No external forces such as voltages, electric fields, magnetic fields or temperature gradient are acting on the semiconductor. All properties of the semiconductor will be independent of time in this case.
The Semiconductor in Equilibrium
Equilibrium or thermal equilibrium:
Electron or hole concentrations at equilibrium:
The number of electrons per unit energy per unit volume in the conduction band is,
)()()( EfEgEn Fc=
)](1)[()( EfEgEp Fv −=For holes,
The total number of electrons per unit volume at thermal equilibrium,
dEEfEgn F
E
E cc
c
)()(0 ∫+
=χ
(A)
E
g(E)
g(E) ∝ (E-Ec)1/2
f(E)
EF
n(E) or p(E)
E E
Forelectrons
For holes
[1-f(E)]
n(E)
p(E)
Ec
EvEv
Ec
0
Ec+χ
EF
VB
CB
(a) (b) (c) (d)
Area = ∫ n(E)dE=n0
Area = p0
Since the Fermi function fF(E) decays rapidly with energy so that gc(E)fF(E) → 0 near the top of the band. Therefore, the upper limit of integration in equation (A) to be Ec+χ→ ∞. We assume that (Ec-EF)> 3kT so that
]/)(exp[)( kTEEEf FF −−≈ [Boltzman approximation]
dEkTEEEE
hmn F
CEn
C ⎥⎦⎤
⎢⎣⎡ −−
−= ∫∝ )(exp)2(4
3
2/3*
0π
This approximation is good for nondegenerate Semiconductors:
Let, ηη kTddEkTEE c =⇒
−=
ηηπ ηdekTEE
hkTmn FCn ∫
∝ −⎥⎦⎤
⎢⎣⎡ −−
=0
2/13
2/32/3*
0)(exp)()2(4
πηη η
21)2/1(
21)12/1(
0
2/1 =Γ=+Γ=∫∝ − de
)()(exp0 ccFc
c EfNkTEENn =⎥⎦
⎤⎢⎣⎡ −−
=∴
For nondegenerate Semiconductors:
232/3
2
*22 ThkTmN n
c ∝⎟⎟⎠
⎞⎜⎜⎝
⎛=
π Nc is the effective density of states in the CB
( )⎥⎦⎤
⎢⎣⎡ −−
=kT
EENp vFv exp0Similarly,
kTEvc
geNNpn /00
−=
Self-Study: Determine the expression of p0
( ) ( ) kTEvc
kTEEvc
vFFcvc
gvc eNNeNNkTEE
kTEENNpn //)(
00 expexp −−− ==⎥⎦⎤
⎢⎣⎡ −−
⎥⎦⎤
⎢⎣⎡ −−
=
I Intrinsic Semiconductors:A pair of an electron and a hole (EHP) is generated due to the thermal excitation.
Conduction electron+4
Si+4Si
+4Si
+4Si
+4Si
+4Si
+4Si
+4Si
+4Si
Eg
Ec
Ev
Ef
Valance hole
Electron and hole pairs (EHP) are generated due to the thermal excitation. no is the electron density (number of electrons / cm3), po is the hole density (number of holes / cm3) . for intrinsic materials no = po = ni . For silicon ni = 1.5 x 1010 and for GaAsni = 106 .
nn
⎥⎦⎤
⎢⎣⎡ −−
=kTEENp vF
v)(exp0
⎥⎦⎤
⎢⎣⎡ −−
=kTEENn Fc
cc)(exp
)()()( EfEgEN F=
)()()( EfEgEN F=
kTEvci
geNNn 2/−=∴
200 inpn = → Mass action law (Valid at thermal equilibrium)
Intrinsic Carrier Concentration:
n0=p0=ni; the concentration of electrons in the conduction band is equal to the concentration of holes in the valence band. The Fermi energy level for intrinsic semiconductor is called the intrinsic Fermi energy or EF=EFi, Therefore,
⎥⎦⎤
⎢⎣⎡ −−
=kTEENn Fic
ci)(exp ( )
⎥⎦⎤
⎢⎣⎡ −−
=kTEENp vFi
vi expand
00/2 pneNNnpn kTE
vciiig ===∴ −
Example 4.1: Given, EF-Ev=0.27eV, Nv=1.04×1019cm-3 at 300°K. Calculate p0 at 400°K.
⎥⎦⎤
⎢⎣⎡ −
=kTEEnn FiF
i)(exp0
⎥⎦⎤
⎢⎣⎡ −
=kTEEnp FFi
i)(exp0
And
⎥⎦⎤
⎢⎣⎡ −−
=kTEENn Fc
c)(exp0 ⎥⎦
⎤⎢⎣⎡ −+−−
=kT
EEEEN FiFFicc
)()(exp
⎥⎦⎤
⎢⎣⎡ −
⎥⎦⎤
⎢⎣⎡ −−
=kTEE
kTEEN FiFFic
c exp)(exp
Example 4.2:Given, Nc=2.8×1019cm-3, Nv=1.04×1019cm-3 at 300°K in Si. Eg=1.1eV in Si.Calculate, (i) ni at 300K
(ii) ni at 450K
Temperature dependency of nifor Si, Ge and GaAs
kTEvci
geNNn 2/−=∴
232/3
2
*22 ThkTmN n
c ∝⎟⎟⎠
⎞⎜⎜⎝
⎛=
π
232/3
2
*22 T
h
kTmN pv ∝
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
π
( ) kTEi
geTTn 2/23 −∝∴
Intrinsic Fermi-level position
For intrinsic semiconductor,
ii pn =( )
( )
( )
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛=+−
=⎥⎦⎤
⎢⎣⎡ +−
⎥⎦⎤
⎢⎣⎡ −−
=⎥⎦⎤
⎢⎣⎡ −−
*
*
2/3
*
*
ln43
2
ln21
2
ln2
2exp
exp)(exp
n
pvcFi
n
pvcFi
c
vvcFi
c
vvcFi
vFiv
Ficc
mm
kTEEE
mm
kTEEE
NNkTEEE
NN
kTEEE
kTEEN
kTEEN
2vc EE + is the midway between Ec
and Ev
⎟⎟⎠
⎞⎜⎜⎝
⎛+=∴ *
*
ln43
n
pmidgapFi m
mkTEE
Example 4.3:
0* 08.1 mmn = 0
* 56.0 mmp =Given and in Si. Calculate EFi at 300°K
Extrinsic Semiconductor
x
As+ As+ As+ As+
EcEd
Ev Distanceinto crystal
~0.05 eV
CB
Electron Energy
N-type
If Nd >> ni, at room temperature, n0 ≈ Nd. 0
2
0 nnp i=
Ref.: S.O. Kasap, Principles of Electronic materials and devices, 3rd Edition, (McGraw-Hill, 2005)
n-type doping:
As+
e-
x
B -
Ev
Ea
Distance intocrystal
~0.05 eV
B - B - B -
h+
VB
Ec
Electron energy
If Na >> ni, at room temperature, p0 ≈ Na.
0
2
0 pnn i=
Ref.: S.O. Kasap, Principles of Electronic materials and devices, 3rd Edition, (McGraw-Hill, 2005)
B-h+
Free
B-
(a)
(b)
p-type doping:
And
Concept of donors, acceptors
Charge neutrality:
nd = Nd – Nd+ ; pa = Na – Na
-
nd & pa are the densities of electrons and holes occupying donor and acceptor levels respectively.
Statistics of donors and acceptors:
The probability function of electrons occupying the donor state is,
⎟⎠⎞
⎜⎝⎛ −
+=
kTEE
EfFd
dd
exp211
1)(
The factor ½ is due to the fact that the electron state at the donor can take an electron with spin either up or down but not both (Both electrons have a chance to occupy the state. Therefore the probability of occupancy is increased).
The density of electrons occupying the donor level Ed,
⎥⎦⎤
⎢⎣⎡ −
+==
kTEE
NEfNnFd
ddddd
exp211
)(
For (Ed-EF)>>kT ⎥⎦
⎤⎢⎣
⎡ −−=
kTEE
Nn Fddd
)(exp2
⎥⎦
⎤⎢⎣
⎡ −−=
kTEE
Nn Fddd
)(exp2
⎥⎦
⎤⎢⎣
⎡ −−=
kTEE
Nn Fcc
)(exp0 → Free electrons in CB
n0+nd → Total number of electrons
nd → No of electrons in the donor state.
( )⎥⎦
⎤⎢⎣
⎡ −−+=+=
+∴
kTEE
NN
nn
nnn dc
d
c
dd
d exp2
11 00
⎥⎦⎤
⎢⎣⎡ −−
+=
+∴
kTEE
NNnn
ndc
d
cd
d
)(exp2
1
1
0→ Determine relative ionization
⎥⎦⎤
⎢⎣⎡ −−
+=
+∴
kTEE
NNpp
pva
d
va
a
)(exp4
1
1
0Similarly for acceptor states,
Example 4.4:Given: (i) Nd = 1016 cm-3, (ii) 1017 cm-3 and (iii) 1018 cm-3 Phosphorus doping in Si at T=300°K. Ec-Ed = 0.045 eV. Calculate the fraction of the total electrons still in the donor states.
⎥⎦⎤
⎢⎣⎡ −−
+=
+∴
kTEE
NNnn
ndc
d
cd
d
)(exp2
1
1
0
A compensated semiconductor is one that contains both donor and acceptor impurity atoms in the same region. In n-type compensated semiconductor Nd> Na and p-type compensated semiconductor Na > Nd. If Na= Nd, the semiconductor is completely compensated and n0 = p0 (like intrinsic SC).
Compensated Semiconductor:
iad nNN >>− )( ad NNn −≈0ad
i
NNnp−
=2
0
ida nNN >>− )( da NNp −≈0da
i
NNnn−
=2
0
At room temperature,
1. n-type non degenerate semiconductor,
and
2. p-type non degenerate semiconductor,,
and
Charge neutrality
Charge neutrality:
In thermal equilibrium, the semiconductor crystal is electrically neutral. This charge-neutrality condition is used to determine the thermal equilibrium electron and hole concentrations as a function of the impurity doping concentration.
If doping level is comparable to ni, then, n0 and p0 can be calculated using charge neutrality:
+− +=+ da NpNn 00
For full ionization aa NN =−dd NN =+and
,
da NpNn +=+ 00
For n-type:0
2
0 nnp i=
di
a NnnNn +=+∴
0
2
0
0)( 20
20 =−−+⇒ ida nnNNn
22
0 22 iadad nNNNNn +⎟
⎠⎞
⎜⎝⎛ −
+−
=∴
dai NpNpn
+=+ 00
2
0)( 20
20 =−−+⇒ iad npNNp
22
0 22 idada nNNNNp +⎟
⎠⎞
⎜⎝⎛ −
+−
=∴
Similarly for p-type:
Example 4.5:At T =300°K in Germanium, ni=2.4×1013 cm-3. Find n0 and p0 for (i) Nd=1016
(ii) Nd=5×1013 cm-3
( ) kTEi
geTTn 2/23 −∝∴
22
0 22 idd nNNn +⎟
⎠⎞
⎜⎝⎛+=
Temperature dependence of carrier concentration:
Example 4.6:At T =550°K in Silicon, the intrinsic carrier concentration must contribute no more than 5 % of the total electron concentration. Determine the minimum donor concentration Nd required to meet this specification.
Position of Fermi energy level:
⎥⎦
⎤⎢⎣
⎡ −−=
kTEE
Nn Fcc
)(exp0 ⎟⎟
⎠
⎞⎜⎜⎝
⎛=−⇒
0lnnNkTEE c
Fc
⎟⎟⎠
⎞⎜⎜⎝
⎛=−⇒⎥⎦
⎤⎢⎣⎡ −
=i
FiFFiF
i nnkTEE
kTEEnn 0
0 lnexp
n-type:
p-type:
⎟⎟⎠
⎞⎜⎜⎝
⎛=−
⎟⎟⎠
⎞⎜⎜⎝
⎛=−
iFFi
vvF
npkTEE
pNkTEE
0
0
ln
ln
Variation of EF with doping:
As the doping levels increase, the Fermi energy level moves closer to the conduction band for the n-type material and closer to the valence band for the p-type material.
Variation of EF with temperature:
⎟⎟⎠
⎞⎜⎜⎝
⎛=−⇒
0lnnNkTEE c
Fc
Example 4.8:Given: p-type Si at 300°K doped with boron. EF-Ea=3kT. Ea-Ev=0.045eV. Calculate: Maximum doping at which the Boltzmann approximation is valid.
Example 4.7:Given: At T = 300°K, Na =1016 cm-3 and Ec-EF=0.2eV in Si.Determine: Nd
Degenerate Semiconductor:
If acceptor or donor concentration in Si is greater than approximately 1017 cm-3, then the Boltzmann approximation is less valid and the semiconductor is called degenerate semiconductor.
CB
g(E)
E
Impurities forminga band
(a) (b)
EFpEv
EcEFn
Ev
Ec
CB
VB
→ Donor electrons interact with each other.→ The single discrete donor energy level will split into a band of energies.→ As the donor concentration further increases, the band of donor states widens and may overlap the bottom of the conduction band. This overlap occurs when the donor concentration becomes comparable with the effective density of states.→ EC is therefore slightly shifted down and Eg becomes slightly narrower.→ The similar phenomena occurs in p-type degenerate semiconductor.
dEEfEgn F
E
E cc
c
)()(0 ∫+
=χ
∫∞
−+
−=
Ec F
cn
kTEEdEEE
mh
n)exp(1
)()2(4 2/1
2/3*30π
kTEE c−
=η kTEE cF
F−
=η
∫∞
−+=
0
2/1
2/1 )exp(1)(
FF
dFηη
ηηη
)(22/10 FcFNn η
π=
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