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UT Arlington EE 5340 - Lecture Notes

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Slide 1Depletion ApproximationDepletion approx. charge distributionInduced E-field in the D.R.Induced E-field in the D.R.Induced E-field in the D.R. (cont.)Soln to Poisson’s Eq in the D.R.Soln to Poisson’s Eq in the D.R. (cont.)Soln to Poisson’s Eq in the D.R. (cont.)Comments on the Ex and VbiSample calculationsSample calculationsSoln to Poisson’s Eq in the D.R.Effect of V  0Slide 15Junction CapacitanceJunction C (cont.)Junction C (cont.)Junction C (cont.)Junction Capacitancedy/dx - Numerical DifferentiationPractical JunctionsPractical Jctns (cont.)Linear graded junctionLinear graded junction (cont.)Linear graded junction (cont.)Linear graded junction (cont.)Linear graded junction, etc.ReferencesEE 5340Semiconductor Device TheoryLecture 12 – Spring 2011Professor Ronald L. [email protected]://www.uta.edu/ronc©rlc L12-01Mar2011DepletionApproximation•Assume p << po = Na for -xp < x < 0, so r = q(Nd-Na+p-n) = -qNa, -xp < x < 0, and p = po = Na for -xpc < x < -xp, so r = q(Nd-Na+p-n) = 0, -xpc < x < -xp•Assume n << no = Nd for 0 < x < xn, so r = q(Nd-Na+p-n) = qNd, 0 < x < xn, and n = no = Nd for xn < x < xnc, so r = q(Nd-Na+p-n) = 0, xn < x < xnc2©rlc L12-01Mar2011Depletion approx.charge distributionxnx-xp-xpcxncr+qNd-qNa+Qn’=qNdxnQp’=-qNaxpDue to Charge neutrality Qp’ + Qn’ = 0, => Naxp = Ndxn [Coul/cm2][Coul/cm2]3©rlc L12-01Mar2011Induced 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 that Ex = 0 for -xpc < x < -xp and Ex = 0 for -xn < x < xnc  QQAdxEAdVdSE'p'nxxxxxVSnp-≈04©rlc L12-01Mar2011Induced E-fieldin the D.R.xnx-xp-xpcxncO-O-O-O+O+O+Depletion region (DR)p-type CNRExExposed Donor ionsExposed Acceptor Ionsn-type chg neutral regp-contactN-contactW05©rlc L12-01Mar2011Induced 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 range6©rlc L12-01Mar2011Soln to Poisson’sEq in the D.R.xnx-xp-xpcxncEx-EmaxdxqNdxdEaxqNdxdE7©rlc L12-01Mar2011Soln to Poisson’sEq in the D.R. (cont.) )VqkT (note ,xNxN2qdxdVE ,dxEVnNNlnqkTthat is D.R. the in P.E. the of solnthe to V of iprelationsh the Now,t2pa2ndxxxxbi2idabinp8©rlc L12-01Mar2011Soln to Poisson’sEq in the D.R. (cont.)WV2N2qVE then,WE21V have also must we Since.NNNNN where ,qNV2Wthen ,xxW let and ,xNxNbiefbimaxmaxbidadaefefbipnpand9©rlc L12-01Mar2011Comments on theEx and Vbi•Vbi is not measurable externally since Ex is zero at both contacts•The efect of Ex does not extend beyond the depletion region•The lever rule [Naxp=Ndxn] was obtained assuming charge neutrality. It could also be obtained by requiring Ex(x=0-dx) = Ex(x=0+dx) = Emax10©rlc L12-01Mar2011Sample calculations•Vt = 25.86 mV at 300K• e = ereo = 11.7*8.85E-14 Fd/cm= 1.035E-12 Fd/cm•If Na=5E17/cm3, and Nd=2E15 /cm3, then for ni=1.4E10/cm3, then what is Vbi = 757 mV11©rlc L12-01Mar2011Sample calculations•What are Nef, W ?Nef, = 1.97E15/cm3W = 0.707 micron•What is xn ? = 0.704 micron•What is Emax ? 2.14E4 V/cm12©rlc L12-01Mar2011Soln to Poisson’sEq in the D.R.xnx-xp-xpcxncEx-Emax(V)dxqNdxdEaxqNdxdE-Emax(V-dV)W(Va)W(Va-dV)13©rlc L12-01Mar2011  efbimaxefbixaabinxpxxNVaV2qEand ,qNVaV2Ware Solutions .E reduce to tends V todue field the since ,VVdxEthat is now change only TheEfect of V  014©rlc L12-01Mar2011JunctionC (cont.)xnx-xp-xpcxncr+qNd-qNa+Qn’=qNdxnQp’=-qNaxpCharge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn dQn’=qNddxndQp’=-qNadxp15©rlc L12-01Mar2011JunctionCapacitance•The junction has +Q’n=qNdxn (exposed donors), and (exposed acceptors) Q’p=-qNaxp = -Q’n, forming a parallel sheet charge capacitor.    2dadaabidadaabiaddndncmCoul ,NNNNVVq2,NqNNNVV2NN1qNxqN'Q16©rlc L12-01Mar2011JunctionC (cont.)•So this definition of the capacitance gives a parallel plate capacitor with charges dQ’n and dQ’p(=-dQ’n), separated by, L (=W), with an area A and the capacitance is then the ideal parallel plate capacitance.•Still non-linear and Q is not zero at Va=0.17©rlc L12-01Mar2011JunctionC (cont.)•The C-V relationship simplifies to ][Fd/cm ,NNV2NqN'C herewequation model a ,VV1'C'C2dabida0j21bia0jj18©rlc L12-01Mar2011JunctionC (cont.)•If one plots [Cj]-2 vs. VaSlope = -[(Cj0)2Vbi]-1vertical axis intercept = [Cj0]-2 horizontal axis intercept = VbiCj-2VbiVaCj0-21M31 VVJ C0CJVJV10CJACC:Equation Modelbi0jMjj,~,~'19©rlc L12-01Mar2011Junction Capacitance•Estimate CJO•Define y  Cj/CJO•Calculate y/(dy/dV) = {d[ln(y)]/dV}-1•A plot of r  y/(dy/dV) vs. V has slope = -1/M, and intercept = VJ/M20©rlc L12-01Mar2011dy/dx - Numerical Diferentiationx y dy/ dx (central dif erence)x(n-1) y(n-1) [y(n) - y(n-2)]/ [x(n) - x(n-2)]x(n) y(n) [y(n+1) - y(n-1)]/ [x(n+1) - x(n-1)]x(n+1) y(n+1) [y(n+2) - y(n)]/ [x(n+2) - x(n)]x(n+2) y(n+2) [y(n+3) - y(n+1)]/ [x(n+3) - x(n+1)]21©rlc L12-01Mar2011Practical Junctions•Junctions are formed by difusion or implantation into a uniform concentration wafer. The profile can be approximated by a step or linear function in the region of the junction.•If a step, then previous models OK.•If linear, let the local charge density r=qax in the region of the junction.22©rlc L12-01Mar2011Practical Jctns (cont.)Shallow (steep) implantNx (depth)Box or step junction approx.Nx (depth)Na(x)xjLinear approx.NdNdNa(x)Uniform wafer con23©rlc L12-01Mar2011Linear gradedjunction•Let the net donor concentration, N(x) = Nd(x) - Na(x) = ax, so r =qax, -xp < x < xn = xp = xo, (chg neu)xo-xorr = qa xQ’n=qaxo2/2Q’p=-qaxo2/2x24©rlc L12-01Mar2011Linear gradedjunction (cont.)•Let Ex(-xo) = 0, since this is the edge of the DR (also true at +xo)2omax2omaxxxox-xox-xx2qaEwhere ,xx1E)x(Eso ,axdxqdE Law, Gauss'


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UT Arlington EE 5340 - Lecture Notes

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