1. Microelectronics I
Chapter 4: Semiconductor in
Equilibrium
Equilibrium; no external forces such as voltages, electrical
fields, magnetic fields, or temperature gradients are acting on
the semiconductor
3. Microelectronics I
How to count number of carriers,n?
If we know
1. No. of energy states
2. Occupied energy states
Density of states (DOS)
The probability that energy states is
Assumption; Pauli exclusion principle
2. Occupied energy states The probability that energy states is
occupied
“Fermi-Dirac distribution function”
n = DOS x “Fermi-Dirac distribution function”
4. e
Ec
Conduction
band
CEE
h
m
Eg −= 3
2/3
*)2(4
)(
π
No of states (seats) above EC for electron
Microelectronics I
Density of state
E
e
Ec
Ev
Valence
band
EE
h
m
Eg v −= 3
2/3
*)2(4
)(
π
No of states (seats) below Ev for hole
g (E)
5. Fermi-Dirac distribution
Microelectronics I
e
Ec
Probability of electron having certain energy
E
Electron (blue line)
−
+
=
kT
EE
Ef
F
F
exp1
1
)(
Electron
having energy
above Ec
e
Ec
Ev
f (E)10.5
Fermi energy, EF
EF; the energy below which all states are filled with electron and above
which all states are empty at 0K
kT
hole (red line)
−
+
−
kT
EE F
exp1
1
1
Hole having
energy below
Ev
6. e
Ec
E
Microelectronics I
No of carrier
No of free electron
e
Ec
free electron
e
Ev
g(E) x f (E)1
No of free hole e
Ev free hole
7. Microelectronics I
Thermal equilibrium concentration of electron, no
∫
∞
=
CE
o dEEfEgn )()(
CEE
h
m
Eg −= 3
2/3
*)2(4
)(
π
−−
≈=
EE )(1
−−
≈
−
+
=
kT
EE
kT
EE
Ef F
F
F
)(
exp
exp1
1
)( Boltzmann approximation
−−
=
−−
=
kT
EE
N
kT
EE
h
kTm
n
FC
C
FCn
o
)(
exp
)(
exp
2
2
2/3
2
*
π
NC; effective density of states
function in conduction band
8. Microelectronics I
Ex. 1
Calculate the thermal equilibrium electron concentration in Si at T= 300K.
Assume that Fermi energy is 0.25 eV below the conduction band. The value of Nc for Si
at T=300 K is 2.8 x 1019 cm-3.
Ec
EF
0.25 eV
)25.0(
)(
exp
+−−
−−
=
EE
kT
EE
Nn FC
Co
Ev
315
19
108.1
0259.0
)25.0(
exp108.2
−
×=
+−−
⋅×=
cm
EE CC
9. Thermal equilibrium concentration of hole, po
Microelectronics I
[ ]∫∞
−=
Ev
o dEEfEgp )(1)(
EE
h
m
Eg v −= 3
2/3
*)2(4
)(
π
−−
≈=−
EE )(1
−−
≈
−
+
=−
kT
EE
kT
EE
Ef F
F
F
)(
exp
exp1
1
)(1 Boltzmann approximation
−−
=
−−
=
kT
EE
N
kT
EE
h
kTm
p
vF
v
vFp
o
)(
exp
)(
exp
2
2
2/3
2
*
π
Nv; effective density of states
function in valence band
10. Microelectronics I
Ex.2
Calculate the thermal equilibrium hole concentration in Si at T= 300K.
Assume that Fermi energy is 0.27 eV below the conduction band. The value of Nc for Si
at T=300 K is 1.04 x 1019 cm-3.
Ec
EF
)27.0(
)(
exp
−+−
−−
=
EE
kT
EE
Np vF
vo
Ev
EF
0.27 eV
314
19
1009.3
0259.0
)27.0(
exp1004.1
−
×=
−+−
⋅×=
cm
EE Vv
11. Microelectronics I
−−
=
−−
=
kT
EE
N
kT
EE
h
kTm
n FC
C
FCn
o
)(
exp
)(
exp
2
2
2/3
2
*
π
−−
=
−−
=
kT
EE
N
kT
EE
h
kTm
p vF
v
vFp
o
)(
exp
)(
exp
2
2
2/3
2
*
π
kTkTh
NC and Nv are constant for a given material (effective mass) and temperature
Position of Fermi energy is important
If EF is closer to EC than to Ev, n>p
If EF is closer to Ev than to EC, n<p
12. Microelectronics I
Consider ex. 1
Ec
Ev
EF
0.25 eV
315
19
108.1
0259.0
)25.0(
exp108.2
)(
exp
−
×=
+−−
⋅×=
−−
=
cm
EE
kT
EE
Nn
CC
FC
Co
Eg-0.25 eV
Hole concentration
Eg=1.12 eV
34
19
1068.2
0259.0
)25.012.1(
exp1004.1
)(
exp
−
×=
−−
⋅×=
−−
=
cm
kT
EE
Np vF
vo
13. Microelectronics I
Intrinsic semiconductor; A pure semiconductor with no impurity atoms
and no lattice defects in crystal
1. Carrier concentration(ni, pi)
2. Position of EFi
1. Intrinsic carrier concentration
Concentration of electron in in conduction band, ni
Concentration of hole in in valence band, pi
−−
=
−−
==
kT
EE
N
kT
EE
Npn vFi
v
FiC
Cii
)(
exp
)(
exp
−
=
−−
=
kT
E
NN
kT
EE
NNn
g
vc
vC
vCi exp
)(
exp
2
Independent of Fermi energy
14. Microelectronics I
Ex. 3; Calculate the intrinsic carrier concentration in gallium arsenide (GaAs)
at room temperature (T=300K). Energy gap, Eg, of GaAs is 1.42 eV. The
value of Nc and Nv at 300 K are 4.7 x 1017 cm-3 and 7.0 x 1018 cm-3,
respectively.
1218172
1009.5
0259.0
42.1
exp)100.7)(107.4( ×=
−
××=ni
36
1026.2
0259.0
−
×=
cmni
15. Microelectronics I
2. Intrinsic Fermi level position, EFi
If EF closer to Ec, n>p
If EF closer to Ev, n<p
Intrinsic; n=p
EF is located near the center of the forbidden bandgap
−− −− )()( EEEE
+=
−−
=
−−
*
*
ln
4
3
)(
exp
)(
exp
n
p
midgapFi
vFi
v
FiC
C
m
m
kTEE
kT
EE
N
kT
EE
N
Ec
Ev
Emidgap
Mp ≠ mn
Mp = mn EFi = Emidgap
EFi shifts slightly from Emidgap
17. Microelectronics I
Dopant atoms and energy levels
adding small, controlled amounts of specific dopant, or impurity, atoms
Increase no. of carrier (either electron or hole)
Alter the conductivity of semiconductor
III IV V
B C
Al Si P
Ga Ge As
In Sb
3 valence
electrons
5 valence
electrons
Consider Phosphorus (P) and boron (B) as
impurity atoms in Silicon (Si)
18. Microelectronics I
1. P as substitutional impurity (group V element; 5 valence electron)
In intrinsic Si, all 4 valence electrons contribute to covalent bonding.
In Si doped with P, 4 valence electron of P contribute to covalent bonding
and 1 electron loosely bound to P atom (Donor electron).
Donor electron
can easily break the bond and freely moves
19. Microelectronics I
Energy to elevate the donor electron into conduction band is less than that for
the electron involved in covalent bonding
Ed(; energy state of the donor electron) is located near Ec
When small energy is added, donor electron is elevated to conduction band,
leaving behind positively charged P ion
P atoms donate electron to conduction band P; donor impurity atom
No. of electron > no. of hole n-type semiconductor (majority carrier is electron)
20. Microelectronics I
2. B as substitutional impurity (group III element; 3 valence electron)
In Si doped with B, all 3 valence electron of B contribute to covalent
bonding and one covalent bonding is empty
When small energy is added, electron that involved in covalent bond will
occupy the empty position leaving behind empty position that associated
with Si atom
Hole is created
21. Microelectronics I
Electron occupying the empty state associated with B atom does not have
sufficient energy to be in the conduction band no free electron is created
Ea (;acceptor energy state) is located near Ev
When electron from valence band elevate to Ea, hole and negatively
charged B are created
B accepts electron from valence band B; acceptor impurity atom
No. of hole > no. of electron p-type material (majority carrier is hole)
22. Microelectronics I
Pure single-crystal semiconductor; intrinsic semiconductor
Semiconductor with dopant atoms; extrinsic semiconductor
p-typen-type
Dopant atom;
Majority carrier;
Donor impurity atom
electron
Acceptor impurity atom
holeMajority carrier; electron hole
Ionization Energy
The energy that required to elevate donor electron into the conduction (in
case of donor impurity atom) or to elevate valence electron into acceptor
state (in case of acceptor impurity atom).
23. Microelectronics I
III-V semiconductors
GaAs
Group III Group V
Dopant atoms;Dopant atoms;
Group II (beryllium, zinc and cadmium) replacing Ga; acceptor
Group VI (selenium, tellurium) replacing As; donor
Group IV (Si and germanium) replacing Ga; donor
As; acceptor
24. Microelectronics I
Carrier concentration of extrinsic semiconductor
When dopant atoms are added, Fermi energy and distribution of electron and hole
will change.
EF>EFi EF<EFi
Electron> hole
n-type
hole> electron
p-type
25. Microelectronics I
−−
=
kT
EE
Nn FC
Co
)(
exp
−−
=
kT
EE
Np vF
vo
)(
exp
Thermal equilibrium concentration of electron
Thermal equilibrium concentration of hole
Ex. 4
Ec
0.25 eV
Band diagram of Si. At T= 300 K,Ec
Ev
EF
1.12 eV
0.25 eV
Band diagram of Si. At T= 300 K,
Nc=2.8x1019cm-3 and Nv=1.04x1019cm-3.
Calculate no and po.
31519
108.1
0259.0
25.0
exp)108.2( −
×=
−
×= cmno
3419
107.2
0259.0
)25.012.1(
exp)1004.1( −
×=
−−
×= cmpo
N-type Si
26. Microelectronics I
Change of Fermi energy causes change of carrier concentration.
no and po equation as function of the change of Fermi energy
−
=
−−
=
kT
EE
n
kT
EE
Nn FiF
i
FC
Co exp
)(
exp
−− −− EEEE )()(
−−
=
−−
=
kT
EE
n
kT
EE
Np FiF
i
vF
vo
)(
exp
)(
exp
ni; intrinsic carrier concentration
Efi; intrinsic Fermi energy
27. Microelectronics I
The nopo product
2
exp
)(
exp
)(
exp
i
g
vC
vFFC
vCoo
n
kT
E
NN
kT
EE
kT
EE
NNpn
=
−
=
−−
−−
=
in=
2
ioo npn =
Product of no and po is always a constant for a given material at a given
temperature.
28. Microelectronics I
Degenerate and Non degenerate semiconductors
Small amount of dopant atoms (impurity atoms)
No interaction between dopant atoms
Discrete, noninteracting energy state.
EF at the bandgap
E
donor acceptor
Nondegenerate semiconductor
EF
EF
29. Large amount of dopant atoms (~effective density of states)
Microelectronics I
Dopant atoms interact with each other
Band of dopant states widens and overlap the allowed band
(conduction @ valence band)
EF lies within conduction @ valence band
e
e
e
Ec
Ev
Filled states
EF
e
Ec
Ev empty states
EF
Degenerate semiconductor
30. Microelectronics I
Statistic donors and acceptors
Discrete donor level
donor
+
−=
−
+
= dd
Fd
d
d NN
kT
EE
N
n
exp
2
1
1Density of electron
occupying the
donor level
Concentration of
donors
Concentration of
ionized donors
31. Microelectronics I
acceptor
Discrete acceptor level
acceptor
−
−=
−
+
= aa
aF
a
a NN
kT
EE
g
N
p
exp
1
1
Concentration of
holes in the
acceptor states
Concentration of
acceptors
Concentration of
ionized acceptor
g; degeneracy factor (Si; 4)
32. Microelectronics I
from the probability function, we can calculate the friction of total electrons still in
the donor states at T=300 K
−−
+
=
+
kT
EE
N
Nnn
n
dC
d
Cod
d
)(
exp
2
1
1
Consider phosphorus doping in Si for T=300K at concentration of 1016 cm-3
ionization energy
Consider phosphorus doping in Si for T=300K at concentration of 10 cm
(NC=2.8 x1019 cm-3, EC-Ed= 0.045 eV)
%41.00041.0
0259.0
045.0
exp
102
108.2
1
1
16
19
==
−
×
×
+
=
+ od
d
nn
n
only 0.4% of donor states contain electron. the donor states are states
are said to be completely ionized
33. Microelectronics I
Complete ionization; The condition when all donor atoms are positively
charged by giving up their donor electrons and all acceptor atoms are negatively
charged by accepting electrons
34. Microelectronics I
At T=0 K, all electron in their lowest possible energy state
Nd
+=0 and Na
-=0
EF
EF
Freeze-out; The condition that occurs in a semiconductor when the temperature
is lowered and the donors and acceptors become neutrally charged. The
electron and hole concentrations become very small.
35. Microelectronics I
Charge neutrality
In thermal equilibrium, semiconductor crystal is electrically neutral
“Negative charges = positive charge”
Determined the carrier concentrations as a function of impurity doping
concentration
Charge-neutrality condition
Compensated semiconductor; A semiconductor that contains both donor and
acceptors at the same region
If Nd > Na n-type compensated semiconductor
If Na > Nd p-type compensated semiconductor
If Nd = Na has the characteristics of an intrinsic semiconductor
concentration
37. Microelectronics I
)()( ddoaao nNppNn −+=−+
If we assume complete ionization (pa=0, nd=0)
doao NpNn +=+
From nopo=ni
2
2
i
N
n
Nn +=+
2
2
0
22
i
adad
o
d
i
ao
n
NNNN
n
N
n
n
Nn
+
−
+
−
=
+=+
Electron concentration is given as function of donors and acceptors concentrations
38. Microelectronics I
Example;
Consider an n-type silicon semiconductor at T=300 K in which Nd=1016 cm-3
and Na=0. The intrinsic carrier concentration is assumed to be ni=1.5x1010
cm-3. Determine the thermal equilibrium electron and hole concentrations.
210
1616
2
2
)105.1(
1010
22
×+
+=
+
−
+
−
= n
NNNN
n i
adad
o
Electron,
316
210
10
)105.1(
2
10
2
10
−
≈
×+
+=
cm
hole, 34
16
2102
1025.2
10
)105.1( −
×=
×
== cm
n
n
p
o
i
o
39. Microelectronics I
Redistribution of electrons when donors are added
+ + + +
- - - - - -
Intrinsic electron
+ + +
Intrinsic hole
When donors are added, no > ni and po < ni
A few donor electron will fall into the empty states in valence band and
hole concentration will decrease
Net electron concentration in conduction band ≠ intrinsic electron +
donor concentration
40. Microelectronics I
Temperature dependence of no
2
2
22
i
adad
o n
NNNN
n +
−
+
−
=
Very strong function of temperature
As temperature increases, n2 term will dominate. Shows intrinsic characteristicsAs temperature increases, ni
2 term will dominate. Shows intrinsic characteristics
0 K Temperature
Freeze-out
Partial ionization
Extrinsic
no=Nd
Intrinsic
no=ni
41. Microelectronics I
Hole concentration
From charge-neutrality condition and nopo product
)()( ddoaao nNppNn −+=−+
2
ioo npn =
2
i
NpN
n
+=+
2
2
22
i
dada
o
doa
o
i
n
NNNN
p
NpN
p
n
+
−
+
−
=
+=+
42. Microelectronics I
Example;
Consider an p-type silicon semiconductor at T=300 K in which Na=1016 cm-3
and Nd=3 x 1015 cm-3. The intrinsic carrier concentration is assumed to be
ni=1.5x1010 cm-3. Determine the thermal equilibrium electron and hole
concentrations.
2
2
22
+
−
+
−
= n
NNNN
p i
dada
o
Hole,
315
210
15161516
107
)105.1(
2
10310
2
10310
−
×≈
×+
×−
+
×−
=
cm
electron,
34
15
2102
1021.3
107
)105.1( −
×=
×
×
== cm
p
n
n
o
i
o
po=Na-Ndapproximation
43. Microelectronics I
Position of Fermi Energy Level
As a function of doping concentration and temperature
Equations for position of Fermi level (n-type)
=−
o
C
FC
n
N
kTEE ln
Compensated semiconductor, n =N -N
−
=−
ad
C
FC
NN
N
kTEE ln
Compensated semiconductor, no=Nd-Na
=−
i
o
FiF
n
n
kTEE ln
44. Microelectronics I
Equations for position of Fermi level (p-type)
=−
o
v
CF
p
N
kTEE ln
Compensated semiconductor, po=Na-Nd
−
=−
da
v
vF
NN
N
kTEE ln
− da NN
=−
i
o
FFi
n
p
kTEE ln
45. Microelectronics I
Example;
Silicon at T=300 K contains an acceptor impurity concentration of Na=1016 cm-3.
Determine the concentration of donor impurity atoms that must be added so that
the Silicon is n-type and Fermi energy is 0.20 eV below the conduction band
edge.
)(
ln
−−
−
=−
EE
NN
N
kTEE
ad
C
FC
31619
1024.1
0259.0
2.0
exp108.2
)(
exp
−
×=
−
×=
−−
=−
cm
kT
EE
NNN FC
Cad
316316
1024.21024.1 −−
×=+×= cmNcmN ad
48. Microelectronics I
Important terms
Intrinsic semiconductor; A pure semiconductor material with no impurity
atoms and no lattice defects in the crystal
Extrinsic semiconductor; A semiconductor in which controlled amounts
of donors and/or acceptors have been added so that the electron and hole
concentrations change from the intrinsic carrier concentration and a
preponderance of either electron (n-type) or hole (p-type) is created.preponderance of either electron (n-type) or hole (p-type) is created.
Acceptor atoms; Impurity atoms added to a semiconductor to create a p-
type material
Donor atoms; Impurity atoms added to a semiconductor to create n-type
material
49. Microelectronics I
Complete ionization; The condition when all donor atoms are positively
charged by giving up their donor electrons and all acceptor atoms are
negatively charged by accepting electrons
Freeze-out; The condition that occurs in a semiconductor when the
temperature is lowered and the donors and acceptors become neutrally
charged. The electron and hole concentrations become very small
Fundamental relationship
2
ioo npn =
50. Microelectronics I
problems
1. The value of po in Silicon at T=300K is 1015 cm-3. Determine (a) Ec-EF and (b) no
2. Determine the equilibrium electron and hole concentrations in Silicon for the
following conditions;
(a) T=300 K, Nd= 2x1015cm-3, Na=0
(b) T=300 K, Nd=Na=1015 cm-3
3. (a) Determine the position of the Fermi level with respect to the intrinsic Fermi3. (a) Determine the position of the Fermi level with respect to the intrinsic Fermi
level in Silicon at T=300K that is doped with phosphorus atoms at a
concentration of 1015cm-3. (b) Repeat part (a) if the Si is doped with boron
atoms at a concentration of 1015cm-3. (c) Calculate the electron concentration in
the Si for part (a) and (b)
