Firstranker's choice Semiconductors (SC)
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Fermi Dirac Distribution Function:- The particles which obey
Pauli's exclusion principle and are indistinguishable from each
other are called Fermions. It has been a well known fact that
all half integral spin particles are fermions. Since electrons and
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holes half integral spins, therefore these particles are Fermions.
In order to: The distribution of fermions among Various energy
levels at a given temperature is governed by a probability
distribution function called Fermi - Dirac distribution function,
which is given by the following expression:-
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f(E)= 1 / 1 + e(E-EF)/RT
This function gives probability that an electron can occupy
energy level E at thermal equilibrium. Here EF is a
reference energy called fermi Energy and Energy level corresponding
to EF is called fermi level. The function f(E) is also known
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as Fermi Factor. Equation ? is also known as Fermi-Dirac equation.
Variation of Fermi Factor:- The fermi-Dirac distribution
function given in equation ? above can take values only between
0 and 1. Thus it represents probability or occupancy of energy
levels by electrons.
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Case I At absolute zero temperature (0K)
consider any energy level corresponding to energy E less than
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reference energy EF (Called Fermi-Energy). Therefore
negative. Since T=0K :: from ? f(E)= 1 / 1 + e-8 = 1 / 1 + 0 = 1
This means that any level below EF is occupied with electrons
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at 0K.
Now consider an energy level having energy E greater than EF.
Now E-EF is positive. Since T= 0K Thus from ? we can write
f(E)= 1 / 1 + e8 = 1 / 1 + 8 = 0
Thus all Energy levels above Ef are completely empty at 0K.
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When E= EF we take lim (Rather than T= 0) Thus
T?0
We get f(EF) = lim 1 / 1 + e(E-EF)/RT
T?0
= lim 1 / 1 + e0
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T?0
= lim 1 / 1 + 1
T?0
= ½
It means that probability of occupancy of Fermi level is ½. That
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is, it may be a filled level or an empty level. In fact when
EF is a virtual level (Just like pure/intrinsic semi-conductor), then
it is completely empty at 0K and when EF is a real level
(Just like conductors), then it is completely filled at 0K. The
Variation of f(E) at 0K for a conductor is shown in Fig(1a) and
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Fig 1b.
f(E)
0.5
(a)
E
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0
0
EF
Fig.1
f(E)
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(b)
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From this discussion, www.FirstRanker.com -
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it is a level at 0K, below which all existing levels
are completely filled with electrons and above which all existing
levels are completely empty"
Case II If temperature is more than absolute zero (T>0K)
When a solid is heated, then electrons start absorbing this heat
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and transfer to excited states. However the energy given
in this manner can be absorbed by electrons which are
near the Fermi level Ef only. The electrons, which are in levels
well below EF are not able to excite to higher states.
Consider a state E>EF, So that E-Ef is positive. Since
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RT is also positive, therefore if E?8 then (E-EF)/KT?8
So that f(E)?0
On the other hand when E< EF then E-EF is
negative. But T>0: (E-EF)/KT is also negative. We assume
that temp. T is such that quantity RT is much greater than
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Value of EF. Under this assumption as E?0 (E-EF)/KT?8
So that f(E) ? 1
For E = EF (and T>0 of Course!), we get
f(E=EF)= f(EF) = 1 / 1 + e0 = ½ = 0.5
This result is same as in case I. Thus Value of
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fermi factor f(E) is 0.5 at any temp. whether
0K or any higher temperature.
with this discussion, we can see that shape of
graph of f(E) vs E at a given temp should be of the form
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as shown in Fig. (2)
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f(E)
1
0.5
temp T
temp=T'
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0
EF-RT
EF
E+kT
E
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Fig.2.
Let us now plot same graph at another temperature T' which is
greater than T (ie. T'>T). Consider the region E>EF. Since
T'>T :: (E-EF)/RT > (E-EF)/RT'
? f(E) at T < f(E) at T' [See equation ?]
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Now consider Now consider region E<EF. In this case
we must remember that E-Ef is negative. Therefore
(E-EF)/RT < (E-EF)/RT'
? f (E) at T < f(E) at T'
Thus graph of f(E) at T' will be above graph of f(E)
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at T in the region E>EF and it will be below graph of
f(E) at T in the region (E<EF), while f (E) = ½ will be a
crossover point as shown in fig 2.
We shall make two changes in Fig. 2 for practical purposes.
i) we will show energy along Yanis and f(E) along X-axis for
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practical purposes (although it appears to be somewhat illogical right
now)
ii) Actually rail of graph extends upto 8, but for practical purposes
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Firstranker's choice SC
f(E) does not chang www.FirstRanker.com we will
show this graph touching the axes within a region of RT.
with thise changes, the variation of f(E) at different
temperatures (including T= 0K case) is shown in fig.3.
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E+RT
EF
EF-kT
E
0
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0
0.5
1
kT
RT
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Fig.3
Effective Mass (m*) :- It is a general perception that mass
of an electron in a solid is same as the mass of a free electron
inside
However experimentally, it is observed that if some solids the
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mass of electron, when in motion, is more than the mass of
a free electron, while for some other solids, it is less than
mass of a free election. We cannot associate this change in
mass with relativistic effects due to two reasons!
i) Speed of electron inside a solid is very small as
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compared to speed of light (Typically of the order of 105 m/s
so relativistic effects are supposed to be NIL .
ii) Due to relativistic effects mass in motion is always more
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while decrease in mass can not take place
decrease in mass is also observed in certain solids.
It is observed that when an electron moves in a
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Solid, it experiences interactions
from other electrons and positive ions
present in its neighbourhood. The
combination of electrons and nearby
+ve ions form a potential U around
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the moving electron due to which
interaction takes place, which results in change in mass
of electron moving inside a solid. Thus "Effective mass of
an electron is the mass in the presence of lattice potential
in a solid". It is devoted by symbol m*. The expression for
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effective mass is given by
m* = h2 / d2E / dk2
Where h = h / 2p
p=hk=linear
momentum
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E = Energy of electron
moving in a solid-
Variation
The Nation of energy of an electron with respect to propagation
constant k in the condition band and Valence band is shown
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i'n Fig. 1. Equation ? can also be written as 1 / m* = 1 / h2 d2E / dk2
Since d2E / dk2 represents curvature of graph between Efk. Thus from
Fig & above, we can make following conclusions
(1). Near the bottom of band (V.B.) m* m (where m = mass of free electron)
(because E and k graph is a parabola here, and for free electron potential
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energy is zero, So that
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E= KE+ PE
= h2k2 / 2m + 0
= h2k2 / 2m
? E? k2
( Note that for free electron
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use Symbol m for mass
and for bound electron
use symbol m* for mass
which means that Efk graph of a free electron is
a parabola.
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(ii) At the points of inflexion A&A', Eck the graph
between E and k changes between Concave to convex
shape so that d2E / dk2 = 0 From ? 1 / m* = 0 or m*8
This means that external potential of lattice cannot
exert any action on the motion of electron in this region.
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(iii) Between points 0 fA (or of A') curlative
of Eand & Curve is +ve so 1 / m* is +ve or effective
mass of electron is +ve.
(iv) Beyond points A and A' (ie. hear the top of
Valence band) the Elk graph has negative curvature
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ie. d2E / dk2 for m* is negative. Thus effective mass
is negative near this region.
The concept of effective mass provides satisfactory description
of the charge carriers in crystals. In crystals like alkali metals,
energy bards are pastrally freeed and conduction takes place mainly
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due to electrons. However for crystals whire energy band is nearly full
the Vacancies having -Ve charge & -Ve mass are equivalent to particle
of +Ve charge and +ve mass called holes.
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Firstranker's choice SC
Calculation of electron www.FirstRanker.com
intrinse semiconductor:-
Consider two narrowly spaced energy
levels E and E+dE, Lit number of
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For doing this article
electrons existing per unit volume of student has to remember
the sample in this energy range are following standard
dn. If density of energy states in results:-
this region is Z(E)de and fermi (1) The density of energy
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function is f (E), then we have States in condition band
in small energy region
dn= Energy states in the x Probability of de around energy value
region E & E+dE a state to be E is given by
Volume filled by 3/2
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? an electron Z(E) dE = 4p / h3 (2m*)3/2 (E-Ec)½ dE
density of states fermi (ii) ? x½ e-x dx = vp / 2a3/2
function 0
= (Z(E)de) X f(E) (Well maths is sometimes
challenging but always very
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or dn= Z(E) f(E) dE - ? exciting! )
The expression for density of states This integration is challenge to
in energy region E & E+dE is given ILATE Concept learned in
by:- 10+2 classes
Z(E) dE = 4p / h3 (2m*)3/2 (E-Ec)½ dE - ?
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Where m* = effective mass of electron in conduction conduction
band, Ec = energy corresponding to bottom edge of condition
band.
The expression for fermi-function is given by
f (E) = 1 / 1 + e(E-EF)/RT ˜ e-(E-EF)/KT - ?
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(Neglecting 1 en comparison to exponential term)
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Put equations ? & ? in ?, we get
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dn = 4p / h3 (2m*)3/2 (E-Ec)½ e-(E-EF)/KT dE
= [4p / h3 (2m*)3/2 (E-Ec)½ x e-(E-Ec)/KT x e-(EC-EF)/KT] dE
Total number of free electrons per unit volume in the conduction
band are given by:
n= ? dn = 4p / h3 (2m*)3/2 e-(EC-EF)/KT ? (E-Ec)½ e-(E-Ec)/KT dE
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Ec Ec
In equation ?, we can put a = 1 / KT
and E- Ec = x (say)
So that dE = dx
When E=Ec then x= E-Ec=Ec-Ec=0
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when E=8 then x = 8 -Ee = 8
? Integration part of equation ? can be calculated as
? (E-Ec)½ e-(E-Ec)/KT dE = ? x½ e-ax dx
Ec 0
= vp / 2a3/2
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= vp (KT)3/2 / 2
Put this value in ?, we get
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n= 4p / h3 (2m*)3/2 e-(EC-EF)/KT x vp (KT)3/2 / 2
? n= 2 / h3 (2 m* p KT)3/2 e-(EC-EF)/KT - ?
or n = Nc e-(EC-EF)/KT - ?
Where NC = 2 / h3 (2 m* p KT)3/2 - ?
Nc is called effective density of (energy) states in
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the condition band. Equation ? shows that population
density of electrons (n) in the conduction band is a function
of temperature and energy of fermi level.
Calculation of hole density (b) in the Valence band (VB) of
an intrinsic Semiconductor:-
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We know that fermi function f (E) = 1 / 1 + e(E-EF)/KT
represents the probability of finding an electron in the energy
State E. However a hole is created when an electron is
taken out of a bond/band. This means means that presence
of hole is equivalent to absence of electron. If Prob (h) is
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probability of a hole, then we can write
Prob(h) = 1- Probability of electron = 1-f(E)
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? Prob(h) = 1- 1 / 1 + e(E-EF)/KT
= 1 + e(E-EF)/KT - 1 / 1 + e(E-EF)/KT
= e(E-EF)/KT / 1 + e(E-EF)/KT = 1 / e-(E-EF)/KT + 1
= 1 / e(EF-E)/KT + 1
˜ 1 / e(EF-E)/KT
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= e-(EF-E)/KT - ?
(°° For Valence band E<EF
So I can be neglected
in comparison to exponential
part
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The density of energy states within the energy range
E and E+dE of valence band is given by:
Z(E) dE = 4p / h3 (2m*)3/2 (Ev-E)½ dE - ?
(where m* = effective mass of hole and Ev is energy
corresponding to top edge of Valence band)
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Let dp = concentration of holes per unit volume (also called
as hole density) in the energy range E to E+de within
Valence band
? dp = Prob(h) XZ(E) dE
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