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Semiconductor Device Modeling and Characterization EE5342, Lecture 2-Spring 2005

Semiconductor Device Modeling and Characterization EE5342, Lecture 2-Spring 2005. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/. 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

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Semiconductor Device Modeling and Characterization EE5342, Lecture 2-Spring 2005

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  1. Semiconductor Device Modeling and CharacterizationEE5342, Lecture 2-Spring 2005 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

  2. 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 • http://www.uta.edu/studentaffairs/judicialaffairs/ • www.uta.edu/ronc/5342/Ethics.htm

  3. 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.

  4. 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)|

  5. 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

  6. 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.

  7. 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

  8. 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.

  9. 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

  10. 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

  11. K-P Potential Function*

  12. 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

  13. 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

  14. K-P Solutions*

  15. K-P E(k) Relationship*

  16. 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

  17. Generalizationsand 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.

  18. 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

  19. Silicon BandStructure** • 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

  20. Si Energy BandStructure 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

  21. Si Bond ModelAbove 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)

  22. Band Model forthermal 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

  23. 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

  24. 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

  25. Band Model fordonor 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

  26. Acceptor: Holedue 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

  27. Hole orbits andacceptor 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

  28. 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

  29. Quantum densityof 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

  30. 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*

  31. Fermi-Diracdistribution 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

  32. Fermi-DiracDF (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

  33. Maxwell-BoltzmanApproximation • 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

  34. Electron Conc. inthe MB approx. • Assuming the MB approx., the equilibrium electron concentration is

  35. 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.

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