Physics of Semiconductor Devices.pptx

Physics of Semiconductor Devices
an Introduction
Dr. Wolfgang Ploss
Texas Instruments Germany
Dr. Wolfgang Ploss, Studien Material, 2022 1
wolfgang.ploss@extern.oth-regensburg.de
My Email OTH
mastermem@oth-regensburg.de
Master Mail Box
For questions, please use the OTH email addresses

Introduction
PART 2b

Band Structure
• Electron Statistics
• Fermi Function, Fermi Energy
• Electrons and Holes in Bands

r
Potential energy
Potential Energy ~ -
1
𝑟
+

+ + +
The nuclei in the lattice are positive charged and we have an attraction force between
the positive nuclei and negative charged electrons. If we move the electrons along the
Electrical field with an external force, which is opposite to the internal field,
we increase the potential energy of the electrons in the field.
E – Field
of nucleus
Direction
In crystal
+
+
+

At Edges of Conduction Band and Valence Band:
Electrons or Holes have no kinetic energy, no velocity
Negative electrons increase energy if we move them up in the conductive band.
Positive charged holes increase the potential energy if we move them down in the Valence band

Fermi Statistic

Some Statistics
Questions:
 What is the probability of an event ?
o Expected event divided by all possibilities of the event
o e.g. dice: what is probability to get number 5 to play a dice ? Dice has 6 areas. On
every area is written a number 1,2,.. 6. All areas are same. The expected event is one
area. We have 6 areas on a dice. Probability = 1 / 6 ( to get a special number, e.g. 5 or 1
or 4 etc.)
 What is the probability to find an electron between E …E +dE
x
probability
x x + dx
dp (x) = p(x) dx = area
Probability to find an electron between x and x + dx.
Probability to find the particle between a and b
should be 1.
p(x)
area
a b 𝑎
𝑏
𝑝 𝑥 𝑑𝑥 = 1 p = probability function
If we have a function f(x) which shows us opportunities , but not all opportunities
are possible or allowed between a and b , which is defined by a probability function p(x),
we need to calculate: 𝑎
𝑏
𝑝 𝑥 𝑓 𝑥 𝑑𝑥

BOLTZMANN – Statistic: Particle are capable of being differentiated
BOSE – Statistic: Particle are NOT capable of being differentiated
Fermi – Statistic: Particle are NOT capable of being differentiated and
only one particle is allowed to take a certain condition
Example (next page) :
2 particles should be distributed to 4 possible states with the same energy.
N= 2 (particle) , g = 4 (states with same energy)
How many possibilities do we get for the 3 different Statistics ?
Nature of particles requires different statistics

Example: 2 particle distributes on 4 possible states with the same energy
𝒈𝑵
N + g -1
N
g
N
( )
( )
N = 2 particle with same energy
g = 4 states (options to be with
same energy)

Energy Ei : gi states , Ni particle are distributes on gi states
Assumptions:
𝑁𝑖 𝐸𝑖 = 𝑐𝑜𝑛𝑠𝑡= E ; 𝑁𝑖 = 𝑐𝑜𝑛𝑠𝑡
Thermal Equilibrium State:
Particle distribution according to the highest probability, P
P  maximum of probability
S = K ln (P ) , Entropy S get maximum because of P
 Calculation of Entropy for different particle types gives statistic distribution
Bose
Boltzmann
Fermi
Probability P (E)

Probability to find an electron within Energy E + dE
Fermi Function

Probability to find an electron within Energy E + dE
Fermi Function
Important about the Fermi Energy : Definition of Fermi Energy
At T= 0 , Fermi Energy is max Energy. All States below Fermi Energy are occupied, f (E<=EF)=1
At T>0, at Fermi Energy the probability to find a Fermion (electron). f(E=EF ) = ½ , 50%
Note, to be exact:
in Physics and Theory of Heat, the Fermi Energy for T > 0 K is called “Chemical Potential”.
In our course we call also for T > 0 the Energy with 50% probability the Fermi Energy.

For E > EF :
For E < EF :
0
)
(
exp
1
1
)
( F 



 E
E
f
1
)
(
exp
1
1
)
( F 



 E
E
f
Fermi-Dirac distribution: Consider T  0 K
0
∞
For Energy E < = Fermi Energy
the probability to find a electron
with this energy is 1.
For Energy > Fermi Energy we
will find no electron.
Probability = 0

15
If E = EF then f(EF) =
𝟏
𝟐
for all temperatures
If then
Thus the following approximation is valid:
i.e., most states at energies 3kT above EF are empty.
If then
Thus the following approximation is valid:
kT
E
E 3
F 
 1
exp F 





 
kT
E
E





 


kT
E
E
E
f
)
(
exp
)
( F
kT
E
E 3
F 
 1
exp F 





 
kT
E
E





 


kT
E
E
E
f F
exp
1
)
(
Fermi-Dirac distribution: Consider T > 0 K
Maxwell - Boltzmann
Mathematics:
1
1+𝑥
= 1 – x, if x << 1, x =
Dr. Wolfgang Ploss, Studien Material, 2022

• kT (at 300 K) = 0.025eV (25 meV) , 3KT = 0.075eV (75 meV)
• In comparison to Eg(Si) = 1.1eV,
• 3kT is very small in comparison to band gap
Fermi Dirac Distribution and Maxwell – Boltzmann Distribution
For E > EF +3KT
Maxwell – Boltzmann
is good enough to get
the probability





 


kT
E
E
E
f
)
(
exp
)
( F

T >0
EF – 3KT
EF + 3KT
For Energies +- 3KT around EF Fermi Dirac Function can be described
with Maxwell Boltzmann Probability Function
3KT = 75meV at 300K

Which statistics must be used depends on where the Fermi Energy level is
Fermi Level in this Section

20
Fermi-Dirac distribution and the Fermi-level
Density of states tells us how many states exist at a given energy E. The Fermi
function f(E) specifies how many of the existing states at the energy E will be
filled with electrons. The function f(E) specifies, under equilibrium conditions,
the probability that an available state at an energy E will be occupied by an
electron. It is a probability distribution function.
EF = Fermi energy or Fermi level
k = Boltzmann constant = 1.38 1023 J/K =
8.6  105 eV/K
T = absolute temperature in K

21
Equilibrium distribution of carriers
Distribution of carriers = DOS probability of occupancy
= g(E) f(E)
(where DOS = Density of states)
Total number of electrons in CB (conduction band) =

 top
C
d
)
(
)
(
C
0
E
E
E
E
f
E
g
n
Total number of holes in VB (valence band) =
 
 
 V
Bottom
d
)
(
1
)
(
V
0
E
E
E
E
f
E
g
p
Ec
Ev
CB
VB
E top
E bottom

K = Boltzmann Constant Charge of electron:
K = 8.617 10 -5 eV / K e = q = 1.6 10 -19 As
KT for T = 300K, Energy at 300K
KT = 25.8 meV , KT /e = 25.8mV
What is 1eV ?  electron with charge q passes potential difference of 1V and get
energy of 1eV.
Energy = q U, unit in eV
Energy per electron at 300K = KT / e = 25.8mV
KT at 300K
K T = 25.8 meV = 25.8 1.6 10 -19 10 -3 As V ~ 4.1 10 -21 Ws
K T NA = R T, in J / mole ; R / NA = K ,
NA = 6 10 +23 particle per mole, Avogadro Const
information

Free Electrons in MET: Fermi Energy of ‘Free Electron Gas’
Fermi Energy
Depends only of
electron number per volume
in MET
Typical Value for Fermi
Energy in Metal ~ 4eV
Fermi Temperature Typical Value ~ 50,000K
Fermi Wave Lengths Typical Value ~ 1 A
Fermi Velocity Typical Value ~ 10 +8 cm / s
information

Fermi Energy moves to lower Values with T > 0
T0 = 0
All states E > Ef are empty
All states E<= Ef occupied
T1 > T0 > 0
2 electrons have higher
Energy. probability
f(E F ) = ½ moves
to lower values.
T2 > T1
3 electrons have higher energy
Fermi Energy gets lower again.
Density of states

Energy state density g(E) in Silicon
Lines (delta function)
from electrons in
Shelves close to nuclei
Energy state density
In valence – and conduction
bands

Density of states g (E) for Silicon – simple Model
Close to Ev and Ec : g(E) ~ 𝑬
Question: Where is Fermi Energy ?
g ( E) ~ 𝑬

Thermal Equilibrium and Fermi Energy
The definition of the Fermi Energy has a useful and important consequence:
In an electronic system in thermal equilibrium, there is an unique Fermi Energy
that is constant throughout the entire system.
In other words, if an electronic system is characterized by Fermi Energy that
varies with location, the system is not in thermal equilibrium. This makes sense:
We have regions with high energy and the high - energy states are occupied.
There are also regions where the Fermi Energy is low and low- energy states are empty.
The hole system can lower its energy by allowing the electrons in high-energy states to
move to low- energy states that are empty. This movement of electrons stops when
the Fermi Energy becomes constant.

Position of Fermi Energy and Consequences
Metal
Above Fermi Energy are
enough free states for
electrons. They can
gain energy and move.
We have current in Metals.
Semiconductor
Fermi Energy is within
The band gap. For T>0 K,
some electrons are in
conduction band and “free”.
Holes are “free”
+ +
Isolator
The band gap in isolator
is big. No ‘free’ electron can be
In the condition band .
No current is possible
valence band
conduction band

29
Equilibrium distribution of carriers
Distribution of carriers = DOS probability of occupancy
= g(E) f(E)
(where DOS = Density of states)
Total number of electrons in CB (conduction band) =

 top
C
d
)
(
)
(
C
0
E
E
E
E
f
E
g
n
Total number of holes in VB (valence band) =
 
 
 V
Bottom
d
)
(
1
)
(
V
0
E
E
E
E
f
E
g
p
Ec
Ev
CB
VB
E top
E bottom

How to calculate the number of electrons in the conduction band ?

 top
C
d
)
(
)
(
C
0
E
E
E
E
f
E
g
n ; gc ( E) ~ 𝐸
Non- degenerate semiconductor: EF is in the gap.
If Ef is > 3KT away from Ec we can use Maxwell – Boltzmann.
Degenerated semiconductor: EF is in conduction band. You must use Fermi Dirac.

Intrinsic Semiconductor

intrinsic

What does it mean ?
Nc and Nv describes all available states in the conduction band for electrons
or holes in the valence band - the effective density of states
Assumptions:
• Maxwell Boltzmann (Fermi Energy within Conduction and Valence band)
• Non - degenerated Silicon
Nc ~ 3 x 10+19 cm-3
Nv ~ 1 x 10+19 cm-3
Silicon at 25C

Intrinsic Semiconductor, no doping
We have electrons and holes in intrinsic semiconductor and call:
Electrons per cm-3 = ni i stands for ‘intrinsic’
Holes per cm-3 = p i
The Fermi Energy (Reference) for this System is EF = E i
Maxwell Boltzmann
for n i
for p i = n i
𝐍𝐜
𝐍𝐯
> 0
Ec – Ev = Egap
Ei =
𝐄𝐜+𝐄𝐯
𝟐
- K T ln (
𝐍𝐜
𝐍𝐯
)
𝟏
𝟐

Intrinsic Semiconductor, no doping
EF = Ei =
𝐄𝐜+𝐄𝐯
𝟐
- K T ln (
𝐍𝐜
𝐍𝐯
)
𝟏
𝟐 ~
𝐄𝐜+𝐄𝐯
𝟐
= Ei
If mass of electron and holes would be equal, EF or Ei would be in the middle of Gap.
Ei is slightly below the middle of the gap, because Nc > Nv,
~
𝐄𝐜+𝐄𝐯
𝟐

Semiconductor, ni and pn Product
ni = 𝐍𝐯 𝐍𝐜 𝒆−
𝟏
𝟐
(𝑬𝒈/𝑲𝑻)

END Part 4

Physics of Semiconductor Devices.pptx

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