EE 5340 Semiconductor Device Theory Lecture 12 – Spring 2011 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

Depletion Approximation 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 < xnc ©rlc L12-01Mar2011

Depletion approx. charge distribution +Qn’=qNdxn +qNd [Coul/cm2] -xp x -xpc xn xnc -qNa Due to Charge neutrality Qp’ + Qn’ = 0, => Naxp = Ndxn Qp’=-qNaxp [Coul/cm2] ©rlc L12-01Mar2011

Induced E-field in 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 ≈0 ©rlc L12-01Mar2011

Induced E-field in the D.R. Ex p-contact N-contact O - O + p-type CNR n-type chg neutral reg O - O + O - O + Exposed Acceptor Ions Depletion region (DR) Exposed Donor ions W x -xpc -xp xn xnc ©rlc L12-01Mar2011

Induced E-field in 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 ©rlc L12-01Mar2011

Soln to Poisson’s Eq in the D.R. Ex -xp xn x -xpc xnc -Emax ©rlc L12-01Mar2011

Soln to Poisson’s Eq in the D.R. (cont.) ©rlc L12-01Mar2011

Soln to Poisson’s Eq in the D.R. (cont.) ©rlc L12-01Mar2011

Comments on the Ex and Vbi Vbi is not measurable externally since Ex is zero at both contacts The effect 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) = Emax ©rlc L12-01Mar2011

Sample 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 mV ©rlc L12-01Mar2011

Sample calculations What are Neff, W ? Neff, = 1.97E15/cm3 W = 0.707 micron What is xn ? = 0.704 micron What is Emax ? 2.14E4 V/cm ©rlc L12-01Mar2011

Soln to Poisson’s Eq in the D.R. Ex W(Va-dV) W(Va) -xp xn x -xpc xnc -Emax(V) -Emax(V-dV) ©rlc L12-01Mar2011

Effect of V  0 ©rlc L12-01Mar2011

Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn Junction C (cont.) r +Qn’=qNdxn +qNd dQn’=qNddxn -xp x -xpc xn xnc -qNa Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn dQp’=-qNadxp Qp’=-qNaxp ©rlc L12-01Mar2011

Junction Capacitance The junction has +Q’n=qNdxn (exposed donors), and (exposed acceptors) Q’p=-qNaxp = -Q’n, forming a parallel sheet charge capacitor. ©rlc L12-01Mar2011

Junction C (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. ©rlc L12-01Mar2011

Junction C (cont.) The C-V relationship simplifies to ©rlc L12-01Mar2011

Junction C (cont.) If one plots [Cj]-2 vs. Va Slope = -[(Cj0)2Vbi]-1 vertical axis intercept = [Cj0]-2 horizontal axis intercept = Vbi Cj-2 Vbi Va Cj0-2 ©rlc L12-01Mar2011

Junction 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/M ©rlc L12-01Mar2011

dy/dx - Numerical Differentiation ©rlc L12-01Mar2011

Practical Junctions Junctions are formed by diffusion 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. ©rlc L12-01Mar2011

Practical Jctns (cont.) Na(x) N N Shallow (steep) implant Na(x) Linear approx. Box or step junction approx. Nd Nd Uniform wafer con x (depth) x (depth) xj ©rlc L12-01Mar2011

Linear graded junction Let the net donor concentration, N(x) = Nd(x) - Na(x) = ax, so r =qax, -xp < x < xn = xp = xo, (chg neu) r = qa x r Q’n=qaxo2/2 -xo x xo Q’p=-qaxo2/2 ©rlc L12-01Mar2011

Linear graded junction (cont.) Let Ex(-xo) = 0, since this is the edge of the DR (also true at +xo) ©rlc L12-01Mar2011

Linear graded junction (cont.) Ex -xo xo x -Emax |area| = Vbi-Va ©rlc L12-01Mar2011

Linear graded junction (cont.) ©rlc L12-01Mar2011

Linear graded junction, etc. ©rlc L13-03Mar2011

References 1 and M&KDevice 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. 3 and **Semiconductor Physics & Devices, 2nd ed., by Neamen, Irwin, Chicago, 1997. Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989. ©rlc L12-01Mar2011