EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2010 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

First Assignment Send e-mail to ronc@uta.edu On the subject line, put “5340 e-mail” In the body of message include email address: ______________________ Your Name*: _______________________ Last four digits of your Student ID: _____ * Your name as it appears in the UTA Record - no more, no less L04 03Sep10

Second Assignment Please print and bring to class a signed copy of the document appearing at http://www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf L04 03Sep10

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* L04 03Sep10

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 L04 03Sep10

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 L04 03Sep10

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 L04 03Sep10

Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is L04 03Sep10

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 = 1.45E10/cm3 L04 03Sep10

Calculating the equilibrium no The idea 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 L04 03Sep10

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 L04 03Sep10

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 L04 03Sep10

Figure 1.10 (a) Fermi-Dirac distribution function describing the probability that an allowed state at energy E is occupied by an electron. (b) The density of allowed states for a semiconductor as a function of energy; note that g(E) is zero in the forbidden gap between Ev and Ec. (c) The product of the distribution function and the density-of-states function. (p. 17 - M&K) L04 03Sep10

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 L04 03Sep10

Figure 1.13 Energy-gap narrowing Eg as a function of electron concentration. [A. Neugroschel, S. C. Pao, and F. A. Lindhold, IEEE Trans. Electr. Devices, ED-29, 894 (May 1982).] taken from p. 25 - M&K) L04 03Sep10

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 L04 03Sep10

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 L04 03Sep10

Figure 1. 9 Electron concentration vs Figure 1.9 Electron concentration vs. temperature for two n-type doped semiconductors: (a) Silicon doped with 1.15 X 1016 arsenic atoms cm-3[1], (b) Germanium doped with 7.5 X 1015 arsenic atoms cm-3[2]. (p.12 in M&K) L04 03Sep10

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 L04 03Sep10

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 L04 03Sep10

Equilibrium conc (cont.) For Nd > Na (taking the + root) no = (Nd-Na)/2 + {[(Nd-Na)/2]2+ni2}1/2 For Nd >> Na and Nd >> ni, can use the binomial expansion, giving no = Nd/2 + Nd/2[1 + 2ni2/Nd2 + … ] So no = Nd, and po = ni2/Nd in the limit of Nd >> Na and Nd >> ni L04 03Sep10

Example calculations For Nd = 3.2E16/cm3, ni = 1.4E10/cm3 no = Nd = 3.2E16/cm3 po = ni2/Nd , (po is always ni2/no) = (1.4E10/cm3)2/3.2E16/cm3 = 6.125E3/cm3 (comp to ~1E23 Si) For po = Na = 4E17/cm3, no = ni2/Na = (1.4E10/cm3)2/4E17/cm3 = 490/cm3 L04 03Sep10

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 L04 03Sep10

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) L04 03Sep10

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) L04 03Sep10

Locating Efi in the bandgap Since Ec - Efi = kT ln(Nc/ni), and Efi - Ev = kT ln(Nv/ni) The 1st equation minus the 2nd 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 L04 03Sep10

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 L04 03Sep10

Equilibrium electron conc. and energies L04 03Sep10

Equilibrium hole conc. and energies L04 03Sep10

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 L04 03Sep10

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. L04 03Sep10

Carrier mobility (cont.) If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is L04 03Sep10

Figure 1.16 (p. 31 M&K) Electron and hole mobilities in silicon at 300 K as functions of the total dopant concentration. The values plotted are the results of curve fitting measurements from several sources. The mobility curves can be generated using Equation 1.2.10 with the following values of the parameters [3] (see table on next slide). L04 03Sep10

Parameter Arsenic Phosphorus Boron μmin 52.2 68.5 44.9 μmax 1417 1414 470.5 Nref 9.68 X 1016 9.20 X 1016 2.23 X 1017 α 0.680 0.711 0.719 Figure 1.16 (cont. M&K) L04 03Sep10

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 L04 03Sep10

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.) L04 03Sep10

Figure 1.15 (p. 29) M&K Dopant density versus resistivity at 23°C (296 K) for silicon doped with phosphorus and with boron. The curves can be used with little error to represent conditions at 300 K. [W. R. Thurber, R. L. Mattis, and Y. M. Liu, National Bureau of Standards Special Publication 400–64, 42 (May 1981).] L04 03Sep10

References *Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989. **Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago. M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003. L04 03Sep10