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Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011

Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/. First Assignment. e-mail to listserv@listserv.uta.edu In the body of the message include subscribe EE5342

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Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011

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  1. Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

  2. 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.

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

  4. Additional University Closure Means More Schedule Changes • Plan to meet until noon some days in the next few weeks. This way we will make up for the lost time. The first extended class will be Monday, 2/14. • The MT changed to Friday 2/18 • The P1 test changed to Friday 3/11. • The P2 test is still Wednesday 4/13 • The Final is still Wednesday 5/11.

  5. MT and P1 Assignment on Friday, 2/18/11 • Quizzes and tests are open book • must have a legally obtained copy-no Xerox copies. • OR one handwritten page of notes. • Calculator allowed. • A cover sheet will be published by Wednesday, 2/16/11.

  6. p-type Ec Ec Ev EFn qfn= kT ln(Nd/ni) EFi Ev Energy bands forp- and n-type s/c n-type EFi qfp= kT ln(ni/Na) EFp

  7. Eo Making contactin a p-n junction • Equate the EF in the p- and n-type materials far from the junction • Eo(the free level), Ec, Efi and Ev must be continuous N.B.: qc = 4.05 eV (Si), and qf = qc + Ec - EF qc(electron affinity) qf (work function) Ec Ef Efi qfF Ev

  8. EfN Band diagram forp+-n jctn* at Va = 0 Ec qVbi = q(fn -fp) qfp < 0 Ec Efi EfP Ev Efi qfn > 0 *Na > Nd -> |fp|> fn Ev p-type for x<0 n-type for x>0 x -xpc xn 0 -xp xnc

  9. Band diagram forp+-n at Va=0 (cont.) • A total band bending of qVbi = q(fn-fp) = kT ln(NdNa/ni2) is necessary to set EfP = EfN • For -xp < x < 0, Efi - EfP < -qfp, = |qfp| so p < Na = po, (depleted of maj. carr.) • For 0 < x < xn, EfN - Efi < qfn, so n < Nd = no, (depleted of maj. carr.) -xp < x < xn is the Depletion Region

  10. DepletionApproximation • Assume p << po = Nafor -xp < x < 0, so r = q(Nd-Na+p-n) = -qNa, -xp < x < 0, and p = po = Nafor -xpc < x < -xp, so r = q(Nd-Na+p-n) = 0, -xpc < x < -xp • Assume n << no = Ndfor 0 < x < xn, so r = q(Nd-Na+p-n) = qNd, 0 < x < xn, and n = no = Ndfor xn < x < xnc, so r = q(Nd-Na+p-n) = 0, xn < x < xnc

  11. Poisson’sEquation • The electric field at (x,y,z) is related to the charge density r=q(Nd-Na-p-n) by the Poisson Equation:

  12. Poisson’sEquation • For n-type material, N = (Nd - Na) > 0, no = N, and (Nd-Na+p-n)=-dn +dp +ni2/N • For p-type material, N = (Nd - Na) < 0, po = -N, and (Nd-Na+p-n) = dp-dn-ni2/N • So neglecting ni2/N, [r=(Nd-Na+p-n)]

  13. Quasi-FermiEnergy

  14. Quasi-FermiEnergy (cont.)

  15. Quasi-FermiEnergy (cont.)

  16. Induced E-fieldin the D.R. • The sheet dipole of charge, due to Qp’ and Qn’ induces an electric field which must satisfy the conditions • Charge neutrality and Gauss’ Law* require thatEx = 0 for -xpc < x < -xp and Ex = 0 for -xn < x < xnc h0

  17. O O O O O O + + + - - - Induced E-fieldin the D.R. Ex N-contact p-contact p-type CNR n-type chg neutral reg Depletion region (DR) Exposed Donor ions Exposed Acceptor Ions W x -xpc -xp xn xnc 0

  18. Depletion approx.charge distribution r +Qn’=qNdxn +qNd [Coul/cm2] -xp x -xpc xn xnc Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn -qNa Qp’=-qNaxp [Coul/cm2]

  19. 1-dim soln. ofGauss’ law Ex -xp xn xnc -xpc x -Emax

  20. Depletion Approxi-mation (Summary) • For the step junction defined by doping Na (p-type) for x < 0 and Nd, (n-type) for x > 0, the depletion width W = {2e(Vbi-Va)/qNeff}1/2, where Vbi = Vt ln{NaNd/ni2}, and Neff=NaNd/(Na+Nd). Since Naxp=Ndxn, xn = W/(1 + Nd/Na), and xp = W/(1 + Na/Nd).

  21. One-sided p+n or n+p jctns • If p+n, then Na >>Nd, and NaNd/(Na +Nd) = Neff --> Nd, and W --> xn, DR is all on lightly d. side • If n+p, then Nd >>Na, and NaNd/(Na +Nd) = Neff --> Na, and W --> xp, DR is all on lightly d. side • The net effect is that Neff --> N-, (- = lightly doped side) and W --> x-

  22. JunctionC (cont.) r +Qn’=qNdxn +qNd dQn’=qNddxn -xp x -xpc xn xnc Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn -qNa dQp’=-qNadxp Qp’=-qNaxp

  23. JunctionC (cont.) • The C-V relationship simplifies to

  24. JunctionC (cont.) • If one plots [C’j]-2vs. Va Slope = -[(C’j0)2Vbi]-1 vertical axis intercept = [C’j0]-2 horizontal axis intercept = Vbi C’j-2 C’j0-2 Va Vbi

  25. Arbitrary dopingprofile • If the net donor conc, N = N(x), then at xn, the extra charge put into the DR when Va->Va+dVa is dQ’=-qN(xn)dxn • The increase in field, dEx =-(qN/e)dxn, by Gauss’ Law (at xn, but also const). • So dVa=-(xn+xp)dEx= (W/e) dQ’ • Further, since N(xn)dxn = N(xp)dxp gives, the dC/dxn as ...

  26. Arbitrary dopingprofile (cont.)

  27. Arbitrary dopingprofile (cont.)

  28. Arbitrary dopingprofile (cont.)

  29. Arbitrary dopingprofile (cont.)

  30. n Nd 0 xn x Debye length • The DA assumes n changes from Nd to 0 discontinuously at xn, likewise, p changes from Na to 0 discontinuously at -xp. • In the region of xn, the 1-dim Poisson equation is dEx/dx = q(Nd - n), and since Ex = -df/dx, the potential is the solution to -d2f/dx2 = q(Nd - n)/e

  31. Debye length (cont) • Since the level EFi is a reference for equil, we set f = Vt ln(n/ni) • In the region of xn, n = ni exp(f/Vt), so d2f/dx2 = -q(Nd - ni ef/Vt), let f = fo + f’, where fo = Vt ln(Nd/ni) so Nd - ni ef/Vt = Nd[1 - ef/Vt-fo/Vt], for f - fo = f’ << fo, the DE becomes d2f’/dx2 = (q2Nd/ekT)f’, f’ << fo

  32. Debye length (cont) • So f’= f’(xn) exp[+(x-xn)/LD]+con. and n = Nd ef’/Vt, x ~ xn, where LD is the “Debye length”

  33. Debye length (cont) • LD estimates the transition length of a step-junction DR (concentrations Na and Nd with Neff = NaNd/(Na +Nd)). Thus, • For Va=0, & 1E13 <Na,Nd< 1E19 cm-3 • 13% <d< 28% => DA is OK

  34. Example • An assymetrical p+ n junction has a lightly doped concentration of 1E16 and with p+ = 1E18. What is W(V=0)? Vbi=0.816 V, Neff=9.9E15, W=0.33mm • What is C’j? = 31.9 nFd/cm2 • What is LD? = 0.04 mm

  35. 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. • 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. • 3 Physics of Semiconductor Devices, Shur, Prentice-Hall, 1990.

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