Bipolar Junction Transistors: Solving Ebers-Moll Problems

Bermel ECE 305 F16

ECE 305: Fall 2016

Bipolar Junction Transistors:Solving Ebers-Moll Problems

Professor Peter BermelElectrical and Computer Engineering

Purdue University, West Lafayette, IN [email protected]

Pierret, Semiconductor Device Fundamentals (SDF)

Chapters 10 and 11 (pp. 371-385, 389-403)

11/30/2016

mailto:[email protected]

Recap: essence of current gain

2

,1BE

p i E q

E

B

V

E

qD ne

NI

W

N+

N

P

2

,1BEi B

B

V

E

qn

B

nqDe

NI

W

Input Response Input

Response

VBE VBC

11/30/2016 2

Ebers Moll Model

0

0

1

1

BE

BC

qV

F F

qV

R R

I I e

I I e

IC=Ic,n+Ic,p

2 2

, ,

0

2

,

0

1 1

1 1

BCBE

BCBE

pi B i qVqi Bn

B B

F F

Cn

B B C C

R

V

qVq

C

V

e eqDn nqD

AW N

nqDA

W N

I e

IW

e

N

I

22

,

0

2

, ,

0

1 1

1 1

BCBE

BCBE

p i E i i Bn

E B

qVqVBn

E E B

qV

R R

V

B

E

F

q

nqe e

e

qD n nqDA

W N W

DA

W

e

N

I

IN

I

IE

IF

IB

IC

E

B

C

IR

αFIF

αRIR

IE=IE,n+IE,p

3

Ebers-Moll model

Bermel ECE 305 F16

IC VBE,VBC( ) = aFIF0 eqVBE kBT -1( )- IR0 e

qVBC kBT -1( )

IE VBE ,VBC( ) = IF0 eqVBE kBT -1( )-aRIR0 e

qVBC kBT -1( )

IB VBE ,VBC( ) = IE VBE ,VBC( )- IC VBE ,VBC( )

See Pierret SDF, Chapter 11, sec. 11.1.4

aFIF0 = aRIR0

11/30/2016 4

Q1) Which is the base transport factor?

n+

emitter

p

base

n

collectorn+

FB RB

IE ICIEn

IEp

ICn

IB

a) b)

c) d)

e)

IEn IEn + IEp( )

ICn IEp

ICn IEn IEp

ICn IEn + IEp( )

11/30/2016 5

Q2) Which is beta_dc?

n+

emitter

p

base

n

collectorn+

FB RB

IE ICIEn

IEp

ICn

IB

a) b)

c) d)

e)

IEn IEn + IEp( )

ICn IEp

ICn IEn IEp

ICn IEn + IEp( )

11/30/2016 6

Q3) Which is alpha_dc?

n+

emitter

p

base

n

collectorn+

FB RB

IE ICIEn

IEp

ICn

IB

a) b)

c) d)

e)

IEn IEn + IEp( )

ICn IEp

ICn IEn IEp

ICn IEn + IEp( )

11/30/2016 7

Q4) Which is the base current?

n+

emitter

p

base

n

collectorn+

FB RB

IE ICIEn

IEp

ICn

IB

a) b)

c) d)

e)

IEn IEn + IEp( )

ICn IEp

ICn IEn IEp

ICn IEn + IEp( )

11/30/2016 8

Q5) Which is the emitter injection efficiency?

n+

emitter

p

base

n

collectorn+

FB RB

IE ICIEn

IEp

ICn

IB

a) b)

c) d)

e)

IEn IEn + IEp( )

ICn IEp

ICn IEn IEp

ICn IEn + IEp( )

11/30/2016 9

active region example (NPN)

Bermel ECE 305 F16

AE =10 mm ´ mm

IC = 50 mA

NDE =1.5 ´1019 cm-3

NAB =1.0 ´1017 cm-3

NDC = 2.0 ´1016 cm-3

WB = 0.25 mm

WB = 0.50 mm

WC =1.50 mm

t pE = 0.1ns

t nB = 75 ns

t pC =150 ns

11/30/2016 10

Step 1: diffusion coefficients and lengths

NDE =1.5 ´1019 cm-3

NAB =1.0 ´1017 cm-3

NDC = 2.0 ´1016 cm-3

WB = 0.25 mm

WE = 0.50 mm

WC =1.50 mm

t pE = 0.1ns

t nB = 75 ns

t pC =150 ns

Pierret, SDF, Fig. 3.5a, p. 80

mpE » 70 cm2 /V-s

mnB » 800 cm2 /V-s

mpC » 420 cm2 /V-s

DpE =kBT

qmpE »1.8 cm2 /s

DnB » 20.8 cm2 /s

DpC »10.9 cm2 /s

LpE = DpEt pE » 0.13 mm

LnB »12.5 mm

LpC »12.8 mm11

find the emitter injection efficiency

g F =1

1+DpE

DnB

WB

WE

NABNDE

g F =1

1+1.1´10-3= 0.9989

Bermel ECE 305 F1611/30/2016 12

find the base transport factor

aT =1

1+1

2

WB

LnB

æ

èçö

ø÷

2

aT =1

1+ 2.0 ´10-4= 0.9998

Bermel ECE 305 F1611/30/2016 13

find beta

bF =g FaT

1-g FaT

=aF

1-aF

bF =0.9989 ´0.9998

1- 0.9989 ´0.9998=

0.9987

1- 0.9987= 768

Bermel ECE 305 F1611/30/2016 14

find the base current

IC = 50 mA

IB = IC bF

bF = 768

IB = IC bF = 65 nA

aF =bF

bF +1= 0.9987

Bermel ECE 305 F1611/30/2016 15

Forward saturation current density

IC VBE ,VBC( ) = aFIF0 eqVBE kBT -1( )- IR0 e

qVBC kBT -1( )


qVBC kBT -1( )

IF0 = qADnB

WB

ni2

NAB+DpE

WE

ni2

NDE

æ

èçö

ø÷® qA

DnB

WB

ni2

NAB+DpE

LpE

ni2

NDE

æ

èç

ö

ø÷

IF0 =1.33´10-16 A

Bermel ECE 305 F1611/30/2016 16

Reverse saturation current density

IC VBE ,VBC( ) = aFIF0 eqVBE kBT -1( )- IR0 e

qVBC kBT -1( )


qVBC kBT -1( )

IR0 =1.91´10-16 A

IR0 = qADnB

WB

ni2

NAB+DpC

WC

ni2

NDC

æ

èçö

ø÷

Bermel ECE 305 F1611/30/2016 17

Check active region

Bermel ECE 305 F16

IC VBE,VBC( ) = aFIF0 eqVBE kBT -1( )- IR0 e

qVBC kBT -1( )®aFIF0eqVBE kBT + IR0


qVBC kBT -1( )® IF0eqVBE kBT +aRIR0

IB VBE ,VBC( ) = IE VBE ,VBC( )- IC VBE ,VBC( )® 1-aF( ) IF0eqVBE kBT - 1-aR( ) IR0

IC »aFIF0eqVBE kBT

IB » 1-aF( ) IF0eqVBE kBT =

1-aF( )aF

IC =1

bdcIC

✔

11/30/2016 18

What is ID?

IE = ID

IB

VBC

VBE

IE = ID

VCE =VD VD

ID

Bermel ECE 305 F16 19

What is ID?

IE = ID

IB

IC = aFIF0 eqVBE kBT -1( )- IR0 e

qVBC kBT -1( )

IE = IF0 eqVBE kBT -1( )-aRIR0 e

qVBC kBT -1( )

VBC = 0

VBC

VBE

IE = ID

VCE =VD

IE = IF0 eqVBE kBT -1( )

ID = IF0 eqVD kBT -1( )

IC


What is IC?

IE

IB = 0

IC = aFIF0 eqVBE kBT -1( )- IR0 e

qVBC kBT -1( )

IE = IF0 eqVBE kBT -1( )-aRIR0 e

qVBC kBT -1( )VBC

VBE

IC

VCE


Common Base Configuration

Bermel ECE 305 F16

P NE

IE IC

C

BB

VEB

(in)

VCB

(out)

How would the model change if this was a Schottky barrier

BJT?

IE

IF

IB

IC

E

B

C

IR

CBE CBC

αRIR

αFIFN+

11/30/2016 22

Common Emitter Configuration

Bermel ECE 305 F16

αRIR

αFIF

CBE

CBC

E

B

IB

IE

IC

IF

IR

1

1

F F F FF F B

FF

F

I II I

F F

F

I

F F R RI I

Cm

Cp

IR

P

+

N

P

C

E E

B

VEB

(in)

VEC

(out)

ICIB

11/30/2016 23

Current Gain

Bermel ECE 305 F16

P

N

P+

C

E E

B

VEB

(in)

ICIB

Common Emitter current gain ..

CDC

BI

I

2 2

( / ), ,/

2

, /

( 1) ( 1)

( 1)

BCB

BE

E qV kTi B i BqV kTn n

B B B B

i E q kTn

E

V

E

n nqD qDe e

W N W N

nqDe

W N

2

,

2

,

i Bn E E

B p i E B

nD W N

W D n N

Common Base current gain ..

P+ N P

E

IE IC

C

BB

VEB

(in)

VCB

(out)

CDC

E

I

I C C

DC

B E C

I I

I I I

1

DC

DC

11/30/2016 24

Gummel Plot and Output Characteristics

2 2

, , //( 1) ( 1)BCBEi B i B q kTq kTn n

B B B B

C VVn nqD qD

e eA W N W N

I

2

, /( 1)BE

p i E V kT

E E

B qqD n

eA W N

I

CDC

BI

I

DCCommon

emitter

Current Gain

VBE11/30/2016 25

=

How to make a Good Silicon Transistor

Bermel ECE 305 F16

Emitter doping higher

than Base doping

~1, same material

Make-Base short …

(few mm in 1950s, 200 A now)

For a given Emitter length

2

,

2

,

i Bn E EDC

B p i E B

nD W N

W D n N

11/30/2016 26

transistor doping

n-collector

n+

n+

IBIB

high base doping lowers the

base resistancehigh sub-collector doping

lowers the collector series

resistance

light collector doping increases

the breakdown voltage

11/30/2016 27

Doping for Gain …

Bermel ECE 305 F16

NE

NB

NC

N+

P

N

2

,

2

,

i Bn E EDC

B p i E B

nD W N

W D n N

11/30/2016 28

BJT doping profiles

29

n+

emitter

p

base

n

collectorn+

FB RB

IEn

IEp

Bermel ECE 305 F16

log10 ND NA( ){ }

NDE NAB

NDC

ND++

11/30/2016

emitter-base doping

30

n+

emitter

p

base

n

collectorn+

FB RB

IEn

IEp

IEn µni

2

NAB

IEp µni

2

NDE

High emitter

doping increases

the emitter

injection

efficiency.

g F =1

1+DpE

DnB

WB

WE

NABNDE

NDE >> NAB

Bermel ECE 305 F16

Outline

11/30/2016 Bermel ECE 305 F16

1) Emitter injection efficiency

2) Base transport factor

3) Early effect

4) Speed (base transit time)

5) Effects of saturation

6) Gummel plots

7) Transconductance

8) HBTs

9) Emitter crowding

31

Conclusions

The Ebers-Moll model can be solved for key

performance parameters quickly and self-

consistently

Current gain can be very substantial, under

the correct circumstances

Performance can be systematically improved

through term-by-term maximization of:

11/30/2016 Bermel ECE 305 F16 32

2

,

2

,

i Bn E EDC

B p i E B

nD W N

W D n N

Documents

Bipolar Junction Transistors: Solving Ebers-Moll Problems