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EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2009

EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2009. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc. Test 1 – Sept. 24, 2009. 8 AM Room 108 Nedderman Hall Open book - 1 legal text or ref., only. You may write notes in your book. Calculator allowed

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EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2009

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  1. EE 5340Semiconductor Device TheoryLecture 8 - Fall 2009 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

  2. Test 1 – Sept. 24, 2009 • 8 AM Room 108 Nedderman Hall • Open book - 1 legal text or ref., only. • You may write notes in your book. • Calculator allowed • A cover sheet will be included with full instructions. See http://www.uta.edu/ronc/5340/tests/ for examples from previous semesters.

  3. Si and Al and model (approx. to scale) metal n-type s/c p-type s/c Eo Eo Eo qcsi~ 4.05eV qcsi~ 4.05eV qfm,Al ~ 4.1 eV qfs,n qfs,p Ec Ec EFm EFn EFi EFi EFp Ev Ev

  4. Eo Making contact be-tween metal & s/c • Equate the EF in the metal and s/c materials far from the junction • Eo(the free level), must be continuous across the jctn. N.B.: qc = 4.05 eV (Si), and qf = qc + Ec - EF qc(electron affinity) qf (work function) Ec EF EFi qfF Ev

  5. Equilibrium Boundary Conditions w/ contact • No discontinuity in the free level, Eo at the metal/semiconductor interface. • EF,metal = EF,semiconductor to bring the electron populations in the metal and semiconductor to thermal equilibrium. • Eo - EC = qcsemiconductor in all of the s/c. • Eo - EF,metal = qfmetal throughout metal.

  6. No disc in Eo Ex=0 in metal ==> Eoflat fBn=fm- cs = elec mtl to s/c barr fi=fBn-fn= fm-fs elect s/c to mtl barr Ideal metal to n-typebarrier diode (fm>fs,Va=0) metal n-type s/c Eo qcs qfm qfi qfs,n qfBn Ec EFm EFn EFi Depl reg Ev qf’n

  7. Metal to n-typenon-rect cont (fm<fs) n-type s/c No disc in Eo Ex=0 in metal ==> Eo flat fB,n=fm - cs = elec mtl to s/c barr fi= fBn-fn< 0 Accumulation region metal Eo qcs qfm qfs,n qfi qfB,n Ec EFm EFn EFi Ev qfn Acc reg

  8. Ideal metal to p-typebarrier diode (fm<fs) p-type s/c No disc in Eo Ex=0 in metal ==> Eoflat fBn= fm- cs = elec mtl to s/c barr fBp= fm- cs + Eg = hole m to s fi = fBp-fs,p = hole s/c to mtl barr metal Eo qcs qfm qfi qfs,p qfBn Ec EFi EFm EFp qfBp Ev qfi qfp<0 Depl reg

  9. Metal to p-typenon-rect cont (fm>fs) metal n-type s/c No disc in Eo Ex=0 in metal ==> Eo flat fB,n=fm- fs,n = elec mtl to s/c barr fBp= fm- cs + Eg = hole m to s Accumulation region Eo qcs qfm q(fi) qfs,n qfBn Ec EFm EFi EfP qfp Ev qfBp qfi Accum reg

  10. Metal/semiconductorsystem types n-type semiconductor • Schottky diode - blocking for fm > fs • contact - conducting for fm < fs p-type semiconductor • contact - conducting for fm > fs • Schottky diode - blocking for fm < fs

  11. Barrier transistion region, d Interface states above fo acc, p neutrl below fo dnr, n neutrl Ditd-> oo, qfBn=Eg- fo Fermi level “pinned” Ditd-> 0, qfBn=fm - c Goes to “ideal” case Real Schottkyband structure1

  12. Fig 8.41 (a) Image charge and electric field at a metal-dielectric interface (b) Distortion of potential barrier at E=0 and (c) E0

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

  14. Poisson’s Equation • n = no + dn, and p = po + dp, in non-equil • 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

  15. No disc in Eo Ex=0 in metal ==> Eoflat fBn=fm- cs = elec mtl to s/c barr fbi=fBn-fn= fm-fs elect s/c to mtl barr Ideal metal to n-typebarrier diode (fm>fs,Va=0) n-type s/c metal 0 xn xnc Eo qcs qfm qfbi qfs,n qfBn Ec EFm EFn EFi Depl reg Ev qf’n

  16. DepletionApproximation • For 0 < x < xn, assume n << no = Nd, sor = q(Nd-Na+p-n) = qNd • For xn < x < xnc, assume n = no = Nd, sor = q(Nd-Na+p-n) = 0 • For x = 0-, there is a pulse of charge balancing the qNdxn in 0 < x < xn

  17. Ideal n-type Schottky depletion width (Va=0) Ex r xn qNd x Q’d = qNdxn x xn -Em d (Sheet of negative charge on metal)= -Q’d

  18. n Nd 0 xn x Debye length • The DA assumes n changes from Nd to 0 discontinuously at xn. • In the region of xn, Poisson’s eq is E = r/e --> dEx/dx = q(Nd - n), and since Ex = -df/dx, we have -d2f/dx2 = q(Nd - n)/e to be solved

  19. 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

  20. 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”

  21. Debye length (cont) • LD estimates the transition length of a step-junction DR. Thus, • For Va = 0, i ~ 1V, Vt ~ 25 mV d< 11%  DA assumption OK

  22. Effect of V  0 • Define an external voltage source, Va, with the +term at the metal contact and the -term at the n-type contact • For Va > 0, the Va induced field tends to oppose Ex caused by the DR • For Va < 0, the Va induced field tends to aid Ex due to DR • Will consider Va < 0 now

  23. Effect of V  0

  24. qVa = Efn - Efm Barrier for electrons from sc to m reduced toq(fbi-Va) qfBn the same DR decr Ideal metal to n-typeSchottky (Va > 0) metal n-type s/c Eo qcs qfm q(fi-Va) qfs,n qfBn Ec EFm EFn EFi Ev Depl reg qf’n

  25. Schottky diodecapacitance r qNd Q’d = qNdxn dQ’ x -Q-dQ xn Ex xn x -Em

  26. Schottky Capacitance(continued) • The junction has +Q’n=qNdxn (exposed donors), and Q’n = - Q’metal (Coul/cm2), forming a parallel sheet charge capacitor.

  27. Schottky Capacitance(continued) • This Q ~ (i-Va)1/2 is clearly non-linear, and Q is not zero at Va = 0. • Redefining the capacitance,

  28. Schottky Capacitance(continued) • So this definition of the capacitance gives a parallel plate capacitor with charges dQ’n and dQ’p(=-dQ’n), separated by, L (=xn), with an area A and the capacitance is then the ideal parallel plate capacitance. • Still non-linear and Q is not zero at Va=0.

  29. Schottky Capacitance(continued) • The C-V relationship simplifies to

  30. Schottky Capacitance(continued) • If one plots [Cj]-2vs. Va Slope = -[(Cj0)2Vbi]-1 vertical axis intercept = [Cj0]-2 horizontal axis intercept = fi Cj-2 Cj0-2 Va fi

  31. References 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. See Semiconductor Device Fundamentals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model. 2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981. 3Semiconductor Physics & Devices, 2nd ed., by Neamen, Irwin, Chicago, 1997.

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