Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ EE5342 – Semiconductor Device Modeling and Characterization Lecture 05-Spring 2010 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

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Direct carrier gen/recomb k Ec Ev (Excitation can be by light) gen rec - + Ev Ec Ef Efi L05 February 01

Direct gen/rec of excess carriers Generation rates, Gn0 = Gp0 Recombination rates, Rn0 = Rp0 In equilibrium: Gn0 = Gp0 = Rn0 = Rp0 In non-equilibrium condition: n = no + dn and p = po + dp, where nopo=ni2 and for dn and dp > 0, the recombination rates increase to R’n and R’p L05 February 01

Direct rec for low-level injection Define low-level injection as dn = dp < no, for n-type, and dn = dp < po, for p-type The recombination rates then are R’n = R’p = dn(t)/tn0, for p-type, and R’n = R’p = dp(t)/tp0, for n-type Where tn0 and tp0 are the minority-carrier lifetimes L05 February 01

Shockley-Read- Hall Recomb Indirect, like Si, so intermediate state Ec Ec ET Ef Efi Ev Ev k L05 February 01

S-R-H trap characteristics1 The Shockley-Read-Hall Theory requires an intermediate “trap” site in order to conserve both E and p If trap neutral when orbited (filled) by an excess electron - “donor-like” Gives up electron with energy Ec - ET “Donor-like” trap which has given up the extra electron is +q and “empty” L05 February 01

S-R-H trap char. (cont.) If trap neutral when orbited (filled) by an excess hole - “acceptor-like” Gives up hole with energy ET - Ev “Acceptor-like” trap which has given up the extra hole is -q and “empty” Balance of 4 processes of electron capture/emission and hole capture/ emission gives the recomb rates L05 February 01

tpo = (Ntvthsn)-1, where sn~p(rBohr)2 S-R-H recombination Recombination rate determined by: Nt (trap conc.), vth (thermal vel of the carriers), sn (capture cross sect for electrons), sp (capture cross sect for holes), with tno = (Ntvthsn)-1, and tpo = (Ntvthsn)-1, where sn~p(rBohr)2 L05 February 01

S-R-H recomb. (cont.) In the special case where tno = tpo = to the net recombination rate, U is L05 February 01

S-R-H “U” function characteristics The numerator, (np-ni2) simplifies in the case of extrinsic material at low level injection (for equil., nopo = ni2) For n-type (no > dn = dp > po = ni2/no): (np-ni2) = (no+dn)(po+dp)-ni2 = nopo - ni2 + nodp + dnpo + dndp ~ nodp (largest term) Similarly, for p-type, (np-ni2) ~ podn L05 February 01

S-R-H “U” function characteristics (cont) For n-type, as above, the denominator = to{no+dn+po+dp+2nicosh[(Et-Ei)kT]}, simplifies to the smallest value for Et~Ei, where the denom is tono, giving U = dp/to as the largest (fastest) For p-type, the same argument gives U = dn/to Rec rate, U, fixed by minority carrier L05 February 01

S-R-H net recom- bination rate, U In the special case where tno = tpo = to = (Ntvthso)-1 the net rec. rate, U is L05 February 01

S-R-H rec for excess min carr For n-type low-level injection and net excess minority carriers, (i.e., no > dn = dp > po = ni2/no), U = dp/to, (prop to exc min carr) For p-type low-level injection and net excess minority carriers, (i.e., po > dn = dp > no = ni2/po), U = dn/to, (prop to exc min carr) L05 February 01

Minority hole lifetimes. Taken from Shur3, (p.101). L05 February 01

Minority electron lifetimes. Taken from Shur3, (p.101). L05 February 01

Parameter example tmin = (45 msec) 1+(7.7E-18cm3)Ni+(4.5E-36cm6)Ni2 For Nd = 1E17cm3, tp = 25 msec Why Nd and tp ? L05 February 01

S-R-H rec for deficient min carr If n < ni and p < pi, then the S-R-H net recomb rate becomes (p < po, n < no): U = R - G = - ni/(2t0cosh[(ET-Efi)/kT]) And with the substitution that the gen lifetime, tg = 2t0cosh[(ET-Efi)/kT], and net gen rate U = R - G = - ni/tg The intrinsic concentration drives the return to equilibrium L05 February 01

The Continuity Equation The chain rule for the total time derivative dn/dt (the net generation rate of electrons) gives L05 February 01

The Continuity Equation (cont.) L05 February 01

The Continuity Equation (cont.) L05 February 01

The Continuity Equation (cont.) L05 February 01

The Continuity Equation (cont.) L05 February 01

The Continuity Equation (cont.) L05 February 01

The Continuity Equation (cont.) L05 February 01

Energy bands for p- and n-type s/c p-type n-type Ec Ev Ec EFi EFn qfn= kT ln(Nd/ni) EFi qfp= kT ln(ni/Na) EFp Ev L05 February 01

Making contact in a p-n junction Equate the EF in the p- and n-type materials far from the junction Eo(the free level), Ec, Efi and Ev must be continuous N.B.: qc = 4.05 eV (Si), and qf = qc + Ec - EF Eo qc (electron affinity) qf (work function) Ec Ef Efi qfF Ev L05 February 01

Band diagram for p+-n jctn* at Va = 0 Ec qVbi = q(fn - fp) qfp < 0 Efi Ec EfP EfN Ev Efi qfn > 0 *Na > Nd -> |fp| > fn Ev p-type for x<0 n-type for x>0 x -xpc -xp xn xnc L05 February 01

Band diagram for p+-n at Va=0 (cont.) A total band bending of qVbi = q(fn-fp) = kT ln(NdNa/ni2) is necessary to set EfP = EfN For -xp < x < 0, Efi - EfP < -qfp, = |qfp| so p < Na = po, (depleted of maj. carr.) For 0 < x < xn, EfN - Efi < qfn, so n < Nd = no, (depleted of maj. carr.) -xp < x < xn is the Depletion Region L05 February 01

Depletion Approximation Assume p << po = Na for -xp < x < 0, so r = q(Nd-Na+p-n) = -qNa, -xp < x < 0, and p = po = Na for -xpc < x < -xp, so r = q(Nd-Na+p-n) = 0, -xpc < x < -xp Assume n << no = Nd for 0 < x < xn, so r = q(Nd-Na+p-n) = qNd, 0 < x < xn, and n = no = Nd for xn < x < xnc, so r = q(Nd-Na+p-n) = 0, xn < x < xnc L05 February 01

Poisson’s Equation The electric field at (x,y,z) is related to the charge density r=q(Nd-Na-p-n) by the Poisson Equation: L05 February 01

Poisson’s Equation For n-type material, N = (Nd - Na) > 0, no = N, and (Nd-Na+p-n)=-dn +dp +ni2/N For p-type material, N = (Nd - Na) < 0, po = -N, and (Nd-Na+p-n) = dp-dn-ni2/N So neglecting ni2/N, [r=(Nd-Na+p-n)] L05 February 01

Quasi-Fermi Energy L05 February 01

Quasi-Fermi Energy (cont.) L05 February 01

Quasi-Fermi Energy (cont.) L05 February 01

Depletion approx. charge distribution +Qn’=qNdxn +qNd [Coul/cm2] -xp x -xpc xn xnc -qNa Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn Qp’=-qNaxp [Coul/cm2] L05 February 01

Induced E-field in the D.R. The sheet dipole of charge, due to Qp’ and Qn’ induces an electric field which must satisfy the conditions Charge neutrality and Gauss’ Law* require that Ex = 0 for -xpc < x < -xp and Ex = 0 for -xn < x < xnc h 0 L05 February 01

Induced E-field in the D.R. Ex p-contact N-contact O - O + p-type CNR n-type chg neutral reg O - O + O - O + Exposed Acceptor Ions Depletion region (DR) Exposed Donor ions W x -xpc -xp xn xnc L05 February 01

1-dim soln. of Gauss’ law Ex -xpc -xp xn xnc x -Emax L05 February 01

Depletion Approxi- mation (Summary) For the step junction defined by doping Na (p-type) for x < 0 and Nd, (n-type) for x > 0, the depletion width W = {2e(Vbi-Va)/qNeff}1/2, where Vbi = Vt ln{NaNd/ni2}, and Neff=NaNd/(Na+Nd). Since Naxp=Ndxn, xn = W/(1 + Nd/Na), and xp = W/(1 + Na/Nd). L05 February 01

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. 3 Physics of Semiconductor Devices, Shur, Prentice-Hall, 1990. L05 February 01