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This lecture by Professor Ronald L. Carter explores the effects of carrier recombination in diode rectifiers (DR) and high-level injection effects. The S-R-H rate's impact on current density equations is discussed, emphasizing how recombination reduces available carriers for ideal diode current. The lecture also addresses reverse bias conditions, junction breakdown mechanisms such as avalanche and Zener breakdown, and provides example calculations pertinent to practical diode operations. A comprehensive overview of equations and models is also presented.
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Semiconductor Device Modeling and CharacterizationEE5342, Lecture 8-Spring 2002 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/
Effect of carrierrecombination in DR • The S-R-H rate (tno = tpo = to) is
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
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)),
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
Plot of typical Va > 0 current density eqns. Vext-Vd=JARs ln J low level injection ln(JKF) Effect ofRs ln[(Js*JKF) 1/2] Effect of high level injection ln(Jsrec) data ln(Js) Vext recomb. current VKF
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.
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
Ecrit for reverse breakdown (M&K**) Taken from p. 198, M&K**
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
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
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
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 = ?
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 = ?
Example • Js,long, = ? • If xnc, = 2 micron, Js,short, = ?
Example(cont.) • Estimate VKF • Estimate IKF
Example(cont.) • Estimate Js,rec • Estimate Rs if xnc is 100 micron
Example(cont.) • Estimate Jgen for 10 V reverse bias • Estimate BV
Diode equivalentcircuit (small sig) ID h is the practical “ideality factor” IQ VD VQ
Small-signal eqcircuit Cdiff and Cdepl are both charged by Va = VQ Va rdiff Cdepl Cdiff
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
Diode switching(cont.) VF,VR >> Va F: t < 0 Sw RF R: t > 0 VF + RR D VR +
Diode chargefor t < 0 pn pno x xn xnc
Diode charge fort >>> 0 (long times) pn pno x xn xnc
Snapshot for tbarely > 0 pn Total charge removed, Qdis=IRt pno x xn xnc
I(t) for diodeswitching ID IF ts ts+trr t - 0.1 IR -IR
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.