Spring 2007EE130 Lecture 23, Slide 1 Lecture #23 QUIZ #3 Results (undergraduate scores only, N = 39) Mean = 22.1; Median = 22; Std. Dev. = 1.995 High =

EE130 Lecture 23, Slide 1Spring 2007

Lecture #23

QUIZ #3 Results (undergraduate scores only, N = 39)

Mean = 22.1; Median = 22; Std. Dev. = 1.995

High = 25; Low = 18

OUTLINE

The Bipolar Junction Transistor– Fundamentals– Ideal Transistor Analysis

Reading: Chapter 10, 11.1

EE130 Lecture 23, Slide 2Spring 2007

The base current consists of majority carriers supplied for1. Recombination of injected minority carriers in the base

2. Injection of carriers into the emitter

3. Reverse saturation current in collector junction• Reduces | IB |

4. Recombination in the base-emitter depletion region

Base Current Components (Active Bias)

EMITTER BASE COLLECTOR

p-type n-type p-type

EE130 Lecture 23, Slide 3Spring 2007

Circuit Configurations

Output Characteristics for Common-Emitter Configuration

EE130 Lecture 23, Slide 4Spring 2007

Modes of OperationCommon-emitter output characteristics

(IC vs. VCE)

Mode Emitter Junction Collector Junction

CUTOFF reverse bias reverse bias

Forward ACTIVE forward bias reverse bias*

Reverse ACTIVE reverse bias* forward bias

SATURATION forward bias forward bias

*or not strongly forward biased

EE130 Lecture 23, Slide 5Spring 2007

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

viewed electrostatically as two independent pn junctions

EE130 Lecture 23, Slide 6Spring 2007

Electrostatic potential, V(x)

Electric field, (x)

Charge density, (x)

EE130 Lecture 23, Slide 7Spring 2007

BJT Performance Parameters (PNP)

• Emitter Efficiency:

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

• Base Transport Factor:

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

Ep

CpT I

I

Tdc • Common-Base d.c. Current Gain:

EnEp

Ep

II

I

EE130 Lecture 23, Slide 8Spring 2007

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

α

α

CEB

dc

CBB

dc

dcC

CBBCdcC

IβI

III

IIII

0α CBEdcC III

• Common-Emitter d.c. Current Gain:

dc

dc

B

Cdc I

I

1

EE130 Lecture 23, Slide 9Spring 2007

Summary: BJT Fundamentals• Notation & conventions:

• Electrostatics:– Under normal operating conditions, the BJT may

be viewed electrostatically as two independent pn junctions

IE = IB + IC

pnp BJT npn BJT

EE130 Lecture 23, Slide 10Spring 2007

• Performance parameters:

– Emitter efficiency

– Base transport factor

– Common base d.c. current gain

– Common emitter d.c. current gain

EnEp

Ep

II

I

E

CpT I

Idc

dc

dc

B

Cdc I

I

1

Ep

CpT I

I

EE130 Lecture 23, Slide 11Spring 2007

Notation (PNP BJT)

NE = NAE

DE = DN

E = n

LE = LN

nE0 = np0 = ni2/NE

NB = NDB

DB = DP

B = p

LB = LP

pB0 = pn0 = ni2/NB

NC = NAC

DC = DN

C = n

LC = LN

nC0 = np0 = ni2/NC

EE130 Lecture 23, Slide 12Spring 2007

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

EE130 Lecture 23, Slide 13Spring 2007

Emitter Region Formulation

• Diffusion equation:

• Boundary Conditions:

E

EE n

dx

ndED

2

2

"0

)1()0"(

0)"(/

0

kTqV

EE

E

EBenxn

xn

EE130 Lecture 23, Slide 14Spring 2007

Base Region Formulation

• Diffusion equation:

• Boundary Conditions:

B

BB p

dx

pdBD

2

2

0

)1()(

)1()0(/

0

/0

kTqV

BB

kTqVBB

CB

EB

epWp

epp

EE130 Lecture 23, Slide 15Spring 2007

Collector Region Formulation

• Diffusion equation:

• Boundary Conditions:

C

CC n

dx

ndCD

2

2

'0

)1()0'(

0)'(/

0

kTqV

CC

C

CBenxn

xn