Chapter4 semiconductor in equilibrium
-
Upload
k-m -
Category
Engineering
-
view
90 -
download
1
Transcript of Chapter4 semiconductor in equilibrium
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
eEc
Ev
Conduction
band
Valence
band
T>0K
eband
�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 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”
e
Ec
Conduction
bandCEE
h
mEg −=
3
2/3*)2(4)(
π
No of states (seats) above EC for electron
Microelectronics I
Density of state
E
e
Ec
EvValence
band
EEh
mEg 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
EEEf
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 Fexp1
11
Hole having
energy below
Ev
e
Ec
E
Microelectronics I
No of carrier
No of free electrone
Ecfree electron
e
Ev
g(E) x f (E)1
No of free holee
Ev free hole
Microelectronics I
Thermal equilibrium concentration of electron, no
∫∞
=
CE
o dEEfEgn )()(
CEEh
mEg −=
3
2/3*)2(4)(
π
−−≈=
EE )(1
−−≈
−+
=kT
EE
kT
EEEf F
F
F
)(exp
exp1
1)( Boltzmann approximation
−−=
−−
=
kT
EEN
kT
EE
h
kTmn
FCC
FCno
)(exp
)(exp
22
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 1019 cm-3.
Ec
EF
0.25 eV
)25.0(
)(exp
+−−
−−=
EE
kT
EENn FC
Co
Ev
315
19
108.1
0259.0
)25.0(exp108.2
−×=
+−−⋅×=
cm
EE CC
Thermal equilibrium concentration of hole, po
Microelectronics I
[ ]∫∞
−=
Ev
o dEEfEgp )(1)(
EEh
mEg v −=
3
2/3*)2(4)(
π
−−≈=−
EE )(1
−−≈
−+
=−kT
EE
kT
EEEf F
F
F
)(exp
exp1
1)(1 Boltzmann approximation
−−=
−−
=
kT
EEN
kT
EE
h
kTmp
vFv
vFp
o
)(exp
)(exp
22
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 1019 cm-3.
Ec
EF)27.0(
)(exp
−+−
−−=
EE
kT
EENp vF
vo
Ev
EF
0.27 eV
314
19
1009.3
0259.0
)27.0(exp1004.1
−×=
−+−⋅×=
cm
EE Vv
Microelectronics I
−−=
−−
=
kT
EEN
kT
EE
h
kTmn FC
CFCn
o
)(exp
)(exp
22
2/3
2
*π
−−=
−−
=
kT
EEN
kT
EE
h
kTmp vF
vvFp
o
)(exp
)(exp
22
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
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
EENn
CC
FCCo
Eg-0.25 eV
Hole concentration
Eg=1.12 eV
34
19
1068.2
0259.0
)25.012.1(exp1004.1
)(exp
−×=
−−⋅×=
−−=
cm
kT
EENp vF
vo
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
EEN
kT
EENpn vFi
vFiC
Cii
)(exp
)(exp
−=
−−=
kT
ENN
kT
EENNn
g
vcvC
vCi exp)(
exp2
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 1009.50259.0
42.1exp)100.7)(107.4( ×=
−××=ni
361026.2
0259.0
−×=
cmni
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
+=
−−=
−−
*
*
ln4
3
)(exp
)(exp
n
p
midgapFi
vFiv
FiCC
m
mkTEE
kT
EEN
kT
EEN
Ec
Ev
Emidgap
Mp ≠ mn
Mp = mn EFi = Emidgap
EFi shifts slightly from Emidgap
Microelectronics I
Efi is located near the center of Eg
no=po
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-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).
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
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-typehole> electron
p-type
Microelectronics I
−−=
kT
EENn FC
Co
)(exp
−−=
kT
EENp vF
vo
)(exp
Thermal equilibrium concentration of electron
Thermal equilibrium concentration of hole
Ex. 4
Ec0.25 eV
Band diagram of Si. At T= 300 K, Ec
Ev
EF
1.12 eV
0.25 eVBand diagram of Si. At T= 300 K,
Nc=2.8x1019cm-3 and Nv=1.04x1019cm-3.
Calculate no and po.
31519 108.10259.0
25.0exp)108.2( −
×=
−×= cmno
3419 107.20259.0
)25.012.1(exp)1004.1( −
×=
−−×= cmpo
N-type Si
Microelectronics I
�Change of Fermi energy causes change of carrier concentration.
no and po equation as function of the change of Fermi energy
−=
−−=
kT
EEn
kT
EENn FiF
iFC
Co exp)(
exp
−− −− EEEE )()(
−−=
−−=
kT
EEn
kT
EENp FiF
ivF
vo
)(exp
)(exp
ni; intrinsic carrier concentration
Efi; intrinsic Fermi energy
Microelectronics I
The nopo product
2
exp
)(exp
)(exp
i
g
vC
vFFCvCoo
n
kT
ENN
kT
EE
kT
EENNpn
=
−=
−−
−−=
in=
2
ioo npn =
Product of no and po 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
EF
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
EFe
Ec
Ev empty states
EF
Degenerate semiconductor
Microelectronics I
Statistic donors and acceptors
Discrete donor level
donor
+−=
−+
= dd
Fd
dd NN
kT
EE
Nn
exp2
11Density of electron
occupying the
donor levelConcentration of
donors
Concentration of
ionized donors
Microelectronics I
acceptor
Discrete acceptor level
acceptor
−−=
−+
= aa
aF
aa NN
kT
EE
g
Np
exp1
1Concentration of
holes in the
acceptor statesConcentration of
acceptorsConcentration 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
Nnn
n
dC
d
Cod
d
)(exp
21
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
045.0exp
102
108.21
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 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.
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
Microelectronics I
Charge-neutrality condition
+−+=+ doao NpNn
Negative charges Positive chargesNegative charges Positive charges
)()( ddoaao nNppNn −+=−+
Microelectronics I
)()( ddoaao nNppNn −+=−+
If we assume complete ionization (pa=0, nd=0)
doao NpNn +=+
From nopo=ni2
2
i Nn
Nn +=+
2
2
0
22i
adado
di
ao
nNNNN
n
Nn
nNn
+
−+
−=
+=+
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 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.
2101616
2
2
)105.1(1010
22
×+
+=
+
−+
−= n
NNNNn i
adado
Electron,
316
210
10
)105.1(2
10
2
10
−≈
×+
+=
cm
hole, 34
16
2102
1025.210
)105.1( −×=
×== cm
n
np
o
io
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
Microelectronics I
Temperature dependence of no
2
2
22i
adado n
NNNNn +
−+
−=
Very strong function of temperature
�As temperature increases, n 2 term will dominate. Shows intrinsic characteristics�As temperature increases, ni2 term will dominate. Shows intrinsic characteristics
0 K Temperature
Freeze-out
Partial ionizationExtrinsic
no=Nd
Intrinsic
no=ni
Microelectronics I
Hole concentration
From charge-neutrality condition and nopo product
)()( ddoaao nNppNn −+=−+
2
ioo npn =
2
i NpNn
+=+
2
2
22i
dadao
doa
o
i
nNNNN
p
NpNp
n
+
−+
−=
+=+
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
NNNNp i
dadao
Hole,
315
21015161516
107
)105.1(2
10310
2
10310
−×≈
×+
×−+
×−=
cm
electron,34
15
2102
1021.3107
)105.1( −×=
×
×== cm
p
nn
o
io
po=Na-Ndapproximation
Microelectronics I
Position of Fermi Energy Level
As a function of doping concentration and temperature
Equations for position of Fermi level (n-type)
=−
o
CFC
n
NkTEE ln
Compensated semiconductor, n =N -N
−=−
ad
CFC
NN
NkTEE ln
Compensated semiconductor, no=Nd-Na
=−
i
oFiF
n
nkTEE ln
Microelectronics I
Equations for position of Fermi level (p-type)
=−
o
vCF
p
NkTEE ln
Compensated semiconductor, po=Na-Nd
−=−
da
vvF
NN
NkTEE ln
− da NN
=−
i
oFFi
n
pkTEE ln
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
NkTEE
ad
CFC
31619 1024.10259.0
2.0exp108.2
)(exp
−×=
−×=
−−=−
cm
kT
EENNN FC
Cad
316316 1024.21024.1 −−×=+×= cmNcmN ad
Microelectronics I
Position of EF as function of donor concentration (n-type) and acceptor
concentration (p-type)
Microelectronics I
Position of EF 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 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 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 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)