chapter 5 carrier transport phenomenaocw.nctu.edu.tw/upload/classbfs120904360159510.pdfw.k. chen...

W.K. Chen Electrophysics, NCTU 1

Chapter 5Carrier transport phenomena


Transport

The net flow of electrons and holes in material is called transport

Two basic transport mechanismsDrift: movement of charged due to electric fields

Diffusion: the flow of charges due to density gradient

We implicitly assume the thermal equilibrium during the carrier transport is not substantially disturbed


Outline

Carrier drift

Carrier diffusion

Graded impurity distribution

The Hall effect


5.1 Carrier driftDrift: the net movement of charge due to electric fields is called drift

Drift current: The net drift of charge gives rise to a drift current

)cmsec

#( densityFlux

)sec

#(

)(Flux

2⋅=

Φ=

=ΔΔ

=Δ

=Δ

=Φ

υφ

υυ

nA

nAt

tnA

t

nA

t

N

l

l l

A

N: total number of flow chargen: volume density of flow chargeA: cross-sectional areaυ: average drift velocityl: traveling length of carrier per Δt

)cm

A( density current Drift

2dqnqJ υφ ==Ampere


mobilityThe average drift velocity for low electric fields is directly proportional to the electric field, similar to the terminal velocity case in “Fundamental Physics”

E)(* eamF p +==

υdp: average drift velocity for holesμp: hole mobility, proportionality factor

crystal

extFr

amFext*=

r

Err

pdp μυ =

The mobility describe how well a particle will move due to an electric field

E)()(, pdpdrfp pepeJ μυ +=+=Drift current due to holes

Hole: E: electric field


E

EeF )(+=EeF )(−=

)cm

A(

2dqnqJ υφ ==

Errndn μυ −=

)()()(, EpeneqnJ ndpdrfn μυυ −−=−==

E, ndrfn enJ μ=Drift current due to electrons

E)(,, pndrfpdrfn epenJJJ μμ +=+=Total drift current

drfpJ ,

drfnJ ,

nφFlux pφFlux


Example 5.1 drift current density

21619

,,

A/cm136)10)(10)(8500)(106.1(

E)(

=×=

+=+=

−dft

pndrfpdrfndft

J

epenJJJ μμ

V/cm 10E field electric applied ifdensity current drift theCalulate

ionization complete Assume

cm 10 ,0 300K,at sample GaAs 316

=⇒

== −da NN

Solution:

V/cm 10E field electric applied ifdensity current drift theCalulate

cm 1024.310

)108.1(

cm 10 ionization complete

3-416

2162

-316

=⇒

×=×

==

=≈⇒

−

n

np

Nn

i

d


5.1.2 Mobility effect

tm

ee

dt

dmamF

ppp *** E

E)( =⇒+=== υυ

Eity peak velocmean *, ⎟⎟⎠

⎞⎜⎜⎝

⎛=

p

cppeakd m

eτυ

F

t

Under thermal equilibrium

Assume the net drift velocity is a small perturbation on the random thermal velocity, so the time between collision will not be altered appreciably

τcp: mean time between collisions


E2

1 velocity average

* ⎟⎟⎠

⎞⎜⎜⎝

⎛=

p

cpd m

eτυ

Due to the statistic nature, the factor of ½ does not appear in a more accurate model

*

*

E

,E

n

cndnn

p

cpdpp

m

e

m

e

τυμ

τυμ

==

==

The average drift velocity is one half the peak value

EE velocity average p*μ

τυ =⎟

⎟⎠

⎞⎜⎜⎝

⎛=

p

cpdp m

e

The less the collisions, the longer the mean collision time and the higher the mobility


Scattering (collision) mechanismsTwo major scattering mechanisms

Phonon (lattice) scattering

A perfect periodic potential in a solid allows electrons to move unimpeded, or with no scattering, through the crystal

The thermal vibrations of lattice atoms cause a disruption in the perfect periodic potential, resulting the interactions between the electrons or holes and the vibrating lattice atoms

Ionized impurity scattering

The impurites in semiconductor at higher temperatures. The coulomb interactions between the electrons or holes and the ionized impurities produce scattering or collisions.

2/3−∝TLμ

II N

T 2/3−

∝μ

lattice

impurity )(

impurity ionized total−+ += adI NNN


Mobilitis versus temperarure

Inserts show the temperature dependence for “almost” intrinsic silicon

The inserts show that the parameter n is not equal to 3/2, but is 2.2, as the first-order scattering theory predicted. However , the mobilites do increase as the temperature increases


Mobilties versus impurity concentrations at 300K

Ge

Si

GaAs


τμτυμτυ

μ ∝⇒==== E

,E **

n

cndnn

p

cpdpp m

e

m

eQ

LI

dtdtdt

τττ+=

LI μμμ111

+=

The probability of a scattering even in the differential time dt is the sum of individual events

Net mobility

The net mobility due to the ionized and lattice scattering processes

High effective mass of carrier results in low mobility

The mobility will increase with the increasing collision time


5.1.3 Conductivity

EE)( σμμ =+= pn epneJ

)( pn epen μμσ += )(

11

pn epne μμσρ

+==

A

L

A

LRIRV

σρ

=== ,

E1E/ σ

σ

=⎟⎠⎞

⎜⎝⎛⋅===

ALA

L

A

RV

A

IJ

Eσ=J

The conductivity and resistivity of an extrinsic semiconductor are a function primarily of the majority carrier parameters, such carrier concentrations and mobilities


Resistivity versus impurity concentration at 300K

Si Ge, GaAs and GaP


In the midtemperature range (extrinsic range)

We have complete ionization, the electron concentration remains essentially constant, However, the mobility decreases with increasing temperature

At higher temperatures

The intrinsic carrier concentration begins to dominate the electron concentration as well as the conductivity

ipni nee )( μμσ +=


Example 5.2 mobility

mobilityelectron andion concentratdonor theDetermine

cm 10a ,cm)-( 16

300K,at Si type-n dCompensate3161

⇒=Ω= −− Nσ

Solution:

)10()106.1(16

)(

)(

300Kat ionization complete andtor semiconduc dcompenstae

1719 −×=

−=≈−≈⇒

−dn

adnn

ad

N

NNene

NNn

μ

μμσ

1-2

3-173-17

cm)-( 16.8 s-/Vcm 510Then

)cm102 .,.( cm103

choose weif exampleFor

error and l with triafigureleft theUse

Ω=⇒=

×=×=−= −+

σμn

dadI NeiNNN


1-2

3-173-17

cm)-( 8.20 s-/Vcm 325Then

)cm105 .,.( cm106

choose weIf

Ω=⇒=

×=×=−= −+

σμn

dadI NeiNNN

e)given valu with (agree cm)-( 16

s-/Vcm 400

)cm105.3

cm105.4

yieldserror and alFurther ti

1-

2

3-17

3-17

Ω=⇒

=

×=

×=+= −+

σ

μn

d

adI

N

NNN


5.1.4 Velocity saturation

kTm th 2

3

2

1 2 =υ

eV03885.0)0259.0(2

3

2

1

K 300Tat

2 ==

=

thmυ

m/s 10)1011.9(

)106.1)(03885.0(2)eV03885.0(2 531

19

≈×

×== −

−

mυth

cm/s 107≈thυ


Figure 5.8

For SiAt low electric fields, there is linear variation of velocity with electric field

At high electric fields, the velocity saturated at approximately 107cm/s


For GaAsDue to low effective mass, the low-electric field electron velocity in GaAs is much larger than in Si.

At high electric fields, negative differential mobility occurs due to the scattering of electrons into upper valley. Because of larger effective mass in the upper valley (0.55 mo vs. 0.067 mo), the intervalley transfer mechanism results in decreasing average drift velocity of electrons with electric field.

onon mmmm 08.1 :Si 067.0 :GaAs ** ==


5.2 Carrier diffusion

)(xn

xox

ox

dx

dnlF thn υ−=

cnthl τυ=(1)(2)

ox lxo +lxo −


During electron travel between collisions

In a mean free time, One half of electrons at segment (1) will move to the right and cross the xoplane into segment (2)

One half of electrons at segment (2) will move to the left and cross the xo plane into (segment (1))

)(2

1)(

)(2

1

2

1

2

1)(

flux)(electron right the toflowelectron of rateNet

21

2121

ldx

dnxF

ldx

dnnn

nnnnxF

thn

thththon

−⋅=⇒

−≈−

−=−=

υ

υυυ

dx

dnlF thn υ

2

1flux electron net −=

cnthl τυ=

(1)(2)

ox lxo +lxo −

l

1n2n


dx

dnleFeJ thnn υ

2

1)( +=−=

dx

dneDJ ndifn =,

lD thn υ2

1

t coefficiendiffusion electron

=

dx

dpeDJ pdifp −=,

lD thp υ2

1

coefficentdiffusion hole

=

cnthl τυ=

cpthl τυ=

Hole diffusion current

Electron diffusion current


Example 5.4

21718

19

,

A/cm 10810.0

107101)225)(106.1( =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ×−××=

ΔΔ

≈=

−

x

neD

dx

dneDJ nndifn

/scm 225t coefficiendiffusion ifcurrent diffusion theCalculate

cm 0.10 of distance

aover cm 107 to101 fromlinearly ion variesconcentratelectron The

300K,at GaAs type-n

2

31818

=⇒

×× −

nD

Solution:


5.2.2 Total current density

dx

dpeD

dx

dneDJepenJ pndifnpn −+++= ,EE μμ

peDneDJepenJ pndifnpn ∇−∇+++= ,EE μμ


5.3 Graded impurity distribution

)(1

FiF EEe

−=⇒φ

dx

dE

edx

d Fix

1E =−=

φ)( e

Ec−

=φ

The electric field is defined as

The electric potential is related to the electric potential energy by the charge ( -e)

In nonuniform doped semiconductor, there will be a diffusion of majority electrons from the region of high concentration to the region of low concentration.

The flow of electrons leave behind positively charged donor ions. The separation of positive and negative charges induces a electric field


The induced electric field due to the nonuniform doping

dx

xdN

xNe

kT d

dx

)(

)(

1E ⎟

⎠⎞

⎜⎝⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=−⇒≈⎥

⎦

⎤⎢⎣

⎡ −=

i

dfifd

fifio n

xNkTEExN

kT

EEnn

)(ln )(exp

dx

xdN

xN

kT

dx

dEd

d

fi )(

)(=−


Example 5.5)1(0 )(cm 1010 )( -31916 mxxxNd μ≤≤−=

V/cm 9.25)0(E

find we,0At

)1010(

)10)(0259.0()(

)(

1E

1916

19

==

=

−−=⎟

⎠⎞

⎜⎝⎛−=

x

x

xdx

xdN

xNe

kT

x

d

dx

Solution:


5.3.2 The Einstein relation

dx

xdNeDenJ

dx

dneDenJ

dnnn

nnn

)(E0

E0

+==

+==

μ

μ

dx

xdNeD

dx

xdN

xNe

kTen d

nd

dn

)()(

)(

10 +

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛−= μ

e

kTD

n

n =μ e

kTDD

p

p

n

n ==μμ

In thermal equilibrium, the individual electron and hole current must be zero

)()( xNxn d≈Q


The hole current must also be zero

e

kTDD

p

p

n

n ==μμ

Einstein relation K 300Tat 0256.0 =

300K Tat 40 =≈D

μ


BqFB ×= υ

zxy Bqqq

BqF

υυ==

=×+=

HEE

0]E[

WV HH E+=

The Hall effect is used

to distinguish whether a semiconductor is n-type or p-type

To measure the majority carrier concentration and

Majority carrier mobility

The induced electric field in the y-direction is called the Hall field

The induced electric field produce a voltage in the y-direction is called the Hall Voltage

5.4 The Hall effect

zxH WBV υ= citydrift velo :xυ


))(( Wdep

I

ep

J xxdx ==υ

For p-type semiconductor, the drift velocity is related to material parameters

H

zx

edV

BIp =

epd

BIV zxH =⇒

ned

BIV zxH =⇒

H

zx

edV

BIn =

Semiconductor type & concentration


xpx epJ Eμ=Q

L

Vep

Wd

I xp

x ⋅= μ

WdepV

LI

x

xp =μ

WdenV

LI

x

xn =μ

mobilityOnce the majority carrier concentration is determined, we can calculate the low-field majority carrier mobility


Example 5.7 carrier concnetration & mobility

mV 25.6 and tesla,105 gauss 005 V, 5.12 mA, 0.1

cm 10 cm, 10 cm, 1.0

H2

zx

32

−=×====

===−

−−

VBVI

dWL

x

Solution:

3153213519

23

cm 105m 105)1025.6)(10)(106.1(

)105)(10(×=×=

×−××−

= −−−

−−

nH

zx

edV

BIn =

WdenV

LI

x

xn =μ

s-/Vcm 1000

s-/Vm 10.0)10)(10)(5.12)(105)(106.1(

)10)(10(

2

2542119

33

=

=××

= −−−

−−

n

n

μ

μ


Figure 5.14 Figure for Problem 5.22

chapter 5 carrier transport phenomenaocw.nctu.edu.tw/upload/classbfs120904360159510.pdfw.k. chen...

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