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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?
Sin
Da
Vacuum
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.
2 4 6 8 10
0.2
0.4
0.6
0.8
In a ultra-thin film, resonant tunnelling is possible
Randomness tends to create local states (P.W. Anderson 1958)
E0
100
AV
E0 - VA
cE
vE
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.
The energy of injected electrons ~
Typical scales of the potential
iE
00 iEU
0 bd
iE AAV
AV
nmVdVU
nmd
eV
eV
A
A
/2.0/)(
15~5
1
77.0
0
0
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.
)6()5()2()10/1()( 0000 xVxVxVxUxU
incidentelectrons electrons
dtransmitte
0 ax
0U
2 5 6
The hamiltonian
n
iii xxVxVixxU
10 )()(}),2,1|{,(
),0( dxi
xeF
UxVr
0)(
dxeFU AA
),0( dx
dx
)(2 2
22
xUdx
d
mH
m
EV 0
0
2 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
Let , ,
one has
02
02
xeF
EUm
r
3
2
2 Fml r
2
02)(
2lEU
m
l
x
0)()(
2
2
d
d
i) Classical region
Let
one has
This is a 1/3 order Bessel equation, the solution is
Two independent wave-function of energy E are
0
)(1
)(,3
2 2/3
uRu
0)()3
1()()(
222 uRuuRuuRu
)(),( 3/13/1 uYuJ
)(2
3)( 3/1
3/1
1 uJu
x
)(2
3)( 3/1
3/1
2 uYu
x
ii) Non-classical region
Let
The Schoedinger equation becomes
It is a 1/3 order modified Bessel equation. The solutions are
Two independent wave-functions are
0
)(2
3)(,
3
23/1
2/3 xv
vSv
0)()3
1()()(
222 vSvvSvvSv
)(),( 3/13/1 vKvI
)(2
3)( 3/1
3/1
1 vIv
x
)(2
3)( 3/1
3/1
2 vKv
x
iii) Matrix representation
In classical region, a general state of fixed energy E can be written as
In the matrix representation,
In the non-classical region,
iv) Connection condition at the transition point
)()()( 22
11 xbxbx
2
1
b
bB
)()()( 22
11 xcxcx
2
1
c
cC
x0
MCB
20
2
31
M
2) Include the random potential
j 0 1 2 …… i x0 i+1 i+2 ...... n d y0 b
C0 C1 Ci Bi+1 Bi+2 Bn Bn+1 Cn+2 Bn+3
At there is a local potential
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:
,
,
jxxV 0jx
)()(
)()()(
21
21
xx
xxxN
0xx
)()(
)()(
21
21
xx
xx
0xx
Connection conditions
i) In non-classical region j
Cj-1 Cj
In matrix representation
Define matrix
0)()0()0( jjj xxx
202
Vm
121
1 )()(
00)()(
j
jjjjjj C
xxCxNCxN
)()()(
)()()(
)(
1)(
2121
2212
jjj
jjj
jj xxx
xxx
xxT
)()()()()( 2121 jjjjj xxxxx
Then the connection condition is
ii) In classical region
Replace by in the matrix , one has
iii) Transmission coefficient
Define
1)( jjj CxTIC
21, 21, )( jxT
1)( jjj BxTIB
i
ll
n
ill NxTMxTdNG
1
1
1
)0()(1)(1)(
11 )()(
bNMdNH
22121222211111211 HGHGkkHGHGz fi
22111221112221122 HGHGkHGHGkz fi
Where ,
with the wave numbers of incident and transmitted electrons respectively, is the longitudinal mass of electron in Si.
The transmission coefficient is given by
3) Average over n
For specification, assume n follows the Poisson distribution
where is the mean value of n.
il
i km
mk
fe
f km
mk
fi kk ,
22
21
2
21
4),,,(
zz
GkkxxxD finn
!n
nep
nn
n
n
lm
For a given n, we generate m positions
In the case of uniform distribution, those positions regularly locate in the range (0, d) with separation . The average transmission coefficient is
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.
),,( 21 nxxx
)1/( md
0
),(n
nni DpFED
n
nnni xxxDpFED ,,),( 21
0
A numerical solution for n=3
2 4 6 8
0.5
1
1.5
2
2.5
2
3
4
5
6
1
1.5
2
2.53
0
0.2
0.4
2
3
4
5
6
1
1.5
2
2.53
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
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
iBF
tii dEdE
TkEE
mdEdETEEN
/)(exp1
1
2),,(
32
E
E
iii
c
dEFEDTEENTFEj ),(),,(),,(
tm
The emitted current is
Integrate with respect to E first, one attains
iB
FiBtii dE
Tk
EETkmdETEN
)exp(1ln
2),(
32
i
E
ii dEFEDTENeTFJc
),(),(),(
cE
dETFEjeTFJ ),,(),(
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