Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2003 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ L3 January 21

First Assignment Send e-mail to ronc@uta.edu On the subject line, put “5342 e-mail” In the body of message include email address: ______________________ Last Name*: _______________________ First Name*: _______________________ Last four digits of your Student ID: _____ * As it appears in the UTA Record - no more, no less L3 January 21

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) L3 January 21

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 L3 January 21

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* L3 January 21

Fermi-Dirac distribution 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 L3 January 21

Fermi-Dirac DF (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 L3 January 21

Maxwell-Boltzman Approximation 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 L3 January 21

Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is L3 January 21

Electron and 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 L3 January 21

Calculating the equilibrium 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 L3 January 21

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 L3 January 21

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 L3 January 21

Degenerate and nondegenerate 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 L3 January 21

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 L3 January 21

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 L3 January 21

Classes of semiconductors 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 L3 January 21

Equilibrium concentrations 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 L3 January 21

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 L3 January 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 L3 January 21

Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is L3 January 21

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 L3 January 21

Position of the Fermi 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 L3 January 21

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) L3 January 21

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) L3 January 21

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 L3 January 21

Sample calculations 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 L3 January 21

Equilibrium electron conc. and energies L3 January 21

Equilibrium hole conc. and energies L3 January 21

vx = axt = (qEx/m*)t, and the displ 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 L3 January 21

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. L3 January 21

Carrier mobility (cont.) If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is L3 January 21

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 L3 January 21

Drift current resistance 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) L3 January 21

Drift current resistance (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) L3 January 21

Drift current resistance (cont.) Note: for an extrinsic semiconductor and multiple scattering mechanisms, since R = l/(nqmnA) or l/(pqmpA), and (mn or p total)-1 = S mi-1, then Rtotal = S Ri (series Rs) The individual scattering mechanisms are: Lattice, ionized impurity, etc. L3 January 21

Exp. mobility model function 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 L3 January 21

Exp. mobility model for P, As and B in Si L3 January 21

Carrier mobility functions (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 L3 January 21

Carrier mobility functions (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 L3 January 21

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 L3 January 21

Ionized impurity mobility 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 L3 January 21

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) L3 January 21

Net silicon extr resistivity (cont.) L3 January 21

Net silicon extr resistivity (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.) L3 January 21

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 L3 January 21

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. L3 January 21