States and state filling
• So far, we saw how to calculate bands for solids
• Kronig-Penny was a simple example
• Real bandstructures more complex
• Often look like free electrons with effective mass m*
• Given E-k, we can calculate ‘density of states’
• High density of conducting states would imply metallicity
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Carrier Statistics
Carrier populations depend on
• number of available energy states (density of states)
• statistical distribution of energies (Fermi-Dirac function)
Assume electrons
act ‘free’ with a
parametrized
effective mass m*
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Labeling states allows us to count them!
GaAs
Si
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Where are the states?
E
dEdE
dk
x x x x x x x xkx
k
For 1D parabolic bands, DOS peaks at edges
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Where are the states?
E
dEdE
dk
x x x x x x x xkx
k
Dk = 2p/L
(# states) = 2(dk/Dk) = Ldk/p
DOS = g = # states/dE = (L/p)(dk/dE)
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Analytical results for simple bands
E
dEdE
dk
x x x x x x x xkx
k
Dk = 2p/L
E = h2k2/2m* + Ec  dk/dE = m/2ħ2(E-Ec)
DOS = Lm*/2p2ħ2(E-Ec) ~ 1/(E-Ec)
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Increasing Dimensions
E
dE
dEdE
....
..
...
...
...
....
....2pkdk
. dk
k
k
# k points increases
due to angular integral
along circumference,
as (E-Ec)
dNs = 2 x 2pkdk/(2p/L)2
g ~ Sq(E-Ec), step fn
In higher dimensions, DOS has complex shapes
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From E-k to Density of States
dNs = g(E)dE = 2
Σ1
k
 2 (dk/[2p/L]) for each dimension
Use E = Ec + ħ2k2/2mc to convert
kddk into dE
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From E-k to DOS for free els
E
E
Ec
1-D
2-D
3-D
DOS
(Wmc/2p2ħ3)[2mc(E-Ec)]1/2
E
DOS
(Smc/2pħ2)q(E-Ec)
DOS
(mcL/pħ)/√2mc[E-Ec]
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Real DOS needs computation
DOS
E (eV)
VB CB
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Keep in mind 3-D Density of States
m 2m( E  Ec )
g (E) 
2 3
p 
In the interest of simplicity, we’ll try to reduce all
g‘s to this form…
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(Ev-E)
lh
(Ev-E)
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What if ellipsoids?
Such that number of states is preserved
b a
E – EC = ħ2k12/2ml*+ ħ2k22/2mt* + ħ2k32/2mt*
1 = k12/a2
+ k22/b2
+ k32/b2
a = 2ml*(E-EC)/ħ2
b = 2mt*(E-EC)/ħ2
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What if ellipsoids?
Total k-space volume of Nel ellipsoids
= (4pab2/3)Nel
where
a = 2ml*(E-EC)/ħ2
b = 2mt*(E-EC)/ħ2
k-space volume of equivalent sphere
= (4pk3/3)
where
k = 2mn*(E-EC)/ħ2
Equating
K-space
Volumes
mn* = (ml*mt*2)1/3(Nel)2/3
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Valence bands more complex
where
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Valence bands more complex
hh
lh
But can try to fit two paraboloids for heavy and light holes and sum
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Valence bands more complex
hh
lh
4pk3/3 = 4pk13/3 + 4pk23/3
k = 2mp*(EV-E)/ħ2
k1 = 2mhh(EV-E)/ħ2
k2 = 2mlh(EV-E)/ħ2
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So map onto 3D isotropic
free-electron DOS
gC(E) = mn*[2mn*(E-EC)]/p2ħ3
gV(E) = mp*[2mp*(EV-E)]/p2ħ3
with the right masses
mn* = (ml*mt*2)1/3(Nel)2/3
mp* = (mhh3/2 + mlh3/2)2/3
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Density of states effective
mass for various solids
mn* = (ml*mt*2)1/3(Nel)2/3
mp* = (mhh3/2 + mlh3/2)2/3
ml* mt*
GaAs 0.067 0.067
Nel
mhh mlh
mn* mp*
1
0.45 0.082 0.067 0.473
Si
0.98 0.19
6
0.49
Ge
1.64 0.082
8
0.28 0.042 0.89
0.16
1.084 0.5492
0.29
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But how do we fill these states?
Fermi-Dirac Function
•
Find number of carriers in CB/VB - need to know
–
–
•
Number of available energy states (g(E))
Probability that a given state is occupied (f(E))
Fermi-Dirac function derived from statistical mechanics of
“free” particles with three assumptions:
1. Pauli Exclusion Principle – each allowed state can accommodate
only one electron
2. The total number of electrons is fixed N=Ni
3. The total energy is fixed ETOT =  EiNi
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Fermi-Dirac Function
f (E ) 
1
1  e E E
F
 kT
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Carrier concentrations
• Can figure out # of electrons in conduction band
Etop
n   g c (E )f (E )dE
Ec
El. Density
State
Density
gc(E)
Occupancy
per state
• And # of holes in valence band
f(E)
Ev
p   gv (E )1  f (E )dE
Ebottom
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Full F-D statistics
n  NC
2
F1 (c )
p 2
F1 ()  
2
1
2
 d
 
0 1 e

c  (E F  EC ) / kT
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Carrier Concentrations
• If EC-EF >> kT the integral simplifies – nondegenerate
semiconductors Fermi level more than ~3kT away from
bottom/top of band
• We then have  << -1, so drop +1 in denominator of F1/2 function
• For electrons in conduction band:
3
2
 2pmn * kT 
n  2
 e
2
h


Or equivalently
n  Nc e
EF  EC
kT
( EC EF )
kT
 2pmn * kT 
NC  2

2
h


3
2
CB lumped density
of states
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Carrier Concentrations
• Can do same thing for holes (nondegenerate approximation)
3
2
E E
 2pm p * kT 
kT
p  2
e

2
h


p  Nv e
V
F
( EF Ev )
kT
 2pm p * kT 
Nv  2

2
h


3
2
VB lumped density
of states
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Intrinsic Semiconductors
•
For every electron in the CB there is a hole in the VB
np
NC e
( EC EF )
kT
 NV e
( EF EV )
kT
(EC  EV ) kT  NV 
EF 

ln

2
2  NC 
3kT  mn
EF  midgap 
ln
4  mp
•
*

* 
Fermi level is in middle of bandgap if effective masses not too
different for e and h
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Intrinsic Semiconductors
• Can plug in Fermi energy to find intrinsic carrier
concentrations
( EC EV )
2 kT
i
C V
C
npn  N N e
 N NV e
EG
2 kT
• Electrical conductivity proportional to n so intrinsic
semiconductors have resistance change with temperature
(thermistor) but not useful for much else.
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Doped Semiconductors
P has 1 more el than
tetrahedral
Boron has 1 less el than
tetrahedral
Extra el loosely tied (why?)
It steals an el from a nb. Si
to form a tetrahedron. The
deficit ‘hole’ p-dopes Si
It n-dopes Si
(1016/cm3 means 1 in 5 million)
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Carrier Concentrations - nondegenerate
n  Nc e
( EC EF )
kT
p  Nv e
np  NC NV e ( E E
v
C
) / kT
( EF Ev )
kT
 NC NV e E
G
/ kT
np  ni2
Independent of Doping
(This is at Equilibrium)
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A more useful form for n and p
For intrinsic semiconductors
ni  N c e
( EC Ei )
kT
n = nie(EF-Ei)/kT
EF
Ei
p i  Nv e
( Ei  Ev )
kT
p = nie(Ei-EF)/kT
Ei
EF
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Charge Neutrality
     / k0
Poisson’s Equation
In equilibrium, E=0 and =0
  q p  n  ND  N A 
-
 p  n  ND  N A-  0
Charge Neutrality
Relationship
Number of ionized donors:
ND
1

N D 1  g D e ( E
F
ED ) / kT
(gD = 2 for e’s, 4 for light holes, 1 for deep traps)
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Charge Neutrality
“1”
ED
“0”
0
PN  e-(EN-EFN)/kT
<N> = 0.P0 + 1.P1 = f(ED-EF) = 1/[1 + e-(EF-ED)/kT]
ND
1

N D 1  g D e ( E
F
ED ) / kT
(gD = 2 for e’s, 4 for light holes, 1 for deep traps)
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Charge Neutrality


2ED + U0

0
PN  e-(EN-EFN)/kT
ED
<N> = 0.P0 + 1.(P + P) + 2.P
ND
1

N D 1  g D e ( E
F
ED ) / kT
(gD = 2 for e’s, 4 for light holes, 1 for deep traps)
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Charge Neutrality
Can equivalently alter ED to account for degeneracy
NV e
( EV EF ) / kT
n
 NC e
( EF EC ) / kT

1
1  gDe
( EF ED ) / kT

1
1  g Ae
( E A EF ) / kT
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0
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Nondegenerate Fully Ionized Extrinsic Semiconductors
• n-type (donor)
ND >> NA, ND >> ni
n  ND
p = ni2/ND
• p-type (acceptor)
NA >> ND, NA >> ni
p  NA
n = ni2/NA
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Altering Fermi Level with doping
(… and later with fields)
• Recall:
n  ni e ( E
F
p  ni e ( E E
i
Take ln
Ei ) / kT
F
) / kT
n
p
kT ln  kT ln  EF  Ei
ni
ni
ND
EF  Ei  kT ln
ni
n-type
NA
Ei  EF  kT ln
ni
p-type
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In summary
• Labeling states with ‘k’ index allows us to count and get
a DOS
• In simple limits, we can get this analytically
• The Fermi-Dirac distribution helps us fill these states
• For non-degenerate semiconductors, we get simple
formulae for n and p at equilibrium in terms of Ei and
EF, with EF determined by doping
• Let’s now go away from equilibrium and see what
happens