Lecture #15 - EECS Instructional Support Group Home Pageee130/sp03/lecture/lecture15.pdf · 2 Spring 2003 EE130 Lecture 15, Slide 3 Introduction • The BJT is a 3-terminal device

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EE130 Lecture 15, Slide 1Spring 2003

Lecture #15

OUTLINE

The Bipolar Junction Transistor– Fundamentals

– Ideal Transistor Analysis

Reading: Chapter 10, 11.1

EE130 Lecture 15, Slide 2Spring 2003

Bipolar Junction Transistors (BJTs)• Over the past 3 decades, the higher layout density and

low-power advantage of CMOS technology has eroded away the BJT’s dominance in integrated-circuit products.

(higher circuit density better system performance)

• BJTs are still preferred in some digital-circuit and analog-circuit applications because of their high speed and superior gain.

faster circuit speedlarger power dissipation

limits integration level to ~104 circuits/chip

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EE130 Lecture 15, Slide 3Spring 2003

Introduction• The BJT is a 3-terminal device

– 2 types: PNP and NPN

VEB = VE – VBVCB = VC – VBVEC = VE – VC

= VEB - VCB

VBE = VB – VEVBC = VB – VCVCE = VC – VE

= VCB - VEB

• The convention used in the textbook does not follow IEEE convention (currents defined as positive flowing into a terminal)

• We will follow the convention used in the textbook

EE130 Lecture 15, Slide 4Spring 2003

Charge Transport in a BJT• Consider a reverse-biased pn junction:

– Reverse saturation current depends on rate of minority-carrier generation near the junction⇒ can increase reverse current by increasing rate of

minority-carrier generation:Optical excitation of carriers

Electrical injection of minority carriers into the neighborhood of the junction

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EE130 Lecture 15, Slide 5Spring 2003

ICp

ICn

PNP BJT Operation (Qualitative)

B

Cdc I

I≅β

“Active Bias”: VEB > 0 (forward bias), VCB < 0 (reverse bias)

“Emitter”

“Base”

“Collector”

EE130 Lecture 15, Slide 6Spring 2003

• Important features of a good transistor:– Injected minority carriers do not recombine in the

neutral base region

– Emitter current is comprised almost entirely of carriers injected into the base (rather than carriers injected into the emitter

BJT Design

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EE130 Lecture 15, Slide 7Spring 2003

The base current consists of majority carriers supplied for1. Recombination of injected minority carriers in the base2. Injection of carriers into the emitter3. Reverse saturation current in collector junction

• Reduces | IB |4. Recombination in the base-emitter depletion region

Base Current Components

EE130 Lecture 15, Slide 8Spring 2003

Circuit Configurations

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EE130 Lecture 15, Slide 9Spring 2003

Modes of OperationCommon-emitter output characteristics

(IC vs. VCE)

EE130 Lecture 15, Slide 10Spring 2003

BJT Electrostatics• Under normal operating conditions, the BJT may be

viewed electrostatically as two independent pn junctions

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EE130 Lecture 15, Slide 11Spring 2003

• Emitter Efficiency:

– Decrease (5) relative to (1+2)to increase efficiency

• Base Transport Factor:

– Decrease (1) relative to (2)to increase transport factor

BJT Performance Parameters (PNP)

EnEp

Ep

III+=γ

Ep

Cp

T II=α

Tdc γαα ≡• Common-Base d.c. Current Gain:

EE130 Lecture 15, Slide 12Spring 2003

Collector Current (PNP)• The collector current is comprised of

• Holes injected from emitter, which do not recombine in the base ← (2)

• Reverse saturation current of collector junction ← (3)

where ICB0 is the collector current which flows when IE = 0

( )

0

0

0

α1α1

αα

CEB

dc

CBB

dc

dcC

CBBCdcC

IβI

III

IIII

+=−

+−

=

++=

0α CBEdcC III +=

• Common-Emitter d.c. Current Gain:

dc

dc1dc ααβ −=

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EE130 Lecture 15, Slide 13Spring 2003

Notation (PNP BJT)

NE = NAEDE = DNτE = τn

LE = LNnE0 = np0 = ni

2/NE

NB = NDBDB = DPτB = τpLB = LP

pB0 = pn0 = ni2/NB

NC = NACDC = DNτC = τn

LC = LNnC0 = np0 = ni

2/NC

EE130 Lecture 15, Slide 14Spring 2003

Ideal Transistor Analysis• Solve the minority-carrier diffusion equation in each quasi-neutral

region to obtain excess minority-carrier profiles– different set of boundary conditions for each region

• Evaluate minority-carrier diffusion currents at edges of depletion regions

• Add hole & electron components together terminal currents

0"" =

∆−=xdx

ndEEn

EqADI0=

∆−=xdx

pdBEp

BqADI

Wxdxpd

BCpBqADI

=

∆−=0'' =

∆=xdx

ndCCn

CqADI

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EE130 Lecture 15, Slide 15Spring 2003

Emitter Region Formulation• Diffusion equation:

• Boundary Conditions:

E

EE ndxnd

ED τ∆∆ −= 2

2

"0

)1()0"(

0)"(/

0 −==∆

=∞→∆kTqV

EE

E

EBenxn

xn

EE130 Lecture 15, Slide 16Spring 2003

Base Region Formulation• Diffusion equation:

• Boundary Conditions:

B

BB pdxpd

BD τ∆∆ −= 2

2

0

)1()(

)1()0(/

0

/0

−=∆

−=∆kTqV

BB

kTqVBB

CB

EB

epWp

epp

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EE130 Lecture 15, Slide 17Spring 2003

Collector Region Formulation• Diffusion equation:

• Boundary Conditions:

C

CC ndxnd

CD τ∆∆ −= 2

2

'0

)1()0'(

0)'(/

0 −==∆

=∞→∆kTqV

CC

C

CBenxn

xn

EE130 Lecture 15, Slide 18Spring 2003

Current Formulation

0"" =

∆−=xdx

ndEEn

EqADI

0=

∆−=xdx

pdBEp

BqADI

Wxdxpd

BCpBqADI

=

∆−=

0'' =

∆=xdx

ndCCn

CqADI

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EE130 Lecture 15, Slide 19Spring 2003

Emitter Region Solution• The solution of is:

• From the boundary conditions:

we have:

and:

E

EE ndxnd

ED τ∆∆ −= 2

2

"0

EE LxLxE eAeAxn /"

2/"

1)"( +=∆ −

)1()0"(

0)"(/

0 −==∆

=∞→∆kTqV

EE

E

EBenxn

xn

EEB LxkTqVEE eenxn /"/

0 )1()"( −−=∆

)1( /0 −= kTqVEL

DEn

EB

E

E enqAI

EE130 Lecture 15, Slide 20Spring 2003

Collector Region Solution• The solution of is:

• From the boundary conditions:

• we have:

and:

CC LxLxC eAeAxn /'

2/'

1)'( +=∆ −

CCB LxkTqVCC eenxn /'/

0 )1()'( −−=∆

)1( /0 −−= kTqV

CLD

CnCB

C

C enqAI

C

CC ndxnd

CD τ∆∆ −= 2

2

'0

)1()0'(

0)'(/

0 −==∆

=∞→∆kTqV

CC

C

CBenxn

xn

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EE130 Lecture 15, Slide 21Spring 2003

Base Region Solution• The solution of is:

• From the boundary conditions:

we have:

BB LxLxB eAeAxp /

2/

1)( +=∆ −

B

BB pdxnd

BD τ∆∆ −= 2

2

0

)1()(

)1()0(/

0

/0

−=∆

−=∆kTqV

BB

kTqVBB

CB

EB

epWp

epp

( )( )BLWBLW

BLxBLxCB

BLWBLWBLxWBLxW

EB

eeeekTqV

B

eeeekTqV

BB

ep

epxp

//

//

//

/)(/)(

)1(

)1()(/

0

/0

−

−

−

−−−

−−

−−

−+

−=∆

EE130 Lecture 15, Slide 22Spring 2003

• Now, we know• Therefore, we can write:

as

( ) 2sinh ξξξ −−= ee

( )( )BLWBLW

BLxBLxCB

BLWBLWBLxWBLxW

EB

eeeekTqV

B

eeeekTqV

BB

ep

epxp

//

//

//

/)(/)(

)1(

)1()(/

0

/0

−

−

−

−−−

−−

−−

−+

−=∆

( )[ ]( )

[ ]( )

B

BCB

B

BEB

LW

Lx

kTqVB

LW

LxW

kTqVBB

ep

epxp

sinhsinh

)1(

sinhsinh

)1()(

/0

/0

−+

−=∆−

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EE130 Lecture 15, Slide 23Spring 2003

• We know• Therefore, we have:

and:

( ) 2cosh ξξξ −+= ee

( )[ ]1)1( /)/sinh(

1/)/sinh(

/cosh(0

) −−−= kTqVLW

kTqVLWLW

BLD

EpCB

B

EB

B

B

B

B eepqAI

( )[ ]1)1( /)/sinh(

/cosh(/)/sinh(

10

) −−−= kTqVLWLWkTqV

LWBLD

CpCB

B

BEB

BB

B eepqAI

EE130 Lecture 15, Slide 24Spring 2003

Terminal Currents• We know:

• Therefore:

( )[ ]1)1( /)/sinh(

1/)/sinh(

/cosh(0

) −−−= kTqVLW

kTqVLWLW

BLD

EpCB

B

EB

B

B

B

B eepqAI

)1( /0 −= kTqVEL

DEn

EB

E

E enqAI

( )[ ]1)1( /)/sinh(

/cosh(/)/sinh(

10

) −−−= kTqVLWLWkTqV

LWBLD

CpCB

B

BEB

BB

B eepqAI

)1( /0 −−= kTqV

CLD

CnCB

C

C enqAI

( ) ( )( )[ ]1)1( /)/sinh(

10

/)/sinh(

/cosh(00

) −−−+= kTqVLWBL

DkTqVLWLW

BLD

ELD

ECB

BB

BEB

B

B

B

B

E

E epepnqAI

( ) ( )( )[ ]1)1( /)/sinh(

/cosh(00

/)/sinh(

10

) −+−−= kTqVLWLW

BLD

CLDkTqV

LWBLD

CCB

B

B

B

B

C

CEB

BB

B epnepqAI

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EE130 Lecture 15, Slide 25Spring 2003

Simplification• In real BJTs, we make W << LB for high gain.

Then, since

we have:

( )( ) 1for 1cosh

1for sinh

2

2

<<+→

<<→

ξξ

ξξξξ

( )( )WxkTqV

B

WxkTqV

BB

CB

EB

ep

epxp

)1(

1)1()(/

0

/0

−+

−−≅∆

EE130 Lecture 15, Slide 26Spring 2003

Performance Parameters (Active Mode)

( )

( )

( )221

221

221

2

2

2

2

2

2

1

11

11

11

BEE

B

B

E

Bi

Ei

BEE

B

B

E

Bi

Ei

B

EE

B

B

E

Bi

Ei

LW

LW

NN

DD

n

dc

LW

LW

NN

DD

n

dc

LWT

LW

NN

DD

n

n

n

n

+=

++=

+=

+=

β

α

α

γ Assumptions: • emitter junction forward biased, collector junction reverse biased• W << LB