A MT Ch20 BJT Basics.pdf

8/8/2019 A MT Ch20 BJT Basics.pdf

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ECSE2210,MicroelectronicsTechnology,Prof.E.F.Schubert

Chapter20page1

Bipolar Junction Transistor BJT - Basics

Introduction

Inventors: Shockley, Bardeen, and Brattain, 1949 Bell Telephone LaboratoriesRevolutionary invention which changed the world The transistor is a three terminal device Applications include the use as amplifierandswitch There are pnp transistor and npn transistors The following materials will focus on the pnp transistor

o Two n regions merge to form a very thin baseo EB junction: Forward biaso BC junction: Reverse bias

Band diagram of pnp transistor


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Chapter20page2

Amplifier definitions

Common base configuration

E

C

I

I (1)

o current amplification of common base circuito 99.0 for state-of-the-art transistors

Common emitterconfiguration

11

11

CE

C

B

C

II

I

I

I(2)

Common collectorconfiguration


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Chapter20page3

B

C

B

E /

I

I

I

I(3)

Nature of bipolar transistor is a current amplifier(not a voltage amplifier)o It is a current-controlled current sourceo We control the collector current with the base current


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Chapter20page4

Qualitative analysis

Basic ideaso EB junction is asymmetric:

EnEp II (4)

Emitter hole current is controlled by EB junction.

o Base width is small:pB LW (5)

Most holes diffusing into the base will reach the collector sinceLp >> WB.Thus the base current controls the collector current.

o EB junction (EB junction is forward biased)(1)Holes diffusing from E into the B(2) Electrons diffusing from the B into the E

o Base(3) Recombination of holes injected into the base

(4) Most holes reach the C since Bp WL

o BC junction (BC junction is reverse biased)(5) Electron minority carrier current from C to B.(6) Hole minority carrier current from B to C.

We know that current (5) and (6) can be neglectedfrom mostpractical purposes.


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Chapter20page5

Quantitative analysis

What fraction of the emitter hole current reaches the collector?EpC IBI (6)

B = Base Transport FactorB = Probability that a hole injected into B reaches C

B 1

What fraction of the total emitter current is the emitterhole current?)( EpEnEEp IIII (7)

= Emitter Efficiency = Ratio ofIEn toIE

1

Ep

En

1

Ep

En

EpEn

Ep11

I

I

I

I

II

I

(8)

Current amplification BI/IBI/I EEpEC (9)

Next, we will calculateB and . We will employ:1.) Approximate calculation2.) Exact calculation

Approximate calculation: Hole Distribution in Base (pnp transistor)o Assume base is long(WB >>Lp)

pn /n e)(

Lxpxp (10)


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Chapter20page6

o Assume base isshort(WB


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Chapter20page7

o Assume that emitter is longElectron current from base into emitter

1e /p0n

nEn

kTeVnL

DAeI (17)

o Base isshortHole current from emitter into base

slopeinincreasetodueCorrection

B

p/n0

p

pEp 1e

W

Lp

L

DAeI kTeV (18)

o Emitter efficiency using Eqns. (8), (17), and (18)

n0B

p

p0n

n

Ep

En 11

pW

D

nLD

I

I (19)

Using A2

i2

ip0 // Nnpnn (20)

and D2

i2

in0 // Nnnnp (21)

One obtains

NLD

NWD

Anp

DBn

1 (22)

o To design a transistor with a high value of, we choose:1. WB very short2. NA >>ND (Emitter doping >> Base doping) (23)

Example: Assume pnp transistor, with Emitter 318A cm101 N ,Base 317

Dcm101 N ,

npDD , nm100

B

W ,Ln = 1 m

Calculate emitter efficiency

99.0100

111

Anp

DBn NLD

NWD


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Chapter20page8

Hole Distribution in the Base (pnp transistor)

o Hole concentration at the emitter side of basekTeVkTeV ppxpp /n0

/n0nE

BEBE e)1e()0( (24)

o Hole concentration at the collector side of basen0/n0BnC )1e()( BE ppWxpp kTeV (25)

note that BCV is negative

o Eqns. (24) and (25) are the boundary conditions for the hole concentration in thebase

o There is no electric field in the neutral region of the base. Therefore, transport isdescribed by the diffusionequation

2p

nn2n

2

)()(dd

Lxpxp

x (26)

General solution

CCxpLxLxn pnp /

2/

1n ee)(

(27)

The constants C1 and C2 will be determined by the boundary conditions. They are:

E21n )0( pCCxp (28)

C/

2/

1BnpBpB ee)( pCCWxp

LWLW (29)


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Chapter20page9

Solving Eqns. (28) and (29) forC1 and C2 yields

pBpB

pB

//

/EC

1ee

e

LWLW

LWpp

C

(30)

pBpB

pB

//C

/E

2ee

e

LWLW

LWpp

C

(31)

Next: Insert the constants C1 and C2 into Eqn. (27).

For 0C p , the hole concentration in the base is given by:

pBpB

pnpBpnpB

//

////

Enee

eee)(

LWLW

LxLWLxLWe

pxp

(32)

This function has an exponentially decreasingpart and an exponentially increasingpart, as shown in the illustration below.

o Recall that the slope,x

xp

d

)(d n , determines the diffusion current

Slope is larger at 0n x as compared to Bn Wx .

Difference: Recombination in the base.


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Chapter20page10

o For pB LW , we can expand the exponential function into a power series:...

!2!11e

2

xxx

Neglecting all quadratic and higher terms in Eqn. (32) yields

B

nEn 1)(

W

xpxp (33)

this is the diffusion triangle for negligible recombination in the base.


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Chapter20page11

Emitter (E), base (B), and collector (C) current

E, B and C currentso We have calculated the hole distribution in the base and can now calculate the

currents of the three terminals E, B and C by using

)(d

dn

np xp

xDAeI (34)

o Emitter current is obtained by using Eqns. (27), (30), (35), (34))()0( 12

p

pnpEp CC

L

DAexII (35)

pBC

pBE

p

pEp cosechcoth L

WpL

WpL

D

AeI (36)

o Collector current:

pBpB /

1/

2p

pBnpC ee)(

LWLWCC

L

DAeWxII (37)

p

BC

p

BE

p

pC cothcosech

L

Wp

L

Wp

L

DAeI (38)

o Base current:CEpCEB IIIII (39)

L

Wpp

L

DAeI

p

BCE

p

pB

2tanh)( (40)

o Eqns. (36), (38), and (40) aregeneral, i.e. they are valid forany bias configuration ofthe transistor

The equation can be simplified for a transistor under regular operating conditions,which are:

VBE = forward bias VCB = reverse bias


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Chapter20page12

Approximate E, B and C currentso VBE = forward bias pE 0o VCB = forward bias pC = 0o From Eqns. (36), (38), and (40):

p

BE

p

pEp coth

L

Wp

L

DAeI (41)

Using 3//1coth xxx , one obtains

p

B

B

pE

p

pEp

3L

W

W

Lp

L

DAeI (42)

p

BE

p

pC cosech

L

Wp

L

DAeI (43)

Using 6//1cosech xxx , one obtains

p

B

B

pE

p

pC

6L

W

W

Lp

L

DAeI (44)

p

B

p

BE

p

pCECEB

6

1

3

1p L

W

L

Wp

L

DAeIIIII (45)

pW

AepWL

DAeI E

p

BEB2

p

pB

22

(46)


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Chapter20page13

Base Transport Factor, Bo Using Eqns. (41) and (43) we calculate

p

B

pB

pB

Ep

C sech)/(coth

)/(cosech

L

W

LW

LW

I

IB (47)

Using2)2/1(1sech xx , one obtains

L

WB

2

p

B

2

11

(48)

We now have an expression forB.

o We now have (Emitter efficiency) Eqn. (22) andB (Base transport factor) Eqn. (48).Since B , we can now give the current amplification of a pnp transistor:

B

L

W

NLD

NWDB

2p

2B

Anp

DBn

211 (49)

Example: Calculate the Base Transport Factor forWB = 0.1 m and(a) Lp = 0.1 mand(b) Lp = 1.0 m

Solution:

Base transport factor:

2

p

B211

L

WB

(a) B = 0.5

(b) B = 0.995


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Chapter20page14

Appendix: Mathematics

Exponentialfunction

e lim 1 1

2.718e 1 1!

2!

3!

Function: e/

Slope:

Integral: e/ d

HyperbolicexponentialfunctionsHyperbolicsinfunction: sinh e e

Hyperboliccosfunction:

cosh

e

e

(chainfunction)

Hyperbolictanfunction: tanh

Hyperboliccotfunction: coth

Hyperbolicsecanfunction: sech

Hyperboliccosecanfunction: cosech

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A MT Ch20 BJT Basics.pdf