Download A simple model for electron emission through a non

Resonant field emission
through diamond thin films
Zhibing Li
1. The problem
2. The picture
3. A simple model
4. Solution
5. Emitted current
6. Discussions
1. The problem
 A model for this system
 The emitted current?
 Can one expect any novel
(useful) feature in amorphous
diamond ultra-thin films?
n   Si
Vacuum
aD
2. The Picture
 experiment hints

The film is an insulator of nano scale.
---- Quantum effects would be important.

The amorphous diamond locally likes a crystal but is disordered in
long-range.
---- The band structure is similar to the diamond crystal but both
valence band and conduction band have band tails of local states
(Mott 1967).

Emission is enhanced by dopants of N and Li etc.

Low threshold voltage is detected in polycrystalline diamond.
 In a ultra-thin film, resonant tunnelling is possible
0.8
0.6
0.4
E0
VA
0.2
2
10
0
4
6
8
10
E0 - VA
 Randomness tends to create local states (P.W.
Anderson 1958)
Ec
Ev
local states (empted)
local states (occupied)
 What create the local states?
 the randomness of amorphous diamond
 the grain boundaries of polycrystalline
 defects, impurities, and stack faults etc
 The scattering mainly is caused by
(i) film boundaries;
(ii) local potentials corresponding to the local states.
Typical scales of the potential
The energy of injected electrons ~
Ei
U 0  Ei  0
VA   A
Ei
0  0.77eV
 A  1eV
VA
0
d  5 ~ 15nm
(U 0  VA ) / d  0.2V / nm
d
b
3. A simple model
One dimension model.
The effect of local states is represented by a series of
delta potentials, each of which has a bound state.
U ( x)  U 0 (1  x / 10)  V0 ( x  2)  V0 ( x  5)  V0 ( x  6)
incident
U0
transmitted
electrons
electrons
0
2
x 5
6
a
The hamiltonian
2 d 2
H 
 U ( x)

2
2m dx
n
U ( x, {xi | i  1,2, })  V ( x)  V0   ( x  xi )
xi  (0, d )
V ( x)  U 0 
eF
r
x
 U A   A  eF x  d 
V0 
2 E0


m
i 1
x  (0, d )
xd
E0 is the difference between the bottom of
conduction band and the energy of the local state
4. The solution
It is an exercise of quantum mechanics.
1) Solution for a linear potential
The Schoedinger equation is
2m 
eF
   2 U 0  E 
 
r
Let l  3
one has

x   0


2
m
2 ,
, 
(
U

E
)
l
0
2m F
2
 r2
d 2 ( )
  ( )  0
2
d
x
  
l
i) Classical region   0
Let
2
1
u   3 / 2 , R(u ) 
 ( )
3

one has
1
) R(u )  0
2
3
This is a 1/3 order Bessel equation, the solution is J1/ 3 (u ), Y1/ 3 (u )
u 2 R(u )  uR(u )  (u 2 
Two independent wave-function of energy E are
1/ 3
 3u 
 1 ( x)    J1/ 3 (u )
 2
1/ 3
 3u 
 2 ( x)    Y1/ 3 (u )
 2
ii) Non-classical region
Let
 0
2
 3v 
3/ 2
v     , S (v)   
3
 2
1 / 3
 ( x)
The Schoedinger equation becomes
1
v S (v)  vS (v)  (v  2 ) S (v)  0
3
2
2
It is a 1/3 order modified Bessel equation. The solutions are I1/ 3 (v), K1/ 3 (v)
Two independent wave-functions are
1/ 3
 3v 
1 ( x )    I 1 / 3 ( v )
 2
1/ 3
 3v 
 2 ( x )    K1 / 3 ( v )
 2
iii) Matrix representation
In classical region, a general state of fixed energy E can be written as
 ( x)  b1 1 ( x)  b 2 2 ( x)
In the matrix representation,
 b1 
B   2 
b 
In the non-classical region,
 ( x)  c11 ( x)  c 2 2 ( x)
 c1 
C   2 
c 
iv) Connection condition at the transition point
B  MC
x0

 1
M 

0

3 

2 
 
 
2 
2) Include the random potential
j
0
1
2 …… i
C0 C1
x0
Ci
i+1
i+2 ...... n
Bi+1 Bi+2
d
y0
b
Bn Bn+1 Cn+2 Bn+3
At x j there is a local potential  V0 x  x j 
At x0 electron crosses from non-classical region to classical region,
at a it enters the non-classical region again and gets out at y0.
Define a matrix:
 1 ( x)  2 ( x) 

N ( x)  
 1( x)  2 ( x)  ,
x  x0
1 ( x)  2 ( x) 

 
1( x)  2 ( x)  ,
x  x0
Connection conditions
i) In non-classical region
j
 ( x j  0)   ( x j  0)   ( x j )  0
2mV0
 
2
Cj-1
In matrix representation
0 
 0

C j 1
N ( x j )C j  N ( x j )C j 1   


(
x
)

(
x
)
1
j
2
j


Define matrix
  22 ( x j ) 
1    2 ( x j )1 ( x j )

T (x j ) 
2

1 ( x j )
1 ( x j ) 2 ( x j ) 
( x j ) 
( x j )  1 ( x j ) 2 ( x j )  1 ( x j ) 2 ( x j )
Cj
Then the connection condition is


C j  I  T ( x j ) C j 1
ii) In classical region
Replace 1 ,  2 by 1 , 2 in the matrix T ( x j ) , one has


B j  I  T ( x j ) B j 1
iii) Transmission coefficient
Define
n
i
l i 1
l 1
G  N (d  )  1  T ( xl ) M  1  T ( xl  ) N (0  ) 1
H  N (d  ) M 1 N (b ) 1
z1  G21H11  G11H 21   ki k f G22 H12  G12 H 22 
z2  ki G12 H 21  G22 H11   k f G21H12  G11H 22 
m
ki 
ki
ml
Where
,
m
kf 
kf
me
with ki , k f the wave numbers of incident and transmitted electrons
respectively, ml is the longitudinal mass of electron in Si.
The transmission coefficient is given by
Dn ( x1 , x2 , , xn ) 
4k i k f G
2
z12  z 22
3) Average over n
For specification, assume n follows the Poisson distribution
pn  e
where
n
n
nn
n!
is the mean value of n.
For a given n, we generate m positions ( x1 , x2 , xn )
In the case of uniform distribution, those positions regularly locate in
the range (0, d) with separation d /( m  1) . The average transmission
coefficient is

D ( Ei , F )   pn Dn
n 0
In the case of random distribution, the positions of delta potentials are
generated by Monte Carlo method. Average over samples of positions should
be done.

D ( Ei , F )   pn Dn  x1 , x2 ,  xn  
n 0
A numerical solution for n=3
3
2.5
2.5
2
2
1.5
1
1.5
1
0.4
0.5
2
4
6
8
0.2
0
2
3
4
5
6
5. Emitted current
The number of electrons with energy between E~E+dE and
with normal energy between Ei~Ei+dEi impinging on the
diamond film from the semiconductor (heavy dopped) is
N ( E , Ei , T )dEdEi 
mt
1
dEdEi
2 3
2  1  exp ( E  EF ) / k BT 
mt is the transversal mass of electron in Si.
The number of electrons emitted per unit area per unit time
with total energy E~E+dE , that is the so-called total energy
distribution (TED) of the emitted electrons, is given by
E
j ( E , F , T )   N ( E , Ei , T ) D ( Ei , F )dEi
Ec
The emitted current is

J ( F , T )  e  j ( E , F , T )dE
Ec
Integrate with respect to E first, one attains

J ( F , T )  e  N ( Ei , T ) D ( Ei , F )dEi
Ec
N ( Ei , T )dEi 
mt k BT 
Ei  EF 
ln
1

exp(

) dEi

2 3
2  
k BT 
6. Discussions
 Replace the  potential by a more realistic one.
 For thick film, defect density should be used instead of
isolate defects.
 3D
 The defects have importance consequences. The simple
model shows the possibility of emission enhancement by
defects (such as doping of nitrogen)
 Resonance transmission is the most idea case for
applications