BJTs: Current-Voltage CharacteristicsBJT_IV_S...BJTs: Current-Voltage Characteristics Professor Mark...

Lundstrom ECE 305 S15

ECE-305: Spring 2015

BJTs: Current-Voltage Characteristics

Professor Mark Lundstrom

Electrical and Computer Engineering Purdue University, West Lafayette, IN USA

[email protected]

4/22/15

Pierret, Semiconductor Device Fundamentals (SDF) pp. 371-399

bipolar transistors


VCE

E: emitter

C: collector

B: base

IC

NPN BJT

IC

VBE1, IB1

IE

IB

(forward) active region EB: FB, BC: RB

saturation region EB: FB, BC: FB

cut-off region EB: RB, BC: RB

2

BJT operation: active region

3 3

n+ emitter

p base

n collector

n+

FB RB

x

To understand this device, we should first draw an Energy Band Diagram.


equilibrium

x

EF EC

EV

E

emitter base collector


qVbi

4

EC

EV

VBE = 0, VCE > 0

x

EC

EV

E

collector


qVbi

5 emitter base

Fn

Fn

EC

EV“off” “cut-off”

VBE > 0, VCE > 0

x

EC

EV

E

collector


q Vbi −VA( )

6 emitter base

Fn

Fn

EC

EV“active”

IEn

ICn =αT IEn = IC

IC = I0eqVBE kBT

IEp

IB = IEp

IB = IC βdc << IC

NPN BJT operation (active region)

7

n+ emitter

p base

n collector

n+

FB RB

IE IC

IB

InE ICn ≈ IEn

I pE

IC ≈ IEn IE = IEn + IEpIB = IEp

WB << Ln( )


Question 1)

1) For an NPN bipolar transistor biased in the forward active region, which of the following is true?

a)  VBE = 0, VCE = 0. b)  VBE > 0, VCE > 0. c)  VBE > 0, VCE < 0. d)  VBE < 0, VCE > 0. e)  VBE < 0, VCE < 0.

Lundstrom ECE 305 S15 8

Question 2)

2) For a PNP bipolar transistor biased in the forward active region, which of the following is true?

a)  VBE = 0, VCE = 0. b)  VBE > 0, VCE > 0. c)  VBE > 0, VCE < 0. d)  VBE < 0, VCE > 0. e)  VBE < 0, VCE < 0.


Question 3)


3) How are the PN junctions biased in the saturation region of an NPN BJT?

a) Emitter-base: forward biased. Base-collector: forward-biased.

b) Emitter-base: forward biased. Base-collector: reverse-biased.

c) Emitter-base: reverse biased. Base-collector: forward-biased.

d) Emitter-base: reverse biased. Base-collector: reverse-biased.

e) Emitter-base: forward biased. Base-collector: biased breakdown.

10

Question 4)


4) How are the PN junctions biased in the saturation region of an PNP BJT?

a) Emitter-base: forward biased. Base-collector: forward-biased.

b) Emitter-base: forward biased. Base-collector: reverse-biased.

c) Emitter-base: reverse biased. Base-collector: forward-biased.

d) Emitter-base: reverse biased. Base-collector: reverse-biased.

e) Emitter-base: forward biased. Base-collector: biased breakdown.

11

outline


1)  Review 2)  Review of PN junctions under bias 3)  IV Characteristics (Active region) 4)  IV characteristics (Saturation region) 5)  CE vs. CB 6)  Wrap-up

12

NP junction in FB


q Vbi −VA( )Fp

Jn Jn = qDn

WP

ni2

NA

eqVA kBT −1( )

Jp Jp = qDp

WN

ni2

ND

eqVA kBT −1( )

WP

WN

13

Fn

quasi-neutral regions


q Vbi −VA( )Fp

WPWN

14

Fn

diffusion in the quasi-neutral regions


0x

Δn 0( ) = ni2

NA

eqVA kBT −1( )

WP << Ln

WP

Δn 0( ) = 0

′x

Δn x( )

′0

Δp ′0( ) = ni2

ND

eqVA kBT −1( )

Δp x( )

WN

WN << Ln

15

NP junction in FB (N-region)


0x

Δn x( ) Δn 0( ) = ni2

NA

eqVA kBT −1( )

WB << Ln

WP

Jn = qDndΔn x( )dx x=0

Jn = −q Dn

WP

Δn 0( )

Jn = q Dn

WP

ni2

NA

eqVA kBT −1( )Δn 0( ) = 0

16

N+P junction in FB


Jn

Jn VA( ) = q Dn

WP

ni2

NA

eqVA kBT −1( )

Jp

Jp VA( ) = q Dp

WN

ni2

ND

eqVA kBT −1( )JD VA( )

ND >> NA

γ ≡Jn VA( )

Jn VA( ) + Jp VA( ) ≤1

“electron injection efficiency”

γ ≡ 1

1+Dp

Dn

WP

WN

NA

ND 17

outline



18

NPN BJT operation (general)

19

n+ emitter

p base

n collector

n+ IE IC

IB

IEn ICn

IEp ICp

IE = IEn + IEp IC = ICn + ICp IB = IE − IC

FB/RB FB/RB


NPN BJT operation (active region)

20

n+ emitter

p base

n collector

n+

FB RB

IE IC

IB

InE ICn

I pE

IC ≈ IEn IE = IEn + IEpIB = IEp

WB << Ln( )


NPN BJT operation (active)

21

n+ emitter

p base

n collector

n+ IE IC

IB

IEn ICn

IEp ICp ≈ 0

IEn VBE( ) = qAE Dn

WB

ni2

NAB

eqVBE kBT −1( ) ???

IEp VBE( ) = qAEDp

WE

ni2

NDE

eqVBE kBT −1( )

ICn VBE( ) = IEn VBE( )

(no recombination in the base) αT = 1

diffusion in the quasi-neutral regions


0x

WB

′x′0

Δp ′0( ) = ni2

NDE

eqVBE kBT −1( )

Δp x( )

WE

WE << Ln

Emitter Base

Δn 0( ) = ni2

NAB

eqVBE kBT −1( )

WB << Ln

Δn x( )

Δn WB( ) = ni2

NAB

eqVBC kBT −1( )

22

diffusion in the quasi-neutral base


0

x

Δn 0( ) = ni2

NAB

eqVBE kBT −1( )

WB << Ln

WB

Δn x( )

Base

Δn WB( ) = ni2

NAB

eqVBC kBT −1( )

IEn VBE( ) = qAE Dn

WB

ni2

NAB

eqVBE kBT −1( ) ???

IEn VBE( ) = qAE Dn

WB

ni2

NAB

eqVBE kBT − eqVBC kBT( )

23

Δn WB( ) = − ni2

NAB

≈ 0

VBC << 0 (active region)



0

x

Δn 0( ) = ni2

NAB

eqVBE kBT −1( )

WB << Ln

WB

Δn x( )

Base

Δn WB( ) ≈ 0

IEn VBE( ) = qAE Dn

WB

ni2

NAB

eqVBE kBT − eqVBC kBT( )VBC << 0

IEn VBE( ) = qAE Dn

WB

ni2

NAB

eqVBE kBT

24


25

n+ emitter

p base

n collector

n+ IE IC

IB

IEn ICn

IEp ICp ≈ 0

IEn VBE( ) = qAE Dn

WB

ni2

NAB

eqVBE kBT

IEp VA( ) = qAEDp

WE

ni2

NDE

eqVBE kBT −1( )

ICn VBE( ) = IEn VBE( ) αT = 1( )

IC = IEn VBE( )

IC VBE( ) = I0eqVBE kBT I0 = qAEDn

WB

ni2

NAB

NPN BJT in active region

26

VCE

IC

VBE ,IB

IC = I0eqVBE kBT

What base current produced this collector current?


NPN BJT (active region base current)

27

n+ emitter

p base

n collector

n+ IE IC

IB

IEn ICn

IEp ICp ≈ 0

IEp VBE( ) = qA Dp

WE

ni2

NDE

eqVBE kBT −1( )

IB VBE( ) = qAEDp

WE

ni2

NDE

eqVBE kBT

(forward) active region


VCE

IC

VBE ,IB

IC = I0eqVBE kBT

IB VBE( ) = qAEDp

WE

ni2

NDE

eqVBE kBT

ICIB

= Dn

Dp

NDE

NAB

WE

WB

= βdc

28

IC VBE( ) = qAEDp

WE

ni2

NDE

eqVBE kBT

(forward) active region summary


VCE

IC

VBE ,IBI0 = qAE

Dn

WB

ni2

NAB

βdc =Dn

Dp

NDE

NAB

WE

WB

IC = I0eqVBE kBT

IB = IC βdc

29

outline



30

bipolar transistors


VCE

E: emitter

C: collector

B: base

IC

NPN BJT

IC

VBE1, IB1

IE

IB

(forward) active region EB: FB, BC: RB

saturation region EB: FB, BC: FB

31



0y

Δn 0( ) = ni2

NAB

eqVBE kBT −1( )

WB << Ln

WB

Δn x( )

Base

Δn WB( ) = ni2

NAB

eqVBC kBT −1( )

IEn VBE( ) = qAE Dn

WB

ni2

NAB

eqVBE kBT − eqVBC kBT( )

32

NPN BJT operation (saturation)

33

n+ emitter

p base

n collector

n+

FB FB

IE IC

IB

IEn ICn

I0 = qAEDn

WB

ni2

NABIC = I0e

qVBE kBT 1− e−qVCE kBT( )

outline



34

NPN bipolar transistor

35

BE: FB BC: RB

VBE > 0VCB = VCE −VBE > 0

IC

IB VCE

IEVBE

Pierret, Fig. 10.4

active saturation

cut-off inverted active

VCE (V )

IC = βdcIB

common base (active region)

36

IC

IB VCE

VBE

VCB

IE

IC = αdcIE

VCB > 0VEB < 0

IE

IB = IC β

IV characteristics

common base (active region)

37

IC

VCBVEB IB

IE

BE: FB BC: RB

VEB < 0VCB > 0

Pierret, Fig. 10.4

active

cut-off

saturation

VCB (V )

IC = αdcIE

outline



38

bipolar transistors


VCE

E: emitter

C: collector

B: base

IC

NPN BJT

IC

VBE1, IB1

IE

IB

39

IC = I0eqVBE kBT 1− e−qVCE kBT( )

I0 = qADn

WB

ni2

NAB

IC = I0eqVBE kBT IB = IC βdc

IB > IC βdc

NPN BJT operation (general)

40

n+ emitter

p base

n collector

n+ IE IC

IB

IEn ICn

IEp ICp

IE = IEn + IEp IC = ICn + ICp IB = IE − IC

FB/RB FB/RB



41

n+ emitter

p base

n collector

n+ IE IC

IB

IEn ICn ≈ IEn

IEp

IC ≈ IEn IB = IEp

FB/RB FB/RB


NPN BJT operation (saturation)

42

n+ emitter

p base

n collector

n+ IE IC

IB

IEn ICn

IEp ICp

IC = IEn − ICn IB = IEp + ICp

FB/RB FB/RB


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