EE 5340 Semiconductor Device Theory Lecture 05 – Spring 2011

1. EE 5340Semiconductor Device TheoryLecture 05 – Spring 2011 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

2. Review the Following • R. L. Carter’s web page: • www.uta.edu/ronc/ • EE 5340 web page and syllabus. (Refresh all EE 5340 pages when downloading to assure the latest version.) All links at: • www.uta.edu/ronc/5340/syllabus.htm • University and College Ethics Policies • www.uta.edu/studentaffairs/conduct/ • Makeup lecture at noon Friday (1/28) in 108 Nedderman Hall. This will be available on the web.

3. 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

4. Second Assignment • Submit a signed copy of the document posted at www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf

5. Schedule Changes Due to University Weather Closings • Make-up class will be held Friday, February 11 at 12 noon in 108 Nedderman Hall. • Additional changes will be announced as necessary. • Syllabus and lecture dates postings will be updated in the next 24 hours. • Project Assignment will be posted in the next 36 hours.

6. Intrinsic carrierconc. (MB limit) • ni2 = no po = Nc Nv e-Eg/kT • Nc = 2{2pm*nkT/h2}3/2 • Nv = 2{2pm*pkT/h2}3/2 • Eg = 1.17 eV - aT2/(T+b) a = 4.73E-4 eV/K b = 636K

7. Classes ofsemiconductors • Intrinsic: no = po = ni, since Na&Nd << ni, ni2 = NcNve-Eg/kT, ~1E-13 dopant level ! • 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

8. 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

9. 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

10. n-type equilibriumconcentrations • N ≡Nd- Na , n type N > 0 • For all N, no = N/2 + {[N/2]2+ni2}1/2 • In most cases, N >> ni, so no = N, and po= ni2/no = ni2/N, (Law of Mass Action is al- ways true in equilibrium)

11. 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

12. p-type equilibriumconcentrations • N ≡Nd - Na , p type  N < 0 • For all N, po = |N|/2 + {[|N|/2]2+ni2}1/2 • In most cases, |N| >> ni, so po = |N|, and no = ni2/po = ni2/|N|, (Law of Mass Action is al- ways true in equilibrium)

13. 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

14. 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)

15. 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)

16. 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

17. 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

18. 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

19. Equilibrium electronconc. and energies

20. Equilibrium hole conc. and energies

21. 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

22. 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.

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

24. 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).

25. Figure 1.16 (cont. M&K)

26. 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

27. Drift currentresistance • Given: a semiconductor resistor with length, l, and cross-section, A. What is the resistance? • As stated previously, the conductivity, s = nqmn + pqmp • So the resistivity, r = 1/s = 1/(nqmn + pqmp)

28. Drift currentresistance (cont.) • Consequently, since R = rl/A R = (nqmn + pqmp)-1(l/A) • For n >> p, (an n-type extrinsic s/c) R = l/(nqmnA) • For p >> n, (a p-type extrinsic s/c) R = l/(pqmpA)

29. References M&K and 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. • See Semiconductor Device Fundamen-tals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model. 2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.

30. 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.