Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

First Assignment e-mail to listserv@listserv.uta.edu In the body of the message include subscribe EE5342 This will subscribe you to the EE5342 list. Will receive all EE5342 messages If you have any questions, send to ronc@uta.edu, with EE5342 in subject line. ©rlc L09-14Feb2011

Second Assignment Submit a signed copy of the document that is posted at www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf ©rlc L09-14Feb2011

Additional University Closure Means More Schedule Changes Plan to meet until noon some days in the next few weeks. This way we will make up for the lost time. The first extended class will be Monday, 2/14. The MT changed to Friday 2/18 The P1 test changed to Friday 3/11. The P2 test is still Wednesday 4/13 The Final is still Wednesday 5/11. ©rlc L09-14Feb2011

MT and P1 Assignment on Friday, 2/18/11 Quizzes and tests are open book must have a legally obtained copy-no Xerox copies. OR one handwritten page of notes. Calculator allowed. ©rlc L09-14Feb2011

Energy bands for p- and n-type s/c p-type n-type Ec Ev Ec EFi EFn qfn= kT ln(Nd/ni) EFi qfp= kT ln(ni/Na) EFp Ev ©rlc L09-14Feb2011

Making contact in a p-n junction Equate the EF in the p- and n-type materials far from the junction Eo(the free level), Ec, Efi and Ev must be continuous N.B.: qc = 4.05 eV (Si), and qf = qc + Ec - EF Eo qc (electron affinity) qf (work function) Ec Ef Efi qfF Ev ©rlc L09-14Feb2011

Band diagram for p+-n jctn* at Va = 0 Ec qVbi = q(fn - fp) qfp < 0 Efi Ec EfP EfN Ev Efi qfn > 0 *Na > Nd -> |fp| > fn Ev p-type for x<0 n-type for x>0 x -xpc -xp xn xnc ©rlc L09-14Feb2011

Band diagram for p+-n at Va=0 (cont.) A total band bending of qVbi = q(fn-fp) = kT ln(NdNa/ni2) is necessary to set EfP = EfN For -xp < x < 0, Efi - EfP < -qfp, = |qfp| so p < Na = po, (depleted of maj. carr.) For 0 < x < xn, EfN - Efi < qfn, so n < Nd = no, (depleted of maj. carr.) -xp < x < xn is the Depletion Region ©rlc L09-14Feb2011

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 L09-14Feb2011

Poisson’s Equation The electric field at (x,y,z) is related to the charge density r=q(Nd-Na-p-n) by the Poisson Equation: ©rlc L09-14Feb2011

Poisson’s Equation For n-type material, N = (Nd - Na) > 0, no = N, and (Nd-Na+p-n)=-dn +dp +ni2/N For p-type material, N = (Nd - Na) < 0, po = -N, and (Nd-Na+p-n) = dp-dn-ni2/N So neglecting ni2/N, [r=(Nd-Na+p-n)] ©rlc L09-14Feb2011

Quasi-Fermi Energy ©rlc L09-14Feb2011

Quasi-Fermi Energy (cont.) ©rlc L09-14Feb2011

Quasi-Fermi Energy (cont.) ©rlc L09-14Feb2011

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 h 0 ©rlc L09-14Feb2011

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 L09-14Feb2011

Depletion approx. charge distribution +Qn’=qNdxn +qNd [Coul/cm2] -xp x -xpc xn xnc -qNa Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn Qp’=-qNaxp [Coul/cm2] ©rlc L09-14Feb2011

1-dim soln. of Gauss’ law Ex -xpc -xp xn xnc x -Emax ©rlc L09-14Feb2011

Depletion Approxi- mation (Summary) For the step junction defined by doping Na (p-type) for x < 0 and Nd, (n-type) for x > 0, the depletion width W = {2e(Vbi-Va)/qNeff}1/2, where Vbi = Vt ln{NaNd/ni2}, and Neff=NaNd/(Na+Nd). Since Naxp=Ndxn, xn = W/(1 + Nd/Na), and xp = W/(1 + Na/Nd). ©rlc L09-14Feb2011

One-sided p+n or n+p jctns If p+n, then Na >> Nd, and NaNd/(Na + Nd) = Neff --> Nd, and W --> xn, DR is all on lightly d. side If n+p, then Nd >> Na, and NaNd/(Na + Nd) = Neff --> Na, and W --> xp, DR is all on lightly d. side The net effect is that Neff --> N-, (- = lightly doped side) and W --> x- ©rlc L09-14Feb2011

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 L09-14Feb2011

Junction C (cont.) The C-V relationship simplifies to ©rlc L09-14Feb2011

Junction C (cont.) If one plots [C’j]-2 vs. Va Slope = -[(C’j0)2Vbi]-1 vertical axis intercept = [C’j0]-2 horizontal axis intercept = Vbi C’j-2 C’j0-2 Va Vbi ©rlc L09-14Feb2011

Arbitrary doping profile If the net donor conc, N = N(x), then at xn, the extra charge put into the DR when Va->Va+dVa is dQ’=-qN(xn)dxn The increase in field, dEx =-(qN/e)dxn, by Gauss’ Law (at xn, but also const). So dVa=-(xn+xp)dEx= (W/e) dQ’ Further, since N(xn)dxn = N(xp)dxp gives, the dC/dxn as ... ©rlc L09-14Feb2011

Arbitrary doping profile (cont.) ©rlc L09-14Feb2011

Arbitrary doping profile (cont.) ©rlc L09-14Feb2011

Arbitrary doping profile (cont.) ©rlc L09-14Feb2011

Arbitrary doping profile (cont.) ©rlc L09-14Feb2011

n x xn Nd Debye length The DA assumes n changes from Nd to 0 discontinuously at xn, likewise, p changes from Na to 0 discontinuously at -xp. In the region of xn, the 1-dim Poisson equation is dEx/dx = q(Nd - n), and since Ex = -df/dx, the potential is the solution to -d2f/dx2 = q(Nd - n)/e ©rlc L09-14Feb2011

Debye length (cont) Since the level EFi is a reference for equil, we set f = Vt ln(n/ni) In the region of xn, n = ni exp(f/Vt), so d2f/dx2 = -q(Nd - ni ef/Vt), let f = fo + f’, where fo = Vt ln(Nd/ni) so Nd - ni ef/Vt = Nd[1 - ef/Vt-fo/Vt], for f - fo = f’ << fo, the DE becomes d2f’/dx2 = (q2Nd/ekT)f’, f’ << fo ©rlc L09-14Feb2011

Debye length (cont) So f’ = f’(xn) exp[+(x-xn)/LD]+con. and n = Nd ef’/Vt, x ~ xn, where LD is the “Debye length” ©rlc L09-14Feb2011

13% < d < 28% => DA is OK Debye length (cont) LD estimates the transition length of a step-junction DR (concentrations Na and Nd with Neff = NaNd/(Na +Nd)). Thus, For Va=0, & 1E13 < Na,Nd < 1E19 cm-3 13% < d < 28% => DA is OK ©rlc L09-14Feb2011

Example An assymetrical p+ n junction has a lightly doped concentration of 1E16 and with p+ = 1E18. What is W(V=0)? Vbi=0.816 V, Neff=9.9E15, W=0.33mm What is C’j? = 31.9 nFd/cm2 What is LD? = 0.04 mm ©rlc L09-14Feb2011

References *Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989. **Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago. M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003. 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. 2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981. 3 Physics of Semiconductor Devices, Shur, Prentice-Hall, 1990. ©rlc L09-14Feb2011