Semiconductor Device Modeling and Characterization EE5342, Lecture 6-Spring 2002 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ L06 31Jan02 1 General Instructions: • All projects should be submitted on 8.5" x 11" paper with a cover sheet attached, or electronically as a single document file which will print as such. If submitted as a paper project report, it should be stapled only in the upper left-hand corner and no other cover or binder or folder should be used. L06 31Jan02 2 Format • The cover sheet – your name, – the project title, – the course name and number, and – your e-mail address. • The report includes – purpose of the project and the theoretical background, L06 31Jan02 – a narrative explaining how you did the project, – answers to all questions asked in the project assignment, and a – list of references used in the order cited in the report (the reference number should appear in the report each time the reference is used). 3 All figures and tables should be clearly marked with a figure or table number and caption. The caption and labels on the figures should make the information in the figure comprehensible without reading further in the text of the report. Circuits used should be shown in the text. Auxiliary information (such as SPICE data outputs, etc.) should be included in appropriate Appendices at the end of the report. Be sure to describe exactly how all results were obtained, giving enough information for anyone who understands EE 5342 to repeat your work. All work submitted must be original. If derived from another source, a full bibliographical citation must be given. (See all of Notes 5 and 6 in the syllabus.) L06 31Jan02 4 • The temperature dependence of the mobility of carriers in silicon (the Arora model - see Arora, Hauser and Roulston, Electron and Hole Mobilities in Silicon as a Function of Concentration and Temperature, IEEE Trans. Electron Devices, ED29, p. 292, ff., 1982) is quoted by Casey (Devices for Integrated Circuits : Silicon and III-V Compound Semiconductors, by H. Craig Casey, John Wiley, New York, 1999, p. 75) and also quoted by Muller and Kamins (Device Electronics for Integrated Circuits, 2nd ed., by Richard S. Muller and Theodore I. Kamins, John Wiley and Sons, New York, 1986, p. 35). L06 31Jan02 5 Question 1: • Careful examination of the form of mn(N,T) and mp(N,T) (N = doping concentration, T = temperature) will reveal that Casey and Muller and Kamins do not agree. Resolve the differences and determine the correct equation for the model. This model will be referred to as mnAHR(N,T) and mpAHR(N,T). L06 31Jan02 6 Question 2: • Determine the values of the model [mnAHR(N,T) and mpAHR(N,T)] for the 3x3 matrix of values of T= 0, 30, and 60C and N=1E15, 3E16, and 1E18 cm-3. Show your results in table format, i.e., one table will be values of mnAHR(N,T) for all nine conditions described in the 3x3 matrix of N,t values, and a similar table will be developed for the mpAHR(N,T) values. L06 31Jan02 7 • Another model is discussed by Mohammad, Bemis, Carter and Renbeck (Temperature, Electric field and Doping Dependent Mobilities of Electrons and Holes in Semiconductors”, Solid-State Electronics, Vol. 36, No. 12, PP. 167783, 1993.) This model will be referred to as mnMBCR(T,E,N) and mpMBCR(T,E,N) L06 31Jan02 8 Question 3: • Determine the same tables defined in Question 2 for the models mnMBCR(T,E,N), mpMBCR(T,E,N)] for the case where E = 0. L06 31Jan02 9 Question 4: • Determine the tables of values for the conditions defined in Question 2 for the relative differences between the models when E = 0. Use the following definitions for the relative differences: rdn |mnMBCR(T,E,N) - mnAHR(N,T)| mnAHR(N,T) and rdp |mpMBCR(T,E,N) - mpAHR(N,T)|/ L06 31Jan02 10 mpAHR(N,T) Questions 5 and 6 • 5: Comment on the results of Question 4. What possible reasons can you give for the differences between the two models? • 6: Comment on the application of a mn(T,N) and a mp(T,N) model to determine a R(T,N) model for an integrated circuit resistor. For one thing, what additional modeling issues would need to be considered? L06 31Jan02 11 Energy bands for p- and n-type s/c p-type n-type Ec EFi EFp Ev qfn= kT ln(Nd/ni) qfp= kT ln(ni/Na) L06 31Jan02 Ec EFn EFi Ev 12 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 L06 31Jan02 qc (electron Eo affinity) qf (work function) Ef qfF Ec Efi Ev 13 Band diagram for p+-n jctn* at Va = 0 Ec qf p < 0 Efi EfP Ev *Na > Nd -> |fp| > fn p-type for x<0 -xpc L06 31Jan02 -xp 0 qVbi = q(fn - fp) EfN Efi qfn > 0 n-type for x>0 xn Ec Ev xnc x 14 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.) L06-x 31Jan02< x < x 15 p n is the Depletion Region 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 L06 31Jan02 16 Depletion approx. charge distribution r +Qn’=qNdxn [Coul/cm2] +qNd -xp -xpc xn -qNa Qp’=-qNaxp [Coul/cm2] L06 31Jan02 xnc x Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn 17 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 x xn E dS rdV A Exdx A Qn' + Qp' x xp S V L06 31Jan02 18 Induced E-field in the D.R. Ex p-contact p-type CNR - O + O n-type chg + O O neutral reg + - O O Depletion Exposed Acceptor Ions region (DR) W -xpc L06 31Jan02 N-contact -xp 0 xn Exposed Donor ions xnc x 19 Review of depletion approximation qVbi EFp Ec EFn EFi Ev -xpc -xp 0 xn L06 31Jan02 xnc x • • • • • Depletion Approx. pp << ppo, -xp < x < 0 nn << nno, 0 < x < xn 0 > Ex > -2Vbi/W, in DR (-xp < x < xn) pp=ppo=Na & np=npo= ni2/Na, -xpc< x < -xp nn=nno=Nd & pn=pno= ni2/Nd, xn < x < xnc 20 Review of D. A. (cont.) -xpc-xp Ex xn xnc 2Vbi Va W , W xp + xn , qNeff x Neff NaNd , Na xp Ndxn , Na + Nd Ex 0, x xp q Ex - Na x + xp , xp x 0, q Ex Na x xn , 0 x xn , Ex 0, x xn -Emax L06 31Jan02 21 Depletion Approximation (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 = {2(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). L06 31Jan02 22 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 --> xL06 31Jan02 23 Debye length Nd 0 n xn x • 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, Poisson’s eq is =*E = r/ --> d Ex/dx = q(Nd n), and since Ex = -df/dx, we have -d2f/dx2 = q(Nd - n)/ to be solved L06 31Jan02 24 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/kT)f’, f’ << fo L06 31Jan02 25 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” Vt kT LD , Vt , a transition length. qn + p q Note : n + p Nd for n - type, Na for p - type and 2ni for intrinsic material. L06 31Jan02 26 Debye length (cont) • LD estimates the transition length of a step-junction DR (concentrations Na and Nd with Neff = NaNd/(Na +Nd)). Thus, LD Na + LD Nd 1 1 VtNeff + WV 0 N N 2Vbi a d a • For Va=0, & 1E13 < Na,Nd < 1E19 cm-3 13% < < 28% => DA is OK L06 31Jan02 27