Carrier Transport Phenomena

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

By Dr. Ghanshyam Kumar Singh

In this section, we will

•Describe the mechanism of carrier drift due to an applied electric field

•Describe the characteristics of carrier mobility

•Describe the mechanism of carrier diffusion

•Describe the effects of a nonuniform impurity doping concentration

•Discuss and analyze the Hall Effect

Understanding of electrical properties (I-V characteristics)

Basic current equation;

EneI

e; electronic charged (constant, 1.6 x 10-19 C); mobility ( figure of merit that reflect the speed)n; carrier concentrationE; Electric field

Carrier Transport

“Drift”The movement of carrier due

to electric field (E)

“Diffusion”The flow of carrier due to density

gradients (dn/dx)

V

+ -E

electron

dividerelectron

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 atomE

hole

Lattice atom

Ionized impurity atomE

Holes accelerates due to E

Involves in collision (“Scattering Process”) Loses most of energy

Gain average drift velocity, vdp

Ev pdp

µp; Hole mobility (unit; cm2/Vs)Describes how well a carrier move due to E

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

dpdrfp epvJ |

pEeJ pdrfp |

Drift current density due to electron

nEeJ ndrfn |

Total drift current;

EpneJ pndrf )(

The sum of the individual electron and hole drift current densities

Mobility effects

*p

cpp m

e

Mobility is important parameter to determine the conductivity of material

*n

cnn m

e

; mean time between collisions

If =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

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 TL

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

Temperature , Scattering Probability , Mobility Temperature , Scattering Probability , Mobility

electron hole

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

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.

Temperature , Thermal velocity, Time around ionized impurity, Mobility

• If the number of ionized impurity centers increases, then the probability of a carrier encountering an ionized impurity centers increases, thus reducing mobility

Ionized Impurity , Scattering Probability , Mobility

IN

TI

2/3

The net mobility is given by

IL 111

Due to phonon scattering Due to ionized ion scattering

Normally, more than one source of scattering is present, for example both impurities and lattice phonons.

It is normally a very good approximation to combine their influences using "Matthiessen's Rule" (developed from work by Augustus Matthiessen in 1864):

Conductivity

EEpneJ pndrf )(

Drift current

σ; conductivity [(Ω.cm)-1]

)( pne pn

electron

hole

Function of electron and hole concentrations and mobolities

Ρ; resistivity [Ω.cm]

)(

11

pne pn

L

+- V

I

Area, A

Bar of semiconductor

Current density,A

IJ Electric field,

L

VE

IRIA

LI

A

LV

L

V

A

I

EJ

Resistance, R is a function of resistivity, or conductivity, as well as the geometry of the semiconductor

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

pepne npn )(

Assume complete ionization

1

an Ne

Function of the majority carrier

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

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

)10(10 1720 dn N

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.

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

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

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

154

2

)(5.01010

105

cm

RA

L

The conductivity of the compensated p-type semiconductor is

)( dapp NNepe

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

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 NNe

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

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

Na=1.25x1016cm-3

Velocity Saturation

Evd Drift velocity increase linearly with applied electric field.

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

At high electric field, vd saturates Constant value

Carrier diffusion

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

dividerCarrier

Ele

ctro

n co

ncen

trat

ion,

n

Position x

Electron diffusion current density

Electron flux

dx

dneDJ

dx

dnDeJ

ndifnx

ndifnx

|

| )(

Dn; electron diffusion coefficient

Hol

e ce

ntra

tion,

p

Position x

Hole diffusion current density

Hole flux

dx

dpeDJ

dx

dpeDJ

pdifpx

pdifpx

|

|

Dp; hole diffusion coefficient

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

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

1-D

3-D

peDneDEepEenJ pnpn

Graded impurity distribution

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”

Non-uniformly doped semiconductor

electron

x

EC

EF

Ev

x

Energy-band diagram

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

dx

xdN

xNe

kTE d

dx

)(

)(

1

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

0)(

)( dx

xdNeDExNeJ d

nxdn

…eq.1

…eq.2Electron current

From eq.1 and 2,

e

kTD

n

n

Hole current must also be zero. We can show that

e

kTD

p

p

e

kTDD

p

p

n

n

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

Exercise 1

Assume the mobility of a particular carrier is 1000 cm2/V-s at T=300K.Determine the diffusion coefficient given the carrier mobility.

Using the Einstein relation we have that

e

kTDD

p

p

n

n

20.0259 1000 25.9 cm /sD kT kT

De e

Exercise 2

Assume that electron diffusion coefficient of a semiconductor at T=300K is Dn=215 cm2/s. Determine the electron mobility.

Using the Einstein relation we have that

e

kTDD

p

p

n

n

22158301cm /V-s

0.0259

D kT DkTee

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

the Lorentz force on electron and hole is in –y directionThere will be buildup of negative charge (n-type) or positive charge (p-type) at y=0As a results, an electrical field 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

WBvVW

VqBqv

(p-type)

Polarity of VH is used to determine the type of semiconductor

For p-type

))(( Wdep

Iv x

x

deV

BIp

epd

BIV

H

zx

zxH

Can calculate the hole concentration in p-type

For n-type

deV

BIn

end

BIV

H

zx

zxH

Note that VH is negative for n-type

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

xpx EepJ

Similar with n-type, mobility is determined from

WdenV

LI

x

xn

WdepV

LIL

Vep

Wd

I

x

xp

xpx

Exercise 3

Silicon at T = 300 K is uniformly doped with phophorus atoms at a concentration of 2 1016 cm. A Hall effect device has been fabricated with the following geometry: d = 10-3 cm, W = 10-2 cm, and L = 10-1 cm. The electrical parameters measured are: Ix = 1.2 mA, and Bz = 500 gauss = 5 10-2 Tesla. Determine a) The Hall voltageb) The Hall field

Exercise 3

Determine a) The Hall voltage

3 2

22 19 5

3

1.2 10 5 10

2 10 1.6 10 10

1.875 10 1.875

X ZH

I BV

ned

V mV

b) The Hall Field

3

2

1.875 100.1875 V cm

10

HH H H

HH

VV W

W

V

W

, n=2 1016 cm, Ix = 1.2 mA, Bz = 500

Quiz 1

Consider a silicon at T = 300 K. A Hall effect device has been fabricated with the following geometry: d = 5 10-3 cm, W = 5 10-2 cm, and L = 0.50 cm. The electrical parameters measured are: Ix = 0.50 mA, Vx = 1.25 V, and Bz = 650 gauss = 6.5 10-2 Tesla. The Hall voltage is EH = -16.5 mV/cm. Determine a) The Hall voltageb) The conductivity typec) The majority carrier concentrationd) The majority carrier mobility

Quiz 1

Determine a) The Hall voltage

3 216.5 10 5 10

0.825 mV

H HV W

b) The conductivity type

negative n-typeHV

c) The majority carrier concentration

3 2

15 -3

19 5 3

0.5 10 6.5 104.924 10 cm

1.6 10 5 10 0.825 10x z

H

I Bn

edV

d) The majority carrier mobility

3 2

19 21 4 5

0.5 10 0.5 10

1.6 10 4.924 10 1.25 5 10 5 10

1015 cm / V-s

xn

x

I L

enV Wd

d = 5 10-3 cm, W = 5 10-2 cm, and L = 0.50 cm. Ix = 0.50 mA, Vx = 1.25 V, Bz = 650 gauss = 6.5 10-2 TeslaEH = -16.5 mV/cm