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5340 EE Semiconductor Device Theory Lecture 4 - Fall 2003 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc L 04 Sept 04 1 Web Pages * If you have not already done so: R. L. Carter s web page www.uta.edu/ronc/ EE 5340 web page and syllabus www.uta.edu/ronc/5340/syllabus.htm University and College Ethics Policies www2.uta.edu/discipline/ www.uta.edu/ronc/5340/COE_EthicsStatement_Fall02.htm L 04 Sept 04 2 Assignment 1: If you have not received response from me ... 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 L 04 Sept 04 3 Assignment 2: 930AM students only* IF you have a class OTHER than 5340 at 8 AM, e-mail to ronc@uta.edu Subject line: 5340 8 AM Class In the body of message include email address: ______________________ Your Enrollment Name*: ______________ Class you are taking at 8:00 AM: ________ * If you don t do this, and don t take Test 1 at 8:00 AM, you will not get credit for Test 1 L 04 Sept 04 4 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, no Nd-Na p-type: no < po, since Nd < Na, po Na-Nd 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 L 04 Sept 04 5 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 L 04 Sept 04 6 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 L 04 Sept 04 7 Equilibrium conc (cont.) For Na > Nd (taking the + root) po = (Na-Nd)/2 + {[(Na-Nd)/2]2+ni2}1/2 For Na >> Nd and Na >> ni, can use the binomial expansion, giving po = Na/2 + Na/2[1 + 2ni2/Na2 + ] So po = Na in the limit of Na >> Nd and Na >> ni L 04 Sept 04 8 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 L 04 Sept 04 9 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 midband L 04 Sept 04 10 EF relative to Ec and Ev Inverting no = Nc exp[-(Ec-EF)/kT] gives Ec - EF = kT ln(Nc/no) For n-type Ec material: - 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) L 04 Sept 04 11 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) L 04 Sept 04 12 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 L 04 Sept 04 13 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 L 04 Sept 04 14 Equilibrium electron conc. and energies no Ef - Ec no = exp , or Ef - Ec = kT ln ; Nc kT Nc no Ef - Efi no = exp , or Ef - Efi = kT ln ; n ni kT i noNv Nv and Ef - Ev = kT ln 2 = kT ln p n o i L 04 Sept 04 15 Equilibrium hole conc. and energies po po Ev - Ef = exp , or Ev - Ef = kT ln ; Nv kT Nv po po Efi - Ef = exp , or Efi - Ef = kT ln ; n kT ni i poNc Nc and Ec - Ef = kT ln 2 = kT ln n no i L 04 Sept 04 16 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 L 04 Sept 04 17 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. L 04 Sept 04 18 Carrier mobility (cont.) If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is 1 = t , and the tcoll collisions i total mobility m is given by 1 = m total collisions mi 1 all 1 all L 04 Sept 04 19 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 r r I = J dS L 04 Sept 04 20 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.) L 04 Sept 04 21

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