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Semiconductor Device Modeling and Characterization EE5342, Lecture 2-Spring 2002

Understand the K-P.E(k) relationship, Silicon Covalent Bond, Band Structure & Thermal Carrier Generation. Learn about Donor & Acceptor states and Impurity Levels in Semiconductor Electronics.

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Semiconductor Device Modeling and Characterization EE5342, Lecture 2-Spring 2002

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  1. Semiconductor Device Modeling and CharacterizationEE5342, Lecture 2-Spring 2002 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

  2. K-P E(k) Relationship*

  3. 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

  4. 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.

  5. 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

  6. 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

  7. 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

  8. 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)

  9. 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 = 1E10/cm3

  10. 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

  11. 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)h]/[Zpm*q2] ~ 2 nm for Z=1, m*~mo/2, n=1, ground state

  12. 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

  13. 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) and has one -q ion • #B atoms = #holes, so neutral • H atom-like orbits

  14. 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

  15. 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

  16. Semiconductor Electronics - concepts thus far • Conduction and Valence states due to symmetry of lattice • “Free-elec.” dynamics near band edge • Band Gap • direct or indirect • effective mass in curvature • Thermal carrier generation • Chemical carrier gen (donors/accept)

  17. Counting carriers - 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

  18. 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*

  19. 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

  20. 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

  21. 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

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

  23. Electron and HoleConc in MB approx • Similarly, the equilibrium hole concentration is po = Nv exp[-(EF-Ev)/kT] • So that nopo = NcNv exp[-Eg/kT] • ni2 = nopo, Nc,v = 2{2pm*n,pkT/h2}3/2 • Nc = 2.8E19/cm3, Nv = 1.04E19/cm3 and ni = 1E10/cm3

  24. Calculating theequilibrium no • The ideal is to calculate the equilibrium electron concentration no for the FD distribution, where fF(E) = {1+exp[(E-EF)/kT]}-1 gc(E) = [4p(2mn*)3/2(E-Ec)1/2]/h3

  25. Equilibrium con-centration for no • Earlier quoted the MB approximation no = Nc exp[-(Ec - EF)/kT],(=Nc exp hF) • The exact solution is no = 2NcF1/2(hF)/p1/2 • Where F1/2(hF) is the Fermi integral of order 1/2, and hF = (EF - Ec)/kT • Error in no, e, is smaller than for the DF: e = 31%, 12%, 5% for -hF = 0, 1, 2

  26. Equilibrium con-centration for po • Earlier quoted the MB approximation po = Nv exp[-(EF - Ev)/kT],(=Nv exp h’F) • The exact solution is po = 2NvF1/2(h’F)/p1/2 • Note: F1/2(0) = 0.678, (p1/2/2) = 0.886 • Where F1/2(h’F) is the Fermi integral of order 1/2, and h’F = (Ev - EF)/kT • Errors are the same as for po

  27. Degenerate andnondegenerate cases • Bohr-like doping model assumes no interaction between dopant sites • If adjacent dopant atoms are within 2 Bohr radii, then orbits overlap • This happens when Nd ~ Nc (EF ~ Ec), or when Na ~ Nv (EF ~ Ev) • The degenerate semiconductor is defined by EF ~/> Ec or EF ~/< Ev

  28. Donor ionization • The density of elec trapped at donors is nd = Nd/{1+[exp((Ed-EF)/kT)/2]} • Similar to FD DF except for factor of 2 due to degeneracy (4 for holes) • Furthermore nd = Nd - Nd+, also • For a shallow donor, can have Ed-EF >> kT AND Ec-EF >> kT: Typically EF-Ed ~ 2kT

  29. Donor ionization(continued) • Further, if Ed - EF > 2kT, then nd~ 2Nd exp[-(Ed-EF)/kT], e < 5% • If the above is true, Ec - EF > 4kT, so no ~ Nc exp[-(Ec-EF)/kT], e < 2% • Consequently the fraction of un-ionized donors is nd/no = 2Nd exp[(Ec-Ed)/kT]/Nc = 0.4% for Nd(P) = 1e16/cm3

  30. Classes ofsemiconductors • Intrinsic: no = po = ni, since Na&Nd << ni =[NcNvexp(Eg/kT)]1/2,(not easy to get) • n-type: no > po, since Nd > Na • p-type: no < po, since Nd < Na • Compensated: no=po=ni, w/ Na- = Nd+ > 0 • Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants

  31. Equilibriumconcentrations • Charge neutrality requires q(po + Nd+) + (-q)(no + Na-) = 0 • Assuming complete ionization, so Nd+ = Nd and Na- = Na • Gives two equations to be solved simultaneously 1. Mass action, no po = ni2, and 2. Neutrality po + Nd = no + Na

  32. Equilibriumconc (cont.) • For Nd > Na (taking the + root) no = (Nd-Na)/2 + {[(Nd-Na)/2]2+ni2}1/2 • For Nd >> Na and Nd >> ni, can use the binomial expansion, giving no = Nd/2 + Nd/2[1 + 2ni2/Nd2 + … ] • So no = Nd, and po = ni2/Nd in the limit of Nd >> Na and Nd >> ni

  33. Equilibriumconc (cont.) • For Na > Nd (taking the + root) po = (Na-Nd)/2 + {[(Na-Nd)/2]2+ni2}1/2 • For Na >> Nd and Na >> ni, can use the binomial expansion, giving po = Na/2 + Na/2[1 + 2ni2/Na2 + … ] • So po = Na in the limit of Na >> Nd and Na >> ni

  34. Examplecalculations • For Nd = 3.2E16/cm3, ni = 1.4E10/cm3 no = Nd = 3.2E16/cm3 po = ni2/Nd , (po is always ni2/no) = (1.4E10/cm3)2/3.2E16/cm3 = 6.125E3/cm3 (comp to ~1E23 Si) • For po = Na = 4E17/cm3, no = ni2/Na = (1.4E10/cm3)2/4E17/cm3 = 490/cm3

  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.

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