Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ Semiconductor Device Modeling and Characterization EE5342, Lecture 7-Spring 2004 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ L7 February 10

MidTerm and Project Tests MidTerm on Thursday 2/12 Cover sheet to be posted at http://www.uta.edu/ronc/5342/tests/ Project 1 draft assignment will be posted 2/13. Project report to be used in doing: Project 1 Test on Thursday 3/11 Cover sheet will be posted as above L7 February 10

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 10

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 10

Diffnt’l, one-sided diode cond. (cont.) L7 February 10

Cap. of a (1-sided) short diode (cont.) L7 February 10

General time- constant L7 February 10

General time- constant (cont.) L7 February 10

General time- constant (cont.) L7 February 10

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 10

Effect of carrier recombination in DR The S-R-H rate (tno = tpo = to) is L7 February 10

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 10

Effect of carrier rec. in DR (cont.) L7 February 10

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 10

High level inj effects (cont.) L7 February 10

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 10

Plot of typical Va > 0 current density equations ln(J) data Effect of Rs Vext VKF L7 February 10

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 10

Reverse bias (Va< 0), carr gen in DR (cont.) L7 February 10

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 10

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 10

Reverse bias junction breakdown L7 February 10

Ecrit for reverse breakdown (M&K**) Taken from p. 198, M&K** Casey Model for Ecrit L7 February 10

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 10

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 10

Gauss’ Law rpc rp rj rn rnc L7 February 10

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 10

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 10

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 10

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 10

Spherical diode Breakdown Voltage L7 February 10

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 10

L7 February 10

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 10

Hole lifetimes, taken from Shur***, p. 101. L7 February 10

Example Js,long, = ? If xnc, = 2 micron, Js,short, = ? L7 February 10

Example (cont.) Estimate VKF Estimate IKF L7 February 10

Example (cont.) Estimate Js,rec Estimate Rs if xnc is 100 micron L7 February 10

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

Diode equivalent circuit (small sig) ID h is the practical “ideality factor” IQ VD VQ L7 February 10

Small-signal eq circuit Cdiff and Cdepl are both charged by Va = VQ Va Cdiff rdiff Cdepl L7 February 10

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 10

Diode switching (cont.) VF,VR >> Va F: t < 0 Sw RF R: t > 0 VF + RR D VR + L7 February 10

Diode charge for t < 0 pn pno x xn xnc L7 February 10

Diode charge for t >>> 0 (long times) pn pno x xn xnc L7 February 10

Equation summary L7 February 10

Snapshot for t barely > 0 pn Total charge removed, Qdis=IRt pno x xn xnc L7 February 10

I(t) for diode switching ID IF ts ts+trr t - 0.1 IR -IR L7 February 10

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. L7 February 10