Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ Semiconductor Device Modeling and Characterization EE5342, Lecture 7-Spring 2005 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ L7 February 08

Injection Conditions L7 February 08

Apply the Continuity Eqn in CNR Ideal Junction Theory (cont.) Apply the Continuity Eqn in CNR L7 February 08

Ideal Junction Theory (cont.) L7 February 08

Ideal Junction Theory (cont.) L7 February 08

Excess minority carrier distr fctn L7 February 08

Carrier Injection ln(carrier conc) ln Na ln Nd ln ni ~Va/Vt ~Va/Vt ln ni2/Nd ln ni2/Na x -xpc -xp xnc xn L7 February 08

Minority carrier currents L7 February 08

Evaluating the diode current L7 February 08

Special cases for the diode current L7 February 08

Ideal diode equation Assumptions: Current dens, Jx = Js expd(Va/Vt) low-level injection Maxwell Boltzman statistics Depletion approximation Neglect gen/rec effects in DR Steady-state solution only Current dens, Jx = Js expd(Va/Vt) where expd(x) = [exp(x) -1] L7 February 08

Ideal diode equation (cont.) Js = Js,p + Js,n = hole curr + ele curr Js,p = qni2Dp coth(Wn/Lp)/(NdLp) = qni2Dp/(NdWn), Wn << Lp, “short” = qni2Dp/(NdLp), Wn >> Lp, “long” Js,n = qni2Dn coth(Wp/Ln)/(NaLn) = qni2Dn/(NaWp), Wp << Ln, “short” = qni2Dn/(NaLn), Wp >> Ln, “long” Js,n << Js,p when Na >> Nd L7 February 08

Diffnt’l, one-sided diode conductance Static (steady-state) diode I-V characteristic IQ Va VQ L7 February 08

Diffnt’l, one-sided diode cond. (cont.) L7 February 08

Charge distr in a (1- sided) short diode dpn Assume Nd << Na The sinh (see L12) excess minority carrier distribution becomes linear for Wn << Lp dpn(xn)=pn0expd(Va/Vt) Total chg = Q’p = Q’p = qdpn(xn)Wn/2 Wn = xnc- xn dpn(xn) Q’p x xn xnc L7 February 08

Charge distr in a 1- sided short diode dpn Assume Quasi-static charge distributions Q’p = Q’p = qdpn(xn)Wn/2 ddpn(xn) = (W/2)* {dpn(xn,Va+dV) - dpn(xn,Va)} dpn(xn,Va+dV) dpn(xn,Va) dQ’p Q’p x xn xnc L7 February 08

Cap. of a (1-sided) short diode (cont.) L7 February 08

Diode equivalent circuit (small sig) ID h is the practical “ideality factor” IQ VD VQ L7 February 08

Small-signal eq circuit Cdiff and Cdepl are both charged by Va = VQ Va Cdiff rdiff Cdepl L7 February 08

General time- constant L7 February 08

General time- constant (cont.) L7 February 08

General time- constant (cont.) L7 February 08

Effect of carrier recombination in DR The S-R-H rate (tno = tpo = to) is L7 February 08

Effect of carrier rec. 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 L7 February 08

Effect of carrier rec. in DR (cont.) L7 February 08

Effect of non- zero E in the CNR This is usually not a factor in a short diode, but when E is finite -> resistor In a long diode, there is an additional ohmic resistance (usually called the parasitic diode series resistance, Rs) Rs = L/(nqmnA) for a p+n long diode. L=Wn-Lp (so the current is diode-like for Lp and the resistive otherwise). L7 February 08

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)), L7 February 08

High level inj effects (cont.) L7 February 08

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 L7 February 08

Plot of typical Va > 0 current density equations ln(J) data Effect of Rs Vext VKF L7 February 08

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. L7 February 08

Reverse bias (Va< 0), carr gen in DR (cont.) L7 February 08

Reverse bias junction 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 L7 February 08

Reverse bias junction 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 L7 February 08

Reverse bias junction breakdown L7 February 08

Ecrit for reverse breakdown (M&K**) Taken from p. 198, M&K** Casey Model for Ecrit L7 February 08

Junction curvature effect 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 L7 February 08

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 L7 February 08

Gauss’ Law rpc rp rj rn rnc L7 February 08

Spherical Diode Fields calculations For rj < ro ≤ rn, Setting Er = 0 at r = rn, we get Note that the equivalent of the lever law for this spherical diode is L7 February 08

Spherical Diode Fields calculations Assume Na >> Nd, so rn – rj  d >> rj – rp. Further, setting the usual definition for the potential difference, and evaluating the potential difference at breakdown, we have PHIi – Va = BV and Emax = Em = Ecrit = Ec. We also define a = 3eEm/qNd[cm]. L7 February 08

Showing the rj  ∞ limit C1. Solve for rn – rj = d as a function of Emax and solve for the value of d in the limit of rj  . The solution for rn is given below. . L7 February 08

Solving for the Breakdown (BV) Solve for BV = [fi – Va]Emax = Ecrit, and solve for the value of BV in the limit of rj  . The solution for BV is given below . L7 February 08

Spherical diode Breakdown Voltage L7 February 08

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 = ? L7 February 08

L7 February 08

Parameters for examples 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 = ? L7 February 08

Hole lifetimes, taken from Shur***, p. 101. L7 February 08

Example Js,long, = ? If xnc, = 2 micron, Js,short, = ? L7 February 08

Example (cont.) Estimate VKF Estimate IKF L7 February 08

Example (cont.) Estimate Js,rec Estimate Rs if xnc is 100 micron L7 February 08

Example (cont.) Estimate Jgen for 10 V reverse bias Estimate BV L7 February 08

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 L7 February 08

Diode switching (cont.) VF,VR >> Va F: t < 0 Sw RF R: t > 0 VF + RR D VR + L7 February 08

Diode charge for t < 0 pn pno x xn xnc L7 February 08

Diode charge for t >>> 0 (long times) pn pno x xn xnc L7 February 08

Equation summary L7 February 08

Snapshot for t barely > 0 pn Total charge removed, Qdis=IRt pno x xn xnc L7 February 08

I(t) for diode switching ID IF ts ts+trr t - 0.1 IR -IR L7 February 08