# Bipolar Junction Transistors: Solving Ebers-Moll Problems

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1 C 305: Fall 016 ipolar Junction Transistors: Solving bers-moll Problems Professor Peter ermel lectrical and Computer ngineering Purdue University, West Lafayette, N USA Pierret, Semiconductor Device Fundamentals (SDF) Chapters 10 and 11 (pp , ) 11/30/016 ermel C 305 F16

2 Recap: essence of current gain N P V V C N + nput Response nput qd n W p i, qv N e 1 qd W n n i, N e qv 1 Response 11/30/016

3 C F0 bers Moll Model qdn i, qd A W F n N A W qv n i, p i, C qvc e 1 e 1 qv qvc e 1 R0 e 1 n N qd W n N C C C = c,n + c,p F R C =,n +,p α R R α F F C F R F0 R0 e e qv qv C 1 1 qdp ni, qd n n i, qd ni A A W N W N W N F 0 qv qvc e 1 R R0 e 1 qv n, qvc e 1 e 1 3

4 bers-moll model C ( V, V C ) = a F ( F0 e qv k T -1) - ( R0 e qv C k T -1) ( V,V C ) = ( F0 e qv k T -1) -a R ( R0 e qv C k T -1) ( V,V C ) = ( V,V C ) - C ( V,V C ) a F F0 = a R R0 See Pierret SDF, Chapter 11, sec /30/016 ermel C 305 F16 4

5 Q1) Which is the base transport factor? F R n n+ emitter p base Cn n collector n+ C p a) b) n ( n + ) p Cn ( n + ) p c) Cn n d) p e) Cn 11/30/016 p 5

6 Q) Which is beta_dc? F R n n+ emitter p base Cn n collector n+ C p a) b) n ( n + ) p Cn ( n + ) p c) Cn n d) p e) Cn 11/30/016 p 6

7 Q3) Which is alpha_dc? F R n n+ emitter p base Cn n collector n+ C p a) b) n ( n + ) p Cn ( n + ) p c) Cn n d) p e) Cn 11/30/016 p 7

8 Q4) Which is the base current? F R n n+ emitter p base Cn n collector n+ C p a) b) n ( n + ) p Cn ( n + ) p c) Cn n d) p e) Cn 11/30/016 p 8

9 Q5) Which is the emitter injection efficiency? F R n n+ emitter p base Cn n collector n+ C p a) b) n ( n + ) p Cn ( n + ) p c) Cn n d) p e) Cn 11/30/016 p 9

10 active region example (NPN) A = 10 mm mm C = 50 ma N D = cm -3 N A = cm -3 N DC = cm -3 W = 0.50 mm W = 0.5 mm W C = 1.50 mm t p = 0.1 ns t n = 75 ns t pc = 150 ns 11/30/016 ermel C 305 F16 10

11 Step 1: diffusion coefficients and lengths N D = cm -3 N A = cm -3 N DC = cm -3 W = 0.50 mm W = 0.5 mm W C = 1.50 mm t p = 0.1 ns t n = 75 ns t pc = 150 ns Pierret, SDF, Fig. 3.5a, p. 80 m p» 70 cm /V-s m n» 800 cm /V-s D p = k T q m p»1.8 cm /s D n» 0.8 cm /s L p = L n»1.5 mm D p t p» 0.13 mm m pc» 40 cm /V-s D pc»10.9 cm /s L pc»1.8 mm 11

12 find the emitter injection efficiency g F = 1 1+ D p D n W W N A N D g F = 1 = /30/016 ermel C 305 F16 1

13 find the base transport factor a T = æ ç è W L n ö ø a T = 1 = /30/016 ermel C 305 F16 13

14 find beta b F = g Fa T = a F 1-g F a T 1-a F b F = = = /30/016 ermel C 305 F16 14

15 find the base current C = 50 ma b F = 768 = C b F = C b F = 65 na a F = b F b F +1 = /30/016 ermel C 305 F16 15

16 Forward saturation current density C ( V,V C ) = a F ( F0 e qv k T -1) - ( R0 e qv C k T -1) ( V,V C ) = ( F0 e qv k T -1) -a R ( R0 e qv C k T -1) æ F0 = qa D n ç è W n i + D p N A W n i N D ö ø qa æ D n ç è W n i + D p N A L p n i N D ö ø F0 = A 11/30/016 ermel C 305 F16 16

17 Reverse saturation current density C ( V,V C ) = a F ( F0 e qv k T -1) - ( R0 e qv C k T -1) ( V,V C ) = ( F0 e qv k T -1) -a R ( R0 e qv C k T -1) æ R0 = qa D n ç è W n i + D pc N A W C n i N DC ö ø R0 = A 11/30/016 ermel C 305 F16 17

18 Check active region C ( V, V C ) = a F ( F0 e qv k T -1) - ( R0 e qv C k T -1) a F F0 e qv k T + R0 ( V,V C ) = ( F0 e qv k T -1) -a R ( R0 e qv C k T -1) F0 e qv k T +a R R0 ( V,V C ) = ( V,V C ) - C ( V,V C ) ( 1-a F ) F0 e qv k T - ( 1-a R ) R0 C» a F F0 e qv k T ( )» ( 1-a F ) F0 e qv k T = 1-a F C = 1 C a F b dc 11/30/016 ermel C 305 F16 18

19 What is D? = D D V C V C = V D V D V = D ermel C 305 F16 19

20 What is D? V C = D C C = a F ( F0 e qv k T -1) - ( R0 e qv C k T -1) = ( F0 e qv k T -1) -a R ( R0 e qv C k T -1) V V C = V D V C = 0 = ( F0 e qv k T -1) D = ( F0 e qv D k T -1) = D ermel C 305 F16 0

21 What is C? C C = a F ( F0 e qv k T -1) - ( R0 e qv C k T -1) = ( F0 e qv k T -1) -a R ( R0 e qv C k T -1) V C V C = 0 V ermel C 305 F16 1

22 Common ase Configuration V (in) C N+ P N C V C (out) F α R R C R α F F C C C C How would the model change if this was a Schottky barrier JT? 11/30/016 ermel C 305 F16

23 Common mitter Configuration C m V (in) P N P + C C V C (out) C C C αrr αff C F R F F F F F F F F C p R F 1 1 F F F R R F F 11/30/016 ermel C 305 F16 3

24 Current Gain C Common mitter current gain.. DC C V (in) P N P+ C qd n n i, qv / kt qd n n i, ( qvc / kt ) ( e 1) ( e 1) W N W N qd n n i, qv / kt W N ( e 1) n n i, N p i, N D W W D n P+ N P C Common ase current gain.. C DC C C DC C DC 1 DC V (in) C VC (out) 11/30/016 ermel C 305 F16 4

25 Gummel Plot and Output Characteristics C qd n n i, qv / kt qd n n i, qvc / kt = ( e 1) ( e 1) A W N W N qd n A W N p i, qv / kt ( e 1) DC DC C Common emitter Current Gain V 11/30/016 5

26 How to make a Good Silicon Transistor For a given mitter length ~1, same material DC n n i, N p i, N D W W D n Make-ase short (few mm in 1950s, 00 A now) mitter doping higher than ase doping 11/30/016 ermel C 305 F16 6

27 transistor doping high base doping lowers the base resistance n+ n-collector n+ light collector doping increases the breakdown voltage high sub-collector doping lowers the collector series resistance 11/30/016 7

28 Doping for Gain DC n i, N p i, N D W n W D n N + N P N N N C 11/30/016 ermel C 305 F16 8

29 JT doping profiles F R n n+ emitter p base n collector n+ p N D ++ log 10 { N D ( N A )} N D N A N DC 11/30/016 ermel C 305 F16 9

30 emitter-base doping n µ n i N A p µ n i N D n n+ emitter F p base p R n collector n+ High emitter doping increases the emitter injection efficiency. ermel C 305 F16 g F = 1 1+ D p D n W W N D >> N A N A N D 30

31 Outline 1) mitter injection efficiency ) ase transport factor 3) arly effect 4) Speed (base transit time) 5) ffects of saturation 6) Gummel plots 7) Transconductance 8) HTs 9) mitter crowding 11/30/016 ermel C 305 F16 31

32 Conclusions The bers-moll model can be solved for key performance parameters quickly and selfconsistently Current gain can be very substantial, under the correct circumstances Performance can be systematically improved through term-by-term maximization of: DC n n i, N p i, N D W W D n 11/30/016 ermel C 305 F16 3

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