Resonant field emission through diamond thin films Zhibing Li

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