}}D_e(E)dE$$
Although this equation was used as an approximation, it is only suitable for describing true semiconductors. It does not reflect the effect of a large number of donor levels.
Therefore, it is necessary to choose the Fermi energy to be equal to the electron density of the actual n-type semiconductor.
This Fermi energy is a convenient one that reflects the donor effect and is called the pseudo-Fermi energy (or the Fermi energy of an n-type semiconductor).
The term “pseudo-Fermi energy” is used only when there is an excess of carriers due to voltage or light irradiation.
If there are many donor levels, there are also many electrons that can go up to the conduction band.
If we try to explain the model of a genuine semiconductor in equation (4) with this equation, we need to increase the probability of the existence of free electrons by moving the suspected Fermi energy \(E’_F\) closer to the bottom of the conduction band as shown in the following figure.
In the case of a true semiconductor, this Fermi energy \(E_F\) is located almost in the middle of the bottom of the conduction band and the top of the valence band.
Since the donor is doped in this case, the suspected Fermi energy of the semiconductor is near the conduction band.
Notice : translate from Japanese to English
フェルミ・ディラック分布 : Fermi Dirac distribution
エネルギー : energy
状態密度 : state density
存在確率 : Existence probability
電子密度 : Electron density
The shaded area on the right of the above figure represents the energy distribution of the electron density n.
This shape appears many times in the pn junction diagram, so just remember the image.
As an example, consider the case where each donor atom in an n-type semiconductor is all ionized.
In other words, we consider the case where electrons from all donors are up to the conduction band.
We obtain the \(E’_F\) for this doubt Fermi energy.
Here, the donor density \(N_D\) is assumed to be sufficiently larger than the density of electrons rising from the valence band to the conduction band.
In other words, since most of the free electrons come up from the donor level, we can write \(n=N_D\), which equals the carrier density \(n\) in the conduction band and the donor density \(N_D\).
(Actually, it is not this convenient. When Si (silicon) is doped with P (phosphorus) or As (arsenic) as a donor, the ionization energy is about 40~50 meV, and \(k_BT\) is relatively small, 25.9 meV at room temperature.
The \(k_BT\) is the thermal energy at room temperature of 300K. In other words, the electrons go up to the conduction band after getting 40~50 meV, so at room temperature 300K (27°C), the energy is slightly insufficient.
So we cannot say that all donor atoms are ionized.
But for now, as a special exception, let us consider the case where all donor atoms are ionized).
$$N_D=n=\int_{E_C}^{∞}f(E)D_e(E)dE$$
$$=\int_{E_C}^{∞}\frac{1}{1+e^{\frac{E-E’_F}{k_BT}}}D_e(E)dE$$
$$≈\frac{8\sqrt{2}π}{h^3}(m_e^*)^{\frac{3}{2}}e^{\frac{E’_F}{k_BT}}\int_{E_C}^{∞}\sqrt{E-E_C}e^{-\frac{E}{k_BT}}dE$$
$$=N_Ce^{-\frac{E_C-E’_F}{k_BT}}$$
can be written as: the third equation ≈ is an approximation.
As we did in a much earlier section, if we compare the energies at room temperature, we get E-E’F>>kBT which seems a bit aggressive. I can’t really say >> because I approached the suspected Fermi energy to the conduction band.
So, think of it as a very rough estimate of the pseudo-Fermi energy with very crude accuracy,
$$f(E)=\frac{1}{1+e^{\frac{E-E’_F}{k_BT}}}≈e^{-\frac{E-E’_F}{k_BT}}$$
The result is The basic flow is the same as in equation (5).
This form is the Maxwell-Boltzmann distribution.
$$∴ N_D=N_Ce^{-\frac{E_C-E’_F}{k_BT}}$$
Take the natural logarithm on both sides and solve for \(E’_F\).
Result.$$E’_F=E_C+k_BT\rm{ln}\frac{N_D}{N_C}$$
is.
We have now obtained the suspect Fermi energy (a very rough approximation).
Let’s consider where this alleged Fermi energy is compared to the conduction band.
Since the donor density \(N_D\) is usually smaller than the effective density of states \(N_C\) in the conduction band, this is a negative term.
(For example, when positive integers A and B are A>B>e=2.718……,
$$\rm{ln}\frac{B}{A}=\rm{ln}(\frac{A}{B})^{-1}=-\rm{ln}\frac{A}{B}$$
and the logarithmic term is marked with -)
Similarly,
$$\rm{ln}\frac{N_D}{N_C}=-\rm{ln}\frac{N_C}{N_D}$$
which is the same as the above. Since the room temperature is 300 K, \(k_BT\) is 25.9 meV, and the logarithmic term on the right side of the suspected Fermi energy E’F is about tens of meV to 100 meV.
This means that the pseudo-Fermi energy \(E’_F\) is about several tens of meV to 100 meV below the energy \(E_C\) at the bottom of the conduction band.
Here we have obtained the pseudo-Fermi energy for n-type semiconductors, but if we follow the same procedure, the pseudo-Fermi energy for p-type semiconductors will also be located about several tens of meV to 100 meV above the top of the valence band.
Of course, in this case, separate pseudo-Fermi energies must be defined and calculated for n-type and p-type semiconductors: for n-type, the pseudo-Fermi energy must be set close to the bottom of the conduction band, and for p-type, the pseudo-Fermi energy must be set close to the top of the valence band.
In the case of p-type, the pseudo-Fermi energy should be set close to the top of the valence band.
These areas differ from the case of a true semiconductor (the Fermi energy is halfway between the conduction band and the valence band).
Now we can determine the number of carriers in the conduction band and valence band.
Hence, the goal of this section has been achieved.
Next, let’s look at moving the carriers around. Specifically, we will look at the case of electricity flowing through a semiconductor.
