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EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2003

EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2003. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc. Ideal n-type Schottky depletion width (V a =0). E x. r. x n. qN d. x. Q’ d = qN d x d. x. x n. -E m. d. (Sheet of negative charge on metal)= -Q’ d.

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EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2003

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  1. EE 5340Semiconductor Device TheoryLecture 8 - Fall 2003 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

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

  3. qVa = Efn - Efm Barrier for electrons from sc to m reduced toq(fbi-Va) qfBn the same (data - p.166) DR smaller 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

  4. Effect of V  0

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

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

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

  8. 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.

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

  10. 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

  11. Profiling dopantsin a Schottky diode r qNd Q’d = qNdxn dQ’ x -Q-dQ xn Ex xn x -Em

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

  13. Arbitrary dopingprofile (cont.)

  14. 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

  15. Ideal m to n s/c Schottky diode curr

  16. 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

  17. Metal to n-typeaccum region (fm<fs)

  18. Metal to s/c contact resistance

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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]

  25. 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

  26. 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

  27. Induced E-fieldin the D.R. (cont.) • Poisson’s Equation E = r/e, has the one-dimensional form, dEx/dx = r/e, which must be satisfied for r = -qNa, -xp < x < 0, and r = +qNd, 0 < x < xn, with Ex = 0 for the remaining range

  28. Soln to Poisson’sEq in the D.R. Ex xn -xp x -xpc xnc -Emax

  29. Test 1 - 25Sept03 • 8 AM Room 206 Activities Building • 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 Fall 2002.

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