Download Document

Solid State Theory
Physics 545
Band Theory III
Each atomic orbital leads to a
band of allowed states in the solid
Band of allowed states
Gap: no allowed states
Band of allowed states
Gap: no allowed states
Band of allowed states
Independent Bloch states
F1
0
Solution of the tight binding model
is periodic in k. Apparently have
an infinite number of k-states
k states for
each allowed energy state.
In fact the different k-states
k states all
equivalent.
Bloch states
kR
ψ (r + R ) ≡ eik.R
ψ (r )
E(k)
-2
α= 10
-4
γ=1
-6
-8
-10
-12
-14
-16
-18
-4
Let k = ḱ′ + G where k′ is in the first Brillouin zone
and
d G is
i a reciprocal
i
l lattice
l i vector.
-2
−π/a
π/a
0
2
4
k [111] direction
ψ(r + R ) ≡ eik′.R eiG.R ψ(r )
But G.R = 2πn, n-integer. Definition of the reciprocal lattice. So
eiG.R = 1
and ψ(r + R ) ≡ eik′.R ψ(r )
e ik.R ≡ e ik ′.R
k′ is exactly equivalent to k.
k
The only independent values of k are those in the first Brillouin zone.
Reduced Brillouin zone scheme
The only
y independent
p
values of k are those in the first Brillouin zone.
Discard for
|k| > π/a
Results of tight binding calculation
2π/a
-2π/a
Displace into
1st B.
B Z
Z.
Results of nearly free electron calculation
Reduced Brillouin zone scheme
Extended, reduced and periodic
B ill i zone schemes
Brillouin
h
P i di Zone
Periodic
Z
Reduced
R d
d Zone
Z
Extended
E
d d Zone
Z
All allowed states correspond to k-vectors in the first Brillouin Zone.
Can draw E(k) in 3 different ways
The number of states in a band
Independent k-states in the first Brillouin zone, i.e. ⏐kx⏐ < π/a etc.
Finite crystal: only discrete k
k-states
states allowed
kx = ±
2πn x
, n x = 0,1,2,.... etc.
L
Monatomic simple cubic crystal, lattice constant a, and volume V.
One allowed k state per volume (2π)3/V in k-space.
Volume of first BZ is (2π/a)3
Total number of allowed k-states in a band is therefore
⎛ 2π ⎞
⎜ ⎟
⎝ a ⎠
3
(2π) = V
V
a
3
=N
Precisely N allowed k-states i.e. 2N electron states (Pauli) per band
This result is true for any lattice:
each primitive unit cell contributes exactly one k-state to each band.
Metals and insulators
In full band containing 2N electrons all states within the first B. Z. are
occupied. The sum of all the k-vectors in the band = 0.
A partially filled band can carry current, a filled band cannot
Insulators have an even integer number
of electrons per primitive unit cell.
E
With an even number of electrons per
unit cell can still have metallic behaviour
due to ban overlap.
Overlap in energy need not occur
in the same k direction
EF
0
π k
a
Metal
M
t l due
d to
t
overlapping bands
E
E
E
EF
π k
a
0
π k
a
0
Empty Band
EF
Energy Gap
π k
a
0
Partially
Filled Band
Full Band
Part Filled Band
Part Filled Band
Energy Gap
Full Band
INSULATOR
or SEMICONDUCTOR
METAL
METAL
or SEMI-METAL
Germanium
Bands in 3D
In 3D the band structure is
much more complicated than
in 1D because crystals
y
do not
have spherical symmetry.
The form of E(k) is dependent
upon the direction as well as
the magnitude of k.
Figure removed to
reduce file size
•
Chemical bonds and electron bands.
a) Number of electrons in any band is finite because the density of
states is finite
finite.
8π 2 m ( E )
ρ3 D ( E ) =
h3
3/ 2
E
1/ 2
8π 2
N=
h3
b) Bands are formed from molecular orbitals.
Etop
3 / 2 1/ 2
m
(
E
)
E dE
∫
Ebottom
Filling of Energy Bands
⇒ An important property of a full band is that it is UNABLE to carry a net current since
for each state in the band we can identify a corresponding state with equal and OPPOSITE
momentum that is filled by an electron. To drive a net current through the crystal it is
necessary to induce an IMBALANCE in the filling of momentum states
⇒ For an energy band that is filled completely however this requires that we excite
electrons ACROSS the forbidden gap.
E
• Situation in which the lowest energy band is
filled completely with electrons
• the
h only
l way in which
h h a net current can flow
l is
to excite electrons across the energy gap
ENERGY GAP
• if th
the energy gap is
i large
l
h
however
excitation
it ti
cannot be achieved and so no net current is
allowed to flow
k
−π/a
π/a
⇒By the same arguments if the energy band is PARTIALLY filled then it should be very
EASY to generate a net current flow in the crystal
⇒ In this situation the forbidden gap lies FAR away from the highest filled electron states
and so it is easyy to use an electric field to ggenerate an imbalance in the fillingg of momentum
states
⇒ A small applied voltage will therefore generate a LARGE current as we discussed
previously for free electrons
E
E
ENERGY GAP
ENERGY GAP
k
−π/a
π/a
NO APPLIED ELECTRIC FIELD
k
−π/a
SMALL ELECTRIC FIELD APPLIED
π/a
⇒Electronic band theory presents us a natural scheme for CLASSIFYING different types of
materials
⇒ METALS should be materials whose uppermost energy band is only PARTIALLY filled
with electrons.
⇒ This explains why these materials are GOOD conductors of electricity
⇒ We expect that insulators on the other hand should be materials whose energy bands
are either COMPLETELY full or empty so that an energy gap BLOCKS current flow in these
materials
E
FORBIDDEN GAP
FILLED STATES
METAL
FORBIDDEN GAP
FILLED STATES
INSULATOR
Band structure of metals
monovalent
l metals
l
multivalent metals, semimetals
⇒What types
yp of elements produce
p
partial
p
or complete
p
fillingg of energy
gy
bands?
⇒ The GROUP I elements should be good METALS since these elements
h
have
only
l ONE valence
l
electron,
l t
whereas
h
coordination
di ti number
b is
i 6-12.
6 12
⇒ If we have a crystal composed of N atoms there will therefore be N
g energy
gy band
valence electrons which will HALF-FILL a single
⇒ The GROUP IV elements should be INSULATORS since these
elements have FOUR valence electrons and so in an N-atom crystal there
will 4N valence electrons that FILL two energy bands completely
E
FILLED STATES
FILLED STATES
FILLED STATES
GROUP I
GROUP IV
FILLING OF ENERGY LEVELS BY THE
VALENCE ELECTRONS OF GROUP I & IV
ELEMENTS
Semiconductors
• In certain materials known as SEMICONDUCTORS however the energy gap that separates
the highest filled band in the ground state from the lowest empty band is SMALL
* Such materials are INSULATORS at zero temperature since their ground state is one in
which the energy bands are either completely full or empty
* Since the forbidden gap is small however electrons can be EXCITED across it at
higher temperatures to PARTIALLY fill the next band
⇒ The material will no longer
g be an insulator at this temperature
p
but will CONDUCT
electricity
E
E
FORBIDDEN GAP
FILLED STATES
FILLED STATES
INSULATOR
T=0
INSULATOR
T>0
FILLED STATES
FILLED STATES
SEMICONDUCTOR
T=0
SEMICONDUCTOR
T>0
Some general COMMENTS on semiconductors
•The
The energy band that holds the valence electrons in the ground state is known as the
VALENCE BAND. It is usually formed by Bonding Orbitals.
•The lowest empty band is known as the CONDUCTION BAND.
BAND It is usually formed
by antibonding orbitals.
⇒ The energy gap that separates these bands is usually denoted as Eg
* Room
R
temperature
t
t
semiconductors
i d t are generally
ll materials
t i l in
i which
hi h Eg is
i a FEW eV
V
(≤ 3 eV)
⇒ This should be compared to a thermal energy of approximately 40 meV that is
available to electrons at room temperature (300 K)
E
SEMICONDUCTOR
Si
Ge
InSb
InAs
InP
GaP
GaAs
GaSb
AlSb
Eg (eV)
0K
300 K
1.17
0.74
0.23
0.43
1.42
2.32
1 52
1.52
0.81
1.65
1.11
0.66
0.17
0.36
1.47
2.25
1
1.43
43
0.68
1.60
CONDUCTION
BAND
Eg
VALENCE BAND
SEMICONDUCTOR
T=0
Concept of a hole
• At higher temperatures electrons in semiconductors may be excited into the
conduction band where they are able carry an electrical current
* Each electron leaves behind an EMPTY state in the valence band and to account
for current flow in semiconductors we must ALSO consider the role of these HOLE
states
* If the valence band is COMPLETELY filled, then the total crystal momentum of
this
hi bband
d iis equall to ZERO since
i
for
f any occupied
i d k-state
k
we can identify
id if an
corresponding filled state with OPPOSITE momentum
E
ENERGY GAP
2
1
k
−π/a
π/a
• The total crystal momentum in a filled energy
band is exactly equal to zero
• to illustrate this consider the total momentum
d to occupation of states 1 & 2
due
• state 1 corresponds to an electron with positive
momentum while state 2 corresponds to one
with
ith equall andd opposite
it momentum
t
• the net crystal momentum of electrons occupying
states 1 & 2 is zero and this pairing can be repeated
for all other states in the band
• When the valence band is completely filled with electrons we can write
∑k
i
=0
(17)
* if we excite AN electron from the state with wavenumber
ke in the valence band into the conduction band equation (17) for the valence band may
now be REWRITTEN as
∑k
ki ≠ k e
i
= − ke
(18)
* The empty state in the valence band may therefore be viewed as a HOLE which has
OPPOSITE momentum to the electron that was excited out of that state
kh = − ke
(19)
• Since the hole corresponds to a missing electron its energy may be written as
Eh ( k h ) = − Ee ( − k e ) = − Ee ( k e )
(20)
* Equation 10.4 shows that electrons and holes have OPPOSITE energy scales
since moving DOWNWARD in the valence band implies INCREASING hole energy
• Moving downwards in the valence band corresponds
t increasing
to
i
i hole
h l energy
E
• in the picture shown here hole 1 therefore has more
energy
gy than hole 2
ENERGY GAP
• this is not too difficult to understand if we think of
the total energy of the electrons left in the band
HOLE 2
HOLE 1
k
−π
π/a
π/a
• since hole 1 corresponds to a missing electron from
a lower energy electron state than hole 2 the total
energy
gy off electrons in the band is higher
g
ffor hole 1
than for hole 2
• While the energy scales are oppositely directed for electrons and holes we can show that
the hole VELOCITY is the SAME as that of the state from which the electron is missing
* To do this we simply replace kh by –ke and Eh(kh) by –Ee(ke)
vh =
=−
1 d
Eh ( k h )
= dkh
1 d
1 d
( − Ee ( ke )) =
E e ( k e ) = ve
= dke
= dke
* Using the same approach we can also show that the effective mass of the hole is a
NEGATIVE quantity
1
1 d
1
d
=
E
(
k
)
=
(
−
(− Ee ( ke )))
h
h
*
2
2
mh = kh dkh
= (− ke ) dke
=−
1 d
1
(
(
))
E
k
=
−
e
e
= 2ke dke
me*
• It is important to appreciate that we do NOT actually have positively-charged carriers in
the
h semiconductor
i d
but
b that
h the
h holes
h l simply
i l behave
b h
AS IF they
h hhad
d positive
i i charge
h
* The basic idea is that EACH of the OCCUPIED electron states in the valence band
responds to externally-applied fields as we would expect for a negatively-charged
carrier.
rri r The
Th NET response
r p n off the
th band
b nd however
h
r LOOKS like
lik the
th response
r p n off a single
in l
particle with a POSITIVE charge!
When we discuss conduction band properties of semiconductors or insulators we refer to
electrons but when we discuss the valence properties,
electrons,
properties we refer to holes.
holes This is because
in the valence band only the missing electrons or holes lead to current flow.
E
E
ENERGY GAP
ELECTRIC
FIELD
ENERGY GAP
k
−π/a
π/a
k
−π/a
π/a
AN ELECTRIC FIELD APPLIED IN THE +x DIRECTION ACCELERATES ELECTRONS IN THE –x DIRECTION
AND SO THE HOLE STATE APPEARS TO BE ACCELERATED IN THE +x DIRECTION
General comments
Thermal vibrations or energy can be used to create a
hole by exciting an electron from the valence band to the
conduction band
In
I an intrinsic
i t i i (undoped)
( d
d) semiconductor,
i
d t the
th
number of holes in the valence band equals the electrons
in the conduction band
Holes can move about the valence band and
recombine with electrons in the conduction band (to
disappear)
tatistics of electrons and holes in semiconductors
One can use Boltzmann statistics for electrons and holes if their energy is small in
comparison with EF
⎡ E − EF ⎤
1
=≈ exp ⎢ −
f (E) =
, if E − EF > 3.5k BT
⎥
exp((
(( E − E F ) / k BT ) + 1
⎣ k BT ⎦
Electron excitation on semiconductors
n = N c e ( E F − Ec ) / k B T
⎡ 2 πme*k BT ⎤
, Nc = 2 ⎢
⎥
2
⎣ h
⎦
p = N v e ( Ev − E F ) / k BT
3/ 2
⎡ 2πmh*k BT ⎤
, Nv = 2 ⎢
⎥
2
⎦
⎣ h
3/ 2
For pure semiconductor EF ≈E
Eg/2
n=p ≈C T2/3exp(-Eg /2kT)
C is a material constant
n p ≈C
n·p
C T2/3exp(-E
exp( Eg /kT)=ni
ni is
i concentration
t ti off intrinsic
i ti i
carriers
Impurities in semiconductors: Doping
n-type
yp Band
Diagram
p yp Band
p-type
Diagram
Impurities create levels in the forbidden gap of semiconductors.
Impurities in semiconductors: concentration of carriers
Ionized Dopant Concentration:
1
n= 1 ( N C N d ) 2 exp ⎛⎜ − E d e ⎞⎟
2
⎝ 2 kT ⎠
1
1
⎛ Eae ⎞
2
(
N
N
)
exp
p= V a
⎜−
⎟
2
⎝ 2 kT ⎠
Nd and Na is the concentration of donors and acceptors respectively
Ed and Ea are in eV
Comments on electron and hole concentration
In pure semiconductor concentration of holes and of
electrons are equal.
equal
Concentration of holes and of electrons depends on
Eg.=>
=>
Insulator are semiconductor with very large Eg
Mobility in semiconductors has exactly the same
2
2 conductivity
meaning as in metals. Total
can be
pe
p
τ
neτe
σ = neμ e + peμ p = * + * p
expressed as:
me
mp
Concentration
C
t ti
off
electrons
l t
and
d
h l
holes
i
in
semiconductors can be tailored by introducing impurities
(doping) In this case concentration of electrons and
Interesting to know
$Band gap increases with increasing the strength of the chemical
bonds in the semiconductor.
Example:
p Diamond >
Silicon
>
Germanium
$In many semiconductor alloys the band gap changes almost
li
linearly
l with
i h composition.
ii
Example: 1) GaAs – AlAs
2) HgTe – CdTe
Chemistry of doping
Doping activity of impurities (dopants) depend on the charge
state of the impurity relative to the charge in unperturbed
lattice.
Doping activity depends on site,
site at which dopant is
incorporated.
Some dopants may be both donors and acceptors
(amphoteric).
Impurities
I
iti may be
b have
h
charge
h
more than
th one.
Some dopants are natural defects that are inherent to the
material.
Chemistry of doping: examples.
1 Elementary semiconductors: C (diamond),
1.
(diamond) Si,
Si Ge
Band gap (eV)
@300 K
conductivity
C (diamond)
5.5
< 10-16 (Ω·cm)-1
Si
1.11
4×10-6 (Ω·cm)-1
Ge
0 67
0.67
22.2×10
2×10-22 (Ω·cm)-11
Si
Si
Si
Si
Si
e–
As
1. Activity of substitutionally incorporated
dopants is defined by the difference in
number of valence electrons.
2. Interstitial dopants (mostly metals) are
2
donors.
B
h+
Li
e–
3. Some interstitial dopants (Au, Ni) may
be amphoteric.
amphoteric
4. Influence of natural defects is negligible.
Chemistry of doping: examples.
2. III-V semiconductors. Structure: cubic
(sphalerite) or hexagonal
Band gaps (eV) @300 K
AlN
6.2
AlAs
2 15 w
2.15
GaN
3.44 w
GaP
2.27 s
GaAs
1.42 s
InP
1.34 s
InAs
0.35 s
InSb
0.23 s
w
Sphalerite
Wurtzite
1. Most
1
M
off the
h d
defects
f
are substitutional.
b i i
l
2. The substitution site is largely defined by ionic radii.
Example: Be substitutes Ga, Te substitute As in GaAs.
Homework: Determine the doping action of Be and Te in GaAs.
Interesting to know: both structures are polar (have a specific directio
Chemistry of doping: examples.
3. II-VI semiconductors. Structure: cubic (sphalerite) or
hexagonal
ZnSe
2.8
w
ZnTe
2.4
w
CdS
24
2.4
w
CdTe
1.4
s
1. Most of the defects are substitutional.
2. The substitution site is largely defined by ionic radii.
Example: In substitutes Cd, As substitute Te in CdTe.
3. Electrical properties of II-VI crystals are stronly
affected by natural defects.
Example of natural defects: 1) Cadmium vacancies in CdTe may work as a
single or double charged acceptor. Therefore, undoped CdTe is always ptype.
yp
Anion vacancies
donors
all II-VI
In2)semiconductors
withare
a very
largeinband
gap semiconductors.
acceptor-like defects are not
thermodynamically stable if the energy of vacancy formation, EV, is smaller than the
recombination energy
energy, Er.
C d i band
Conduction
C d i band
Conduction
Donor level
Valence band
Donor level
Recombination
energy, Er
Acceptor level
Valence band