Semiconductor Device Modeling and Characterization EE5342, Lecture 15 -Sp 2002 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ L15 05Mar02 1
Charge components in the BJT From Getreau, Modeling the Bipolar Transistor, Tektronix, Inc. L15 05Mar02 2
Gummel-Poon Static npn Circuit Model C R C B R BB I LC B I LE I BF I BR I CC - I EC = IS(exp(v BE /NFV t - exp(v BC /NRV t )/Q B R E E L15 05Mar02 3
Gummel-Poon Static npn Circuit Model C R C B R BB I LC B I LE I BF I BR I CC - I EC = IS(exp(v BE /NFV t - exp(v BC /NRV t )/Q B R E E L15 05Mar02 4
Gummel-Poon Static Par. NAME PARAMETER UNIT DEFAULT IS transport saturation current A 1.0e-16 BF ideal maximum forward beta - 100 NF forward current emission coef. - 1.0 VAF forward Early voltage V infinite ISE B-E leakage saturation current A 0 NE B-E leakage emission coefficient- 1.5 BR ideal maximum reverse beta - 1 NR reverse current emission coefficient - 1 VAR reverse Early voltage V infinite ISC B-C leakage saturation current A 0 NC B-C leakage emission coefficient- 2 EG energy gap (IS dep on T) ev 1.11 XTI temperature exponent for IS - 3 L15 05Mar02 5
Gummel-Poon Static Model Parameters NAME PARAMETER UNIT DEFAULT IKF corner for forward beta A infinite high current roll-off IKR corner for reverse beta A infinite high current roll-off RB zero bias base resistance W 0 IRB current where base resistance A infinite falls halfway to its min value RBM minimum base resistance W RB at high currents RE emitter resistance W 0 RC collector resistance W 0 TNOM parameter - meas. temperature C 27 L15 05Mar02 6
Gummel Poon npn Model Equations I BF = IS expf(v BE /NFV t )/BF I LE = ISE expf(v BE /NEV t ) I BR = IS expf(v BC /NRV t )/BR I LC = ISC expf(v BC /NCV t ) Q B = (1 + v BC /VAF + v BE /VAR ) { + + (BF IBF/IKF + BR IBR/IKR) } L15 05Mar02 7
Gummel Poon Base Resistance If IRB = 0, R BB = R BM +(R B -R BM )/Q B If IRB > 0 R B = R BM + 3(R B -R BM ) (tan(z)-z)/(ztan 2 (z)) z = [ + i B /( IRB)] 1/2 - ( / )(i B /IRB) 1/2 Regarding (i) R BB and (x) R Th on slide 22, R B = R BM + R/(1+i B /I RB ) RB, R = R B - R BM L15 05Mar02 8
BJT Characterization Forward Gummel v BCx = 0 = v BC + i B R B - i C R C i C R C v BEx = v BE +i B R B +(i B +i C )R E i B = I BF + I LE = IS expf(v BE /NFV t )/BF + ISE expf(v BE /NEV t ) v BEx + i B R B v BC + + v BE - - i C = F I BF /Q B = IS expf(v BE /NFV t )/Q B - R E L15 05Mar02 9
1.E-02 1.E-03 1.E-04 1.E-05 1.E-06 1.E-07 1.E-08 1.E-09 1.E-10 1.E-11 1.E-12 1.E-13 1.E-14 1.E-15 1.E-16 Ideal F-G Data i C and i B (A) vs. v BE (V) N = 1 1/slope = 59.5 mv/dec N = 2 1/slope = 119 mv/dec L15 05Mar02 0.0 0.2 0.4 0.6 0.8 10 Ic Ib
BJT Characterization Reverse Gummel v BEx = 0 = v BE + i B R B - i E R E v BCx = v BC +i B R B +(i B +i E )R C i B = I BR + I LC = IS expf(v BC /NRV t )/BR + ISC expf(v BC /NCV t ) i E = R I BR /Q B = - + v BCx i B R B i E v BC + + v BE R C - - R E IS expf(v BC /NRV t )/Q B L15 05Mar02 11
1.E-02 1.E-03 1.E-04 1.E-05 1.E-06 1.E-07 1.E-08 1.E-09 1.E-10 1.E-11 1.E-12 1.E-13 1.E-14 1.E-15 1.E-16 Ideal R-G Data i E and i B (A) vs. v BE (V) N = 1 1/slope = 59.5 mv/dec N = 2 1/slope = 119 mv/dec L15 05Mar02 0.0 0.2 0.4 0.6 0.8 12 Ic Ie Ib
Distributed resistance in a planar BJT coll. base & emitter contact regions reg 1 reg 2 reg 3 reg 4 emitter base collector The base current must flow lateral to the wafer surface Assume E & C currents perpendicular Each region of the base adds a term of lateral res. v BE diminishes as current flows L15 05Mar02 13
Simulation of 2- dim. current flow Q VCC V ib1 = ib R R Q1 Q2 R Qn Distributed device is repr. by Q 1, Q 2, Q n Area of Q is same as the total area of the distributed device. Both devices have the same v CE = VCC Both sources have same current i B1 = i B. The effective value of the 2-dim. base resistance is R bb (i B ) = V/i B = R BBTh L15 05Mar02 14
Analytical solution for distributed Rbb dib x dx vbe vbe JS L exp JSE L exp NFVt NEVt dvbe x rbi ib x dx L Analytical solution and SPICE simulation both fit R BB = R bmin + R bmax /(1 + i B /I RB ) RB L15 05Mar02 15
Distributed base resistance function Normalized base resistance vs. current. (i) R BB /R Bmax, (ii) R BBSPICE /R Bmax, after fitting R BB and R BBTh = R BM + R/(1+i B /I RB ) RB ( R = R B - R BM ) R BBSPICE to R BBTh (x) R BBTh /R Bmax. FromAn Accurate Mathematical Model for the Intrinsic Base Resistance of Bipolar Transistors, by Ciubotaru and Carter, Sol.- St.Electr. 41, pp. 655-658, 1997. L15 05Mar02 16
Gummel Poon Base Resistance If IRB = 0, R BB = R BM +(R B -R BM )/Q B If IRB > 0 R B = R BM + 3(R B -R BM ) (tan(z)-z)/(ztan 2 (z)) z = [ + i B /( IRB)] 1/2 - ( / )(i B /IRB) 1/2 Regarding (i) R BB and (x) R Th on previous slide, R BB = R bmin + R bmax /(1 + i B /I RB ) RB L15 05Mar02 17
Gummel-Poon Static npn Circuit Model C R C B R BB I LC B I LE I BF I BR I CC - I EC = IS(exp(v BE /NFV t - exp(v BC /NRV t )/Q B R E E L15 05Mar02 18
Gummel Poon npn Model Equations I BF = IS expf(v BE /NFV t )/BF I LE = ISE expf(v BE /NEV t ) I BR = IS expf(v BC /NRV t )/BR I LC = ISC expf(v BC /NCV t ) I CC - I EC = IS(exp(v BE /NFV t - exp(v BC /NRV t )/Q B Q B = { + + (BF IBF/IKF + BR IBR/IKR) 1/2 } (1 - v BC /VAF - v BE /VAR ) -1 L15 05Mar02 19
VAR Parameter Extraction (rearly) i E = - I EC = (IS/Q B )exp(v BC /NRV t ), where I CC = 0, and Q -1 B = (1-v BC /VAF-v BE /VAR ) {IKR terms } -1, so since v BE = v BC - v EC, VAR = i E /[ i E / v BE ] vbc Reverse Active Operation L15 05Mar02 20 i B + - v BC i E v EC 0.2 < v EC < 5.0 0.7 < v BC < 0.9 + -
Reverse Early Data for VAR At a particular data point, an effective VAR value can be calculated 0.0006 0.0004 0.0002 v BC = 0.85 V v BC = 0.75 V VAR eff = i E /[ i E / v BE ] vbc The most accurate is at v BE = 0 (why?) 0.0000 0 1 2 3 4 5 i E (A) vs. v EC (V) L15 05Mar02 21
VAF Parameter Extraction (fearly) i C = I CC = (IS/Q B )exp(v BE /NFV t ), where I CE = 0, and Q -1 B = (1-v BC /VAF-v BE /VAR ) {IKF terms } -1, so since v BC = v BE - v CE, VAF = i C /[ i C / v BC ] vbe Forward Active Operation L15 05Mar02 22 + - i B v BE i C v CE 0.2 < v CE < 5.0 0.7 < v BE < 0.9 + -
Forward Early Data for VAF At a particular data point, an effective VAF value can be calculated VAF eff = i C /[ i C / v BC ] vbe 0.003 0.002 0.001 v BE = 0.85 V v BE = 0.75 V The most accurate is at v BC = 0 (why?) 0.000 0 1 2 3 4 5 i C (A) vs. v CE (V) L15 05Mar02 23
BJT Characterization Forward Gummel v BCx = 0 = v BC + i B R B - i C R C v BEx = v BE +i B R B +(i B +i C )R E i B = I BF + I LE = IS exp(v BE /NFV t )/BF + ISE expf(v BE /NEV t ) i C = F I BF /Q B = IS exp(v BE /NFV t ) (1-v BC /VAF-v BE /VAR ) {IKF terms } -1 v BEx L15 05Mar02 24 + - i B i C R B v BC + + v BE R C - - R E
Forward Gummel Data Sensitivities 1.E-02 1.E-04 1.E-06 1.E-08 1.E-10 1.E-12 v BCx = 0 e b i C i B d 0.1 0.3 0.5 0.7 0.9 i C (A),i B (A) vs. v BE (V) a c 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 L15 05Mar02 25
Region (a) fg Data Sensitivities Region a - IKF IS, RB, RE, NF, VAR i C = F I BF /Q B = IS exp(v BE /NFV t ) (1-v BC /VAF-v BE /VAR ) {IKF terms } -1 L15 05Mar02 26
Region (b) fg Data Sensitivities Region b - IS, NF, VAR, RB, RE i C = F I BF /Q B = IS exp(v BE /NFV t ) (1-v BC /VAF-v BE /VAR ) {IKF terms } -1 L15 05Mar02 27
Region (c) fg Data Sensitivities Region c - IS/BF, NF, RB, RE i B = I BF + I LE = (IS/BF) expf(v BE /NFV t ) + ISE expf(v BE /NEV t ) L15 05Mar02 28
Region (d) fg Data Sensitivities Region d - IS/BF, NF i B = I BF + I LE = (IS/BF) expf(v BE /NFV t ) + ISE expf(v BE /NEV t ) L15 05Mar02 29
Region (e) fg Data Sensitivities Region e - ISE, NE i B = I BF + I LE = (IS/BF) expf(v BE /NFV t ) + ISE expf(v BE /NEV t ) L15 05Mar02 30