Transcript Lecture 2 - web page for staff

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
F1 
0.5n(l )  l
c
 0.5n(l )  vth
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
F2  0.5n(l )  vth
• Then, the net rate of electron flow from left to right is
given by F  F  F  0.5v [n(l )  n(l )]
1
2
th
dn  
dn  

= 0.5vth   n(0)  l    n(0)  l  
dx  
dx  

dn
=  vthl
dx
Diffusion Process
• Therefore, the net rate F can be written as
dn
F   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
(1)
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.
Diffusion Process
Soln
J e,diff
dn
n
 eDn
 eDn
dx
x
8
17


1

10

7

10
19
 (1.6  10 )(22.5) 

0.1


 10.8 A/cm 2
Einstein Relation
• From (1) and the equipartition of energy for onedimensional case
1 2 1
mvth  kT
2
2
we can write
Dn  vthl
 vth  vth c 
 m 
 vth2  e e 
 e 
 kT

 me
  e me 


e


Einstein Relation
• Therefore,
 kT
Dn  
 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
Soln
1 cm
4

10
cm/s
6
100  10 s
vh 104
h  
 200 cm 2 / V-s
E 50
 kT 
D p    h  0.0259  200  5.18 cm 2 / s
 e 
vh 
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
dn 

J e  e  e nE   Dn 
dx 

where n = electron density
Diffusion Process
The total conduction current density is given by
Jtotal  J e  J h
where Jh is the hole current =
dp
J h  eh pE  eDh
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.

where
h
h

p mv
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
Soln
(a)
1 eV  1.6  1019 J
1 2
mv  10 eV
2
2  10  1.6  1019
v
9.1 1031
v  1.88  106 m/s
This implies that electron is not inside a solid since
v > vth (~105 m/s)
Electron as a wave
Soln
(b)
h

mv
6.62  1034

31
6
9.1  10  1.88  10
 3.87  1010 m
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


  (r , t )  V (r ) (r , t )  i  (r , t )
2m
t
2
2
where
ħ = h/2
 2 = Laplacian operator in rectangular coordinate
=
2
2
2
 2 2
2
x
y
z
 = 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
d (t )

 (t )  (r )  V (r ) (r ) (t )  i  (r )
2m
dt
2
2
Divide both sides by (r).(t), we have
2 (r )
1 d (t )

 V (r )  i
2m  ( r )
 (t ) dt
2
Schrödinger’s equation
Both can be equal only if they are separately equal to a
constant E as
• Time dependent case:
i
d (t )
 E (t )
dt
• Position dependent: (Time independent)

2
2m
 2 (r )  V (r ) (r )  E (r )
Schrödinger’s equation
Consider time dependent part, we can solve the
equation as
 iE  
t
 

 (t )  exp  
where E = h = h/2 = ħ
Therefore
 (t )  exp  it 
Schrödinger’s equation
For position dependent, to make it simple, assume
that electrons can move in only one dimension (xdirection).
d 2 ( x)

 V ( x) ( x)  E ( x)
2
2m dx
2
d 2 ( x) 2m
 2  E  V ( x) ( x)  0
2
dx
(2)
where  ( x) is probability of findng electron.
2
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 ( x) 2m
 2 E ( x)  0
2
dx
General solution of this equation is
 ( x)  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
2
k2 1 2
E
 mv
2m 2
Schrödinger’s equation
2
2
k
1 2
E
 mv
2m 2
2 2
2
k
1p

2m 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 ( x)  Ae
k 
2
1
 ik1x
 Be
 ik1x
2mE
2
For region 2, an equation (2) becomes
d 2 2 ( x) 2m
 2  E  V2  2 ( x)  0
2
dx
(3)
Schrödinger’s equation
General solution:
 2 ( x)  Ce
ik2 x
k 
2
2
 De
ik2 x
(4)
2m( E  V2 )
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) d 2 (0)
dx
From (3) and (4), we have
A B C
ik1 ( A  B )  ik2C

dx
B k1  k2

A k1  k2
C
2k1

A k1  k2
Schrödinger’s equation
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.
Schrödinger’s equation
Case 3: Finite width potential barrier (1 dimensional)
Schrödinger’s equation
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
k 
2
3
2mE
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.
Energyin transmitted wave in region 3
T
Energyin incident wave in region1

2mV2  E  
T  2  exp 2w

2

A


F
2
This is called “Tunneling probability”.
Schrödinger’s equation
• k3 is real, so that  3 is not zero.
2
• 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”.
Schrödinger’s equation
Case 4: Finite potential well (1 dimensional)
x
Schrödinger’s equation
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
k 
2
1
2m(V  E )
2
(5)
Schrödinger’s equation
Region 2 (V = 0); Free particle case
 2 ( x)  B sin(k2 x)  C cos(k2 x)
k 
2
2
2mE
2
(6)
Schrödinger’s equation
Region 3 (x > 0, V3 = V1); Solution is again decaying.
 3 ( x)  Dek x
1
k 
2
3
2m(V  E )
2
 k12
Schrödinger’s equation
Applying boundary conditions:
at x = -L/2
at x = L/2
 1 ( x)   2 ( x)
d 1 ( x) d 2 ( x)
dx

dx
 2 ( x)   3 ( x)
d 2 ( x) d 3 ( x)
dx

dx
Schrödinger’s equation
It is enough to solve only at x = L/2 due to its
symmetrical.
k L
 3
 k2 L 
2
C cos 
  De
 2 
k L
 3
 k2 L 
2
Ck sin 


Dk
e

3
 2 
By solving (5), this leads to
 k2 L 
k2 tan 
  k3
 2 
(7)
(8)
Schrödinger’s equation
Substitute (8) into (5) and (6), we have
 mL2 E 
E tan 
  V1  E
 2 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
k2 L  n ; where n = integer
n
k2 
L
(9)
We can solve for energy E by putting (3) = (6) and we
get
n2h2
E
8mL2
The uncertainty relationship
If we know the  and use
A 
*

 A dV
V

V
 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.

-L/2
V(x)
0

L/2
x