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EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2003

EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2003. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc. 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

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EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2003

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  1. EE 5340Semiconductor Device TheoryLecture 4 - Fall 2003 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

  2. 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

  3. 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

  4. Assignment 2: 930AM stu-dents 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

  5. 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, 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

  6. 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

  7. Equilibriumconc (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

  8. Equilibriumconc (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

  9. Examplecalculations • 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

  10. 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

  11. 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)

  12. 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)

  13. 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

  14. 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

  15. Equilibrium electronconc. and energies

  16. Equilibrium hole conc. and energies

  17. 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

  18. 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.

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

  20. 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

  21. 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.)

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