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Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2004

Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2004. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/. Web Pages. You should be aware of information at 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 3-Spring 2004

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  1. Semiconductor Device Modeling and CharacterizationEE5342, Lecture 3-Spring 2004 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

  2. Web Pages • You should be aware of information at • 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 • Submit a signed copy to Dr. Carter

  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. 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)

  5. 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

  6. 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*

  7. 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

  8. 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

  9. 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

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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. Equilibrium conc n-type • For Nd > Na • Let N = Nd-Na, and (taking the + root) no = (N)/2 + {[N/2]2+ni2}1/2 • For Nd+= Nd >> ni >> Na we have • no = Nd, and • po = ni2/Nd

  21. Equilibrium conc p-type • For Na > Nd • Let N = Nd-Na, and (taking the + root) po = (-N)/2 + {[-N/2]2+ni2}1/2 • For Na-= Na >> ni >> Nd we have • po = Na, and • no = ni2/Na

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

  23. Hole Conc 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. Position of theFermi Level • Efi is the Fermi level when no = po • Ef shown is a Fermi level for no > po • Ef < Efi when no < po • Efi < (Ec + Ev)/2, which is the mid-band

  25. EF relative to Ec and Ev • Inverting no = Nc exp[-(Ec-EF)/kT] gives Ec - EF = kT ln(Nc/no) For n-type material: Ec - EF =kTln(Nc/Nd)=kTln[(NcPo)/ni2] • Inverting po = Nv exp[-(EF-Ev)/kT] gives EF - Ev = kT ln(Nv/po) For p-type material: EF - Ev = kT ln(Nv/Na)

  26. EF relative to Efi • Letting ni = no gives  Ef = Efi ni = Nc exp[-(Ec-Efi)/kT], so Ec - Efi = kT ln(Nc/ni). Thus EF - Efi = kT ln(no/ni) and for n-type EF - Efi = kT ln(Nd/ni) • Likewise Efi - EF = kT ln(po/ni) and for p-type Efi - EF = kT ln(Na/ni)

  27. Locating Efi in the bandgap • Since Ec - Efi = kT ln(Nc/ni), and Efi - Ev = kT ln(Nv/ni) • The sum of the two equations gives Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv) • Since Nc = 2.8E19cm-3 > 1.04E19cm-3 = Nv, the intrinsic Fermi level lies below the middle of the band gap

  28. Samplecalculations • Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv), so at 300K, kT = 25.86 meV and Nc/Nv = 2.8/1.04, Efi is 12.8 meV or 1.1% below mid-band • For Nd = 3E17cm-3, given that Ec - EF = kT ln(Nc/Nd), we have Ec - EF = 25.86 meV ln(280/3), Ec - EF = 0.117 eV =117meV ~3x(Ec - ED) what Nd gives Ec-EF =Ec/3

  29. Equilibrium electronconc. and energies

  30. Equilibrium hole conc. and energies

  31. Carrier Mobility • In an electric field, Ex, the velocity (since ax = Fx/m* = qEx/m*) is vx = axt = (qEx/m*)t, and the displ x = (qEx/m*)t2/2 • If every tcoll, a collision occurs which “resets” the velocity to <vx(tcoll)> = 0, then <vx> = qExtcoll/m* = mEx

  32. Carrier mobility (cont.) • The response function m is the mobility. • The mean time between collisions, tcoll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few. • Hence mthermal = qtthermal/m*, etc.

  33. Carrier mobility (cont.) • If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is

  34. Drift Current • The drift current density (amp/cm2) is given by the point form of Ohm Law J = (nqmn+pqmp)(Exi+ Eyj+ Ezk), so J = (sn + sp)E =sE, where s = nqmn+pqmp defines the conductivity • The net current is

  35. Drift currentresistance • Given: a semiconductor resistor with length, l, and cross-section, A. What is the resistance? • As stated previously, the conductivity, s = nqmn + pqmp • So the resistivity, r = 1/s = 1/(nqmn + pqmp)

  36. Drift currentresistance (cont.) • Consequently, since R = rl/A R = (nqmn + pqmp)-1(l/A) • For n >> p, (an n-type extrinsic s/c) R = l/(nqmnA) • For p >> n, (a p-type extrinsic s/c) R = l/(pqmpA)

  37. Drift currentresistance (cont.) • Note: for an extrinsic semiconductor and multiple scattering mechanisms, since R = l/(nqmnA) or l/(pqmpA), and (mn or p total)-1 = Smi-1, then Rtotal = S Ri (series Rs) • The individual scattering mechanisms are: Lattice, ionized impurity, etc.

  38. Exp. mobility modelfunction for Si1 Parameter As P B mmin 52.2 68.5 44.9 mmax 1417 1414 470.5 Nref 9.68e16 9.20e16 2.23e17 a 0.680 0.711 0.719

  39. Exp. mobility modelfor P, As and B in Si

  40. Carrier mobilityfunctions (cont.) • The parameter mmax models 1/tlattice the thermal collision rate • The parameters mmin, Nref and a model 1/timpur the impurity collision rate • The function is approximately of the ideal theoretical form: 1/mtotal = 1/mthermal + 1/mimpurity

  41. Carrier mobilityfunctions (ex.) • Let Nd= 1.78E17/cm3 of phosphorous, so mmin = 68.5, mmax = 1414, Nref = 9.20e16 and a = 0.711. Thus mn = 586 cm2/V-s • Let Na= 5.62E17/cm3 of boron, so mmin = 44.9, mmax = 470.5, Nref = 9.68e16 and a = 0.680. Thus mn = 189 cm2/V-s

  42. Lattice mobility • The mlattice is the lattice scattering mobility due to thermal vibrations • Simple theory gives mlattice ~ T-3/2 • Experimentally mn,lattice ~ T-n where n = 2.42 for electrons and 2.2 for holes • Consequently, the model equation is mlattice(T) = mlattice(300)(T/300)-n

  43. Ionized impuritymobility function • The mimpur is the scattering mobility due to ionized impurities • Simple theory gives mimpur ~ T3/2/Nimpur • Consequently, the model equation is mimpur(T) = mimpur(300)(T/300)3/2

  44. Net silicon (ex-trinsic) resistivity • Since r = s-1 = (nqmn + pqmp)-1 • The net conductivity can be obtained by using the model equation for the mobilities as functions of doping concentrations. • The model function gives agreement with the measured s(Nimpur)

  45. Net silicon extrresistivity (cont.)

  46. Net silicon extrresistivity (cont.) • Since r = (nqmn + pqmp)-1, and mn > mp, (m = qt/m*) we have rp > rn • Note that since 1.6(high conc.) < rp/rn < 3(low conc.), so 1.6(high conc.) < mn/mp < 3(low conc.)

  47. Net silicon (com-pensated) res. • For an n-type (n >> p) compensated semiconductor, r = (nqmn)-1 • But now n = N = Nd - Na, and the mobility must be considered to be determined by the total ionized impurity scattering Nd + Na= NI • Consequently, a good estimate is r = (nqmn)-1 = [Nqmn(NI)]-1

  48. References • 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. • 2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.

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