Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2005 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ L3 January 25

Web Pages You should be aware of information at R. L. Carter’s web page www.uta.edu/ronc/ EE 5342 web page and syllabus www.uta.edu/ronc/5342/syllabus.htm University and College Ethics Policies http://www.uta.edu/studentaffairs/judicialaffairs/ www.uta.edu/ronc/5342/Ethics.htm Submit a signed copy to Dr. Carter L3 January 25

First Assignment e-mail to listserv@listserv.uta.edu In the body of the message include subscribe EE5342 This will subscribe you to the EE5342 list. Will receive all EE5342 messages If you have any questions, send to ronc@uta.edu, with EE5342 in subject line. L3 January 25

Silicon Band Structure** Indirect Bandgap Curvature (hence m*) is function of direction and band. [100] is x-dir, [111] is cube diagonal Eg = 1.17-aT2/(T+b) a = 4.73E-4 eV/K b = 636K L3 January 25

Analogy: a nearly -free electr. model Solutions can be displaced by ka = 2np Allowed and forbidden energies Infinite well approximation by replacing the free electron mass with an “effective” mass (noting E = p2/2m = h2k2/2m) of L3 January 25

Si Energy Band Structure at 0 K Every valence site is occupied by an electron No electrons allowed in band gap No electrons with enough energy to populate the conduction band L3 January 25

Silicon Covalent Bond (2D Repr) Each Si atom has 4 nearest neighbors Si atom: 4 valence elec and 4+ ion core 8 bond sites / atom All bond sites filled Bonding electrons shared 50/50 _ = Bonding electron L3 January 25

Si Bond Model Above Zero Kelvin Enough therm energy ~kT(k=8.62E-5eV/K) to break some bonds Free electron and broken bond separate One electron for every “hole” (absent electron of broken bond) L3 January 25

Band Model for thermal carriers Thermal energy ~kT generates electron-hole pairs At 300K Eg(Si) = 1.124 eV >> kT = 25.86 meV, Nc = 2.8E19/cm3 > Nv = 1.04E19/cm3 >> ni = 1.45E10/cm3 L3 January 25

Donor: cond. electr. due to phosphorous P atom: 5 valence elec and 5+ ion core 5th valence electr has no avail bond Each extra free el, -q, has one +q ion # P atoms = # free elect, so neutral H atom-like orbits L3 January 25

Bohr model H atom- like orbits at donor Electron (-q) rev. around proton (+q) Coulomb force, F=q2/4peSieo,q=1.6E-19 Coul, eSi=11.7, eo=8.854E-14 Fd/cm Quantization L = mvr = nh/2p En= -(Z2m*q4)/[8(eoeSi)2h2n2] ~-40meV rn= [n2(eoeSi)h2]/[Zpm*q2] ~ 2 nm for Z=1, m*~mo/2, n=1, ground state L3 January 25

Band Model for donor electrons Ionization energy of donor Ei = Ec-Ed ~ 40 meV Since Ec-Ed ~ kT, all donors are ionized, so ND ~ n Electron “freeze-out” when kT is too small L3 January 25

Acceptor: Hole due to boron B atom: 3 valence elec and 3+ ion core 4th bond site has no avail el (=> hole) Each hole, adds --q, has one -q ion #B atoms = #holes, so neutral H atom-like orbits L3 January 25

Hole orbits and acceptor states Similar to free electrons and donor sites, there are hole orbits at acceptor sites The ionization energy of these states is EA - EV ~ 40 meV, so NA ~ p and there is a hole “freeze-out” at low temperatures L3 January 25

Impurity Levels in Si: EG = 1,124 meV Phosphorous, P: EC - ED = 44 meV Arsenic, As: EC - ED = 49 meV Boron, B: EA - EV = 45 meV Aluminum, Al: EA - EV = 57 meV Gallium, Ga: EA - EV = 65meV Gold, Au: EA - EV = 584 meV EC - ED = 774 meV L3 January 25

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 25

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 25

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 25

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 25

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 25

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 25

Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is L3 January 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

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 25

Equilibrium electron conc. and energies L3 January 25

Equilibrium hole conc. and energies L3 January 25

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. L3 January 25

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 25