1 / 29

Semiconductor Device Modeling and Characterization EE5342, Lecture 19 Spring 2003

Semiconductor Device Modeling and Characterization EE5342, Lecture 19 Spring 2003. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/. The base current must flow lateral to the wafer surface Assume E & C cur-rents perpendicular.

darin
Download Presentation

Semiconductor Device Modeling and Characterization EE5342, Lecture 19 Spring 2003

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Semiconductor Device Modeling and CharacterizationEE5342, Lecture 19Spring 2003 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

  2. The base current must flow lateral to the wafer surface Assume E & C cur-rents perpendicular Each region of the base adds a term of lateral res. vBE diminishes as current flows coll. base & emitter contact regions reg 1 reg 2 reg 3 reg 4 emitter base collector Distributed resis-tance in a planar BJT

  3. Distributed device is repr. by Q1, Q2, … Qn Area of Q is same as the total area of the distributed device. Both devices have the same vCE = VCC Both sources have same current iB1 = iB. The effective value of the 2-dim. base resistance is Rbb’(iB) = DV/iB = RBBTh  DV  = Simulation of 2-dim. current flow

  4. Analytical solutionfor distributed Rbb • Analytical solution and SPICE simulation both fit RBB = Rbmin + Rbmax/(1 + iB/IRB)aRB

  5. Distributed baseresistance function Normalized base resis-tance vs. current. (i) RBB/RBmax, (ii) RBBSPICE/RBmax, after fitting RBB and RBBSPICE to RBBTh (x) RBBTh/RBmax. FromAn Accurate Mathematical Model for the Intrinsic Base Resistance of Bipolar Transistors, by Ciubotaru and Carter, Sol.-St.Electr. 41, pp. 655-658, 1997. RBBTh = RBM + DR/(1+iB/IRB)aRB (DR = RB - RBM )

  6. Gummel PoonBase Resistance If IRB = 0, RBB = RBM+(RB-RBM)/QB If IRB > 0 RB = RBM + 3(RB-RBM)(tan(z)-z)/(ztan2(z)) [1+144iB/(p2IRB)]1/2-1 z = (24/p2)(iB/IRB)1/2 Regarding (i) RBB and (x) RTh on previous slide, RBB = Rbmin + Rbmax/(1 + iB/IRB)aRB

  7. Gummel-Poon Staticnpn Circuit Model Intrinsic Transistor C RC IBR B RBB ILC ICC -IEC = {IS/QB}* {exp(vBE/NFVt)-exp(vBC/NRVt)} IBF B’ ILE RE E

  8. IBF = IS expf(vBE/NFVt)/BF ILE = ISE expf(vBE/NEVt) IBR = IS expf(vBC/NRVt)/BR ILC = ISC expf(vBC/NCVt) ICC -IEC = IS(exp(vBE/NFVt - exp(vBC/NRVt)/QB QB= {+ [+ (BFIBF/IKF + BRIBR/IKR)]1/2}  (1 - vBC/VAF - vBE/VAR )-1 Gummel Poon npnModel Equations

  9. Reverse Active Operation iE iB vEC vBC 0.2 < vEC < 5.0 0.7 < vBC < 0.9 VAR ParameterExtraction (rEarly) iE = -IEC= (IS/QB)exp(vBC/NRVt), where ICC= 0, and QB-1= (1-vBC/VAF-vBE/VAR ) {IKR terms}-1, so since vBE = vBC - vEC, VAR ~ iE/[iE/vBE]vBC

  10. Reverse EarlyData for VAR • At a particular data point, an effective VAR value can be calculated VAReff = iE/[iE/vBE]vBC • The most accurate is at vBE = 0 (why?) vBC = 0.85 V vBC = 0.75 V iE(A) vs. vEC (V)

  11. Reverse EarlyVAR extraction VAReff = iE/[iE/vBE]vBC • VAR was set at 200V for this data • When vBE = 0 vBC = 0.75VAR=200.5 vBC = 0.85VAR=200.2 vBC = 0.75 V vBC = 0.85 V VAReff(V) vs. vEC (V)

  12. VAF ParameterExtraction (fEarly) Forward Active Operation iC = ICC= (IS/QB)exp(vBE/NFVt), where ICE= 0, and QB-1= (1-vBC/VAF-vBE/VAR )* {IKF terms}-1, so since vBC = vBE - vCE, VAF ~ iC/[iC/vBC]vBE iC iB vCE vBE 0.2 < vCE < 5.0 0.7 < vBE < 0.9

  13. Forward EarlyData for VAF • At a particular data point, an effective VAF value can be calculated VAFeff = iC/[iC/vBC]vBE • The most accurate is at vBC = 0 (why?) vBE = 0.85 V vBE = 0.75 V iC(A) vs. vCE (V)

  14. Forward EarlyVAf extraction VAFeff = iC/[iC/vBC]vBE • VAF was set at 100V for this data • When vBC = 0 vBE = 0.75VAF=101.2 vBE = 0.85VAF=101.0 vBE = 0.75 V vBE = 0.85 V VAFeff(V) vs. vCE (V)

  15. iC RC vBC - iB + + RB vBE - vBEx RE BJT CharacterizationForward Gummel vBCx= 0 = vBC+ iBRB- iCRC vBEx = vBE+iBRB+(iB+iC)RE iB = IBF + ILE = ISexp(vBE/NFVt)/BF + ISEexpf(vBE/NEVt) iC = bFIBF/QB = ISexp(vBE/NFVt)  (1-vBC/VAF-vBE/VAR ) {IKF terms}-1

  16. Sample fg data forparameter extraction • IS = 10f • NF = 1 • BF = 100 • Ise = 10E-14 • Ne = 2 • Ikf = .1m • Var = 200 • Re = 1 • Rb = 100 iC data iB data iC, iB vs. vBEext

  17. Definitions ofNeff and ISeff • In a region where iC or iB is approxi-mately a single exponential term, then iC or iB ~ ISeffexp (vBEext /(NFeffVt) where Neff={dvBEext/d[ln(i)]}/Vt, and ISeff = exp[ln(i) - vBEext/(NeffVt)]

  18. Forward GummelData Sensitivities a Region a - IKFIS, RB, RE, NF, VAR Region b - IS, NF, VAR, RB, RE Region c - IS/BF, NF, RB, RE Region d - IS/BF, NF Region e - ISE, NE vBCx = 0 c iC b d iB e iC(A),iB(A) vs. vBE(V)

  19. Region (b) fgData Sensitivities Region b - IS, NF, VAR, RB, RE iC = bFIBF/QB = ISexp(vBE/NFVt)  (1-vBC/VAF-vBE/VAR ){IKF terms}-1

  20. Region (e) fgData Sensitivities Region e - ISE, NE iB = IBF + ILE = (IS/BF)expf(vBE/NFVt) + ISEexpf(vBE/NEVt)

  21. Simple extractionof IS, ISE from data Data set used • IS = 10f • ISE = 10E-14 Flat ISeff for iC data = 9.99E-15 for 0.230 < vD < 0.255 Max ISeff value for iB data is 8.94E-14 for vD = 0.180 iC data iB data ISeff vs. vBEext

  22. Simple extraction of NF, NE from fg data Data set used NF=1 NE=2 Flat Neff region from iC data = 1.00 for 0.195 < vD < 0.390 Max Neff value from iB data is 1.881 for 0.180 < vD < 0.181 iB data iC data NEeff vs. vBEext

  23. Region (d) fgData Sensitivities Region d - IS/BF, NF iB = IBF + ILE = (IS/BF)expf(vBE/NFVt) + ISEexpf(vBE/NEVt)

  24. Simple extractionof BF from data • Data set used BF = 100 • Extraction gives max iC/iB = 92 for 0.50 V < vD < 0.51 V 2.42A< iD < 3.53A • Minimum value of Neff =1 for slightly lower vD and iD iC/iB vs. iC

  25. Region (a) fgData Sensitivities Region a - IKFIS, RB, RE, NF, VAR iC = bFIBF/QB = ISexp(vBE/NFVt)  (1-vBC/VAF-vBE/VAR ){IKF terms}-1 If iC > IKF, then iC ~ [IS*IKF]1/2exp(vBE/2NFVt)  (1-vBC/VAF-vBE/VAR )

  26. Region (c) fgData Sensitivities Region c - IS/BF, NF, RB, RE iB = IBF + ILE = (IS/BF)expf(vBE/NFVt) + ISEexpf(vBE/NEVt)

  27. RC vBCx vBC - iB + + RB vBE - RE iE BJT CharacterizationReverse Gummel vBEx= 0 = vBE+ iBRB- iERE vBCx = vBC+iBRB+(iB+iE)RC iB = IBR + ILC = (IS/BR)expf(vBC/NRVt) + ISCexpf(vBC/NCVt) iE = bRIBR/QB = ISexpf(vBC/NRVt) (1-vBC/VAF-vBE/VAR ) {IKR terms}-1

  28. Sample rg data forparameter extraction • IS=10f • Nr=1 • Br=2 • Isc=10p • Nc=2 • Ikr=.1m • Vaf=100 • Rc=5 • Rb=100 iB data iE data iE, iB vs. vBCext

  29. Reverse GummelData Sensitivities Region a - IKRIS, RB, RC, NR, VAF Region b - IS, NR, VAF, RB, RC Region c - IS/BR, NR, RB, RC Region d - IS/BR, NR Region e - ISC, NC c vBCx = 0 a d e b iB iE iE(A),iB(A) vs. vBC(V)

More Related