1 / 48

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

nituna
Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.

More Related