Lecture 13: After the band theory and band structures
where do we go from here?
Flat Band Diagrams simplified representation for
energy
However, some
functions cannot be seen, such as direct or indirect band
gap.
Insulators (Diamond):
 Each s-band containing N atoms has space for 2N
electrons. If the highest s- band is occupied by 2
electrons, thus the band is completely filled
(insulator) when you apply an electric field,
electrons cannot absorb energy and they can’t drift.
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Alkali Metals (one valence electron per atom):
 The valence band is half- filled. There are available
states for electrons. Application of an electric field
causes electron drift.
2
Bivalent Metals:
 They should be insulators, but they are not, why?
Because the upper band partially overlap (look to
the diagram) and electrons flow in the lower
portion of the next higher band (almost partially
filled bands). Conductors
Intrinsic Semiconductors:
 Valence bands are filled. They have a narrow
forbidden gap. Energy can excite electrons from the
completely filled valence band into the empty
conduction band.
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Semiconductors:
 Valence and conduction bands consist of mixed
(hybrid) S- and p- states
 8 highest states (2s, 6p) split into two separate
(s+p) bands.
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 As you see from the graph each band consists o one
s-state and 3 p-states.
 The valence band can have 4 electrons.
 The conduction band can have 4 electrons
 Intrinsic semiconductor (Si, Ge) have 4 valence
electrons
∴ The valence bands are completely filled and the
conduction bands are empty.
 There is no gap bet? The 1-s state and the 3-p
states in the valence band.
 Important Note: All covalent band elements have
the same bands shown in the figure.
 The energy gap causes the difference
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The loser atomic number⟶ highest energy gap in IV
elements
Fermi energy in intrinsic semiconductor:
Intrinsic semiconductor:
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Fermi energy is defined as the energy where Fermi
distribution function F(E) equals ½.
In Intrinsic semiconductor:
At T=0
F(E) = 1 for E < Ev
F(E) = 0 for E > Ec
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At higher temperatures
Some of the electrons have been excited into the
conduction band.
F(E) is slightly reduced at the top of valence band
F(E) is not zero at the bottom of the conduction band.
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EF is the center of the gap.
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Next, let’s evaluate the electrical conductivity:
Metals:
In classical theory, we assumed that all electrons drift
under the influence of an electric field. All electrons
contribute to conductivity. Is it right?
Lets investigate that from the following diagram for
electron velocities
(a) equilibrium
(b) when an electric field is applied
DRAWING
Only electrons near the Fermi energy contribute to the
conductivity
Derivation of Conductivity:
Current density j
j = N e vf
n*: numer of electrons displaced by electric field
vf : Fermi velocity (velocity at Fermi surface)
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The effect of an applied electric field:
Increase ∆𝐸 at the Fermi energy to higher levels
DRAWING
n*= N(Ef )∆E
j = evf N(Ef )∆E
dE
= evf N(Ef )
∙ dK
dK
Let me find
𝑑𝐸
𝑑𝐾
h2 k 2
E=
for free electrons
2m
dE h2 k h2 P hmvf
=
=
=
= hvf
dk 2m mh
m
j = evf 2 N(Ef )h dk
Let’s find a relevant expression for dk
ℇ = electric field
dv
F=m
dt
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𝑒ℇ =
𝑑𝑝
𝑑𝑘
=ℎ
𝑑𝑡
𝑑𝑡
eℇ
dk = dt
h
2 2 ( )
j = e vf N Ef ℇζ
where ζ = dt
(the relaxation time: time between collisions)
j = e2 vf 2 ζ N(Ef )ℰ
For 3 dimensions:
1
j = e2 vf 2 ζN(Ef )ℰ
3
1 2 2
j = e vf ζN(Ef )ℰ
3
Depends on
 Fermi velocity
 Occipital density of states at the Fermi level
 Relaxation time
Not all free electrons are responsible for conduction
DRAWING
-monovalent metals (Cu, Ag, Au): have partially filled
valence bands, population density is high near Fermi
energy
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→ high conductivity
-Bivalent metals: overlap of the upper band electron
population near Fermi energy is small
→ less conductivity
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