Chapter 4: Semiconductor in Equilibrium

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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
Microelectronics I
e
e
Ec
Ev
Conduction
band
Valence
band
T>0K
e
band
￿Particles that can freely move and contribute to the current flow (conduction)
1.Electron in conduction band
2.Hole in valence band
carrier
Microelectronics I
￿How to count numberof 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”
e
Ec
Conduction
band
C
EE
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)
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
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
Microelectronics I
Thermal equilibrium concentration of electron, n
o


=
C
E
o
dEEfEgn)()(
C
EE
h
m
Eg−=
3
2/3
*)2(4
)(
π





=
E
E
)
(
1












+
=
kT
E
E
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
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 10
19
cm
-3
.
Ec
EF
0.25 eV
)
25
.
0
(
)(
exp


+






−−
=
E
E
kT
EE
Nn
FC
Co
Ev
315
19
10
8
.
1
0259.0
)
25
.
0
(
exp108.2

×
=




+


⋅×=
cm
E
E
CC
Thermal equilibrium concentration of hole, p
o
Microelectronics I
[]


−=
Ev
o
dEEfEgp)(1)(
EE
h
m
Eg
v
−=
3
2/3
*)2(4
)(
π





=

E
E
)
(
1












+
=

kT
E
E
kT
EE
Ef
F
F
F
)
(
exp
exp1
1
)(1
Boltzmann approximation




−−
=




−−






=
kT
EE
N
kT
EE
h
kTm
p
vF
v
vF
p
o
)(
exp
)(
exp
2
2
2/3
2
*
π
Nv; effective density of states
function in valence band
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 10
19
cm
-3
.
Ec
E
F
)
27
.
0
(
)(
exp



+





−−
=
E
E
kT
EE
Np
vF
vo
Ev
E
F
0.27 eV
314
19
10
09
.
3
0259.0
)
27
.
0
(
exp1004.1

×
=





+

⋅×=
cm
E
E
Vv
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
vF
p
o
)(
exp
)(
exp
2
2
2/3
2
*
π






kT
kT
h
￿NC
and N
v
are constant for a given material (effective mass) and temperature
￿Position of Fermi energy is important
If E
F
is closer to E
C
than to E
v
, n>p
If E
F
is closer to E
v
than to E
C, n<p
Microelectronics I
Consider ex. 1
Ec
Ev
EF
0.25 eV
315
19
10
8
.
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
10
68
.
2
0259.0
)25.012.1(
exp1004.1
)(
exp

×
=




−−
⋅×=




−−
=
cm
kT
EE
Np
vF
vo
Microelectronics I
Intrinsic semiconductor; A pure semiconductor with no impurity atoms
and no lattice defects in crystal
1.Carrier concentration(n
i, p
i)
2.Position of E
Fi
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
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(×=







××=n
i
36
1026.2
0259
.
0

×=




cmn
i
Microelectronics I
2. Intrinsic Fermi level position, E
Fi
If E
F
closer to Ec, n>p
If E
F
closer to Ev, n<p
Intrinsic; n=p
EF
is located near the center of the forbidden bandgap








)
(
)
(
E
E
E
E






+=






=






*
*
ln
4
3
)
(
exp
)
(
exp
n
p
midgapFi
vFi
v
FiC
C
m
m
kTEE
kT
E
E
N
kT
E
E
N
Ec
Ev
Emidgap
Mp
≠ m
n
Mp
= m
n
EFi
= E
midgap
EFi
shifts slightly from E
midgap
Microelectronics I
Efi is located near the center of Eg
no=p
o
Efi is located near the center of Eg
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)
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
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)
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
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)
Microelectronics I
￿Pure single-crystal semiconductor; intrinsic semiconductor
￿Semiconductor with dopant atoms; extrinsic semiconductor
p-type
n-type
Dopant atom;
Majority carrier;
￿Donor impurity atom
￿
electron
￿Acceptor impurity atom
￿
hole
Majority 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).
Microelectronics I
III-V semiconductors
GaAs
Group IIIGroup 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
Microelectronics I
Carrier concentration of extrinsic semiconductor
When dopant atoms are added, Fermi energy and distribution of electron and hole
will change.
E
F>E
Fi
EF<E
Fi
Electron> hole
n-type
hole> electron
p-type
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
EvEF
1.12 eV
0.25 eV
Band diagram of Si. At T= 300 K,
Nc=2.8x10
19
cm
-3
and Nv=1.04x10
19
cm
-3
.
Calculate n
o
and p
o.
31519
108.1
0259.0
25.0
exp)108.2(

×=





×=cmn
o
3419
107.2
0259.0
)25.012.1(
exp)1004.1(

×=




−−
×=cmpo
N-type Si
Microelectronics I
￿Change of Fermi energy causes change of carrier concentration.
no
and p
o
equation as function of the change of Fermi energy





=




−−
=
kT
EE
n
kT
EE
Nn
FiF
i
FC
Co
exp
)(
exp








E
E
E
E
)
(
)
(






=






=
kT
E
E
n
kT
E
E
Np
FiF
i
vF
vo
)
(
exp
)
(
exp
ni; intrinsic carrier concentration
Efi
; intrinsic Fermi energy
Microelectronics I
The n
o
p
o
product
2
exp
)(
exp
)(
exp
i
g
vC
vFFC
vCoo
n
kT
E
NN
kT
EE
kT
EE
NNpn
=





=




−−




−−
=
i
n
=
2
ioo
npn=
Product of n
o
and p
o
is always a constant for a given material at a given
temperature.
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
E
F
EF
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
Microelectronics I
Statistic donors and acceptors
Discrete donor level
donor
+
−=





+
=
dd
Fd
d
d
NN
kT
EE
N
n
exp
2
1
1
Density of electron
occupying the
donor level
Concentration of
donors
Concentration of
ionized donors
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)
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
N
nn
n
dC
d
C
od
d
)(
exp
2
1
1
Consider phosphorus doping in Si for T=300K at concentration of 10
16
cm
-3
ionization energy
Consider phosphorus doping in Si for T=300K at concentration of 10
cm
(N
C=2.8 x10
19
cm
-3
, E
C-E
d= 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
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
Microelectronics I
At T=0 K, all electron in their lowest possible energy state
Nd+=0 and N
a-=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.
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 N
d
> N
a
￿n-type compensated semiconductor
If N
a
> N
d
￿p-type compensated semiconductor
If N
d
= N
a
￿has the characteristics of an intrinsic semiconductor
concentration
Microelectronics I
Charge-neutrality condition
+−
+=+
doao
NpNn
Negative charges
Positive charges
Negative charges
Positive charges
)()(
ddoaao
nNppNn

+
=

+
Microelectronics I
)()(
ddoaao
nNppNn

+
=

+
If we assume complete ionization (p
a=0, n
d=0)
doao
NpNn
+
=
+
From n
opo=n
i2
2
i
N
n
N
n
+
=
+
2
2
0
22
i
adad
o
d
i
ao
n
NNNN
n
N
n
n
N
n
+





+

=
+
=
+
Electron concentration is given as function of donors and acceptors concentrations
Microelectronics I
Example;
Consider an n-type silicon semiconductor at T=300 K in which N
d=10
16
cm
-3
and N
a=0. The intrinsic carrier concentration is assumed to be n
i=1.5x10
10
cm
-3
. Determine the thermal equilibrium electron and hole concentrations.
2
10
1616
2
2
)
10
5
.
1
(
10
10
22
×
+




+
=
+





+

=n
NNNN
n
i
adad
o
Electron,
316
2
10
10
)
10
5
.
1
(
2
10
2
10


×
+








+
=
cm
hole,
34
16
210
2
1025.2
10
)105.1(

×=
×
==cm
n
n
p
o
i
o
Microelectronics I
Redistribution of electrons when donors are added
++++
-
-
-
-
-
-
Intrinsic electron
+
+
+
Intrinsic hole
￿When donors are added, n
o
> n
i
and p
o
< n
i
￿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
Microelectronics I
Temperature dependence of n
o
2
2
22
i
adad
o
n
NNNN
n+





+

=
Very strong function of temperature
￿
As temperature increases, n
2
term will dominate. Shows intrinsic characteristics
￿
As temperature increases, n
i
2
term will dominate. Shows intrinsic characteristics
0 K
Temperature
Freeze-out
Partial ionization
Extrinsicno=N
d
Intrinsicno=n
i
Microelectronics I
Hole concentration
From charge-neutrality condition and n
opo
product
)()(
ddoaao
nNppNn

+
=

+
2
ioo
npn=
2
i
N
p
N
n
+
=
+
2
2
22
i
dada
o
doa
o
i
n
NNNN
p
N
p
N
p
n
+





+

=
+
=
+
Microelectronics I
Example;
Consider an p-type silicon semiconductor at T=300 K in which N
a=10
16
cm
-3
and N
d=3 x 10
15
cm
-3
. The intrinsic carrier concentration is assumed to be
n
i=1.5x10
10
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=N
a-N
d
approximation
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
kTEEln
Compensated semiconductor, n
=N
-
N







=−
ad
C
FC
NN
N
kTEEln
Compensated semiconductor, n
o
=N
d
-
N
a






=−
i
o
FiF
n
n
kTEEln
Microelectronics I
Equations for position of Fermi level (p-type)






=−
o
v
CF
p
N
kTEEln
Compensated semiconductor, p
o
=N
a-N
d









=−
d
a
v
vF
N
N
N
kTEEln





d
a
N
N






=−
i
o
FFi
n
p
kTEEln
Microelectronics I
Example;
Silicon at T=300 K contains an acceptor impurity concentration of N
a=10
16
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











=−
E
E
NN
N
kTEE
ad
C
FC
31619
1024.1
0259.0
2.0
exp108.2
)
(
exp

×=





×=






=−
cm
kT
E
E
NNN
FC
Cad
316316
1024.21024.1
−−
×=+×=cmNcmN
ad
Microelectronics I
Position of E
F
as function of donor concentration (n-type) and acceptor
concentration (p-type)
Microelectronics I
Position of E
F
as function of temperature for various doping concentration
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
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=
Microelectronics I
problems
1. The value of p
o
in Silicon at T=300K is 10
15
cm
-3
. Determine (a) E
c-E
F
and (b) n
o
2. Determine the equilibrium electron and hole concentrations in Silicon for the
following conditions;
(a)T=300 K, N
d= 2x10
15
cm
-3
, N
a=0
(b)T=300 K, N
d=Na=10
15
cm
-3
3. (a) Determine
the
position of the Fermi level with respect to the intrinsic Fermi
3. (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 10
15
cm
-3
. (b) Repeat part (a) if the Si is doped with boron
atoms at a concentration of 10
15
cm
-3
. (c) Calculate the electron concentration in
the Si for part (a) and (b)