ENE 311

• 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 v th 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 F
1 of electron crossing plane x = 0 from the left is expressed as
F
1

0.5 (
  l
 c

0.5 (
  v th

Diffusion Process
• Likewise, the average of electron flow per unit area F
2 of electron at x
= l crossing plan at x = 0 from the right is
F
2

0.5 ( )
 v th
• Then, the net rate of electron flow from left to right is given by
F

F
1

F
2

0.5 [ ( th l ) ( )]
= 0.5
v th 

 n (0)
 l dn dx n (0)
 l dn dx

 
=
 dn dx

Diffusion Process
• Therefore, the net rate F can be written as
F
 
D n dn dx where D n
(diffusivity)
= v th
.l
= the diffusion coefficient
• A current density caused by the electron diffusion is given by
(1)
J e
  eF
 eD n dn dx

Diffusion Process
• Ex.
An n-type semiconductor at T = 300 K, the electron concentration varies linearly from
1 x 10 8 to 7 x 10
7 cm
-3 over a distance of 0.1 cm. Calculate the diffusion current density if the electron diffusion coefficient D n cm
2
/s.
is 22.5

Diffusion Process
Sol n
J
 eD n dn dx
 eD n
 n
 x

(1.6 10

19
)(22.5) 



10.8 A/cm
2
 8   17
0.1




Einstein Relation
• From (1) and the equipartition of energy for onedimensional case
1
2 mv th
2 
1
2 kT we can write
D n
 v l th
 v
 v
 th th c

 2 v th
 m e e e
 kT
  e m e
 m e
 e


Einstein Relation
• Therefore,
D n
 kT

 e 
 e
• 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.

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
Sol n v h

1 cm

6 s

 h
D p v
E h
10
4
  
50

 kT

 e

 h
  
5.18 cm / s

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
J e
 e


  e nE
 
D n dn dx

 where n = electron density

Diffusion Process
The total conduction current density is given by
J total

J e

J h where J h is the hole current =
J h
 e
 h pE
 eD h dp dx p = hole density

Diffusion Process
• At a very high electric field, the drift velocity will saturated 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 .
   h p mv where h
= Planck’s constant = 6.62 x 10 -34
J.s
m = electron mass v = speed of electron

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 wave
Sol n
(a)

19 1 eV 1.6 10 J
1
2 mv 2 
10 eV

19 v


31 v
  6
1.88 10 m/s
This implies that electron is not inside a solid since v > v th
(~10 5 m/s)

Electron as a wave
Sol n
(b)
  h mv

34

 
31   6

3.87 10

10
m

Electron as a wave
Sol n
(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
2
 
2 m
2

( , )
   i

 t
 where ħ = h /2 
 2 = Laplacian operator in rectangular coordinate
= 
2
 x
2


 y
2
2


 z
2
2

= wave function
V = potential term

Schrödinger’s equation
To solve the equation, we assume

(r,t) =

(r).

(t) and substitute it in Schrödinger’s equation.
We have

2
2 m
 t
 2

( )
    i

( ) d

( ) dt
Divide both sides by

(r).

(t) , we have

2  2

2 m

( )

( )
 i
1 d

( )

( ) dt

Schrödinger’s equation
Both can be equal only if they are separately equal to a constant E as
• Time dependent case: i d

( )

E

( ) dt
• Position dependent: (Time independent)
2
 
2 m
2

( )
  
E

( )

Schrödinger’s equation
Consider time dependent part, we can solve the equation as

( )
 exp




 iE

 t
 where E = h

= h
 /2  = ħ 
Therefore

( )
 exp
   

Schrödinger’s equation
For position dependent, to make it simple, assume that electrons can move in only one dimension (xdirection).

2 2 d

2 m dx
2
( )
  
E

( ) d
2
 dx
2
( )

2 m
2
 
( )
 
( )

0
(2)
 2 where is probability of findng electron.

Schrödinger’s equation
Now let us consider the 4 different cases for V(x)
Case 1: The electron as a free particle ( V = 0)
Equation (2) becomes d
2
 dx
( )
2

2 m
2
E

( )

0
General solution of this equation is

( )

Ae
 ikx 
Be
 ikx

Schrödinger’s equation
From

(r,t) =

(r).

(t) , the general solution of
Schrodinger’s equation in this case is

( x , t )


Ae
 ikx 
B
 ikx
 e
 i
 t where A and B are amplitudes of forward and backward propagating waves and k is related to E by
E

2
2 2 k m

1
2 mv
2

Schrödinger’s equation
E

2 2 k

1
2 m 2 mv
2
2
2 2 k m

1
2 p
2 m p
 k
This equation is called “de Broglie’s relation”.

Schrödinger’s equation
Case 2: Step potential barrier (1 dimensional) x

Schrödinger’s equation
For region 1, the solution is already known as

1
( )

Ae
 k
2 
1
2 mE
2

Be

For region 2, an equation (2) becomes d
2
 dx
2
2 x

2 m
2


2
 
2
( )

0
(3)

Schrödinger’s equation
General solution:

2
( )

Ce

De

(4) k
2
2 
2 (

2
)
2
Since there is no wave incident from region 2, D must be zero.

Schrödinger’s equation
By applying a boundary condition, solutions must be continuous at x = 0 with

1
(0)
 
2
(0) d

1
(0) dx
 d

2
(0 ) dx
From (3) and (4), we have
A B C ik A
1

B )
 ik C
2
B
A
 k k
1
1

 k k
2
2
C

2
 k
1
A k k
1 2

Schrödinger’s equation
Now let us consider 2 cases:
1. E > V
2
: In this case k
2 is real and k
2 and k
1 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 < V
2
: In this case k
2 is imaginary. The solution shows that an electron decays exponentially in region
2. An electron may penetrate the potential barrier.

Schrödinger’s equation
Case 3: Finite width potential barrier (1 dimensional)

Schrödinger’s equation
We are interested in the case of E < V
2
. 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
,

3
( )

Fe k
2 
3
2 mE
2

Schrödinger’s equation
Consider transmittance, T is a ratio of energy in transmitted wave in region 3 to energy in incident wave in region 1.
T
T

Energy in transmi
 tted wave in region 3
Energy in incident w
F
2
A
2
 exp 



2 w ave in region 1
2 m

V
2

E

 2



This is called “ Tunneling probability ”.

Schrödinger’s equation
• k
3

3
2 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 < V
2
, this mechanism that particle penetrates the barrier is called “ tunneling
”.

Schrödinger’s equation
Case 4: Finite potential well (1 dimensional) x

Schrödinger’s equation
Consider 2 cases
1. E > V
1
: Solutions in all regions are similar to those previous cases that is particle travels oscillatory everywhere.
2. E < V
1
: Region 1 ( x < 0, V = V
1
) ; Solution must be exponentially decaying

1
( )

Ae k
1
2 
2 (

E )
2
(5)

Schrödinger’s equation
Region 2 (V = 0); Free particle case

2
( )

B sin( k x
2
)

C cos( k x
2
) k
2 
2
2 mE
2
(6)

Schrödinger’s equation
Region 3 (x > 0, V
3
= V
1
); Solution is again decaying.

3
( )

De k
3
2 
2 (

E )
2
 k
1
2

Schrödinger’s equation
Applying boundary conditions: at x = -L/2

1 x
 
2 x d

1 dx
( )
 d

2 dx
( ) at x = L/2

2 x
 
3 x d

2 dx
( )
 d

3 dx
( )

Schrödinger’s equation
It is enough to solve only at x = L/2 due to its symmetrical.
C cos


 k L
2
2 

 
De

2

Ck sin


 k L
2
2



 
Dk e
3

2
By solving (5), this leads to k
2 tan

 k L
2
2
 k
3
(7)
(8)

Schrödinger’s equation
Substitute (8) into (5) and (6), we have
E tan


2 mL E
2

 
V
1

E
2

Schrödinger’s equation
If we consider in the case of V
  or infinite potential well

Schrödinger’s equation
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 k L
2
 n
 n
 k

2
L
; where n = integer
(9)
We can solve for energy E by putting (3) = (6) and we get
E

2 2 n h
8 mL
2

The uncertainty relationship
If we know the
 and use
A


V

V
A dV

2 d dV where <A> is momentum or position, we can find the average and RMS values for both electron’s position and momentum.

The uncertainty relationship x p h where  x = uncertainty in position of electrons
 p = uncertainty in momentum of electrons
The uncertainty relationship also includes
E t h

Ex. Find the allowed energy levels for an electron that is trapped in the infinite one-dimensional potential well as in the figure below.


)
V(x
-L/2 0
L/2 x