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This lecture covers fundamental aspects of semiconductor device modeling and characterization, focusing on intrinsic, n-type, and p-type semiconductors. Topics include charge neutrality, equilibrium concentrations, Fermi level positioning, carrier mobility, and drift current. Detailed mathematical equations and approximations are presented, detailing the relationships between carrier concentrations and energy levels in semiconductors. Real-world implications are also discussed, providing insights into the behavior of semiconductor materials under various conditions and their significance in electronic applications.
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Semiconductor Device Modeling and CharacterizationEE5342, Lecture 3-Spring 2002 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/
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
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
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
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
Electron Conc. inthe MB approx. • Assuming the MB approx., the equilibrium electron concentration is
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
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
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)
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)
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
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
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
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.
Carrier mobility (cont.) • If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is
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
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)
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)
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.
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
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
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
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
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
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)
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.)
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
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.