ENE 311 Lecture 2. Diffusion Process The drift current is the transport of carriers when an electric...

ENE 311 Lecture 2

Diffusion Process

• The drift current is the transport of carriers when an electric field is applied.

• There is another important carrier current called “diffusion current”.

• This diffusion current happens as there is a variation of carrier concentration in the material.

• The carriers will move from a region of high concentration to a region of low concentration.

• This kind of movement is called “diffusion process”.

Diffusion Process

• An electron density varies in the x-direction under uniform temperature.

• The average thermal energy of electrons will not vary with x, but the electron density n(x)

• The electrons have an average thermal velocity vth and a mean free path l.

Diffusion Process• The electron at x = -l

have an equal chances of moving left or right.

• the average of electron flow per unit area F1 of electron crossing plane x = 0 from the left is expressed as

1

0.5 ( )0.5 ( ) th

c

n l lF n l v

Diffusion Process

• Likewise, the average of electron flow per unit area F2 of electron at x = l crossing plan at x = 0 from the right is

• Then, the net rate of electron flow from left to right is given by

2 0.5 ( ) thF n l v

1 2 0.5 [ ( ) ( )]

= 0.5 (0) (0)

=

th

th

th

F F F v n l n l

dn dnv n l n l

dx dx

dnv l

dx

Diffusion Process• Therefore, the net rate F can be written as

where Dn = vth.l = the diffusion coefficient (diffusivity)

• A current density caused by the electron diffusion is given by

(1)

n

dnF D

dx

e n

dnJ eF eD

dx

Diffusion Process

• Ex. An n-type semiconductor at T = 300 K, the electron concentration varies linearly from 1 x 108 to 7 x 107 cm-3 over a distance of 0.1 cm. Calculate the diffusion current density if the electron diffusion coefficient Dn is 22.5 cm2/s.

,

8 1719

2

1 10 7 10(1.6 10 )(22.5)

0.1

10.8 A/cm

e diff n n

dn nJ eD eD

dx x

Soln

Diffusion Process

Einstein Relation• From (1) and the equipartition of energy for one-

dimensional case

we can write

21 1

2 2thmv kT

2

n th

th th c

e eth

e e

e

D v l

v v

mv

e

kT m

m e

Einstein Relation

• Therefore,

• This is called “Einstein relation” since it relates the two constants that describe diffusion and drift transport of carriers, diffusivity and mobility, respectively.

• This Einstein relation can be used with holes as well.

n e

kTD

e

Value of diffusivities

Diffusion Process

Ex. Minority carriers (holes) are injected into a homogeneous n-type semiconductor sample at one point. An electric field of 50 V/cm is applied across the sample, and the field moves these minority carriers a distance of 1 cm in 100 μs. Find the drift velocity and the diffusivity of the minority carriers.

Diffusion Process

46

42

2

1 cm10 cm/s

100 10

10200 cm / V-s

50

0.0259 200 5.18 cm /s

h

hh

p h

vs

v

EkT

De

Soln

Diffusion Process

• For conclusion, when an electric field is applied in addition to a concentration gradient, both drift current and diffusion current will flow.

• The total current density due to electron movement can be written as

where n = electron density

e e n

dnJ e nE D

dx

Diffusion Process

The total conduction current density is given by

where Jh is the hole current =

p = hole density

total e hJ J J

h h h

dpJ e pE eD

dx

Diffusion Process• At a very high electric field, the drift velocity will sat

urated where it approaches the thermal velocity.

Electron as a wave

L. de Broglie said electrons of momentum p exhibits wavelike properties that are characterized by wavelength λ.

where h = Planck’s constant = 6.62 x 10-34 J.s

m = electron mass

v = speed of electron

h h

p mv

Electron as a wave

Ex. An electron is accelerated to a kinetic energy of 10 eV. Find

• velocity v• wavelength λ• accelerating voltage

Electron as a waveSoln

(a)

This implies that electron is not inside a solid since

v > vth (~105 m/s)

19

2

19

31

6

1 eV 1.6 10 J

110 eV

2

2 10 1.6 10

9.1 10

1.88 10 m/s

mv

v

v

Electron as a wave

Soln

(b)

34

31 6

10

6.62 10

9.1 10 1.88 10

3.87 10 m

h

mv

Electron as a wave

Soln

(c)

10 eV means an electron is accelerated by a voltage of 10 volts.

Therefore, accelerating voltage is 10 V.

Schrödinger’s equation• Kinetic energy + Potential energy = Total energy

• Schrodinger’s equation describes a wave equation in 3 dimensions is written as

where ħ = h/2

= Laplacian operator in rectangular coordinate

=

= wave function

V = potential term

22 ( , ) ( ) ( , ) ( , )

2r t V r r t i r t

m t

22 2 2

2 2 2x y z

Schrödinger’s equation

To solve the equation, we assume (r,t) = (r).(t) and substitute it in Schrödinger’s equation. We have

Divide both sides by (r).(t), we have

22 ( )

( ) ( ) ( ) ( ) ( ) ( )2

d tt r V r r t i r

m dt

2 2 ( ) 1 ( )( )

2 ( ) ( )

r d tV r i

m r t dt


Both can be equal only if they are separately equal to a constant E as

• Time dependent case:

• Position dependent: (Time independent)

( )( )

d ti E t

dt

22 ( ) ( ) ( ) ( )

2r V r r E r

m


Consider time dependent part, we can solve the equation as

where E = h = h/2 = ħ

Therefore

( ) expiE

t t

( ) expt i t


For position dependent, to make it simple, assume that electrons can move in only one dimension (x-direction).

(2)

where is probability of findng electron.

2 2

2

( )( ) ( ) ( )

2

d xV x x E x

m dx

2

2 2

( ) 2( ) ( ) 0

d x mE V x x

dx

2( )x


Now let us consider the 4 different cases for V(x)

Case 1: The electron as a free particle (V = 0)Equation (2) becomes

General solution of this equation is

2

2 2

( ) 2( ) 0

d x mE x

dx

( ) ikx ikxx Ae Be


From (r,t) = (r).(t), the general solution of Schrodinger’s equation in this case is

where A and B are amplitudes of forward and backward propagating waves and k is related to E by

2 221

2 2

kE mv

m

tiikxikx eBAetx ),(


This equation is called “de Broglie’s relation”.

2 2 21

2 2

k p

m m

p k

2 221

2 2

kE mv

m


Case 2: Step potential barrier (1 dimensional)

x


For region 1, the solution is already known as

(3)

For region 2, an equation (2) becomes

1 11( ) ik x ik xx Ae Be

21 2

2mEk

2

22 22 2

( ) 2( ) 0

d x mE V x

dx


General solution:

(4)

Since there is no wave incident from region 2, D must be zero.

2 22 ( ) ik x ik xx Ce De

2 22 2

2 ( )m E Vk


By applying a boundary condition, solutions must be continuous at x = 0 with

From (3) and (4), we have

1 2

1 2

(0) (0)

(0) )

(0d d

dx dx

1 2( )

A B C

ik A B ik C

1 2

1 2

1

1 2

2

B k k

A k k

C k

A k k


Now let us consider 2 cases:

1. E > V2: In this case k2 is real and k2 and k1 are different leading B/A finite. This means an electron could be seen in both region 1 and 2. It goes over the barrier and solution is oscillatory in space in both regions.

2. E < V2: In this case k2 is imaginary. The solution shows that an electron decays exponentially in region 2. An electron may penetrate the potential barrier.


Case 3: Finite width potential barrier (1 dimensional)


We are interested in the case of E < V2. In this case, solutions for region 1 and region 2 are the same as previous case. Now we turn our interest to region 3. At region 3,

33( ) ik xx Fe

23 2

2mEk


Consider transmittance, T is a ratio of energy in transmitted wave in region 3 to energy in incident wave in region 1.

This is called “Tunneling probability”.

22

2

22

2exp

1region in aveincident win Energy

3region in wavettedin transmiEnergy

EVm

wA

FT

T


• k3 is real, so that is not zero.

• Thus, there is a probability that electron crosses the barrier and appear at other side.

• Since the particle does not go over the barrier due to E < V2, this mechanism that particle penetrates the barrier is called “tunneling”.

2

3


Case 4: Finite potential well (1 dimensional)

x


Consider 2 cases

1. E > V1: Solutions in all regions are similar to those previous cases that is particle travels oscillatory everywhere.

2. E < V1: Region 1 ( x < 0, V = V1) ; Solution must be exponentially decaying

(5)

11( ) k xx Ae

21 2

2 ( )m V Ek


Region 2 (V = 0); Free particle case

(6)

2 2 2( ) sin( ) cos( )x B k x C k x

22 2

2mEk


Region 3 (x > 0, V3 = V1); Solution is again decaying.

13( ) k xx De

2 23 12

2 ( )m V Ek k


Applying boundary conditions:

at x = -L/2 at x = L/2

1 2

1 2

( ) ( )

( ) ( )

x x

d x d x

dx dx

2 3

2 3

( ) ( )

( ) ( )

x x

d x d x

dx dx


It is enough to solve only at x = L/2 due to its symmetrical.

(7)

By solving (5), this leads to

(8)

3

3

2 2

2 23

cos2

sin2

k L

k L

k LC De

k LCk Dk e

22 3tan

2

k Lk k


Substitute (8) into (5) and (6), we have

2

12tan

2

mL EE V E


If we consider in the case of V or infinite potential well


There is impossible that a particle can penetrate an infinite barrier. This brings (x) = 0 at x =0 and x = L. Applying the boundary conditions, we first have B = 0 and

(9)

We can solve for energy E by putting (3) = (6) and we get

2 ; where n = integerk L n

2

nk

L

2 2

28

n hE

mL

The uncertainty relationship

If we know the and use

where <A> is momentum or position, we can find the average and RMS values for both electron’s position and momentum.

*

2V

d

V

A dVA

dV

The uncertainty relationship

where x = uncertainty in position ofelectrons p = uncertainty in momentum of electrons

The uncertainty relationship also includes

x p h

E t h

Ex . Find the allowed energy levels for an elect -ron that is trapped in the infinite one dimen

sional potential well as in the figure below.

V(x)

-L/2 0 L/2

x

ENE 311 Lecture 2. Diffusion Process The drift current is the transport of carriers when an electric...

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