Microelectronics I

Chapter 5: Carrier Transport Phenomena

Transport; the process by which charged particles (electrons and holes) move

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Understanding of electrical properties ( I-V characteristics)

Basic current equation;

EneI ⋅⋅⋅∝ µ

e; electronic charged (constant, 1.6 x 10-19 C)

u; mobility ( figure of merit that reflect the speed)

n; carrier concentration

E; Electric field

Carrier concentration (electron, no and hole, po)

Chapter 4

Carrier transport (current)

This chapter

E; Electric field

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Carrier Transport

“Drift”

The movement of carrier due

to electric field (E)

“Diffusion”

The flow of carrier due to density

gradients (dn/dx)

dividerelectron

V

+ -

E

electron

dividerelectron

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5.1 Carrier Drift

Drift current density

Consider a positively charged hole,

When electric field, E, is applied, the hole accelerates

eEamF p == *

m*p; effective mass of hole, a; acceleration, e; electronic charge

However, hole collides with ionized impurity atoms and with thermally vibrating

lattice atom

hole

Lattice atom

Ionized impurity atom

E

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hole

Lattice atom

Ionized impurity atom

E

Holes accelerates

due to E

Involves in collision

(“Scattering Process”)

Loses most of energy

Gain average drift velocity, vdp

Ev pdp µ=

µp; Hole mobility (unit; cm2/Vs)

Describes how well a carrier move due to E

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Drift current density, Jdrf (unit; A/cm2) due to hole

dpdrfp epvJ =|

pEeJ pdrfp µ=|

Drift current density due to electron

nEeJ ndrfn µ=|

Total drift current;

EpneJ pndrf )( µµ +=

The sum of the individual electron and hole drift current densities

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Mobility effects

*

p

cp

pm

eτµ =

Mobility is important parameter to determine the conductivity of material

*

n

cnn

m

eτµ =

τ; mean time between collisions

If τ=10-15 s, in average, every 10-15 s, carrier involves in collision @ scattering

Two dominant scattering mechanism

1. Phonon or lattice scattering

2. Ionized scattering

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1. Lattice scattering or phonon scattering

At temperature, T > 0 K, atoms randomly vibrate. This thermal vibrations cause a

disruption of the periodic potential function. This resulting in an interaction

between carrier and the vibrating lattice atoms.

Mobility due to lattice scattering, µL

2/3−∝ Tµ 2/3−∝ TLµ

As temperature decreases, the probability of a scattering event decreases. Thus,

mobility increases

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electron hole

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2. Ionized Ion scattering

Coulomb interaction between carriers and ionized impurities produces scattering

or collusion. This alter the velocity characteristics of the carriers.

Mobility due to ionized ion scattering, µI

LN

T2/3

∝µTotal ionized impurity concentrationIN Total ionized impurity concentration

If temperature increases, the random thermal velocity of a carrier increases,

reducing the time the carrier spends in the vicinity of the ionized impurity center.

This causes the scattering effect decreases and mobility increases.

If the number of ionized impurity centers increases, then the probability of a

carrier encountering an ionized impurity centers increases, thus reducing

mobility

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The net mobility is given by

IL µµµ

111+=

Due to phonon scattering Due to ionized ion scattering

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Conductivity

EEpneJ pndrf σµµ =+= )(

Drift current

σ; conductivity [(Ω.cm)-1]

)( pne pn µµσ +=

electron

hole

Function of electron and hole concentrations and mobolities

Ρ; resistivity [Ω.cm]

)(

11

pne pn µµσρ

+==

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L

+- V

I

Area, A

Bar of semiconductor

I VCurrent density,

A

IJ = Electric field,

L

VE =

IRIA

LI

A

LV

L

V

A

I

EJ

=

=

=

=

=

ρ

σ

σ

σResistance, R is a function of resistivity, or

conductivity, as well as the geometry of the

semiconductor

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Consider p-type semiconductor with an acceptor doping Na (Nd=0) in which Na>>ni

pepne npn µµµσ ≈+= )(

Assume complete ionization

ρµσ

1≈≈ an Ne

Function of the majority carrier

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ex.;

Consider compensated n-type Silicon at T=300 K with a conductivity of σ=16 (Ωcm)-1 and an acceptor doping concentration of 1017 cm-3. Determine the donor concentration and the electron mobility.

Solution;

At T=300 K, we can assume complete ionization. (Nd-Na >>ni)

)10()106.1(16

)(

1719 −×=

−=≈

−

dn

adnn

N

NNene

µ

µµσ

To determine µn and Nd, we can use figure mobility vs. impurity concentration with trial and error

)10(101720 −= dn Nµ

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If Nd=2 x 1017 cm-3, impurity concentration, NI= Nd

++Na-=3 x 1017

cm-3. from the figure, µn= 510 cm2/Vs. so σ=8.16 (Ωcm)-1.

If Nd=5 x 1017 cm-3, impurity concentration, NI= Nd

++Na-=6x 1017

cm-3. from the figure, µn= 325 cm2/Vs. so σ=20.8 (Ωcm)-1.cm2/Vs. so σ=20.8 (Ωcm)-1.

Nd should be between 2 x 1017 and 5 x 1017 cm-3. after trial and error.

Nd= 3.5 x 1017 cm-3

µn=400 cm2/Vsσ= 16 (Ωcm)-1

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Ex 2.; designing a semiconductor resistor with a specified resistance to handle a given current density

A Si semiconductor at T=300 K is initially doped with donors at a concentration of Nd=5 x 1015 cm-3. Acceptors are to be added to form a compensated p-type material. The resistor is to have a resistance of 10 kΩ and handle a current density of 50 A/cm2 when 5 V is applied.

Solution;

When 5 V is applied to 10 kΩ resistor, the current, I

mAR

VI 5.0

10

54

===

If the current density, J is limited to 50 A/cm2, the cross-sectional area, A is

253

1050

105.0cm

J

IA

−−

=×

==

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Consider that electric field, E is limited to 100 V/cm. Then the length of the resistor, L is

The conductivity, σ of the semiconductor is

cmE

VL

2105

100

5 −×===

1

54

2

)(5.01010

105 −

−

−

Ω=×

×== cm

RA

Lσ

The conductivity of the compensated p-type semiconductor is

)( dapp NNepe −=≈ µµσ

Here, the mobility is function of total ionized impurity concentration Na+Nd

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Using trial and error, if Na=1.25x1016cm-3 , then Na+Nd=1.75x1016cm-3, and the hole mobility, from figure mobility versus impurity concentration, is approximately µp=410 cm2/Vs. The conductivity is then,

492.0)10)55.12((410106.1)(1519 =×−×××=−= −

dap NNeµσ

This is very close to the value we need. From the calculation

L=5x10-2 cmA=10-5cm2

Na=1.25x1016cm-3

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Velocity Saturation

Evd µ=

Drift velocity increase linearly with applied electric field.

At low electric field, vd increase linearly with applied E.with applied E.slope=mobility

At high electric field, vd saturates Constant value

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Carrier diffusion

Diffusion; process whereby particles from a region of high concentration toward a region of low concentration.

dividerCarrier

Ele

ctr

on c

oncentr

ation,

n

Position x

Electron diffusion

current density

Electron flux

dx

dneDJ

dx

dnDeJ

ndifnx

ndifnx

=

−−=

|

| )(

Dn; electron diffusion coefficient

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Hole

centr

ation,

p

Hole diffusion

current density

Hole flux

dx

dpeDJ

dx

dpeDJ

pdifpx

pdifpx

−=

−=

|

|H

ole

centr

ation,

p

Position x

current density

Dp; hole diffusion coefficient

Diffusion coefficient; indicates how well carrier move as a result of density gradient.

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Total Current Density

Total Current Density

Electron drift current

hole drift current

Electron diffusion current

hole diffusion current

difpxdrfpdifnxdrfn JJJJJ |||| +++=

dx

dpeD

dx

dneDEepEenJ pnxpxn −++= µµ

1-D

3-D

peDneDEepEenJ pnpn ∇−∇++= µµ

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Mobility,µ; indicates how well carrier moves due to electrical fieldDiffusion coefficient, D; how well carrier moves due to density gradient

Here, we derive relationship between mobility and diffusion coefficient using nonuniformly doped semiconductor model

“Einstein relation”

Graded impurity distribution

nonuniformly doped semiconductor

electron

x

EC

EF

Ev

x

Energy-band diagram

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EC

EF

Ev

x

Doping concentration decreases as x increases Electron diffuse in +x directionThe flow of electron leaves behind positively charged donor

Induce electrical field, Ex, given by

xdNkTE d )(1

−= …eq.1

dx

xdN

xNe

kTE d

d

x

)(

)(

1

−=

Since there are no electrical connections, there is no current(J=0)

0)(

)( =+=dx

xdNeDExNeJ d

nxdnµ

…eq.1

…eq.2Electron current

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From eq.1 and 2,

e

kTD

n

n =µ

Hole current must also be zero. We can show that

e

kTDp=

µ epµ

e

kTDD

p

p

n

n ==µµ

Diffusion coefficient and mobility are not independent parameters.The relationship between this 2 parameter “Einstein relation”

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The Hall effect

Using the effect, we can determine

The type of semiconductorCarrier concentrationmobility

Magnetic field

Applied electrical field

Force on charged particle

in magnetic field (“Lorentz

force”)

BqvF ×=

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the Lorentz force on electron

and hole is in –y direction

There will be buildup of negative

charge (n-type) or positive charge

(p-type) at y=0

As a results, an electrical field

called “Hall field, EH” is induced. called “Hall field, EH” is induced.

Hall field produces “Hall voltage,

VH”

In y-direction, Lorentz force will be balanced by force due to Hall field

zxH

Hzx

WBvV

W

VqBqv

=

=× (p-type)

Polarity of VH is used to determine the type of semiconductor

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For p-type

))(( Wdep

Iv x

x =

deV

BIp

epd

BIV

H

zx

zxH

=

=

Can calculate the hole concentration in p-typedeVH

For n-type

deV

BIn

end

BIV

H

zx

zxH

−=

−=

Note that VH is negative for n-type

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When we know the carrier concentration, we can calculate carrier mobility

xpx EepJ µ=

WdepV

LI

L

Eep

Wd

I

x

xp

xpx

=

=

µ

µ

Similar with n-type, mobility is determined from

WdenV

LI

x

xn =µ