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Donors and acceptors in
Semiconductors
Prof. Dr. Mohammed GulamAhamad
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CARRIER CONCENTRATIONS INSEMICONDUCTORS
Donors and Acceptors
Fermi level , Ef Carrier concentration equations
Donors and acceptors both present
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Donors and Acceptors
The conductivity of a pure(intrinsic) s/c is low due tothe low number of freecarriers.
For an intrinsic semiconductor
n = p = ni
n = concentration of electrons per unit volume
p = concentration of holes per unit volume
ni = the intrinsic carrier concentration of the semiconductor under consideration.
The number of carriers aregenerated by thermally orelectromagnetic radiationfor a pure s/c.
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n.p = ni2
n = p
number of e-s in CB = number of holes in VB
This is due to the fact that when an e- makes
a transition to the CB, it leaves a hole behindin VB. We have a bipolar (two carrier)
conduction and the number of holes and e- sare equal.
n.p = ni2This equation is called as mass-action
law.
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The intrinsic carrier concentration ni depends on;
the semiconductor material, and
the temperature.
n.p = ni2
For silicon at 300 K, ni has a value of 1.4 x 1010 cm-3. Clearly , equation (n = p = ni) can be written as
n.p = ni2
This equation is valid for extrinsic as well as
intrinsic material.
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To increase the conductivity, one can
dope pure s/c with atoms from column lllor V of periodic table. This process iscalled as doping and the added atomsare called as dopantsimpurities.
What is doping and dopants impurities ?
There are two types of doped or extrinsic s/cs;
n-type
p-type
Addition of different atoms modify the conductivity ofthe intrinsic semiconductor.
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p-type doped semiconductor
Si + Column lll impurity atoms
Boron (B)has three valance
e-s
Havefour
valance
e-
s
Si
Si Si
Si
B
Electron
Hole
Bond
with
missingelectron
Normal
bond with
two
electrons
Boron bonding in Silicon
Boron sits on a lattice side
p >> n
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Boron(column III) atoms have three valance
electrons, there is a deficiency of electron ormissing electron to complete the outer shell.
This means that each added or doped boron
atom introduces a single holein the crystal.
There are two ways of producing hole
1) Promote e
-
sfrom VB to CB,2) Add column lll impurities to the s/c.
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Energy Diagram for a p-type s/c
Ec = CB edge energy level
Ev = VB edge energy level
EA= Acceptor energ level
Eg
CB
VB
acceptor(Column lll) atoms
The energy gap is forbidden only for pure material, i.e. Intrinsic material.
Electron
Hole
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The impurity atoms from column lll occupy at an energy level
within Eg . These levels can be1. Shallow levels which is close to the band edge,
2. Deep levels which lies almost at the mid of the band gap.
If the EA
level is shallow i.e. close to the VB edge, each addedboron atom accepts an e- from VB and have a fullconfiguration of e-sat the outer shell.
These atoms are called as acceptor atoms since they acceptan e- from VB to complete its bonding. So each acceptoratom gives rise a hole in VB.
The current is mostly due to holes since the number of holes aremade greater than e-s.
p-type semiconductor
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Holes = p = majority carriers
Electrons = n = minority carriers
Majority and minority carriers in a p-type semiconductor
t2
t1
t3
Electric field direction
Holes movement as afunction of appliedelectric field
Hole movement direction
Electron movementdirection
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Ec
Ev
Ea
Eg
Electron
Hole
Shallow acceptor in silicon
Si
Si Si
Si
P
Electron
Weakly bound
electron
Normal
bond with
two
electrons
Phosporus bonding in silicon
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Ec
Ev
Ed
Eg
Electron
Valance band
Conduction band
Band gap is 1.1 eV for silicon
Shallow donor in silicon
Donor and acceptor charge states
Electron
Hole
Neutral donorcentre
onized (+ve)donor centre
Neutralacceptorcentre
onized (-ve)acceptorcentre
Ec
Ev
Ea
Ec
Ev
Ea
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Extra e- of column V atom is weakly attached to its host atom
n-type semiconductor
Si
Si
Si
As
Si
Si + column V (with five valance e- )
ionized (+ve)donor centre
Ec
Ev
ED = Donor energy level (shallow)
Band gap is 1.1 eV for silicon
Hole
Electron
Eg
n - type semiconductor
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np, pn
n-type , n >> p; n is the majority carrier
concentration nnp is the minority carrierconcentration pn
p-type , p >> n; p is the majority carrierconcentration pp
n is the minority carrierconcentration np
np
pn
Type of semiconductor
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calculation
Calculate the hole and electron densities in a piece of p-type siliconthat has been doped with 5 x 1016acceptor atoms per cm3 .
ni= 1.4 x 10
10
cm
-3
(
at room temperature)
Undoped
n = p = ni
p-type ; p >> n
n.p = ni2 NA= 5 x 10
16 p = NA = 5 x 1016 cm-3
3
316
23102
109.3105
)104.1(xcmx
cmx
p
nn i
electrons per cm3
p >> ni and n
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Fermi level , EF
This is a reference energy level at which the probability of occupationby an electron is .
Since Ef is a reference level therefore it can appear anywhere in theenergy level diagram of a S/C .
Fermi energy level is not fixed.
Occupation probability of an electron and hole can be determinedby Fermi-Dirac distribution function, FFD ;
EF = Fermi energy level
kB = Boltzman constant
T = Temperature
)exp(1
1
Tk
EEF
B
FFD
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E is the energy level under investigation.
FFDdetermines the probability of the energy level E beingoccupied by electron.
determines the probability of not finding anelectron at an energy level E; the probability of finding ahole .
)exp(1
1
Tk
EEF
B
FFD
FD
FDF
f
fEEif
1
2
1
0exp1
1
Fermi level , EF
C i t ti ti
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Carrier concentration equations
The number density, i.e., the number of electronsavailable for conduction in CB is
3/ 2*
2
22 exp ( )
exp ( ) exp( )
p F V
F V i FV i
m kT E Ep
h kT
E E E Ep N p n
kT kT
3 / 2*
2
22 exp ( )
exp ( ) exp( )
n C F
C F F iC i
m kT E E n
h kT
E E E En N n nkT kT
The number density, i.e., the number of holes availablefor conduction in VB is
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Donors and acceptors both present
Both donors and acceptors present in a s/c in
general. However one will outnumber the otherone.
In an n-type material the number of donorconcentration is significantly greater than that
of the acceptor concentration. Similarly, in a p-type material the number of
acceptor concentration is significantly greaterthan that of the donor concentration.
A p-type material can be converted to an n-type material or vice versa by means of addingproper type of dopant atoms. This is in fact howp-n junction diodes are actually fabricated.
W k d l
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How does the position of the Fermi Level change with
(a) increasing donor concentration, and
(b) increasing acceptor concentration?
Worked example
(a) We shall use equation
fn is increasing then the quantity EC-EF must be decreasing
i.e. as the donor concentration goes up the Fermi level movestowards the conduction band edge Ec.
exp ( )C FCE E
n NkT
W k d l
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Worked example
But the carrier density equations such as;
arent valid for all doping concentrations! As the fermi-level comes
to within about 3kT of either band edge the equations are no longer
valid, because they were derived by assuming the simpler Maxwell
Boltzmann statics rather than the proper Fermi-Dirac statistic.
kT
EEnp
andkT
EE
h
kTmn
Fii
Fcn
exp
exp2
223
2
*
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Worked example
Eg/2
Eg/2
Eg/2
Eg/2
Eg/2
Eg/2
EC
EV
EF1
EC
EV
EF1EF2
EF2
EF3
EF3
n3n1 n2
p1 p3p2
p3 > p2 > p1
n3 > n2 > n1
W k d l
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Worked example
(b) Considering the density of holes in valence band;
It is seen that as the acceptor concentration increases, Fermi-level
moves towards the valance band edge. These results will be used in
the construction of device (energy) band diagrams.
kT
EENp VFvexp
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Donors and acceptor both present
n D An N N 2) Similarly, when the number of shallow acceptor concentration is signicantly
greater than the shallow donor concentration in a piece of a s/c, it can be
considered as a p-type s/c and
In general, both donors and acceptors are present in a piece of a
semiconductor although one will outnumber the other one.
The impurities are incorporated unintentionally during the growth of the
semiconductor crystal causing both types of impurities being present in a piece
of a semiconductor.
How do we handle such a piece of s/c?
1) Assume that the shallow donor concentration is significantly greater thanthat of the shallow acceptor concentration. In this case the material behaves as
an n-type material and
D d t b th t
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For the case NA>ND , i.e. for p-type material
Donors and acceptor both present
2
2
2 2
.
0
0 ( ) 0
p p i
p A D P p D p A
ip p D A p D A p i
p
n p n
n N N p p N n N
np p N N p N N p n
p
D d t b th t
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Donors and acceptor both present
2
2 21,2
12 22
2
4( ) 0 , solving for ;2
1
42
p D A p i p
p A D A D i
i
pp
b b acp N N p n p xa
p N N N N n
n
n p
majority
minority
D d t b th t
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Donors and acceptor both present
For the case ND>NA, i.e. n-type material
22
22 2
2
1,2
12 22
2
.
0
0 ( ) 0
4solving for n ;
2
14
2
i
n n i nn
n A D n n A n D
i
n n A D n A D n in
n
n D A D A i
in
nn p n p
n
n N N P n N p N
n
n n N N n N N n nn
b b acx
a
n N N N N n
np