Session One
Theory of Solids and the Semiconductor in Equilibrium
To understand the Current–Voltage characteristics of semiconductors, we need some knowledge of electron behavior in crystal when the electron is subjected to various potential functions.
To do this we have:
· To determine the properties of electrons in a crystal lattice.
· To determine the statistical characteristics of a very large number of electrons in a crystal.
Before this we need also a brief introduction to quantum mechanics to predict the behavior of electrons in solids.
Next we will use the density of quantum states in conduction band along with Fermi – Dirac probability function to determine the concentration of electrons and holes in conduction and valence band. Then we apply the concept of the Fermi energy to the semiconductor material. The Equilibrium, or thermal equilibrium, implies that no external forces such as voltages, electric fields, magnetic fields, or temperature gradients are acting on the semiconductor. That means all properties of the semiconductor are independent of the time in this case. In fact the equilibrium is our starting point for developing the physics of the semiconductor. Then we will be able to determine the characteristics from equilibrium deviations, e.g. applied voltage to a semiconductor material.
After this session we are able to:
· Discuss a few basic principles of quantum mechanics that apply to semiconductor materials and devices physics.
· List a few results of quantum mechanics, including energy quantization and probability concepts, in order to make understandable some properties of electrons in crystals.
· Develop the energy–band theory of semiconductors and the two types of charged carriers in semiconductors.
· Discuss and develop the density of quantum states as a function of electron energy.
· Develop the Fermi –Dirac distribution function, which describes the probability of an electron occupying a quantum state as a function of electron energy.
Principles of Quantum Mechanics
The electrical properties of semiconductor materials are directly related to the behavior of electrons in the crystal lattice. This behavior and characterization of these electrons can be described by the formulation of quantum mechanics called wave mechanics. We consider two principles of quantum mechanics as below:
Energy Quanta
If monochromatic light is incident on a clean surface of a material, then under certain conditions, electrons (photoelectrons) are emitted from the surface (see the picture below).
According to the classic physics, if the intensity of the light is large enough, the work function (the energy required to pull an electron away from the surface) of the material will overcome and an electron will be emitted from the surface independent of the incident energy. This predicted result is not observed. The observed effect is that, at a constant incident intensity, the maximum kinetic energy of photoelectron varies linearly with frequency more than . Below this, no photoelectron is produced (see below figure). If incident intensity varies at a constant frequency, the rate of photoelectron emission changes, but the maximum kinetic energy remains constant.
Planck presumed that the thermal radiation is emitted from a heated surface is discrete packets of energy called quanta. The energy of these quanta is given by , where is the frequency of the radiation and is a constant now known as Planck’s constant equal to . Then Einstein suggested that the energy in a light wave is contained of discrete packets known as photon, with same energy of quanta. The minimum energy required to remove an electron is called the work function of the material any extra photon energy goes into the kinetic energy of the photoelectron. The maximum kinetic energy of the photoelectron can be written as
(for )
where is the incident photon energy and is the minimum energy, or work function, required to remove an electron from the surface.
Example 1: Consider an x-ray with a wavelength of , calculate the photon energy corresponding to it in electron-volts.
Hint: One electron-volt (eV) is equal to .
Example 2: Determine the energy (in eV) of a photon having wavelengths of (a) and (b)
(a) 1.24 eV , (b) 1.24x103 eV
Hint: One angstrom is equal to 10-8 cm = 10-10 m.
Note: Reciprocal relation between photon energy and wavelength; a large energy correspond to a short wavelength.
Wave-Particle Duality Principle
In photoelectric effect, the light wave behaves in form of particles. Then it has to show wave-like properties. The momentum of a photon is given by so the wavelength of a particle can be expressed as , where p is the momentum of the particle and is known as de Broglie wavelength of the matter wave, who suggested a wave-particle duality principle which can be explained in summary: the observation that electrons, photons and other very small entities behave like particles in some experiments and like waves in others [in Malay dualitas zarah-gelombang].
In some cases, the electromagnetic waves behave as if they are particles (photons) and, in some cases, particles behave as if they are waves. The wave-particle duality principle is the basis on which we use wave theory to describe the motion and behavior of electrons in a crystal.
Example 3: Determine the de Broglie wavelength of an electron traveling at a velocity of 107 cm/s (=105 m/s).
Example 4: (a) Find the momentum and energy of a particle with mass 5x10-31 kg and a de Broglie wavelength of 180 Å. (b) An electron has a kinetic energy of 20 meV and determine the de Broglie wavelength.
(a) p=3.68x10-26 kg.m/s, E=4.65x10-3 eV
(b) p=7.64x10-26 kg.m/s, λ=86.8 Å.
Energy Quantization and Probability Concepts
Physical meaning of the wave function
Based on the wave-particle duality principle, we can describe the motion of electrons in a crystal by wave theory. This wave theory is described by Schrodinger’s wave equation, which is independent of the time:
[read as “psi” this is small letter of the Greek alphabet of with same pronunciation]
where E is the total energy of the particle and is assumed to be a constant, V(x) is the potential energy of the particle, m is the mass of particle, and is called a modified Plank’s constant defined as .
For your knowledge: From a classical description of the total energy, E, equals the sum of kinetic energy, KE, and potential energy, V, or
To convert this expression into a wave equation by defining a wave function, , we get
To incorporate the de Broglie wavelength of the particle we now introduce the operator and apply it to a plane wave
Where k is the wavenumber equals to . By replacing the momentum squared, , in equation of wave function we get
Or
Now we are going to understand what relation is between the wave function and the electron. The total wave function of an electron is the product of where the electron is (position of x) and when the electron is in that place. This means that the total wave function of electron is the product of the function and time function of [read it as phi of t]. The time function is expressed by
So the total wave function is given as
which means where and when an electron in crystal lattice is.
The probability of finding the particle between x and x+dx at a given time then is expressed as
Where is the complex conjugate function, and
Therefore
This is the probability density function and is independent of time. The major difference between classical physics and quantum mechanics to find the position and energy of a particle or body is that:
· In classical physics: it can be determined precisely
· In quantum mechanics: it is found in term of probability
So we will be concerned to find the probability of an electron having a particular energy.
One-Electron Atom
Consider the one electron (or hydrogen) atom potential problem:
Nucleus: heavy and positively charged proton
Electron: light and negatively charged particle
In the classical Bohr theory, the electron revolves around the nucleus. The potential energy V(r) is due to the coulomb attraction between the proton and the electron and is given by
where e is the magnitude of the electronic charge, is the permittivity of free space and r is the radial distance from the proton to the electron.
The following results can be derived from the analysis of this potential problem:
1. Quantized Energies
The total energy for the electron then is given by
where n is a positive constant called the principle quantum number and m0 is the mass of the electron. The negative energy indicates that the electron is bound to the nucleus. Since n is an integer, the energy of the bound electron can take only discrete values or the energy is said to be quantized.
Example 5: Determine the first three allowed electron energies in the hydrogen atom.
The mass of an electron is m0=9.11x10-31 kg and the permittivity of free space is ε0=8.85x10-12 F/m. The modified Planck’s constant is
6.625x10-34/2x3.14=1.054x10-34 J.s
So for n=1 we get E1=-2.17x10-18 J = -13.6 eV
n=2 àE2= - 3.39 eV, n=3 à E3 = - 1.51 eV.
Figure Example 5: The energy of the electron increases (becomes less negative) as the orbit of the electron becomes larger.
Question 3:
For the one-electron atom, determine the value of n such that the difference in energies En+1 – En is less than 0.20 eV.
Ans. n=5.
2. Quantum Numbers
The second result from the potential problem of one-electron atom is that how to show the multidimensional aspect of this problem. The Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. Only the size of the orbit was important, which was described by the n quantum number. Schrodinger described an atomic model with electrons in three dimensions. This model required three coordinates, or three quantum numbers, to describe where electrons could be found.
The three coordinates from Schrodinger's wave equations are the principal (n), angular (l), and magnetic (m) quantum numbers. These quantum numbers describe the size, shape, and orientation in space of the orbitals on an atom. In terms of the spherical coordinates, based on Schrodinger's wave equations, are n, l, and m respectively.
These quantum numbers are not independent, but are related by
Geometry of Hydrogen Atom Solution
Now we can solve the Schrodinger’s wave equation for the one-electron potential function which is shown by where n, l, and m are quantum numbers. For the lowest energy state (n=1, l=0, m=0) the wave function is given by
This function is spherically symmetric, and the parameter of is given by
[Angstrom]
The probability density function or the probability of finding the electron at a particular distance from the nucleus, for the given function is
This radial probability density function for one-electron atom at the lowest energy state and next higher energy state has been given in following figure:
As it can be seen in the figure, the most probability distance from the nucleus is at , which is the same as the Bohr Theory. The radial probability function for the next higher spherically symmetric wave function, corresponding to n=2 and 3, red and blue lines respectively, shows the idea of the next higher energy shell of the electron. The second and third energy shell are at a greater distance from the nucleus than the first energy shell, but still a small probability that the electron will exist at the smaller radius.
The above results must be taken into account with two more concepts:
1. The first concept is the electron spin. The electron has an intrinsic angular momentum, or spin, that is quantized and may take on one of two possible value of or . Now, we have four basic quantum numbers: n, l, m, and s.
2. The second concept is the Pauli Exclusion Principle [asas larangan Pauli], which states, in any given system (an atom, molecule, or crystal), no two electrons can occupy the same quantum state.
Exercise 1: Go through the Periodic Table and find the following elements. Discuss how they are notated, and what values are n, l, m, and s for each element: Hydrogen, Helium, Lithium, Beryllium, Boron, Carbon, Nitrogen, Oxygen, Fluorine, and Neon (e.g. Hydrogen: 1s1, n=1, l=0, m=0, and s=+1/2 or -1/2)
Energy-Band Theory
Energy bands consisting of a large number of closely spaced energy levels exist in crystalline materials. The bands can be thought of as the collection of the individual energy levels of electrons surrounding each atom. The wave functions of the individual electrons, however, overlap with those of electrons confined to neighboring atoms. The Pauli Exclusion Principle does not allow the electron energy levels to be the same so that one obtains a set of closely spaced energy levels, forming an energy band. The energy band model is crucial to any detailed treatment of semiconductor devices. It provides the framework needed to understand the concept of an energy band gap and that of conduction in an almost filled band as described by the empty states.
Any solid has a large number of bands. In theory, it can be said to have infinitely many bands (just as an atom has infinitely many energy levels). However, all but a few lie at energies so high that any electron that reaches those energies escapes from the solid. These bands are usually disregarded.
The electron goes to lower band and give up energy or you give energy to atom to bring the electron to outer band
The uppermost occupied band in an insulator or semiconductor is called the valence band by analogy to the valence electrons of individual atoms. The lowermost unoccupied band is called the conduction band because only when electrons are excited to the conduction band can current flow in these materials. The difference between insulators and semiconductors is only that the forbidden band gap between the valence band and conduction band is larger in an insulator, so that fewer electrons are found there and the electrical conductivity is less. Because one of the main mechanisms for electrons to be excited to the conduction band is due to thermal energy, the conductivity of semiconductors is strongly dependent on the temperature of the material.
Actually by breaking of a covalent band in valence band, a negative and a positive charge is generated. The semiconductor as a whole is neutrally charged, which means as the negatively charged electron breaks away from its covalent bonding position, a positively charged empty state is created in the original covalent bonding position in the valence band. As temperature further increases, more covalent bands are broken, more electrons jump to the conduction band, and more positive empty states are created in the valence band.
Charged Carriers – Electrons and Holes
Carriers: we are interested in determining the I–V characteristics of semiconductor devices. Current is result of the flow of charge, and can move when forces are applied. These charges are referred to as carriers.
Electrons: one type of charge is the electron, a negatively charged particle. If a force is applied to a particle and the particle moves, it must gain energy. This effect is expressed as
where F is the applied force, dx is the differential distance the particle moves, v is the velocity, and dE is the increase in energy.
We can write the drift current density due to the motion of electrons as
where e is the magnitude of the electronic charge and N is the number of electrons per unit volume in the conduction band.
Holes: when a valence electron was elevated into the conduction band a positively charged empty state is created. The movement of a valence electron into the empty state is equivalent to the movement of the positively charged empty state itself. Now the crystal has a second important charge carrier that can give rise to a current. This charge carrier is called hole.
Now an important step in the process of describing V–I characteristics is to determine the number of electrons and holes in the semiconductor that will be available for conduction. We know that the number of carriers that can contribute to the conduction process is a function of the number of available energy or quantum states. The density of quantum states [rapat keadaan] per unit of energy and per unit volume is given by
The density of quantum states is a function of energy. As the energy of the electron becomes small, the number of available quantum states decreases.
From the above equation, the generalized density of allowed electronic energy states in the conduction band can be written as
Also the density of states function to apply to the valence band can be generalized as
Where the m* is effective mass of particle. When electron is moving inside the solid material, the force between other atoms will affect its movement and it will not be described by Newton's law. So we introduce the concept of effective mass to describe the movement of electron in Newton's law. The effective mass can be negative or different due to circumstances. The electron effective mass of silicon is 1.08 of its electron mass (1.08me) while its hole effective mass is 0.56me.
And finally the volume density of quantum states can be found as
Example 5: Calculate the density of states per unit volume with energies between zero and 1 eV for an electron.
The mass of electron is 9.11x10-31 kg, the Planck’s constant is 6.625x10-34, and the energy of an electron is 1.6x10-19 coulomb, so the density of states is now
or
Example 6: Calculate the density of states per unit volume for a free electron over the range of energies between 1 eV and 2 eV. [Answer: N=8.29x1021 states/cm3].
Statistical Mechanics
When we deal with large numbers of particles, we are interested only in the statistical behavior of the group as a whole rather than in the behavior of each individual particle. To determine the statistical behavior of particles, we must consider the laws that the particles obey. There are three distribution laws determining the distribution of particles among available energy states:
· Maxwell–Boltzmann probability function: Particles are distinguished by numbers, e.g. from 1 to N, with no limit to the number of particles allowed in each energy. The Maxwell–Boltzmann approximation is given by
It is valid when
· Bose–Einstein function: Particles are indistinguishable and again no limit to the number of particles permitted in each quantum state.
· Fermi–Dirac probability function: Particles are indistinguishable, but only one particle is permitted in each quantum state. Electrons in a crystal obey this law and the Fermi–Dirac distribution function is given by
where is the number of particles per unit volume per unit energy and is the number of quantum states per unit volume per unit energy. The function is called the Fermi–Dirac distribution function or probability function and gives the probability that an allowed quantum state at the energy E is occupied by an electron. The parameter in this equation is called Fermi energy. The probability of state being occupied at E=EF is ½.
The Pauli Exclusion Principle supposes that only one particle can occupy a single quantum state. Therefore, as particles are added to an energy band, they will fill the available states in an energy band just like water fills a bucket. The states with the lowest energy are filled first, followed by the next higher ones. At absolute zero temperature (T = 0 K), the energy levels are all filled up to a maximum energy, which we call the Fermi level. No states above the Fermi level are filled. At higher temperature, one finds that the transition between completely filled states and completely empty states is gradual and not steep.
The Fermi function at three different temperatures
Probability of occupancy versus energy of the Fermi-Dirac, the Bose-Einstein and the Maxwell-Boltzmann distribution. The Fermi energy, EF, is assumed to be zero.
Example 7: Determine the probability that an energy level 3kT above the Fermi energy level is occupied by an electron. Let T=300K. We assumed that such an energy level is allowed.
Exercise 2: Assume the Fermi energy level is 0.30 eV below the condition band energy. (a) Determine the probability of a state being occupied by an electron at E=Ec+kT. (b) Repeat part (a) for an energy state at E=Ec+2kT. Assume T=300K.
Example 8: Find the temperature at which there is a 1 percent probability that an energy state 0.30 eV is empty (not contain an electron) below the Fermi energy level of a particular material equals to 6.25 eV.
Note: The Fermi probability function is a strong function of temperature.
Exercise 3: Assume that the electrons in a material follow the Fermi-Dirac distribution function and assume that the Fermi level is 5.50 eV. Determine the temperature at which there is a 0.5 percent probability that a state 0.20 eV above the Fermi level is occupied by an electron.
Example 9: Calculate the energy, in terms of kT and EF, at which the difference between the Boltzmann approximation and the Fermi-Dirac function is 5 percent of the Fermi function.
Questions:
Please answer to all questions and bring the handwriting with yourself on the next session.
1. Assume that the work function of gold is 4.90 eV and that of cesium is 1.90 eV. Calculate the maximum wavelength of light for the photoelectric emission of electrons for gold and cesium.
2. Calculate the de Broglie wavelength for: (a) an electron with kinetic energy of 1.0 eV and 100 eV. (b) A proton with kinetic energy of 1.0 eV.
3. Why do energy levels split as atoms are brought close to each other?
4. Consider the following figure, which shows the energy-band splitting of silicon. If the equilibrium lattice spacing were to charge by small amount, discuss how you would expect the electrical properties of silicon to change. Determine at what point the material would behave like an insulator or like a metal.
5. Find the ratio of the effective density of states in the conduction band at Ec+kT to the effective density of states in the valence band at Ev–kT.
6. The Fermi energy level for a particular material at T=300K is 6.25 eV. The electrons in this material follow the Fermi – Dirac distribution function. (a) Find the probability of an energy level at 6.50 eV being occupied by an electron. (b) Repeat part (a) if the temperature is increased to T=950K. (Assume that EF is a constant.) (c) Calculate the temperature at which there is a 1 percent probability that a state 0.30 eV below the Fermi level will be empty of an electron.
7. Consider the energy levels shown below. Let T=300K. (a) If E1–EF = 0.30 eV, determine the probability that an energy state at E=E1 is occupied by an electron and the probability that an energy state at E=E2 is empty. (b) Repeat part (a) if EF–E2 = 0.40 eV.
Monday, January 7, 2008
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