ece 340 lecture 8 : fermi level and equilibrium carrier...
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ECE 340Lecture 8 : Fermi Level and
Equilibrium Carrier Distributions
Class Outline:
•Density of States•Fermi Level•Equilibrium Carrier Distributions
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Things you should know when you leave…
Key Questions• What is the density of states?• What does the Fermi function tell
us?• What is the Fermi level?• How do I find the equilibrium
electron and hole concentrations?• When can I use the simplified
equations for the concentrations?
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Extrinsic MaterialsHow tightly bound is the extra electron or hole?
• We can use the Bohr’s hydrogen model to get an idea.
• Electrons move in Si and not in a vacuum.– Different relative
permittivity.• The electron mass must be
represented by the effective mass
r
Donor Acceptor
e-
h+
( ) 220
2
4*
32 r
nB
qmEεεπ
−=
Donor in Si P As SbBinding energy (eV) 0.045 0.054 0.039
Acceptor in Si B Al Ga InBinding energy (eV) 0.045 0.067 0.072 0.16
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Extrinsic MaterialsVisualizing donors on the band diagram…
Ed
EaEv
Ec
Ev
Ec
Δx
Let’s take a look at Silicon with Phosphorus impurity atoms:
Ed
Ev
Ec
Eg = 1.12 eV
0.045 eV
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Density of StatesWe are counting states…
kykx
kz
K
K + dK
Δk
( ) kKN D ∆= 33 82π
The total number of states in a small region of volume Δk can be written as (states / volume):
Consider free electrons, the volume of the sphere between the two spheres:
dkkk 24π=∆
The separation dK corresponds to an energy difference (for free electrons):
dEE
mdk
Emk
12
1
2
*
2
*
=
=
Above:•Constant energy surfaces in 3D k-space.•Δk is the volume of the sphere contained between the two spheres of radius k and k +dk.
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Density of StatesBut it is in k-space…The number of states in the shell is given in terms of the density of states in energy by :
( )dEEN
We now equate the two expressions for the total number of states:
( ) ( )dkkdEEN d2
33 48
2 ππ
=
Plug in what we know about how the free electron model changes in k as we change E then integrate to get the final result:
( ) EmmEN d*
32
*
3 2π
=En
ergy
Ec
Ev
Available Valence Band States
Available Conduction Band States
Eg
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Density of StatesThree things to know about the density of states (DOS)…
1. DOS is like seating at a stadium with the number of seats a given distance from the field being equivalent to the number of states available in the range of energy E to E + dE.
2. In general, the number of states available in the conduction and valence band will not be equal.
• Differences in effective masses.
3. Considering closely spaced energies E and E + dE in the respective bands, we can state:
• gc(E)dE is the number of conduction band states lying in the energy range E to E + dE (if E > Ec ).
• gv(E)dE is the number of valence band states lying in the energy range E to E + dE (if E < Ev ).
Ener
gy
Ec
Ev
Eg
( ) ( )
( )( )
32
**
32
**
2
2
π
π
EEmmEg
EEmmEg
vppv
cnnc
−=
−= E ≥ Ec
E ≤ Ev
Units are typically states/cm3-eV
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Fermi Level Fine, but how many states are ACTUALLY filled?• The Fermi function tells us the probability of how many of the existing states
at any given energy will be filled with an electron.• It was derived based on three assumptions
– Each allowed state has a maximum of one electron (Pauli Exclusion Principle).– The probability of each allowed quantum state is the same.– All electrons are indistinguishable.
• When is it valid?– Under equilibrium conditions.– In ANY material (semiconductor, metal, insulator, etc…)
f(E)
1
0Ef E
Mathematically speaking the end result is a probability distribution function:
( ) ( )TkEE
b
f
e
Ef −
+
=
1
1½
T = 0 K
Fermi Energy
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Fermi LevelWhat about the temperature dependence?
• What observations can we make about temperature?
– At E = Ef, f(E) = ½.
– At E ≥ Ef + 3kbT, f(E) decays rapidly to zero and most states will be empty.
– At E ≤ Ef - 3kbT, 1-f(E) rapidly decays to zero. Therefore, most states will be filled.
Ef + 3kbT
Ef - 3kbT
T = 300 KEf = 0.5 eV
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Fermi LevelLet’s look at a quick example…
• Let’s say that the probability that a state is filled at the conduction band edge is equal to the probability that a state is empty in the valence band. Where is the Fermi level?f(E)
1 - f(E)Probability that a state is filled
Probability that a state is emptyf(Ec) = 1 – f(Ev)
( ) ( )TkEEc
b
fc
e
Ef −
+
=
1
1 ( ) ( )
( )TkEE
TkEEv
b
vf
b
fv
e
e
Ef
−
−
+
+
−=−
1
11
111
TkEE
TkEE
b
vf
b
fc −=
−2
vcf
EEE +=
Ec
Ev
Ef
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Fermi LevelVisualizing the Fermi level in intrinsic material…
• In intrinsic material:– Concentration of holes in valence band is equal
to the number of electrons in the conduction band.
– Electron probability tail, f(E), is symmetric with hole probability tail, 1 - f(E).
– No states in Eg despite non-zero occupation probability.
– Fermi level (Ef) lies in the middle of the energy gap.
Ec
Ev
Ef
Ener
gy
Ec
Ev
EgEf
Ener
gy
Ec
Ev
Eg
Carrier Distributions:
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Fermi LevelVisualizing the Fermi level in n-type material…
• In n-type material:– The Fermi level now lies closer to
the conduction band.– There are many more electrons that
there are holes.– The difference between Ef and Ev
provides a measure of the strength of the doping.
Ev
Ef
Ener
gy
Ec
Ev
Eg
Ef
Ec
Ener
gy
Ec
Ev
Eg
Carrier Distributions:
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Fermi LevelVisualizing the Fermi level in p-type material…
• In p-type material:– The Fermi level now lies
closer to the valence band.– There are many more holes
that there are electrons.
Ev
Ef
Ener
gy
Ec
Ev
Eg Ef
Ec
Ener
gy
Ec
Ev
Eg
Carrier Distributions:
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Equilibrium Carrier DistributionsLet’s put together our knowledge of the density of states and the Fermi function…
We want to know how many carriers we have in our semiconductor for ANY energy or temperature.
For electrons
For holes
( ) ( )∫=top
c
E
Ec dEEfEgn
( ) ( )[ ]∫ −=v
bottom
E
Ev dEEfEgp 1
Focus on the electron relation.
∫ −
+
−=
top
c b
f
E
E TkEE
cnn dE
e
EEmmn
1
232
**
πPlug in the relationships for the density of states and the Fermi function:
Make some substitutions and bring the constants out:
( )TkEE
b
c−=η
( )Tk
EE
b
cfc
−=η Etop ~ ∞ ( )
ηηπ ηη d
eTkmm
nc
bnn ∫∞
−+=
0
21
32
23
**
1
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Equilibrium Carrier DistributionsThat integral is awful! Isn’t there an easier way?
( )ηη
π ηη de
Tkmmn
c
bnn ∫∞
−+=
0
21
32
23
**
1
( )ηη
π ηη de
Tkmmp
v
bpp∫∞
−+=
0
21
32
23
**
1
Fermi-Dirac integral of order ½.
Fermi-Dirac integral of order ½.Nc Nv
23
2
*
23
2
*
22
22
=
=
π
π
TkmN
TkmN
bpv
bnc
Effective conduction band density of states.
Effective valence band density of states.
23
0
*,
319
,11051.2
×=
mm
cmN pn
vc
3kbT
3kbT
Non-degenerate Semiconductor
Degenerate Semiconductor
Degenerate Semiconductor
( )
( )TkEE
v
TkEE
c
b
fv
b
cf
eNp
eNn−
−
=
=Non-degenerate Semiconductor
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Equilibrium Carrier ConcentrationsThese are simple closed form expressions, but are they the simplest?
• There are other forms which you may find more useful in homeworks or tests.
• We can find the carrier concentrations relative to Ei , the Fermi level for an intrinsic semiconductor.
For intrinsic semiconductors, we know the following: n = p = ni and Ei = Ef
Then the relations for n and p become:
TkEiE
vi
TkEE
ci
b
v
b
ci
eNn
eNn−
−
=
= Solve for Nc and Nv
TkEE
iv
TkEE
ic
b
vi
b
ic
enN
enN−
−
=
=
By combining equations, we arrive at two very important and useful results:
TkEE
i
TkEE
i
b
fi
b
if
enp
enn−
−
=
= Solve for NiTk
E
vcib
g
eNNn 2−
=
M. J. Gilbert ECE 340 – Lecture 8 9/09/1 1
Equilibrium Carrier ConcentrationsA note on the effective mass…
kx
kz
ky
kx
ky
kz
ml
mt
mt
----- -- ----
•In silicon, most of the electrons reside in the X valley as it is lowest in energy.
•The X direction has six equivalent valleys with different masses in each of the possible directions.
•We use the effective mass which is the geometric mean of the masses (density of states effective mass).
•GaAs conduction band does not have this complication.
ml = 0.98 m0mt = 0.19 m0
( ) 031
232
* 6 mmmm tln =