Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ Semiconductor Device Modeling and Characterization EE5342, Lecture 2-Spring 2004 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ L2 January 22

Web Pages Bring the following to the first class R. L. Carter’s web page www.uta.edu/ronc/ EE 5342 web page and syllabus www.uta.edu/ronc/5342/syllabus.htm University and College Ethics Policies www2.uta.edu/discipline/ www.uta.edu/ronc/5342/Ethics.htm L2 January 22

First Assignment e-mail to listserv@listserv.uta.edu In the body of the message include subscribe EE5342 This will subscribe you to the EE5342 list. Will receive all EE5342 messages If you have any questions, send to ronc@uta.edu, with EE5342 in subject line. L2 January 22

Quantum Mechanics Schrodinger’s wave equation developed to maintain consistence with wave-particle duality and other “quantum” effects Position, mass, etc. of a particle replaced by a “wave function”, Y(x,t) Prob. density = |Y(x,t)• Y*(x,t)| L2 January 22

Schrodinger Equation Separation of variables gives Y(x,t) = y(x)• f(t) The time-independent part of the Schrodinger equation for a single particle with Total E = E and PE = V. The Kinetic Energy, KE = E - V L2 January 22

Solutions for the Schrodinger Equation Solutions of the form of y(x) = A exp(jKx) + B exp (-jKx) K = [8p2m(E-V)/h2]1/2 Subj. to boundary conds. and norm. y(x) is finite, single-valued, conts. dy(x)/dx is finite, s-v, and conts. L2 January 22

Infinite Potential Well V = 0, 0 < x < a V --> inf. for x < 0 and x > a Assume E is finite, so y(x) = 0 outside of well L2 January 22

Step Potential V = 0, x < 0 (region 1) V = Vo, x > 0 (region 2) Region 1 has free particle solutions Region 2 has free particle soln. for E > Vo , and evanescent solutions for E < Vo A reflection coefficient can be def. L2 January 22

Finite Potential Barrier Region 1: x < 0, V = 0 Region 1: 0 < x < a, V = Vo Region 3: x > a, V = 0 Regions 1 and 3 are free particle solutions Region 2 is evanescent for E < Vo Reflection and Transmission coeffs. For all E L2 January 22

Kronig-Penney Model A simple one-dimensional model of a crystalline solid V = 0, 0 < x < a, the ionic region V = Vo, a < x < (a + b) = L, between ions V(x+nL) = V(x), n = 0, +1, +2, +3, …, representing the symmetry of the assemblage of ions and requiring that y(x+L) = y(x) exp(jkL), Bloch’s Thm L2 January 22

K-P Potential Function* L2 January 22

K-P Static Wavefunctions Inside the ions, 0 < x < a y(x) = A exp(jbx) + B exp (-jbx) b = [8p2mE/h]1/2 Between ions region, a < x < (a + b) = L y(x) = C exp(ax) + D exp (-ax) a = [8p2m(Vo-E)/h2]1/2 L2 January 22

K-P Impulse Solution Limiting case of Vo-> inf. and b -> 0, while a2b = 2P/a is finite In this way a2b2 = 2Pb/a < 1, giving sinh(ab) ~ ab and cosh(ab) ~ 1 The solution is expressed by P sin(ba)/(ba) + cos(ba) = cos(ka) Allowed valued of LHS bounded by +1 k = free electron wave # = 2p/l L2 January 22

K-P Solutions* L2 January 22

K-P E(k) Relationship* L2 January 22

Analogy: a nearly -free electr. model Solutions can be displaced by ka = 2np Allowed and forbidden energies Infinite well approximation by replacing the free electron mass with an “effective” mass (noting E = p2/2m = h2k2/2m) of L2 January 22

Generalizations and Conclusions The symm. of the crystal struct. gives “allowed” and “forbidden” energies (sim to pass- and stop-band) The curvature at band-edge (where k = (n+1)p) gives an “effective” mass. L2 January 22

Silicon Covalent Bond (2D Repr) Each Si atom has 4 nearest neighbors Si atom: 4 valence elec and 4+ ion core 8 bond sites / atom All bond sites filled Bonding electrons shared 50/50 _ = Bonding electron L2 January 22

Silicon Band Structure** Indirect Bandgap Curvature (hence m*) is function of direction and band. [100] is x-dir, [111] is cube diagonal Eg = 1.17-aT2/(T+b) a = 4.73E-4 eV/K b = 636K L2 January 22

Si Energy Band Structure at 0 K Every valence site is occupied by an electron No electrons allowed in band gap No electrons with enough energy to populate the conduction band L2 January 22

Si Bond Model Above Zero Kelvin Enough therm energy ~kT(k=8.62E-5eV/K) to break some bonds Free electron and broken bond separate One electron for every “hole” (absent electron of broken bond) L2 January 22

Band Model for thermal carriers Thermal energy ~kT generates electron-hole pairs At 300K Eg(Si) = 1.124 eV >> kT = 25.86 meV, Nc = 2.8E19/cm3 > Nv = 1.04E19/cm3 >> ni = 1.45E10/cm3 L2 January 22

Donor: cond. electr. due to phosphorous P atom: 5 valence elec and 5+ ion core 5th valence electr has no avail bond Each extra free el, -q, has one +q ion # P atoms = # free elect, so neutral H atom-like orbits L2 January 22

Bohr model H atom- like orbits at donor Electron (-q) rev. around proton (+q) Coulomb force, F=q2/4peSieo,q=1.6E-19 Coul, eSi=11.7, eo=8.854E-14 Fd/cm Quantization L = mvr = nh/2p En= -(Z2m*q4)/[8(eoeSi)2h2n2] ~-40meV rn= [n2(eoeSi)h2]/[Zpm*q2] ~ 2 nm for Z=1, m*~mo/2, n=1, ground state L2 January 22

Band Model for donor electrons Ionization energy of donor Ei = Ec-Ed ~ 40 meV Since Ec-Ed ~ kT, all donors are ionized, so ND ~ n Electron “freeze-out” when kT is too small L2 January 22

Acceptor: Hole due to boron B atom: 3 valence elec and 3+ ion core 4th bond site has no avail el (=> hole) Each hole, adds --q, has one -q ion #B atoms = #holes, so neutral H atom-like orbits L2 January 22

Hole orbits and acceptor states Similar to free electrons and donor sites, there are hole orbits at acceptor sites The ionization energy of these states is EA - EV ~ 40 meV, so NA ~ p and there is a hole “freeze-out” at low temperatures L2 January 22

Impurity Levels in Si: EG = 1,124 meV Phosphorous, P: EC - ED = 44 meV Arsenic, As: EC - ED = 49 meV Boron, B: EA - EV = 45 meV Aluminum, Al: EA - EV = 57 meV Gallium, Ga: EA - EV = 65meV Gold, Au: EA - EV = 584 meV EC - ED = 774 meV L2 January 22

Quantum density of states function 1 dim electron wave #s range for n+1 “atoms” is 2p/L < k < 2p/a where a is “interatomic” distance and L = na is the length of the assembly (k = 2p/l) Shorter ls, would “oversample” if n increases by 1, dp is h/L Extn 3D: E = p2/2m = h2k2/2m so a vol of p-space of 4pp2dp has h3/LxLyLz L2 January 22

QM density of states (cont.) So density of states, gc(E) is (Vol in p-sp)/(Vol per state*V) = 4pp2dp/[(h3/LxLyLz)*V] Noting that p2 = 2mE, this becomes gc(E) = {4p(2mn*)3/2/h3}(E-Ec)1/2 and E - Ec = h2k2/2mn* Similar for the hole states where Ev - E = h2k2/2mp* L2 January 22

Fermi-Dirac distribution fctn The probability of an electron having an energy, E, is given by the F-D distr fF(E) = {1+exp[(E-EF)/kT]}-1 Note: fF (EF) = 1/2 EF is the equilibrium energy of the system The sum of the hole probability and the electron probability is 1 L2 January 22

Fermi-Dirac DF (continued) So the probability of a hole having energy E is 1 - fF(E) At T = 0 K, fF (E) becomes a step function and 0 probability of E > EF At T >> 0 K, there is a finite probability of E >> EF L2 January 22

Maxwell-Boltzman Approximation fF(E) = {1+exp[(E-EF)/kT]}-1 For E - EF > 3 kT, the exp > 20, so within a 5% error, fF(E) ~ exp[-(E-EF)/kT] This is the MB distribution function MB used when E-EF>75 meV (T=300K) For electrons when Ec - EF > 75 meV and for holes when EF - Ev > 75 meV L2 January 22

Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is L2 January 22

References *Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989. **Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago. M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003. L2 January 22