# Semiconductor for beginner #3 ## Introduction

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.

Given the potentially extensive content, feel free to explore based on your specific areas of interest. And most importantly, enjoy the read!

## What determines the magnitude of the current

Semiconductor devices operate by the flow of electricity.

This flow of electricity refers to the electrons and holes that are carriers.

How much electricity flows through a device has a large effect on its performance.

What do you think determines the magnitude of the current?

It is the “number of carriers” and “speed.

If there are many carriers and the speed is fast, a large current will flow. If there are few carriers and the speed is slow, the current will be very small.

In Chapter 3, we will learn about the number of carriers.

In Chapters 1-2, we learned about bands.

We talked about how the valence band is packed with electrons and there are electrons that can move freely in the conduction band.

Holes are in the valence band because they can move using the gaps between the jammed electrons.

You will gain a better understanding of what state of electrons you are thinking about by looking at it from multiple perspectives.

## Density of states of electrons

Suppose an electron in the valence band receives light energy and goes across the band gap to the conduction band. At this time, a hole is created in the valence band.

Notice : translate from Japanese to English

エネルギー : Energy

バンドギャップ : bandgap

ホール : Hole

In this case, we find that the number of free electrons and holes are the same.

And if this number of free electrons and holes is large, it is possible to make the device emit light (free electrons in a conductor emit light and migrate to the valence band) or something.

In other words, knowing the number of electrons in the conduction band (as well as holes in the valence band) is very important for device construction.

The density of states of these electrons is called the density of states.

The motivation is to understand this thing called the electron state and to know how many there are.

## Number of electron states as waves

What we will be considering is the electrons in the conduction band, or free electrons.

What we needed to know was the state of the electron. Have you ever heard of quantum mechanics? In quantum mechanics, electrons have two states, one as a wave and the other as a particle.

What we are going to consider is the state as a wave.

Although it is not strictly true, as an image, imagine a penalty kick in soccer.

After I kick the ball, the result is already decided. This is the state as a particle.

The state as a wave is the state before I kick the ball. I have countless choices. Waves also fluctuate and have countless states.

To understand waves with a firsthand sense, let us consider, as an example, the acoustic waves in a concert hall.

If you can grasp the image of a wave here, you can say that you understand the state of an electron wave.

Imagine a cube with a side of 34 meters. Let this be a concert hall (too complicated would be tedious, so let’s laugh).

The speed of sound is 340 m/s. Therefore, 34 is a convenient number.

Suppose a wave travels from the left end and comes to the boundary of the second cube.

At this point, it oscillates in the same way as the wave that came from the first left end.

This means that when we consider the oscillation of a wave in one cube, we can also instantly see the movement of countless cubes side by side.

Considering periodic motion in this way, separated by boundaries, is called periodic boundary conditions. For example, consider the time it takes for a Ferris wheel to complete one lap.

Once this is known, the periodic boundary condition can be used to determine the time it takes for the Ferris wheel to make 100 laps.

The wavelength that resonates at an open end with a side of 34 meters is 34 meters (it travels 34 meters and returns to its original position). This wave would vibrate 10 times per second (10 Hz).

This frequency can be obtained simply by using the relationship between the sound wave 340 m/s and the frequency (Hz) = 1/period (s).

$$340[\frac{\rm{m}}{\rm{s}}]×\frac{1}{34[\rm{m}]}=10\rm{Hz}$$

This stable and resonant wave is called a standing wave. It is one of the states of a stable sound wave.

So what if we consider a wave that oscillates 20 times? It is a wave with two humps in 34m. The wavelength at this point is 17 meters (after traveling 17 meters, it will be at its original position).

This sound is also a standing wave because it resonates and propagates. In other words, it is one of the stable sound wave states.

The next question to consider here is what regularity there is in the number of this stable state called a standing wave.

We can find out by looking at the number of waves. And let us define it as the number of waves.

We can write $$k=\frac{2π}{λ}$$.

This allows us to express the standing wave condition in a simple way.

Let’s express the wavelengths 34m, 17m, and 11.3m in this order,

$$k_y=2π×\frac{1}{34[\rm{m}]}$$

$$k_y=2π×\frac{2}{34[\rm{m}]}$$

$$k_y=2π×\frac{3}{34[\rm{m}]}$$

This will be the case. Of course, there are many more standing waves than these three.

Therefore, the conditions for standing waves can be summarized in one equation

$$k_y=\frac{2πn}{L}$$ (n is an integer)

(n is an integer). The subscript “y” means y direction. This can be expressed as a graph as follows.

Of course, waves in the x and z directions must also be considered.

Considering in the same way, the graph in the xy-plane becomes

Furthermore, when considering waves in all xyz directions, the wavenumber can be expressed as follows

Just to review, the wavenumber points in these diagrams represent the points where stable sound (standing waves) resonate.

In other words, there is no stable sound between these points. Note that we are now considering the walls of a concert hall as the open ends.

To further visualize the number of states of the electron as a wave, I will make the story a bit more mathematical in order to make it easier to visualize.

As the wavelength changes, the momentum and energy change.

These can be expressed using the wave number k.

For example, in quantum mechanics, the momentum of both light and electrons is.

$$p=\hbar k$$

can be expressed as $$\hbar$$ is Planck’s constant.

If you multiply $$\hbar$$ times, you get momentum, so you can almost always consider the wavenumber as momentum when considering quantitative things.

Since the wavenumber is 2π divided by the wavelength, the larger the wavelength, the smaller the wavenumber (momentum), and the smaller the wavelength, the larger the wavenumber (momentum).

Therefore, the space of wavenumbers in xyz in the above figure can also be thought of as the space of momentum. If we consider a sphere of radius |k|(or |p|), each of its lattice points is a possible number of states.

If you think of it in terms of momentum in the xyz direction, it becomes a little easier to imagine.

## Number of electronic states in semiconductors

So far we have considered a concert hall as an example.

Next, we will consider the case of a solid block of semiconductors (called bulk) instead of a concert hall.

As before, let us assume one side L (we will consider a cubic space with one side L).

What we consider is the property of electrons as waves. In other words, we will consider standing waves that can resonate and exist stably in the bulk.

Specifically, we will count the types of standing waves.

However, the story up to this point is the same as the case of sound waves described earlier, so we will skip this part (please refer to the previous figure as you read on).

Let’s look at the actual number of states of electrons.

First, let us consider the number of states of an electron in the range from a certain energy E to E + ΔE. The kinetic energy of an electron is

$$E=\frac{p^2}{2m}$$. In quantum mechanics, the momentum of the electron is $$p=\hbar k$$, so substituting this, we get $$E=\frac{\hbar^2k^2}{2m}$$. Waves have magnitude and direction.

Since we are now considering a three-dimensional space, we can write $$k^2=k^2_x+k^2_y+k^2_z$$. This is just the inner product of the vector k.

Here the radius of the sphere in the previous figure is $$|k|=\sqrt{k^2_x+k^2_y+k^2_z}$$. In other words, $$E=\frac{\hbar^2k^2}{2m}$$ shows that the smaller the radius $$|k|$$ of the sphere, the smaller the energy and the larger the radius, the larger the energy.

In other words, if a lattice point is on the surface of a sphere of this radius, we can say that it has $$E=\frac{\hbar^2k^2}{2m}$$.

By the way, we are now considering “the number of states of an electron in the range from some energy E to E+ΔE”, which is called the density of states.

In diagrammatic terms, imagine a sphere. Make that sphere just a little bit larger.

The number of lattice points between the original sphere and the enlarged sphere is the density of states. In other words, counting the number of states in this region leads to the density of states.

This is the calculation. The method is simple: subtract the number of states of the sphere E from the sphere E + ΔE to obtain the density of states. The radius of the sphere is from the energy E,

$$k=\frac{\sqrt {2mE}}{\hbar}$$

which is $$k=\frac{\sqrt {2mE}}{\hbar}$$.

Therefore, we can find the number of states by finding the volume of a sphere of this radius.

For a sphere of radius r, the volume is $$\frac{4πr^3}{3}$$, so the number of states of a sphere of energy E is,

$$\frac{4πk^3}{3}=\frac{4π(\frac{\sqrt {2mE}}{\hbar})^3}{3}$$

is.

One cube with one side $$\frac{2π}{L}$$ corresponds to one state,

so dividing the volume of the sphere obtained earlier by $$(\frac{2π}{L})^3$$gives the number of states contained in this volume.

$$\frac{\frac{4π(\frac{\sqrt {2mE}}{\hbar})^3}{3}}{(\frac{2π}{L})^3}$$

can be written as $(\frac{2π}{L})^3}{3}}$$. Similarly, the number of states of E+ΔE is,$$\frac{\frac{4π(\frac{\sqrt {2m(E+ΔE)}}{\hbar})^3}{3}}{(\frac{2π}{L})^3}$$which means that the difference of these two equations is (\frac{2π}{L})^3}{3}$$. In other words, if we take the difference of these two equations, we can find the number of states sandwiched between the two spheres. $$\frac{\frac{4π(\frac{\sqrt {2m(E+ΔE)}}{\hbar})^3}{3}}{(\frac{2π}{L})^3}-\frac{\frac{4π(\frac{\sqrt {2mE}}{\hbar})^3}{3}}{(\frac{2π}{L})^3}$$ $$=\frac {L^3}{6π^2}(\frac{\sqrt{2m}}{\hbar})^3(\sqrt{E+ΔE}^3-\sqrt{E}^3)$$ $$=\frac {L^3}{6π^2}(\frac{\sqrt{2m}}{\hbar})^3((E+ΔE)\sqrt{E+ΔE}-E\sqrt{E})$$ $$=\frac {L^3}{6π^2}(\frac{\sqrt{2m}}{\hbar})^3((E+ΔE)\sqrt{E}\sqrt{1+\frac{ΔE}{E}}-E\sqrt{E})=①$$ Here, using the approximation $$\sqrt{1+a}≈1+\frac{1}{2}a$$ (first order Taylor expansion) which holds for a<<1 ki, $$①≈\frac {L^3}{6π^2}(\frac{\sqrt{2m}}{\hbar})^3((E+ΔE)\sqrt{E}(1+\frac{ΔE}{2E}-E\sqrt{E})=②$$ Furthermore, the ΔE × ΔE term is a multiplication of two tiny ones, so if we ignore it and approximate it, $$②≈\frac {L^3}{6π^2}(\frac{\sqrt{2m}}{\hbar})^3(E\sqrt{E}((1+\frac{ΔE}{2E})+ΔE\sqrt{E}-E\sqrt{E})$$ $$=\frac {L^3}{6π^2}(\frac{\sqrt{2m}}{\hbar})^3\frac{3}{2}\sqrt{E}ΔE$$ $$=\frac {L^3}{4π^2}(\frac{\sqrt{2m}}{\hbar})^3\sqrt{E}ΔE$$ can be calculated. This is the number of states between the two spheres (between energy E and E+ΔE). Let N(E) be a function of the number of electronic states at a given energy E. Then the number of states between energy widths ΔE is expressed as N(E) ΔE, $$N(E)ΔE=\frac {L^3}{4π^2}(\frac{\sqrt{2m}}{\hbar})^3\sqrt{E}ΔE$$ can be written as Dividing both sides by ΔE, $$N(E)=\frac {L^3}{4π^2}(\frac{\sqrt{2m}}{\hbar})^3\sqrt{E}$$ which is the same as the length of one side, L. We now know the number of states contained in a cube of length L per side, i.e., the number of standing wave types. To summarize what we have done so far, we have subtracted the number of states between E and E+ΔE and derived a function N(E) of the number of states, with energy E as the variable. Looking at the above equation, the larger L, the larger the number of states. Therefore, we need to adjust this per unit volume. For example, if a car runs 24 km in 8 hours, it can be expressed as 24 km/8h=3 km/h or 3 km/h per hour. This can be expressed per unit of time. When we ask how fast a car is going, we say 180 km/h (180 km/h) or 100 km/h. We would not say 360 km/2h or 200 km/2h. So, per-unit expressions are useful at times and in some cases to make things easier to understand. That’s why, to fix the number per unit volume, we just divide by $$L^3{/latex], D(E)≡\frac{N(E)}{L^3}=\frac{1}{4π^2}(\frac{\sqrt{2m}}{\hbar})^3\sqrt{E} *≡ means to define. We now have the density of states per unit volume, D(E). This is the number of states of the electron as a wave. As you may know if you have studied quantum mechanics, two more states of the electron correspond to one wave. The states called up-spin and down-spin correspond to the states of one wave, using the expression “spin. You may be able to understand spin by studying quantum mechanics, so I will skip the detailed explanation here. From a quantum mechanical point of view, there can be two states in one lattice point. So far we have only considered the number of states of an electron in one wave (up-spin or down-spin); to correspond to the existence of two states for one electron wave, we can double the density of states D(E), and that doubles the number of states, which means that we have considered up and down spin. D_e(E)=\frac{1}{2π^2}(\frac{\sqrt{2m}}{\hbar})^3\sqrt{E} We have finally derived an expression for the density of states of electrons that also corresponds to quantum mechanics. This can be written graphically as follows. Diagram of density of states vs. energy Notice : translate from Japanese to English エネルギー : Energy 状態密度 : state density It is easier to understand visually if it is drawn graphically. The greater the energy, the greater the number of states. However, as you can see from the graph, as you gradually increase the number of states, a considerably larger amount of energy is required. It is good to keep in mind the shape of this density of states, and although we have focused on a three-dimensional semiconductor (bulk), we can change the shape of this density of states by creating a structure called a quantum well (we will do this later). For now, we can express the number of states of electrons as waves in the semiconductor (bulk) in the form of a spin-aware energy function. ## Quantum statistics The number of states of an electron could be expressed as a function of energy. This means that the distribution of electrons is determined by energy. The next thing to consider is how the electrons are distributed among these states. The “concept of the relationship between energy and electron distribution” in which there are many electrons at high energy and many at low energy is called energy distribution. This is not a spatial distribution. The variable of the function is E. In a spatial distribution, it is a function of position x. To think about this energy distribution, we use our knowledge of statistical mechanics. In the energy of electrons, the states are filled in order of energy, starting from the lowest energy state. However, under the influence of heat or light energy, electrons can receive that energy and exist at higher energies. Recall the story of the move from the valence band to the conduction band. Electrons receive heat and light energy and go to places of higher energy. Notice : translate from Japanese to English エネルギー : Energy 伝導帯 : Conduction band 自由電子 : Free Electrons 光 : light バンドギャップ : bandgap 価電子帯 : valence band ホール : Hole Let us consider the energy distribution. First, as a preparation, consider a particle that is larger than an electron and subject to Newtonian mechanics. As an example, we will deal with the energy distribution of gas molecules. This is called the Maxwell-Boltzmann distribution because Maxwell and Boltzmann revealed it. The probability f(E) of the existence of a particle with a certain energy E is, f(E)=e^{-\frac{E}{k_BT}} can be expressed. k_B=1.3806×10^{-23} \rm{J/K} is called Boltzmann’s constant, where K is absolute temperature, not the Celsius we normally use. -273C absolute zero is the standard for 0K. Therefore, 0°C is 273K. A room temperature of 27°C gives a crisp figure of 300 K. This room temperature of 27°C is called “room temperature. This room temperature of 27°C is called RT, which is an acronym for Room Temperature, and is used in solid state physics papers. The characteristic of this distribution is that the probability of existence decreases as the energy increases. Next, let us consider the statistics of electrons. Electrons are a little more complicated than in the case of gas molecules. Electrons follow the Fermi-Dirac distribution because it was created by Fermi and Dirac. They were born about 70 years later than Maxwell and Boltzmann. They were both born about 70 years later than Maxwell and Boltzmann, and played a major role in the construction of quantum mechanics. Particles that follow the Fermi-Dirac distribution are called Fermi particles (fermions). Electrons are classified as Fermi particles. Incidentally, particles can be divided into two main categories: Bose particles and Fermi particles, with the former allowing a number of particles to exist in the same state. In the latter, no more than two can exist in the same state. For example, look at the figure below again. Wavenumber in xyz plane(Sphere of radius k) This lattice point represents the state of the electron as a wave. And there can be two states at that lattice point: up and down. In other words, there cannot be three or four up-spin particles at a given lattice point. At a given lattice point, there can only be (number of up-spin electrons, number of down-spin electrons)=(1,0),(0,1),(0,0),(1,1). When expressed using energy levels, the image is shown in the following figure. Fermi Particles and Energy Levels Notice : translate from Japanese to English エネルギー準位 : energy level フェルミ粒子 : Fermi Particle アップスピン : Up-spin ダウンスピン: downspin 電子 : electron On the other hand, a Bose particle can have many particles at a single lattice point. In other words, the energy level can be expressed as shown in the figure below. Energy level is an axis that represents energy, but energy does not exist continuously, but discretely. It is represented step by step like a jumping box. Bose Particles and Energy Levels Notice : translate from Japanese to English エネルギー準位 : energy level ボーズ粒子 : Bose Particle Since we are going to consider electrons, we will consider Fermi particles. Then, from the Fermi-Dirac distribution that the Fermi particle follows, we understand the existence probability of an electron with a certain energy E. The existence probability f(E) is expressed as The existence probability f(E) is expressed as follows. f(E)=\frac{1}{1+e^{\frac{E-E_F}{k_BT}}} The EF in the denominator is a quantity called the Fermi energy, Fermi level, chemical potential, etc. F is the F of Fermi. Here \($$E_F\)(← displayed by bug)$$E_F$$ means that when $$E=E_F$$, the probability of electron existence is $$1/2$$ and at absolute zero at 0K (-273°C), the most electron is occupied in the valence band at absolute zero at 0 K (-273 °C). At absolute zero, all molecular motion ceases (of course, there are no living organisms). This means that electrons reside in the conduction band and do not move as free electrons. Let us consider the above and more using actual mathematical equations. First of all, let us consider the case T→0K, which is as close to absolute zero as possible (we would like to substitute T=0K into the equation, but the mathematical rules do not allow us to substitute 0 into the denominator. So we consider the case of being as close as possible). There are two cases to consider. $$E-E_F>0$$ and $$E-E_F<0$$. That is, with the former, the exponential part of e diverges positively and f(E) converges to 0. In the latter case, the exponential part of e diverges negatively and f(E) converges to 1. This can be plotted graphically as follows. Fermi-Dirac distribution (0K) Notice : translate from Japanese to English 絶対零度 : absolute zero エネルギー : energy 必ず1/2を通る : Always through 1/2 存在確率 : Probability of existence Next, consider the room temperature condition at T=300K. This is difficult to calculate by hand, so first refer to the following graph. Fermi-Dirac distribution (300K) Notice : translate from Japanese to English 室温 : Room temperature エネルギー : energy 必ず1/2を通る : Always through 1/2 存在確率 : Probability of existence This graph shows that electrons are also present at $$E-E_F>0$$ compared to 0K. In other words, it is a probabilistic representation of the presence of electrons with higher energy as the temperature is increased. It means that electrons move from the valence band to the conduction band upon receiving thermal energy, and free electrons are present. At higher temperatures, of course, we would expect to see more free electrons in the conduction band due to the receipt of thermal energy. Also, as a final note, when $$E-E_F=0$$, the probability is expressed as $$1/2$$, independent of temperature. We can see that $$E_F$$ is the boundary representing 1/2 of the probability. ## The energy distribution of electrons and holes In the conduction band there are free electrons that can move freely, and in the valence band there are holes that can move freely. First, let us consider electrons. If the bottom of the conduction band is $$E_C$$, the graph looks like this. This is the electron density of states vs. energy diagram derived in the previous section. The origin is $$E_C$$. And the density of states of electrons in the conduction band is $$D_e(E)=\frac{1}{2π^2}(\frac{\sqrt{2m}}{\hbar})^3\sqrt{E}… ③$$$

can be expressed as follows.

By the way, do you remember the term “true semiconductor”?

A true semiconductor is a semiconductor made of a single element.

On the other hand, a compound semiconductor is a semiconductor made of two or more elements.

The Fermi energy of a true semiconductor is approximately halfway between the bottom of the conduction band and the top of the valence band.

The band gap of a semiconductor is about 1 eV, so half of that is about 0.5 eV.

Therefore, the Fermi energy (chemical potential) comes at about 0.5 eV in the relationship between the density of states $$D_e(E)$$ and energy.

The probability of an electron entering the conduction band is determined by the Fermi-Dirac distribution $$f(E)$$.

At absolute zero temperature, there are no electrons in the conduction band, but at room temperature (300 K), free electrons exist in the conduction band due to thermal effects.

This probability $$f(E)$$ multiplied by the number of electron states $$D_e(E)$$ gives the graph below.

Notice : translate from Japanese to English

エネルギー : energy

Written in mathematical form, the electron density n in the conduction band is,

$$n=\int_{E_C}^{∞}f(E)D_e(E)dE・・・④$$

can be written as follows. This is obtained by adding up all the numbers of electron states, taking into account the probability of existence of electrons at temperature.

Similarly, we can consider a hole in the valence band.

The energy of a hole is in contrast to that of an electron since its charge is positive.

In addition, in the Fermi-Dirac distribution,

“the probability of the existence of a hole” = “the probability of the absence of an electron”,

so $$1-f(E)$$ is the probability that a hole exists.

Therefore, it can be considered as shown in the following figure.

Notice : translate from Japanese to English

エネルギー : energy

ホール密度 : Hole density

Similarly mathematically the hole density p in the valence band is,

$$p=\int_{-∞}^{E_V}(1-f(E))D_h(E)dE$$

can be expressed as

Of course, at absolute zero (0 K), there are no holes in the valence band, but at room temperature (300 K), holes exist in the valence band due to thermal effects. In other words, if you understand the role of electrons, you also understand the role of holes.