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Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2002

Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2002. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/. Classes of semiconductors. Intrinsic : n o = p o = n i , since N a &N d << n i =[N c N v exp(E g /kT)] 1/2 ,(not easy to get)

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Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2002

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  1. Semiconductor Device Modeling and CharacterizationEE5342, Lecture 3-Spring 2002 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

  2. 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

  3. 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

  4. 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

  5. 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

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

  7. 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

  8. 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

  9. 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)

  10. 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)

  11. 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

  12. 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

  13. Equilibrium electronconc. and energies

  14. Equilibrium hole conc. and energies

  15. 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

  16. 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.

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

  18. 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

  19. 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)

  20. 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)

  21. 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.

  22. 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

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

  24. 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

  25. 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

  26. 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

  27. 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

  28. 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)

  29. Net silicon extrresistivity (cont.)

  30. 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.)

  31. 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

  32. 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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