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EE 5340 Semiconductor Device Theory Lecture 29 - Fall 2003

EE 5340 Semiconductor Device Theory Lecture 29 - Fall 2003. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc. Notes in Progress. This is a draft of L29. A final copy will appear soon Bring a copy of -http://www.uta.edu/ronc/5340/project/40ProjectSolution03.pdf to class.

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EE 5340 Semiconductor Device Theory Lecture 29 - Fall 2003

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  1. EE 5340Semiconductor Device TheoryLecture 29 - Fall 2003 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

  2. Notes in Progress • This is a draft of L29. A final copy will appear soon • Bring a copy of-http://www.uta.edu/ronc/5340/project/40ProjectSolution03.pdfto class. • Final copy of L29 will include “Problems of interest” from Chs. 8, 9, and 10 in Mueller and Kamins, 3rd edition.

  3. Part A Data First, the range for fitting the data was chosen as there was considerable “scat-ter” in the data. This was noted by plotting C^-1/M vs. Va for M = ½ and 1/3. This is shown in Figure A1.1. Figure A1.1. C^-1/M vs. Va for M = ½ and 1/3 for the data given

  4. Part A Data re-plotted The data “scatter” seemed to be most pronounced for the small values of C corresponding to the most negative values of V. The same data was plotted in the range -1 V < V < 0.4 V for the same M values. This is shown in Figure A1.2. Figure A1.2. C^-1/M vs. Va for M = ½ and 1/3 for the data given in Table 1 over the range-1 V < V < 0.4 V.

  5. Determining Values The M value of 1/3 seems to fit the data better, as demonstrated by the R^2 value of 0.9995. The value of M, etc., will be determined using the approach outlined in Lectures 11 and 14. In order to apply the theory developed in Lectures 11 and 14, it was decided to use the range-1 V ≤ V ≤ 0.4 V to calculate the central and forward derivatives. The range should be chosen by determining the minimum (1 - R2) value (as in Figure A1.2). One could try to use 5 points (rather than the 6 chosen here) or 4 (using the highest voltages only), omit the highest and use the next 4 highest, and so on.

  6. Estimating Junction Capacitance Parameters • Notation different than L14 • If CJ = CJO {1 – Va/VJ}-M • Define y  {d[ln(CJ)]/dV}-1 • A plot of y = yi vs. Va = vi has slope = -1/M, and intercept = VJ/M

  7. Derivatives Defined The central derivative is defined as (following Lecture 14 and 11) yi,Central = (vi+1 – vi-1)/(lnCi+1 – lnCi-1), with vi = (vi+1 + vi-1)/2 Equation A1.1 The Forward derivative (as applied to the theory in L11 and L14) is defined in this case as yi,Forward = (vi+1 – vi)/(lnCi+1 – lnCi), with vi,eff = (vi+1 + vi-1)/2 Equation A1.2

  8. Data calculations Table A1.1. Calculations of yi and vi for the Central and Forward derivatives for the data in Table 1. The yi and vi are defined in Equations A1.1 and A1.2.

  9. y vs. Va plots Figure A1.3. The yi and vi values from the theory in L11 and L14 with associa-ted trend lines and slope, intercept and R^2 values.

  10. It is clear the Central derivative gives the more reliable data as the R^2 value is larger. M is the reciprocal of the magnitude of the slope obtained by a least squares fit (linear) plot of yi vs. Vi VJ is the horizontal axis intercept (computed as the vertical axis intercept divided by the slope) Cj0 is the vertical axis intercept of a least squares fit of Cj-1/M vs. V (must use the value of V for which the Cj was computed). The computations will be shown later. The results of plotting Cj-1/M vs. V for the M value quoted below are shown in Figure A1.4 Comments on thedata interpretation

  11. M = 1/2.551 = 0.392 (the data were generated using M = 0.389, thus we have a 0.77% error). VJ = yi(vi=0)/slope =1.6326/2.551 = 0.640 (the data were generated using fi = 0.648, thus we have a 1.24% error). Cj0 = 1.539E30^-.392 = 1.467 pF (the data were generated using Cj0 = 1.68 pF, thus we have a 12.6% error) Calculating theparameters

  12. Linearized C-V plot Figure A1.4. A plot of the data for Cj^-1/M vs. Va using the M value determined for this data (M = 0.392).

  13. Doping Profile The data were equal-ly spaced (DV=0.1V), the central differ-ence was used, for -7.4V ≤ V ≤ 0.4V, which for Cj = e/x, corresponds to a range of 2.81E-5 cm ≤ x ≤ 8.99E-5 cm. These data are shown. The trend line is also shown for a linear fit. Since R^2 = 1.000, a linear N(x) relationship can be assumed.

  14. Parameter values • B2. Using the same procedure, as for part A (central derivatives), one finds M = 0.3448. The data were generated using M = 0.345, thus we have less than 0.1% error. • B3. Since we found a linear fit in B1, and M ~ 0.333 (the value for linear doping), the two observations are in agreement that the doping profile is linear.

  15. Part C Use the depletion approximation for a spherical diode to find the breakdown voltage as a function N- and junction radius, rj. Place the heavily doped side of the junction (Na = N+) in the range r+c < r < rj and the lightly doped side of the junction (Nd = N-) in the range rj < r < r-c (r+c and r-c are the respective contacts). Use the Gauss Law form of the Poisson Equation to solve the E(r) relationship and then solve for Emax = E(rj) as a function of Va = V(r+) - V(r-). In this (one-dimensional) case, the Gauss Law form of the Poisson Equation is

  16. Reverse biasjunction breakdown

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

  18. rp rpc rj rn rnc Gauss’ Law

  19. Spherical DiodeFields 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

  20. Spherical DiodeFields 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].

  21. Showing therj  ∞ 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. .

  22. Solving for theBreakdown (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 .

  23. Spherical diodeBreakdown Voltage

  24. Final Exam • For BOTH sections, 051 and 001 • 8:00 to 10:30 AM • Tuesday, December 9 in • 206 ACT • Cover sheet on web page at http://www.uta.edu/ronc/5340/tests/ • The Final is comprehensive • 20% to 25% on Test 1 material • 20% to 25% on Test 2 material • Balance of final on material since Test 2

  25. Emphasis • Chapters 1 through 7 • Lecture notes • Problems assigned • Previous tests • Chapters 8 through 10 • As above • Some key problems in 9 and 10 are P9:1,3,5,7 and P10:8

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