Unit 1 - Electronic Materials PDF

P age |1 UNIT – 1 (Electronic Materials) 1.1 Introduction Conducting materials are low resistivity materials, which conduct heat as well as electricity. Electrical conduction is due to free electrons, whereas normal conduction is due to free...

P age |1 UNIT – 1 (Electronic Materials) 1.1 Introduction Conducting materials are low resistivity materials, which conduct heat as well as electricity. Electrical conduction is due to free electrons, whereas normal conduction is due to free electrons as well as phonons. 1.1.1 Basic Terminology Conductors: Experimental measurements showed that the metals and their alloys exhibit large electrical conductivity in the order of 108 Ω-1 m-1. Hence they are known as conductors conducting materials are the materials having high electrical and thermal conductivities. Low resistive materials are also generally known as conducting materials. Bound electrons: All the valence electrons in an isolated atom are bound to their parent nuclei which are called as ‘bound electrons’. Free electrons: In a solid, due to the boundaries of neighboring atoms overlap each other, the valence electrons find continuity from atom to atom. Therefore, they can move easily throughout the solid. All such valence electrons of its constituent atoms in a solid are called free electrons. Difference between ordinary gas and free electron gas: The molecules of ordinary gas are neutral. But, the free electron gas is charged. The density of molecules is smaller than the density of free electrons. Electric field (E): The electric field (E) of a conductor having uniform cross section is defined as the potential drop (V) per unit length (l). E = V/ l Vm-1  P age |2 Current density (j): Current density (j) is defined as the current per unit area of cross section of an imaginary plane hold normal to the direction of flow of current in a current carrying conductor. If ‘I’ is the current, and ‘A’ is the area of cross section, then current density is given by, J = I / A Am-2 1.2 Conducting Materials Conducting materials are classified in to three major categories based on the conductivity. (i) Zero resistive materials (ii) Low resistive materials (iii) High resistive materials (i) Zero resistive materials The super conductors like alloys of aluminium, zinc, gallium, niobium, etc., are a special class of materials. These materials conduct electricity almost with zero resistance blow transition temperature. Thus, they are called zero resistive materials. These materials are used for saving energy in the power systems, super conducting magnets, memory storage elements etc., (ii) Low resistive materials The metals like silver, aluminium and alloys have high electrical conductivity. These materials are called low resistive materials. They are used as conductors, electrical conduct etc., in electrical devices and electrical power transmission and distribution, winding wires in motorsand transformers. (iii) High resistive materials The materials like tungsten, platinum, nichrome etc., have high resistive and low temperature co-efficient of resistance. These materials are called high resistive materials. Such a metals and alloys are used in the manufacturing of resistors, heating elements, resistance thermometers. The conducting properties of solid do not depend on the total number of the electrons available because only the valance electrons of the atoms take part in the conduction. When these valance electrons detached from the orbit they are called free electrons or conduction electrons.  P age |3 In a metal, the number of free electrons available is proportional to its electrical conductivity. Hence, electronic structure of a metal determines its electrical conductivity. 1.3 Electron theory of solids We know that the electrons in the outermost orbit of the atom determine the electrical properties in the solid. The free electron theory of solids explains the structure and properties of solids through their electronic structure. This theory is applicable to all solids, both metals and non-metals. It explains (a) The behavior of conductors, semiconductors, and insulators. (b) The electrical, thermal and magnetic properties of solids. So far three electron theories have been proposed. (i) Classical free electron theory This theory was developed by Drude and Lorentz. According to this theory, a metal consists of electrons which are free to move about in the crystal molecules of a gas it contains mutual repulsion between electrons is ignored and hence potential energy is taken as zero. Therefore the total energy of the electron is equal to its kinetic energy. (ii) Quantum free electron theory Classical free electron theory could not explain many physical properties. In classical free electron theory, we use Maxwell-Boltzmann statics which permits all free electrons to gain energy. In Somerfield developed a new theory, in which he retained some of the features of classical free electron theory included quantum mechanical concepts and Fermi-Dirac statistics to the free electrons in the metals. This theory is called quantum free electron theory. Quantum free electron theory permits only a few electrons to gain energy. (iii) Zone theory or band theory of solids Bloch developed the theory in which the electrons move in a periodic field provided by the Lattice concept of holes, origin of Band gap and effective mass of electrons are the special features of this theory of solids. This theory also explains the mechanism of super conductivity based on band theory.  P age |4 1.4 Classical free electron theory Free electron theory of metals was proposed by P. Drude in the year 1900 to explain electrical conduction in metal. This theory was further extended by H.A. Lorentz in the year 1909. 1.4.1 Postulated of free electron theory  Metal structure consists positive ion core, with valence electrons moves freely among these ion cores.  These valence electrons moves inside metal just like molecules in gas.  In the absence of electric field (E = 0), free electrons moves in random directions and collide with each other.  In the collisions, no loss of energy will observed because collisions are elastic.  When electric filed is applied (E ≠ 0), the free electrons are accelerated in direction opposite to field.  Free electrons obey Maxwell-Boltzmann statistics. 1.4.2 Postulates of free electron theory Drift Velocity (Vd): The drift velocity is defined as the average velocity acquired by the free electron in particular direction, due to the applied electric field. 𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐭𝐫𝐚𝐯𝐞𝐥𝐥𝐞𝐝 𝐛𝐲 𝐭𝐡𝐞 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐧 𝐃𝐫𝐢𝐟𝐭 𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 = 𝐓𝐢𝐦𝐞 𝐭𝐚𝐤𝐞𝐧 Vd = λ/t ms-1  P age |5 Mobility (μ): The mobility is defined as the drift velocity (Vd) acquired by the electron per unit electric field (E). μ = Vd/E m2V-1s-1 Mean free path (λ): The average distance travelled by an electron between two successive collisions is called mean free path. Mean collision time (τc) (or) Collision time It is the time taken by the free electron between two successive collisions. τc = λ/Vd sec Relaxation time (τ): It is the time taken by the electron to reach equilibrium position from disturbed position in the presence of electric field. τ = l/Vd sec Where l is the distance travelled by the electron. The value of relaxation time is of the order of 10–14 sec. Band gap (Eg): Band gap is the energy difference between the minimum energy of conduction band and the maximum energy of valence band. Current density (J) It is defined as the current per unit area of cross section of an imaginary plane holded normal to the direction of the flow of current in a current carrying conductor. J = I/A Am-2  P age |6 1.4.3 Expression for electrical conductivity  Electrical conductivity of a semiconductor is different from that of a conductor. The charge carriers in a conductor are electrons. In the case of a semiconductor, both electron and holes are charge carriers.  In the absence of an electric field, the motion of these free electrons is completely random like those of molecules in a gas.  But when an electric field is applied, the electrons drift in opposite direction to that of applied field with an average velocity called the “drift velocity” (Vd).  When electric field is applied, electrons experience a force eE, due to which they are accelerated. 𝐅 = 𝐞𝐄 𝐁𝐮𝐭 𝐅 = 𝐦𝐚 ∴ 𝐦𝐚 = 𝐞𝐄 ∴ 𝐚 = (𝐞𝐄)/𝐦 … … … (𝟏) As, electrons collision occurs during motion, the electrons will get accelerated momentarily so, the drift velocity will be ∴ 𝐕𝐝 = 𝐚 τ 𝐞𝐄 ∴ 𝐕𝐝 = τ … … … (𝟐) 𝐦 𝐈 𝐂𝐮𝐫𝐫𝐞𝐧𝐭 𝐝𝐞𝐧𝐬𝐢𝐭𝐲 𝐉 = = 𝐧𝐞𝐕𝐝 … … … (𝟑) 𝐀 𝐞𝐄 𝐧𝐞𝟐𝐄 ∴ 𝐉 = 𝐧𝐞 τ= τ … … … (𝟒) 𝐦 𝐦 𝐅𝐫𝐨𝐦, 𝐞𝐥𝐞𝐜𝐭𝐫𝐢𝐜𝐚𝐥 𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐯𝐢𝐭𝐲 𝐉 = 𝜎 𝐄 … … … (𝟓) 𝐧𝐞𝟐𝐄 τ ∴𝜎𝐄 = 𝐦 τ𝐧𝐞𝟐 ∴𝜎= m … … … (𝟔) ≫≫ 𝐄𝐱𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝐟𝐨𝐫 𝐞𝐥𝐞𝐜𝐭𝐫𝐢𝐜𝐚𝐥 𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐢𝐯𝐢𝐭𝐲  P age |7 Conductivity in terms of kBT. The relaxation time (τ) in terms of mean free path (λ) and average thermal velocity is given by: λ ∴τ= … … … (7) V Now, based on kinetic theory of gases, 1 3 ∴ mV2 = kT 2 2 B 3kBT ∴m= … … … (8) V2 Using equation (8) & (7) in (6) λ ne2 ( ) ne2 λV2 ∴σ= V = 3KBT 3KBTV V2 ne2λV ∴σ = … … … (9) (This equation says that electrical conductivity decreases 3KBT with temperature) Q.1 A current of 5 A is passing through a metallic wire of cross sectional area 4×10-6 m2. If the density of the charge carriers in the wire is 5×1026 /m3, find the drift speed of the electrons. (e = 1.6×10-19) Ans. I = 5 A, A = 4 ×10-6 m2, n = 5×1026 /m3, Vd = ? The relation between current (I) and drift velocity (Vd) of electron in wire of area (A) is ∴ I = neAVd  P age |8 I ∴ Vd = neA 5 ∴ Vd = (5×1026)(1.6×10−19)(4×10−6) ∴ Vd = 1.5×10−2 m/s Q.2 A conductor has an electron concentration of 5.9×1028 /m3. What density in the conductor corresponds to a drift velocity of 0.625 m/sec. Calculate the mobility of charge carriers. Given σ = 6.22×107 Ʊm-1. Ans. n = 5.9×1028 /m3, Vd = 0.625 m/sec, e = 1.6×10-19, J =? & μ =? ∴ J = neVd ∴ J = (5.9×1028) (1.6×10-19) (0.625) ∴ J = 5.9×109 A/m2 ∴ σ = neμ σ ∴μ= ne (6.22×107) ∴μ= (5.9×1028)(1.6×10−19) ∴ μ = 6.588×10-3 m2/V·sec Q. 3 Calculate drift velocity of free electrons with a mobility of 3.5×10-3 m2 /V·sec in copper for an electric field strength of 0.5 V/m. Ans. μ = 3.5×10-3 m2/V·sec, E = 0.5 V/m, Vd = ?  P age |9 Vd ∴μ= E ∴ Vd = μE ∴ Vd = (3.5×10−3) (0.5) ∴ Vd = 1.75×10−3 m/sec Q. 4 The density of silver 10.5 × 103 kg–3 assuming that each silver atom provides one conduction electron. The conductivity of silver at 20°C is 6.8 × 107 Ω–1 m–1. Calculate the density and mobility of electron in silver with atomic weight 107.9. Ans. Density of silver d = 10.5 × 103 kg m–3 Conductivity of silver at 20°C = 6.8 × 107 Ω–1 m–1 Atomic weight A = 107.9 Avagadro Number × Density We know the carrier concentration (n) = Atomic weight 6.023 × 1023 × 10.5 × 103 ∴n= 107.9 ∴ n = 5.86 × 1025 m3 We know, the conductivity is given by  = ne σ ∴μ= ne 6.8 × 107 ∴μ= 5.86 × 1025 × 1.6 × 10−19 ∴ μ = 7.25 × 10−3 m2V−1s−1  P a g e | 10 1.4.4 Expression for Thermal conductivity The thermal conductivity is defined as the amount of heat flowing through a unit area per unit temperature gradient. ∆Q Watt ∴K=− dT meter ∙ Kelvin A (dx) The negative sign indicates that heat flows hot end to cold end. ΔQ = rate of heat flow (thermal energy) A = Cross-sectional area of conductor (dT/dx)=temperature gradient Average kinetic energy at A = 3 KB (T) 2 Average kinetic energy at B = 3 KB (T − dt) 2 The excess energy carried by electrons in traveling from point A to B 3 3 KB T = KB (T − dT) 2 2 3 = KB dT 2 The electrons travel in all directions crossing unit area from A to B  P a g e | 11 1 i. e. nv 6 So, excess energy transferred from A to B per unit area in unit time is: 1 3 nv × KB dT 6 2 1 = nvKBdT 4 Similarly, deficiency of energy from B to A per unit time is: 1 =− nvKBdT 4 The rate of heat flow can be defined as the net energy transferred from A to B per unit area per unit time: 1 1 ∴ ∆Q = nvKBdT − (− nvKBdT) 4 4 1 1 ∴ ∆Q = nvKBdT ( + ) 4 4 1 ∴ ∆Q = nvKBdT 2 We know thermal conductivity k is defined as: ∆Q ∴K= dT ( dx ) Here, dx can be written as mean free path (λ) 1 nvK dT ∴K= 2 B dT (λ ) 1 Thermal conductivity, K = nvKBλ 2  P a g e | 12 1.4.5 Wiedemann – Franz law This law states that the ratio of thermal conductivity to electrical conductivity is directly prepositional to the absolute temperature. K i. e. 𝖺T σ K ∴ =LT σ L = Lorentz number 1 ∴K= 2 nVKBλ ne2λV ∴σ= 3KBT By, taking ratio 1 K nVK Bλ ∴ = 2 σ ne2λV 3KBT K 3 KB 2 ∴ = ( ) T σ 2 e K ∴ =LT σ Theoretical value of L: 1.12 × 10-8 WΩ/K2 Experimental value of L: 2.44 × 10-8 WΩ/K2 Theoretical and experimental values of L doesn’t match with each other. So, Assumption is wrong that all electrons contributes in thermal conductivity. 1.4.6 Success of free electron theory  It verifies ohm’s law.  It explains the thermal and electrical conductivities of a metal.  It helps to deduce Wiedemann – Franz law.  It explains optical properties of metals.  P a g e | 13 1.4.7 Drawbacks of classical free electron theory  Theoretical value of specific heat and electronic specific heat of metal based is ( 3R ) is 2 not in agreement with the experimental value (10-4 RT).  The classical free electron theory can’t explain electrical conductivity of insulators and semiconductors.  According to classical theory, K = T is constant at all temperature. But, this is not σ constant at low temperature.  It failed to explain the superconductivity and magnetic susceptibility (χ).  The phenomena such as photoelectric effect, Compton Effect and black body radiation can’t be explained.  According to classical free electron theory, the electrical conductivity (σ) proportional to the free electron density (n). But it failed to explain for different metals. Q.1 Find the thermal conductivity of copper 20°C with a free electron density of 8.48 × 1028/m3 The thermal velocity of copper is 1.1536 × 105 m/sec at 20°C, with a mean free path of 2.813 nm. Ans. n = 8.48 × 1028/m3, λ = 2.813 nm = 2.183 × 10-9 m v = 1.1536 × 105 m/sec K =? We know, 1 ∴K= nVKBλ 2 1 ∴K= (8.48 × 1028) × (1.1536 × 105) × (1.38 × 10−23) × (2.183 × 10−9) 2 ∴ K = 189.92 W/m ∙ K  P a g e | 14 1.5 Quantum free electron theory The failure of classical free electron theory paved this way for Quantum free electron theory. It was introduced by Sommerfield in 1928. This theory is based on making small concepts. This theory was proposed by making small changes in the classical free electron theory and by retaining most of the postulates of the classical free electron theory. 1.5.1 Assumptions (Postulates) of Quantum free electron 1. In a metal the available free electrons are fully responsible for electrical conduction. 2. The electrons move in a constant potential inside the metal. They cannot come out from the metal surface have very high potential barrier. 3. Electrons have wave nature, the velocity and energy distribution of the electron is given by Fermi – Dirac distribution function. 4. The loss of energy due to interaction of the free electron with the other free electron. 5. Electron’s distributed in various energy levels according to Pauli Exclusion Principle. Quantum free electron theory  Each metal contains a very large number of electrons and these electrons can move freely inside the metal.  Inside of the metal material uniform potential is applied: ⮩ Constantly moving electrons has only kinetic energy (K.E.) inside the metal while the potential energy (P.E.) is zero.  P.E. = 0 can be only possible when, ⮩ Interaction of free electron with other electron and metallic core of the metal is negligible.  P a g e | 15 K.E. inside the material is not sufficient to overcome metal boundary.  In quantum approach electrons moves in wave form and wave nature of particle is given by Schrodinger.  Schrodinger wave equation for 3-dimension: d 2ψ d 2ψ d 2ψ 2m ∴ + + + [E − V]ψ = 0 dx2 dy 2 dz 2 ℏ2  Schrodinger wave equation for 1-dimension: d 2ψ 2m ∴ + [E − V]ψ = 0 … … … (1) dx2 ℏ2 E = total energy, V = potential energy, m = mass of electron h ℏ = modified Planck’s constant (ℏ= ) 2π In our case P.E. = 0, so, total energy (E) of the system is because of only K.E. d2ψ 2m ∴ +( E) ψ = 0 dx2 ℏ2 2m Consider, ( E) = k2 2 ℏ d 2ψ ∴ + k2ψ = 0 … … … (2) dx2  To identify complementary function of ψ: ∴ ψ = [A coskx + B sinkx] … … … (3) (remember) Where, k = wave vector, A & B are constants In order to find the values of A & B we can use two boundary condition at X = 0 and X = L. For, (1) X = 0 → ψ = 0 condition ∴ 0 = [A cosk0 + B sink0] ∴ 0 = [A (1) + 0] ∴A=0 Putting this value in equation (3)  P a g e | 16 ∴ ψ = B sinkx … … … (4) For, (2) X = L → ψ = 0 condition ∴ ψ = B sinkL Here, B can’t be zero because than whole equation will be zero. ∴ ψ = [A coskx + B sinkx] … … … (3) If ψ becomes zero than probability of having electron inside the metal becomes zero, which is not possible. This problem can be satisfied by taking sinkL zero instead of B. sinkL = 0 only possible at (sin n𝜋) Where, n = 0, 1, 2, 3….. So, if we compare kL = nπ then: nπ ∴k= L Therefore, the wave function will be valid only if: nπ ∴ ψ(x) = B sin ( x) … … … (5) L n2π2 ∴ k2 = L2 By, comparing the values of k2 2m n2π2 E = L2 ℏ2 n2π2ℏ2 ∴ En = L22m n2π2 h2 ∴ En = L22m 4π2 n 2h 2 ∴ En = … … … (6) 8 mL2 According to this equation, electrons will have nth level of energy (En) inside the metal.  P a g e | 17 Value of constant B in equ. 5 obtained by applying the normalization condition to get the value of wave function ψ.  Normalizing the wave function means you find the exact form of probability (ψ) of the particle which found somewhere in space is 1. ∞ ∴ ∫ Ψ2 dx = 1 −∞ L nπx ∴ ∫ B2 sin2 ( ) dx = 1 0 L 2 L 2 πnx ∴ B ∫ 1 − cos ( ) dx = 1 2 0 L 2 L L 2 πnx ∴ B ∫ 1 dx − ∫ cos ( ) dx = 1 2 0 0 L nπ ∴ ψ(x) = B sin ( x) … … … (5) L 2 2πnx L B ∴ [x]L0 − [sin L ] = 1 2 2πn L 0 B2 L 2πnx L ∴ [x]L0 − [sin ] =1 2 2πn L 0 B2 L 2πnL 2πn0 L ∴ [L] − [sin − sin ] =1 2 2πn L L 0 B2 ∴ [L] = 1 2 Putting this value in equation (5)  P a g e | 18 2 nπ ∴ ψ(x) = √ sin ( x) … … … (7) L L This equation gives the value of probability of electron in a confined system. The wave functions corresponding to four lowest states for different values of n and probability of finding the particle are shown in fig. 1.5.2 Advantages of quantum free electron theory  It explains the specific heat of materials.  It explains Compton Effect, photoelectric effect, blackbody radiation, Zeeman Effect.  It gives the correct mathematical expression for the thermal conductivity of metals.  It explains the superconductivity. 1.5.3 Drawbacks of quantum free electron theory  This theory fails to explain the positive value of Hall coefficients.  It failed to distinguish metal, semiconductor and insulator.  It failed to explain lower conductivities of divalent metals than monovalent metals. 1.6 Band Theory or (Energy Band Diagrams) (a) Formation of energy bands in solids:  In a single isolated atom, the electrons in each orbit have definite energy associated with it.  P a g e | 19  In case of solids, the atoms are close to each other, so the energy levels of outermost orbit electrons are affected by the neighbouring atom.  When two single or isolated atoms are brought close to each other, the outermost orbit electrons of the two atoms interacts with each other.  I.e. The electrons in the outermost orbit of one atom experiences an attractive force from the neighbouring atom nucleus.  As a result of this, the energy of electrons is no more the same, the energy levels are changed to a value which is higher or lower than the original energy level.  The Pauli's exclusive principle allows each energy level to contain only two electrons. “This grouping of different energy levels is called energy bands”.  The energy level of the inner orbit electrons are not much affected by presence of neighbouring atoms. (b) Energy bands in solids: Valence band:  “The energy band formed by grouping the energy levels of valence electrons is called valence band”.  Electrons in valence band have lower energy than electrons in conduction band.  P a g e | 20  The electrons in valence band are loosely bound to nucleus of atom, they can be moved to conduction band by applying energy. Forbidden band:  “The energy gap that is present between the valence band and conduction band by separating the two energy band is called the forbidden band or forbidden gap.”  Electrons can’t stay in this band as there is no allowed energy state in this gap.  The energy associated with the forbidden gap is called the bandgap energy. Conduction band:  “The energy gap or band formed by grouping the energy levels of free electrons is called conduction band.”  It is the uppermost band with free electrons, electrons have higher energy than electron is valence band.  The band is completely empty for insulators and partially filled for conductors.  The conduction band electrons are not bound to the nucleus of atom. 1.6.1 Types of electronic materials: conductor, Insulators, and semiconductors Conductors:  “The materials which easily allow the flow of electric current are called conductors.” e.g. copper, iron, silver, etc.  The resistivity is of an order of 10-9 Ωm at room temperature.  In a conductor, the conduction and valence band overlap each other as shown in figure.  A small amount of energy provides enough energy for valence band electrons to move in conduction band.  P a g e | 21 Insulators:  “The materials which do not allow the flow of electric current through them are called insulators.”  The valence band is completely filled with electrons. The forbidden energy gap is greater than 5 eV.  The electrons in valence band can’t move to conduction as they are locked up between the atoms.  It requires large amount of external energy to move the electrons to conduction band.  The resistivity is of approx. 103 to 1017 Ωm at room temp. Semi-conductors:  “The materials which have electrical conductivity between that of conductors and insulators are called semiconductors.” e.g. Silicon, germanium.  The forbidden gap in semiconductors is very small (approx. 1.1 eV).  P a g e | 22  At low temp., valence band is completely filled with electrons and conduction band is completely empty as electrons don’t have sufficient energy to move in conduction band.  As temp. Increases, electrons gain energy and move to conduction band. I.e. electrical conductivity increases with temperature.  So, they have negative temperature coefficient of resistance.  P a g e | 23 1.6.2 Direct and Indirect bandgap semiconductors as per E – K Diagram Direct bandgap semiconductor:  P a g e | 24  A direct band gap semi–conductor is the one in which maximum energy level of valence band aligns with the minimum energy level of conduction band with respect to momentum. e.g. GaAs  In a semi–conductor, the minimum energy state in conduction band and maximum energy state in valance band is characterized by crystal momentum and wave vector ‘k’ (propagation constant).  In a direct bandgap semiconductor, the k-vectors are same for conduction band minima and valence band maxima.  Whenever an electron from conduction band recombines with a hole from valence band, the energy and momentum, both should be conserved.  The energy difference between conduction band and valence band is released in the form of a photon.  The k-vectors of electron and hole are same. So, we say that momentum is also conserved.  We say the energy is conserved by emitting a photon. Indirect bandgap semiconductor:  An indirect bandgap semiconductor is the one in which maximum energy level of valence band and minimum energy level of valence band and minimum energy level of conduction band are not aligned with respect to momentum. e.g., Si, Ge.  In an indirect bandgap semiconductor, the k-vectors are different for conduction band minima and valence band maxima.  In an indirect bandgap semiconductor, there is a difference in momentum. The recombination process can occur only after the momentum align.  P a g e | 25  Due to crystal imperfections in certain semiconductors, the electron in the conduction band loses energy and momentum at the same time.  The electron actually passes through an intermediate state and transfers its momentum to the crystal lattice.  In this case, the energy is emitted in the form of heat. The transition is non-radiative in nature. 1.6.3 Difference between direct and Indirect bandgap semiconductors DBG semiconductor IBG semiconductor It is one in which maximum energy level of It is the one in which maximum energy level of valence band aligns with the minimum energy valence band and minimum energy level of level of conduction band with respect to valence band and minimum energy level of momentum. conduction band are not aligned with respect to momentum. In this direct recombination takes place with In this due to a difference in momentum, first energy equal to the difference between energy momentum is conserved by release of energy of recombining particles. and only when the two momenta are aligned, recombination occurs. The probability of radiative recombination is The probability of radiative recombination is very high. almost negligible. Efficiency factor is high. Efficiency factor is low. They are preferred for making optical devices Cannot be used to make optical device. like LED’s  P a g e | 26 1.6.4 Fermi level and fermi function Fermi level: “ The highest energy level which an electron can occupy at absolute zero (0 K) or (– 273°C) temperature is called fermi level.” Fermi energy: “ The energy possessed by electrons in the fermi level at absolute zero temperature is called fermi energy.”  P a g e | 27 Fermi function: Also known as Fermi – Dirac distribution function. Fermi distribution function is used for half integer spin particles (fermions) like electrons, which obey Pauli’s exclusive principle. Statement: It is an expression for the distribution of electrons among the energy levels as a function of temperature, the probability of finding an electron in a particular energy state of energy E is given by: 𝟏 𝐟(𝐄) = 𝐄 − 𝐄𝐟 ( ) 𝟏+𝐞 𝐤𝐁 𝐓 Where, f(E) = Fermi function EF = Fermi energy T = Absolute temperature (0 K) KB = Boltzmann constant  From above equation, we can say that a system is characterized by its temperature and fermi energy (Ef).  For a filled energy level f(E) = 1 and for unfilled level f(E) = 0. Case I: At T = 0 K  At T = 0 K, Electron occupy lowest energy first, followed by next higher levels as per Pauli’s exclusive principle. (a) At T = 0 K, E < Ef For energy level E, lying below Ef, (E – Ef) makes negative value. 1 1 ∴ f(E) = −∞ = =1 1+e 1+0 ∴ f(E) = 1  This implies that all energy levels below Ef are completely filled.  P a g e | 28 (b) At T = 0 K, E > Ef For energy level E, lying above Ef, (E – Ef) makes positive value. 1 ∴ f(E) = E − E𝐹 ( ) 1+e kB T 1 1 ∴ f(E) = ∞= =0 1+e 1+∞ ∴ f(E) = 0  This implies that all energy levels above Ef are completely Vacant.  There is no possibility of electron occupying an energy level above Ef. Variation of f (E) at T = 0 K Case II: At T=0 K, E=Ef  At room temperature, the probability starts reducing from 1 for values of E close to Ef, but larger than Ef.  At T > 0 K, for E = Ef, exponential function becomes zero. 1 ∴ f(E) = E − E𝐹 ( ) 1 + e kB T 1 1 1 ∴ f(E) = = = 1 + e0 1+1 2 ∴ f(E) = 0.5  f(E) = 0.5 implies that probability of occupancy of an electron is 50% at any temperature above 0 K.  P a g e | 29  For E = Ef, the value of f (E) falls off to zero rapidly as shown in the figure above. Case III: At T = very high temperature  At room temperature, it can be seen from the figure that the transition between completely filled states and completely empty states is rather gradual than abrupt.  i.e. f(E) changes from 1 to 0 more gradually.  As temperature (T) increases, electron may get an energy of an order kBT and go to the higher vacant state.  Relation between fermi energy Ef, fermi velocity Vf temperature Tf and mean free path λ is given by:  P a g e | 30 2Ef Fermi velocity: Vf = √ m Ef Fermi temperature: Tf = kB The mean free path: λ = τ Vf Q.1 Evaluate the Fermi function for energy kBT above the Fermi energy. Ans. E– EF = kBT, f (E) =? 1 We know Fermi Function ∴ f(E) = (E − E𝐹 ) 1 + e kB T For an energy KBT above Fermi energy: E – EF = kBT 1 1 ∴ f(E) = = 1 + e1 1 + 2.78 Fermi distribution function f(E) = 0.269 Q.2 Calculate the fermi velocity and mean free path for conduction electrons, given that its fermi energy is 11.63 eV and relaxation time for electrons is 7.3 × 10–15 sec. Ans. Ef = 11.63 eV = 11.63 ×1.6 × 10–19 Joule, τ = 7.3 × 10–15 sec, Vf = ?, λ=? 2 Ef ∴ Vf = √ m  P a g e | 31 2 × (11.63 × 1.6 × 10–19) ∴ Vf = √ 9.11 × 10–31 ∴ Vf = √4.085 × 1012 ∴ Vf = 2.02 × 106 m/sec The mean free path: 𝜆 = τ Vf ∴ λ = τ Vf ∴ λ = (7.3 × 10–15) (2.02 × 106) ∴ λ = 1.47 × 10−8 m ∴ λ = 14.75 nm Q.3 Calculate the fermi energy and fermi temperature in a metal. The fermi velocity of electrons in the metal is 0.86 × 106 m/sec. Ans. Vf = 0.86 × 106 m/sec Ef = ? Tf = ? 1 Ef = m Vf2 2 1 ∴ Ef = (9.11 × 10–31)(0.86 × 106)2 2 ∴ Ef = 3.36 × 10−19 Joule 3.36 × 10−19 ∴ Ef = 1.6 × 10−19 ∴ Ef = 2.105 eV  P a g e | 32 Ef Fermi temperature: Tf = kB 3.36 × 10−19 ∴ Tf = 1.38 × 10−23 ∴ Tf = 24.41 × 103 K Q.4 Using fermi function, evaluate the temperature at which there is 1% probability that an electron in a solid will have an energy 0.5 eV above Ef of 5 eV. Ans. E = 5.5 eV Ef = 5 eV f(E) = 1% = 0.01 T =? ∴ E − Ef = 5.5 − 5 = 0.5 eV ∴ E − Ef = 0.5 × (1.6 × 10−19) J 1 ∴ f(E) = ( E − E𝐹 ) 1 + e kB T E − E𝐹 ( ) ∴ f(E) ( 1 + e kB T )=1 E − E𝐹 ( ) ∴ f(E) + f(E) e kB T =1 E − E𝐹 ( ) ∴ f(E) e kB T = 1 − f(E) E − E𝐹 1 − f(E) ( k T ) ∴e B = f(E) E − E𝐹 1 − f(E) ( k T ) ∴ e B = f(E) Taking logarithm on both side  P a g e | 33 E − Ef 1 − f(E) ∴ = ln [ ] kB T f(E) E − Ef ∴ = ln (1 − f(E)) − ln (f(E)) kB T 1 ln(1 − f(E)) − ln(f(E)) ∴ = kB T E − Ef E − Ef ∴ kB T = ln(1 − f(E)) − ln(f(E)) E − Ef ∴T= kB [ln(1 − f(E)) − ln(f(E))] 0.5 × (1.6 × 10−19) ∴T= 1.38 × 10−23 [ln(1 − 0.01) − ln(0.01)] 8 × 10−20 ∴T= (1.38 × 10−23) [−0.01005 − (−4.6051)] 8 × 10−20 ∴T= 6.341 × 10−23 ∴ T = 1261.597 ∴ T = 1.261 × 103 K Q.5 Find the probability with which an energy level 0.02 eV above fermi level will be occupied at room temperature of 300 K and at 1000 K. Ans. E − Ef = 0.02 eV = 0.02 × 1.6 × 10−19 J Probability of occupancy at 300 K  P a g e | 34 1 ∴ f(E) = ( E − E𝐹 ) 1+e kB T 1 ∴ f(E) = 0.02×1.6×10−19 ( ) 1+e 1.38×10−23×300 1 ∴ f(E) = 1 + e(0.7729) 1 ∴ f(E) = 1 + 2.166 ∴ f(E) = 0.315 Probability of occupancy at 1000 K 1 ∴ f(E) = ( E − E𝐹 ) 1+e kB T 1 ∴ f(E) = 0.02×1.6×10−19 ( ) 1+e 1.38×10−23×1000 1 ∴ f(E) = 1 + e(0.2318) 1 ∴ f(E) = 1 + 1.2609 ∴ f(E) = 0.442 