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Semiconductor Device Modeling and Characterization EE5342, Lecture 8-Spring 2003

Semiconductor Device Modeling and Characterization EE5342, Lecture 8-Spring 2003. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/. Effect of carrier recombination in DR. The S-R-H rate ( t no = t po = t o ) is. Effect of carrier rec. in DR (cont.).

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Semiconductor Device Modeling and Characterization EE5342, Lecture 8-Spring 2003

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  1. Semiconductor Device Modeling and CharacterizationEE5342, Lecture 8-Spring 2003 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

  2. Effect of carrierrecombination in DR • The S-R-H rate (tno = tpo = to) is

  3. Effect of carrierrec. in DR (cont.) • For low Va ~ 10 Vt • In DR, n and p are still > ni • The net recombination rate, U, is still finite so there is net carrier recomb. • reduces the carriers available for the ideal diode current • adds an additional current component

  4. Effect of carrierrec. in DR (cont.)

  5. High level injection effects • Law of the junction remains in the same form, [pnnn]xn=ni2exp(Va/Vt), etc. • However, now dpn = dnn become >> nno = Nd, etc. • Consequently, the l.o.t.j. reaches the limiting form dpndnn = ni2exp(Va/Vt) • Giving, dpn(xn) = niexp(Va/(2Vt)), or dnp(-xp) = niexp(Va/(2Vt)),

  6. High level injeffects (cont.)

  7. Summary of Va > 0 current density eqns. • Ideal diode, Jsexpd(Va/(hVt)) • ideality factor, h • Recombination, Js,recexp(Va/(2hVt)) • appears in parallel with ideal term • High-level injection, (Js*JKF)1/2exp(Va/(2hVt)) • SPICE model by modulating ideal Js term • Va = Vext - J*A*Rs = Vext - Idiode*Rs

  8. ln(J) Plot of typical Va > 0 current density equations data Effect of Rs Vext VKF

  9. Reverse bias (Va<0)=> carrier gen in DR • Va< 0 gives the net rec rate, U = -ni/2t0, t0 = mean min carr g/r l.t.

  10. Reverse bias (Va< 0),carr gen in DR (cont.)

  11. Reverse biasjunction breakdown • Avalanche breakdown • Electric field accelerates electrons to sufficient energy to initiate multiplication of impact ionization of valence bonding electrons • field dependence shown on next slide • Heavily doped narrow junction will allow tunneling - see Neamen*, p. 274 • Zener breakdown

  12. Ecrit for reverse breakdown (M&K**) Taken from p. 198, M&K**

  13. Reverse biasjunction breakdown • Assume-Va = VR >> Vbi, so Vbi-Va-->VR • Since Emax~ 2VR/W = (2qN-VR/(e))1/2, and VR = BV when Emax = Ecrit (N- is doping of lightly doped side ~ Neff) BV = e (Ecrit )2/(2qN-) • Remember, this is a 1-dim calculation

  14. Junction curvatureeffect on breakdown • The field due to a sphere, R, with charge, Q is Er = Q/(4per2) for (r > R) • V(R) = Q/(4peR), (V at the surface) • So, for constant potential, V, the field, Er(R) = V/R (E field at surface increases for smaller spheres) Note: corners of a jctn of depth xj are like 1/8 spheres of radius ~ xj

  15. BV for reverse breakdown (M&K**) Taken from Figure 4.13, p. 198, M&K** Breakdown voltage of a one-sided, plan, silicon step junction showing the effect of junction curvature.4,5

  16. Example calculations • Assume throughout that p+n jctn with Na = 3e19cm-3 and Nd = 1e17cm-3 • From graph of Pierret mobility model, mp = 331 cm2/V-sec and Dp = Vtmp = ? • Why mp and Dp? • Neff = ? • Vbi = ?

  17. Parameters forexamples • Get tmin from the model used in Project 2 tmin = (45 msec) 1+(7.7E-18cm3)Ni+(4.5E-36cm6)Ni2 • For Nd = 1E17cm3, tp = 25 msec • Why Nd and tp ? • Lp = ?

  18. Hole lifetimes, taken from Shur***, p. 101.

  19. Example • Js,long, = ? • If xnc, = 2 micron, Js,short, = ?

  20. Example(cont.) • Estimate VKF • Estimate IKF

  21. Example(cont.) • Estimate Js,rec • Estimate Rs if xnc is 100 micron

  22. Example(cont.) • Estimate Jgen for 10 V reverse bias • Estimate BV

  23. Diode equivalentcircuit (small sig) ID h is the practical “ideality factor” IQ VD VQ

  24. Small-signal eqcircuit Cdiff and Cdepl are both charged by Va = VQ Va rdiff Cdepl Cdiff

  25. Diode Switching • Consider the charging and discharging of a Pn diode • (Na > Nd) • Wd << Lp • For t < 0, apply the Thevenin pair VF and RF, so that in steady state • IF = (VF - Va)/RF, VF >> Va , so current source • For t > 0, apply VR and RR • IR = (VR + Va)/RR, VR >> Va, so current source

  26. Diode switching(cont.) VF,VR >> Va F: t < 0 Sw RF R: t > 0 VF + RR D VR +

  27. Diode chargefor t < 0 pn pno x xn xnc

  28. Diode charge fort >>> 0 (long times) pn pno x xn xnc

  29. Equationsummary

  30. Snapshot for tbarely > 0 pn Total charge removed, Qdis=IRt pno x xn xnc

  31. I(t) for diodeswitching ID IF ts ts+trr t - 0.1 IR -IR

  32. References * Semiconductor Physics and Devices, 2nd ed., by Neamen, Irwin, Boston, 1997. **Device Electronics for Integrated Circuits, 2nd ed., by Muller and Kamins, John Wiley, New York, 1986. ***Physics of Semiconductor Devices, Shur, Prentice-Hall, 1990.

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