M,W,F 12:00-12:50 (X), 2015 ECEB Professor John Dallesasse Department of Electrical and Computer Engineering 2114 Micro and Nanotechnology Laboratory Tel: (217) 333-8416 E-mail: [email protected] Office Hours: Wednesday 13:00 – 14:00 Please Review e-Mail Sent Today Today’s Discussion • Distribution Functions • Carrier Concentration Calculation • Assignments • Topics for Next Lecture 3 Tentative Schedule [1] JAN 17 JAN 19 Course overview Intro to semiconductor electronics JAN 22 JAN 24 JAN 26 Materials and crystal Bonding forces and energy Metals, semiconductors, structures bands in solids insulators, electrons, holes JAN 29 JAN 31 FEB 2 Intrinsic and extrinsic Distribution functions and Distribution functions and material carrier concentrations carrier concentrations FEB 5 FEB 7 FEB 9 Temperature dependence, Conductivity and mobility Resistance, temperature, compensation impurity concentration FEB 12 FEB 14 FEB 16 Invariance of Fermi level at Optical absorption and Generation and equilibrium luminescence recombination **Subject to Change** 4 Brief Comments Extrinsic Material: Temperature Effects ni n, p n0 ni 2 n0 p0 = ni 500 K 100 K Hot Cold Important Definitions • Intrinsic Material: For a “perfect” crystal in equilibrium, electron and hole densities are equal – The value of this density ni is called the intrinsic concentration • Extrinsic Material: The crystal is “doped” with – donors, where the material is n-type and the electron density exceeds the hole density, or – acceptors, where the material is p-type and the hole density exceeds the electron density – An amphoteric dopant can act as either a donor or acceptor, depending upon what atom it is replacing in the crystal – In extrinsic material, the carrier with the higher concentration is the majority carrier and the carrier with lower concentration is the minority carrier • We will find out later that the product of the minority carrier density and majority carrier density is constant for a given temperature (in equilibrium), 2 and equal to ni • In p-type material, holes are the majority carrier, electrons the minority carrier • In n-type material, electrons are the majority carrier, holes the minority carrier 7 Determining Conductivity Answer: • For electrons, we integrate the product of the number (density) of states at a given energy times the probability a state is full at that energy – Conduction band only – Because the valence band is full, electrons in the valence band do not contribute to the conduction process ∞ n0 = f (E)Nc (E)dE ∫Ec • For holes, we integrate the product of the number (density) of states at a given energy times the probability a state is empty at that energy – Valence band only Ev p0 = ⎡1− f (E)⎤ Nv (E)dE ∫−∞ ⎣ ⎦ But how do we do that? 9 Determining the Probability of State Occupancy Fermions and Bosons • Fermion: Only a single fermion can occupy any given quantum state – they are subject to the Pauli exclusion principle. Electrons are fermions, and follow Fermi-Dirac statistics. • Boson: There is no limit to the number of bosons that can occupy any given quantum state – they are not subject to the Pauli exclusion principle. Photons are bosons, and follow Bose-Einstein statistics. 11 Fermi-Dirac Distribution • The probability that an available energy state at E will be occupied by an electron at temperature T f E ( ) T2>T1 (temperature in Kelvin) is given T=0K by the Fermi-Dirac Distribution 1 Function T=T1 • Fermi-Dirac distribution function: 1/2 1 T=T2 f (E) = (E E )/kT 1 e − f + E f E E : Fermi Level f 1 1 (a) E = E f E = E = = k : Boltzmann Constant f ( f ) 1+ e0 2 k = 8.62x10−5 eV / K (b) T = 0K 1 f E < E = = 1 ( f ) 1+ e−∞ 1 f E > E = = 0 ( f ) 1+ e+∞ 12 What About Holes? Hole Probability: • A state is either empty or full, so the probability a state is empty is f (E) T=0K f (E f − ΔE) 1-P(full) 1 • P(full) is given by the Fermi f (E f + ΔE) distribution f(E) 1/2 • P(empty) = 1 - f(E) T>>0K • f(Ef+ΔE) is the probability of finding an electron at E+ΔE E f E At T = T : • 1-f(Ef-ΔE) is the probability of 1 finding a hole at E-ΔE 1 f (E) = (E E )/kT Note, for a given ΔE : 1+ e − f 1 f E + ΔE = 1− f E − ΔE (E−E f )/kT1 ( f ) ( f ) e 1 1− f (E) = (E E )/kT = (E E)/kT 1+ e − f 1 1+ e f − 1 13 What Do They Mean? Practical Points • The Fermi Level or Fermi Energy is a number that we can use to calculate the density of free carriers in a semiconductor • The Fermi level has units of energy (typically eV) • The Fermi level, used with the Fermi Distribution, tells us the probability a state is filled if we know the energy of the state relative to the Fermi Level • The Fermi Level is generally referenced to some other energy – make sure you know what it is referenced to when doing your calculations 15 Fermi Level in Metals • At absolute zero, all states above the Fermi level are empty and all states below the Fermi level are full ✓ Ef • Above absolute zero, the Fermi level defines the energy where there is a 50% probability a state is occupied • For metals, the Fermi level is the energy of the topmost filled level in the ground state of an N electron system (implies the temperature is absolute zero, 0K) – Related to the energy needed to remove an electron from a metal at absolute zero ? – This definition gives useful insight, but is E problematic for a semiconductor where f the Fermi level is typically in the forbidden gap where there are no energy states 16 Metal Fermi Energy Maximum Kinetic Energy Vacuum Em = hv − qΦ Level • In a metal, the Fermi Energy referenced to the vacuum level is the metal work function http://www.sci.ccny.cuny.edu/~rstein/papers/jcmst00/jcmst00.html Fermi Level: Definitions • There are several equivalent definitions which can be used: – The change in free energy of the crystal when an electron is added or taken away • Effectively, a measure of how difficult it is to add or remove electrons – The sum of the chemical potential and internal electrostatic potential energy, generally equal to the electrochemical potential • Note: most texts fail to mention the electrostatic component • Comment 1: If material with different Fermi levels are brought together, the Fermi levels must align. This has certain implications which will be discussed soon. • Comment 2: Detailed understanding of the Fermi level and Fermi energy requires a background in statistical mechanics, which is beyond the scope of this course 18 Electrochemical Potential • Electrochemical Potential: a measure of the amount of energy needed to add or remove an incremental amount of a given charged species from a defined locus that may be under the influence of an electric field, incorporating both chemical potential and electrostatics • The Fermi Energy is equal to the electrochemical potential in a semiconductor Other Definitions • Chemical Potential: The incremental change in the Gibbs Free Energy of the system per particle (fixed temperature, pressure, etc.). ∂G µ = ∂n • Gibbs Free Energy: In a chemical system, the enthalpy of the system minus the product of the temperature and entropy (at constant pressure/volume) G = H − TS = U + PV − TS U = internal energy, S = entropy, H = enthalpy Fermi Level: Experimental Determination • In practice, it is relatively straight forward to measure carrier concentration • The Fermi level is typically calculated from the measured carrier concentration using the relationships discussed last time (non- degenerate case), or slightly more complex relationships for degenerate material 21 Fermi Level Versus Doping Type n-Type Intrinsic p-Type Additional Information Conduction Band Density of States Consider a cube of material having sides of length "L" A free electron in this material has a wavefunction given by: j(kxx+kyy+kzz) ψ = Ae where 2D 2π 2π 2π k = n , k = n , k = n x x L y y L z z L We also know: p2 2k 2 2π 22 2π 22 E = = = n2 + n2 + n2 = n2 2m* 2m* m*L2 ( x y z ) m*L2 L m*E We can also express n as a function of E: n = π 22 The constant energy surface is therefore the 3D the surface of a sphere of radius "n" where: 1 2 2 2 2 n = (nx + ny + nz ) The total number of states within the sphere is: 4 N′ = πn3 3 Note: Other variants of this derivation get the same result 24 Conduction Band Density of States (2) Including spin, the total number of states up to a given energy is therefore: 4 8 N′(n) = 2 × πn3 = πn3 3 3 Using the expression for n: L m*E n = so: π 22 3/2 8 ⎛ m*E ⎞ N′(E) = L3 3π 2 ⎝⎜ 22 ⎠⎟ 3/2 dN′(E) L3 ⎛ 2m* ⎞ = E dE 2π 2 ⎝⎜ 2 ⎠⎟ The Density of States is defined as the number of states per incremental unit of energy near E per unit volume: 3/2 dN′(E) 1 1 ⎛ 2m* ⎞ N (E) = = E dE L3 2π 2 ⎝⎜ 2 ⎠⎟ 25 Other Cases • Valence Band – Need to include both light holes and heavy holes in the density of states • Near the Direct-Indirect Crossover – Need to include both valleys • Indirect Materials – Multiple valleys due to symmetry and energy degeneracy – This is included in the density of states effective http://www.ioffe.rssi.ru/SVA/NSM/Semicond/Ge/Figs/221.gif mass 26 Carrier Concentration at Thermal Equilibrium The Fermi Function Multiplied by the Density of States ∞ ni = f (E)Nc (E)dE ∫Ec EV = pi = ⎡1− f (E)⎤ NV (E)dE ∫−∞ ⎣ ⎦ ∞ n0 = f (E)Nc (E)dE ∫Ec Ev p0 = ⎡1− f (E)⎤ Nv (E)dE ∫−∞ ⎣ ⎦ 28 DOS: Appendix IV Approximating the Fermi Function 1 E f (E) = (E−E )/kT 1+ e f T=T ⎛ −5 eV ⎞ 2 kT = ⎜ 8.62x10 ⎟ (300K ) 0.0259eV EC ⎝ K ⎠ T=T 1 In the case where E = E E f C 1 f (EC ) = (E −E )/kT 1+ e C f Note: e1 2.7, e2 7.4, e3 20 EV 1 1 1 1 = = 0.048 ~ 0.05 = = = e−3 1+ e3 21 20 e3 1 (E − E ) f E C f ( ) 1/2 So, for > 3, or