handout 16 electrical conduction in energy bands · 2012. 3. 12. · handout 16 electrical...
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Handout 16
Electrical Conduction in Energy Bands
In this lecture you will learn:
• The conductivity of electrons in energy bands • The electron-hole transformation• The conductivity tensor• Examples• Bloch oscillations
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Inversion Symmetry of Energy Bands
Recall that because of time reversal symmetry:
rr knkn
,,
* kEkE nn
We know that:
kEkv nkn
1
Now let go to in the above equation:k
k
kvkv
kv
kE
kE
kEkv
nn
n
nk
nk
nkn
1
1
1
xk
a
a
Energy
xk
a
a
Energy
2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Current Density for Energy Bands
For a free electron gas the current density was given as:
kvkfkd
ekvkfV
eJk
3
3
all 22
2
In Drude model, the electron current density was given as:
venJ
Now we want to find the current density due to electrons in energy bands
The current density due to electrons in the n-th band can be written in a manner similar to the free-electron case:
kvkf
kde
kvkfV
eJ
nn
nk
nn
FBZ3
3
FBZ in
22
2
xk
a
a
Energy
xk
a
a
Energy
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Current Density for a Completely Filled or Empty Bands
xk
a
a
Energy
xk
a
a
Energy
Completely filled bands do not contribute to electrical current or to electrical conductivity
where I have used the fact:
kvkv nn
Ef
Completely empty bands do not contribute to electrical current or to electrical conductivity
Only partially filled bands contribute to electrical current and to electrical conductivity
Of course, if for all in FBZ: 0kfn
k
0
22
FBZ3
3
kvkfkd
eJ nnn
Consider a completely filled band for which for all in FBZ
Application of an external field will not change anything!
1kfn
k
0
22
22
FBZ3
3
FBZ3
3
kvkd
ekvkfkd
eJ nnnn
3
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Current Density and Electron-Hole Transformation
kvkfkd
e
kvkfkd
ekvkd
e
kvkfkd
e
kvkfkd
eJ
nn
nnn
nn
nnn
12
2
12
22
2
112
2
22
FBZ3
3
FBZ3
3
FBZ3
3
FBZ3
3
FBZ3
3
Consider the expression for the current density for a partially filled band:
The final result implies that since the current density of a filled band is zero, the current density for any band can always be expressed in two equivalent ways:
a) As an integral over all the occupied states assuming negatively charged particles (as in (1) above)
a) As an integral over all the unoccupied states assuming positively charged particles (as in (2) above)
(1)
(2)
0
xk
a
a
Energy
xk
a
a
Energy
Ef
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Current Density and Electron-Hole Transformation
The Electron Choice:The current density is given by:
The Hole Choice:The current density is given by:
xk
a
a
Energy
xk
a
a
Energy
Ef
kvkfkd
eJ nnn
FBZ3
3
22
kvkfkd
eJ nnn
1
22
FBZ3
3
Current is understood to be due to negatively charged electrons This choice is better when the electron number is smaller than the hole number
Current is understood to be due to positively charged fictitious particles called “holes”
This choice is better when the hole number is smaller than the electron number
One has two choices when calculating current from a partially filled band:
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Metals, Semiconductors, and InsulatorsMaterials can be classified into three main categories w.r.t. their electrical properties: Metals: In metals, the highest filled band is partially filled (usually half-filled)Semiconductors: In semiconductors, the highest filled band is completely filled (at least at zero temperature) Insulators: Insulators are like semiconductors but usually have a much larger bandgap
xk
Energy
FBZ
Metal
Ef
xk
Energy
FBZ
Semiconductor
Ef
xk
Energy
FBZ
Insulator
Ef
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Inclusion of Scattering in the Dynamical Equation
In the presence of a uniform electric field the crystal momentum satisfies the dynamical equation:
Ee
dttkd
Now we need to add the effect of electron scattering.As in the free-electron case, we assume that scattering adds damping:
ktkEe
dttkd
Steady State Solution:
The boundary condition is that: ktk
0
Note: the damping term ensures that when the field is turned off, the crystal momentum of the electron goes back to its original value
Ee
ktk
In the presence of an electric field, the crystal momentum of every electron is shifted by an equal amount that is determined by the scattering time and the field strength
Energy
xk
Ex
Conduction band
hh valence band
ℓh valence band
Energy
xk
Ex
Conduction band
hh valence band
ℓh valence band
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity: Conduction Band
oToocc kkMkkkEkE
..
21
2
oc kkMkv
.1
The velocity of electrons is:
Consider a solid in which the energy dispersion for conduction band near a band minimum is given by:
Energy
k
Conduction band
ok
The current density is:
kvkf
kdeJ cc
kc
o
near3
3
22
In equilibrium, for every state with crystal momentum that is occupied, the state is also occupied and these two states have opposite velocities.
Therefore in equilibrium:
okk
okk
0
22
near3
3
kvkfkd
eJ cck
co
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity: Conduction Band
Conduction band
Energy
k
ok
Now assume that an electric field is present that shifts the crystal momentum of all electrons:
xk
yk
Electron distribution in k-space when E-field is zero
xk
yk
Electron distribution is shifted in k-space when E-field is not zero
xEE x ˆ
Ee
ktk
Distribution function: kfc
Distribution function:
E
ekfc
Ee
Since the wavevector of each electron is shifted by the same amount in the presence of the E-field, the net effect in k-space is that the entire electron distribution is shifted as shown
kfc
ok
ok
E
6
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Current Density:
Do a shift in the integration variable:
E
EMenJ
EMkfkd
eJ
Ee
kkMkfkd
eJ
Ee
kvkfkd
eJ
c
ck
c
ock
c
cck
c
o
o
o
.
.
.2
2
.2
2
22
12
1
near3
32
1
near3
3
near3
3
Where the conductivity is now a tensor given by: 12 Men
xk
ykE
e
ok
Electrical Conductivity: Conduction Band
xEE x ˆ
kvE
ekf
kdeJ cc
kc
o
near
3
3
22
E
ekfc
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity Example: Conduction Band of GaAs
Consider the conduction band of GaAs near the -point:
e
e
e
m
m
m
M
100
010
0011 Isotropic!
This implies:
e
z
y
x
e
z
y
x
e
e
e
cz
cy
cx
c
men
E
E
E
E
men
E
E
E
m
m
m
en
J
J
J
EMenJ
2
2
2
,
,
,
12
100
010
001
.
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity Example: Conduction Band of Silicon
t
t
m
m
m
M
100
010
0011
In Silicon there are six conduction band minima (valleys) that occur along the six -X directions. For the one that occurs along the -X(2/a,0,0) direction:
0,0,
285.0
ako
Not isotropic!
mℓ = 0.92 mmt = 0.19 m
This implies that for this valley:
z
y
x
t
t
cz
cy
cx
c
E
E
E
m
m
m
en
J
J
J
EMen
J
100
010
001
6
.6
2
,
,
,
12
The factor of 6 is there because only 1/6th of the total conduction electron density in Silicon is in one valley
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity Example: Conduction Band of Silicon
E
E
E
E
men
E
E
E
mm
mm
mm
en
J
J
J
z
y
x
e
z
y
x
t
t
t
cz
cy
cx
2
2
,
,
,
4200
0420
0042
6
After adding the current density contributions from all six valleys, the resulting conductivity tensor in Silicon is isotropic and described by a conductivity effective mass
To find the conductivity tensor for Silicon one needs to sum over the current density contributions from all six valleys:
Isotropic!
te mmm21
311
Conductivity effective mass
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity: Valence Band
oToovv kkMkkkEkE
..
21
2
ov kkMkv
.1
The velocity of electrons is:
Consider a solid in which the energy dispersion for valence band near a band maximum is given by: Energy
k
Valence band
ok
The current density is (using the electron-hole transformation):
kvkf
kdekvkf
kdeJ vv
kvv
kv
oo
1
22
22
near3
3
near3
3
In equilibrium, for every state with crystal momentum that is unoccupied, the state is also unoccupied and these two states have opposite velocities.
Therefore in equilibrium:
okk
okk
01
22
near3
3
kvkfkd
eJ vvk
vo
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity: Valence Band
Valence band
Energy
k
ok
Now assume that an electric field is present that shifts the crystal momentum of all electrons in the valence band:
xk
yk
Hole distribution in k-space when E-field is zero
xk
yk
Hole distribution is shifted in k-space when E-field is not zero
xEE x ˆ
Ee
ktk
Distribution function: kfv
1 Distribution function:
E
ekfv
1
Ee
Since the wavevector of each electron is shifted by the same amount in the presence of the E-field, the net effect in k-space is that the entire electron distribution (and hole distribution) is shifted as shown
kfv
1
ok
ok
E
9
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Current Density:
Do a shift in the integration variable:
E
EMepJ
EMkfkd
eJ
Ee
kkMkfkd
eJ
Ee
kvkfkd
eJ
v
vk
v
ovk
v
vvk
v
o
o
o
.
.
.12
2
.12
2
12
2
12
1
near3
32
1
near3
3
near3
3
Where the conductivity is now a tensor given by: 12 Mep
xk
ykE
e
ok
Electrical Conductivity: Valence Band
xEE x ˆ
kvE
ekf
kdeJ vv
kv
o
12
2 near
3
3
E
ekfv
1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity Example: Heavy-Hole Band of GaAs
Consider the heavy-hole band of GaAs near the -point:
hh
hh
hh
m
m
m
M
100
010
0011 Isotropic!
This implies:
hh
hh
z
y
x
hh
hh
z
y
x
hh
hh
hh
hh
hhz
hhy
hhx
hhhh
mep
E
E
E
E
mep
E
E
E
m
m
m
ep
J
J
J
EMepJ
2
2
2
,
,
,
12
100
010
001
.
10
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrical Conductivity Example: Light-Hole Band of GaAs
Consider the light-hole band of GaAs near the -point:
h
h
h
m
m
m
M
100
010
0011 Isotropic!
This implies:
h
h
hh
mep
EEMepJ
2
12 .
The total valence band conductivity of GaAs can be written as the sum of the contributions from the heavy-hole and the light-hole bands:
h
h
hh
hh
mep
mep
22
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Phenomenology Of Transport
The presence of external fields, and scattering, the following relations work for electrons in any energy band near the band edge (assuming parabolic bands):
ktkEe
dttkd
on ktkMtkv
.1
tkvkfkd
etkvkfkd
etJ nnnnn
1
22
22
FBZ3
3
FBZ3
3
The first two can also be written as:
kvtkvMEe
dtkvtkvd
M nnnn
..
Problem: One needs simple models for current transport so that non-specialists, like circuit designers, can understand devices and circuits without having to understand energy bands
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Drift Velocity and Mobility for ElectronsWe define the drift velocity for the electrons in the conduction band (for parabolic bands) as:
kvtkvtv cce
The drift velocity is independent of wavevector for parabolic bands and satisfies:
tvM
Eedttvd
M ee
..
Once the drift velocity is calculated, the electron current density is:
tventvkf
kde
kvkvtkvkfkd
etkvkfkd
etJ
eec
cccccce
FBZ3
3
FBZ3
3
FBZ3
3
22
22
22
(1)
(2)
0
EEMekvtkvtv ecce
..1
In steady state:
e = mobility tensor
Electrons in the conduction band are to be thought of as negatively charged particles. In case of multiple electron pockets, current density contributions are calculated separately for each and added in the end.
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
We define the drift velocity for the “holes” in the valence band (assuming parabolic bands) as:
kvtkvtv vvh
The drift velocity is independent of wavevector and satisfies the equation:
tvMEe
dttvd
M hh
..
Once the drift velocity is calculated, the hole current density is:
tveptkvkfkd
etJ hvvh
12
2FBZ
3
3
Holes in the valence band are to be thought of as positively charged particles. In case of degenerate valence band maxima, the heavy and light hole current density contributions are calculated separately and added in the end.
(1)
(2)
Where realizing that the inverse effective mass tensor will have negative diagonal terms for valence band, I have multiplied throughout by a negative sign, with the result that the charge “-e” becomes “+e”
EEMetv hh
..1 In steady state: h = mobility tensor
Drift Velocity and Mobility for Holes
12
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
xk
Energy
FBZ
a
a
The Case of No Scattering: Bloch Oscillations
Consider an electron in a 1D crystal subjected to a uniform electric field. The energy band dispersion and velocity are:
xEE o ˆ
In the absence of scattering, the crystal momentum satisfies the dynamical equation:
0
tkteE
tk
eEdt
tkd
xo
x
ox
akVEkE xsssxn cos2
The time-dependent velocity of the electron is:
atktEae
Va
atkVatv
xo
ss
xssn
0sin2
sin2
akVadkkdE
kv xssx
xnxn sin2
1
Periodic!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Case of No Scattering: Bloch Oscillations
atkt
EaeVatv
dttdx
xo
ssn 0sin2
A periodic velocity means that the electron motion in real space is also periodic:
00 txTtxdtdttdxT
o
where the period T is:oEae
T2
xk
a
a
xk
a
a
xk
a
a
xk
a
a
xk
a
a
0t4T
t 2T
t 4
3Tt Tt
x0
0t 4T
t 2T
t
43T
t Tt
Reciprocal space:
Real space:
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Conductivity of Electrons in Graphene
yk
xk
22
K
oyyoxxpc
o
kkkkvEkE
k
Conduction band dispersion
k
kv
kk
kkv
kkkk
ykkxkkvkEkv
o
o
oyyoxx
oyyoxxckc
ˆˆ122
The dynamical equation for the crystal momentum still works:
ktkEe
dttkd
Ee
ktk
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Conductivity of Electrons in Graphene
xk
yk
Electron distribution in k-space when E-field is zero
xk
yk
Electron distribution is shifted in k-space when E-field is not zero
xEE x ˆ
Ee
ktk
Distribution function: kfc
Distribution function:
E
ekfc
Ee
kfc
ok
ok
Energy
k
ok
kvEe
kfkd
eJok
near2
2
222
Current density can be obtained by the familiar expression:2 spins2 pockets or valleys
o
oc
kEek
kEekvtkv
Velocity magnitude remains the same but the velocity direction changes