ENE 311
Lecture II
Diffusion Process
Now we already know the drift current, that 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”.
Assume an electron density varies in the x-direction, as shown in the figure
below, and assume this is under uniform temperature. As uniform temperature, the
average thermal energy of electrons will not vary with x, but the electron density Ne(x) =
n(x) will. The electrons have an average thermal velocity vth and a mean free path l. The
electron at x = -l have an equal chances of moving left or right. Therefore, the average of
electron flow per unit area F1 of electron crossing plane x = 0 from the left is expressed as
F1 
0.5n(l )  l
c
 0.5n(l )  vth
2
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
F  F1  F2  0.5vth [n(l )  n(l )]
dn  
dn  

= 0.5vth   n(0)  l    n(0)  l  
dx  
dx  

dn
=  vthl
dx
Therefore, the net rate F can be written as
F   Dn
dn
dx
where Dn = vth.l = the diffusion coefficient (diffusivity)
A current density caused by the electron diffusion is given by
J e  eF  eDn
dn
dx
(1)
Ex. An n-type semiconductor at T = 300 K, the electron concentration varies linearly
from 1 x 1018 to 7 x 1017 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.
Einstein Relation
From (1) and the equipartition of energy for one-dimensional case (
1 2 1
mvth  kT ), we can write
2
2
3
Dn  vthl
 vth  vth c 
 m 
 vth2  e e 
 e 
 kT   e me 



 me   e 
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.
4
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.
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
dn 

J e  e  e nE   eDn 
dx 

The total conduction current density is given by
where Jh is the hole current = J h  eh pE  eDh
dp
; p = hole density
dx
Note: At a very high electric field, the drift velocity will saturate where it approaches the
thermal velocity.
5
Electron as a wave
L. de Broglie said electrons of momentum P exhibits wavelike properties that are
characterized by wavelength λ.

h
h

p mv
where h = Planck’s constant = 6.62 x 10-34 J.s
m = electron mass
v = speed of electron
Ex. An electron is accelerated to a kinetic energy of 10 eV. Find
(a) velocity v
(b) wavelength λ
(c) accelerating voltage
6
Schrödinger’s equation
Kinetic energy + Potential energy = Total energy
Schrodinger’s equation describes a wave equation in 3 dimensions is written as

2
2m
2 (r , t )  V (r ) (r , t )  i

 (r , t )
t
where ħ = h/2
 2 = Laplacian operator in rectangular coordinate
2
2
2


=
x 2 y 2 z 2
 = wave function
V = potential term
To solve the equation, we assume (r,t) = (r).(t) and substitute it in
Schrödinger’s equation. We have

2
2m
 (t ) 2 (r )  V (r ) (r ) (t )  i  (r )
d (t )
dt
Divide both sides by (r).(t), we have

2 (r )
1 d (t )
 V (r )  i
2m  ( r )
 (t ) dt
2
Both can be equal only if they are separately equal to a constant E as
-
Time dependent case:
-
Position dependent: (Time independent)
Consider time dependent part, we can solve the equation as
 iE  
t
 

 (t )  exp  
where E = h = h/2 = ħ
7
Therefore
 (t )  exp  it 
For position dependent, to make it simple, assume that electrons can move in only
one dimension (x-direction).

d 2 ( x)
 V ( x) ( x)  E ( x)
2m dx 2
2
d 2 ( x) 2m
 2  E  V ( x) ( x)  0
dx 2
where
(2)
 ( x) is probability of finding electron.
2
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 ( x) 2m
 2 E ( x)  0
dx 2
General solution of this equation is
 ( x)  Aeikx  Beikx
From (r,t) = (r).(t), the general solution of Schrodinger’s equation in this
case is
 ( x, t )   Ae  ikx  Be  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
k
1
 mv 2
2m 2
2 2
k
1 p2

2m 2 m
p k
The last equation is called “de Broglie’s relation”.
8
Case 2: Step potential barrier (1 dimensional)
For region 1, the solution is already known as
 1 ( x)  Aeik x  Beik x
1
where k1 
2
1
(3)
2mE
2
For region 2, an equation (2) becomes
General solution:
 2 ( x)  Ceik x  De ik x
2
where k2 
2
2
(4)
2m( E  V2 )
2
Since there is no wave incident from region 2, D must be zero. By applying a
boundary condition, solutions must be continuous at x = 0 with
 1 ( x)   2 ( x)
d 1 ( x) d 2 ( x)
dx

dx
From (3) and (4), we have
A B C
ik1 ( A  B)  ik2C
9
B k1  k2

A k1  k2
C
2k1

A k1  k2
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,
 3 ( x)  Feik x
3
where
k32 
2mE
2
Consider transmittance, T is a ratio of energy in transmitted wave in region 2 to
energy in incident wave in region 1.
This is called “Tunneling probability”.
10
k3 is real, so that
 3 is not zero. Thus, there is a probability that electron crosses
2
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”.
Case 4: Finite potential well (1 dimensional)
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
 1 ( x)  Aek x
1
where
k12 
2m(V  E )
2
(5)
Region 2 (V = 0); Free particle case
 2 ( x)  B sin(k2 x)  C cos(k2 x)
where
k22 
2mE
(6)
2
Region 3 (x > 0, V3 = V1); Solution is again decaying.
 3 ( x)  Dek x
1
where
k32 
2m(V  E )
2
 k12
11
Applying boundary conditions:
at x = -L/2
 1 ( x)   2 ( x)
d 1 ( x) d 2 ( x)
dx

dx
at x = L/2
 2 ( x)   3 ( x)
d 2 ( x) d 3 ( x)
dx

dx
It is enough to solve only at x = L/2 due to its symmetrical.

k L
C cos  2   De
 2 
k3 L
2
k L
 3
 k2 L 
2
Ck2 sin 
   Dk3e
 2 
(7)
By solving (5), this leads to
k L
k2 tan  2   k3
 2 
Substitute (8) into (5) and (6), we have
 mL2 E
E tan 
 2 2


  V1  E


If we consider in the case of V   or infinite potential well
(8)
12
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
k2 L  n ; where n = integer
where
k2 
n
L
(9)
We can solve for energy E by putting (3) = (6) and we get
E
n2h2
8mL2
Ex. Find the allowed energy levels for an electron that is trapped in the infinite onedimensional potential well as in the figure below.
V(x)


-L/2
Soln
0
L/2
x
Time independent SchrÖdinger Equation
d 2 2m
 2  E  V   0
dx 2
(1)
Electron is trapped inside the potential well (V = 0). Eq. (1) becomes
d 2 2m
 2 E  0
dx 2
General solution of (2) is
 ( x)  A sin kx  B cos kx
Applying boundary conditions for (x) at x =  L/2
  

 kL 
 kL 
  A sin 
  B cos  2   0
2
 2 


L
(2)
13
  
L
 
L

 kL 
 kL 
   A sin    B cos    0
2
 2 
 2 
(3)
 kL 
 kL 
  A sin    B cos    0
2
2
 
 2 
 
(4)
(3) + (4):
 kL 
2 B cos    0
 2 

kL n

; n  1,3,5,...
2
2
(3) - (4):
 kL 
2 A sin    0
 2 

kL n

; n  2, 4, 6,...
2
2
Therefore,
kL n

for n  1, 2,3,...
2
2
n
k
for n  1, 2,3,...
L
Then we can find E by using (2) and its general solution.
 Ak 2 sin kx  Bk 2 sin kx   A sin kx  b cos kx 
2mE
 2mE
 2mE
2
2
 2  k  A sin kx   2  k  B cos kx  0




L
L
 2mE
2
 2  k   0 for -  x 
2
2


2 2
nh
En 
for n  1, 2,3,...
8mL2
2
0