Microelectronics I
Chapter 4: Semiconductor in Equilibrium
Equilibrium; Equilibrium ; no external forces forces such as voltages, voltages, electrical fields, magnetic fields, or temperature gradients are acting on the semiconductor semiconductor
Microelectronics I
T>0K e Ec
Conduction band
Ev e
Valence
Particles that can freely move and contribute to the current flow (conduction)
carrier 1. Elec Electr tron on in cond conduc ucti tion on ban band d 2. Hole Hole in valen alencce ban band
Microelectronics I
T>0K e Ec
Conduction band
Ev e
Valence
Particles that can freely move and contribute to the current flow (conduction)
carrier 1. Elec Electr tron on in cond conduc ucti tion on ban band d 2. Hole Hole in valen alencce ban band
Microelectronics I
How to count number of carriers,n?
Assumption; Pauli exclusion principle If we know 1. No. No. of ener energy gy states ates
Density of states (DOS)
2. Occu Occupi pied ed ener energy gy stat states es occupied “Fermi-Dirac “Fermi-Dirac distribution function”
n = DOS x “Fermi-Dirac “Fermi-Dirac distribution function”
Microelectronics I
Density of state E No of states states (seats) above EC for electron
e
Conduction band
g ( E ) =
4π (2m*)3 / 2 h
3
E − E C
Ev e
Valence band
No of states (seats) below Ev for hole
g ( E ) = g (E)
4π (2m*)3 / 2 h
3
E v − E
Microelectronics I
Fermi-Dirac distribution Probability of electron having certain energy
E Electron (blue line) Electron having energy above Ec
f F ( E ) =
e c
1 E − E F 1 + exp kT
Fermi energy, EF Ev Hole having energy below Ev
hole (red line) e
0.5
1−
1
f (E)
1 E − E F 1 + exp kT
EF; the energy below which which all states states are filled with electron electron and above which all states are empty at 0K
Microelectronics I
No of carrier E
No of free electron
e
Ec
e free electron
Ec
Ev No of free hole
Ev
free hole e
e
1
g(E) x f (E)
Microelectronics I
Thermal equilibrium concentration of electron, no ∞
no
∫ g ( E ) f ( E )dE
=
E C
g ( E ) = f F ( E ) =
4π ( 2m*)3 / 2 h
3
E − E C
E − E F 1 + exp kT 3 / 2
≈
exp
−
−
kT
F
Boltzmann approximation
2π mn* kT − ( E C − E F ) no = 2 exp 2 h kT − ( E C − E F ) NC; effective density of states = N C exp kT 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 Ev
0.25 eV
no
=
− ( E C − E F ) kT
N C exp 19
=
2.8 ×10
=
1.8 ×1015 cm
⋅
exp
−
3
−
(
C −
C +
0.0259
0.2 )
Microelectronics I
Thermal equilibrium concentration of hole, po Ev
po
=
∫ g ( E )[1
−
f ( E )]dE
∞
g ( E ) =
4π (2m*)3 / 2
1 − f F ( E ) =
h
3
E v − E
E − E 1 + exp F kT 3 / 2
≈
exp
−
F −
kT
Boltzmann approximation
2 m kT − ( E F − E v ) po = 2 exp h2 kT − ( E F − E v ) Nv; effective density of states = N v exp function in valence band kT * π p
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.
po
Ec
=
− ( E F − E v ) kT
N v exp
E Ev
0.27 eV
19
=
1.04 ×10
=
3.09 ×1014 cm
⋅
exp
−
3
−
(
v +
0.27 −
0.0259
V
)
Microelectronics I
2π mn* kT no = 2 2 h
3 / 2
2π m p* kT po = 2 2
3 / 2
− ( E C − E F ) − ( E C − E F ) = N C exp kT kT
exp
− ( E F − E v ) − ( E F − E v ) = N v exp
exp
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
Microelectronics I
Consider ex. 1
− ( E C − E F ) kT − ( E C − E C + 0.25) 19 = 2.8 × 10 ⋅ exp 0.0259 no
Ec EF
0.25 eV Eg-0.25 eV
Ev =
Hole concentration Eg=1.12 eV
=
N C exp
1.8 ×1015 cm
−
3
− ( E F − E v ) kT − (1.12 − 0.25) 19 = 1.04 × 10 ⋅ exp 0.0259
po
=
=
N v exp
2.68 ×10 4 cm
−
3
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
ni
=
pi
ni
2
=
− ( E C − E Fi ) − ( E Fi − E v ) = N v exp kT kT
N C exp
=
− E g − ( E C − E v ) = N c N v exp kT kT
N C N v exp
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.
2
ni ni
=
=
− 1.42 12 = 5.09 × 10 .
(4.7 × 1017 )(7.0 ×1018 ) exp 2.26 × 106 cm
−
3
Microelectronics I
2. Intrinsic Fermi level position, EFi If EF closer to Ec, n>p If EF closer to Ev, n
N C exp
−
( E C − E Fi )
N v exp
m p* 3 E Fi = E midgap + kT ln * m 4 n
kT
=
−
( E Fi − E v ) kT
Ec Emidgap Ev
Mp = mn Mp ≠ mn
EFi = Emidgap EFi shifts slightly from Emidgap
Microelectronics I
no=po
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 3 valence electrons
5 valence electrons
III
IV
B Al Ga In
C Si Ge
V
P As Sb
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)
Donor 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).
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
n-type Dopant atom;
Donor
impurity atom
p-type Acceptor
impurity atom
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
opan a oms; 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.
Electron> hole n-type
EF>EFi
hole> electron p-type
EF
Microelectronics I
− ( E C − E F ) kT
no
=
N C exp
po
=
N v exp
− ( E F − E v ) kT
Thermal equilibrium concentration of electron
Thermal equilibrium concentration of hole
Ex. 4 .
Ec
e
EF 1.12 eV
no
po
=
=
Band dia ram of Si. At T= 300 K Nc=2.8x1019cm-3 and Nv=1.04x1019cm-3. Calculate no and po.
Ev
− 0.25 15 = 1.8 ×10 cm 0.0259
(2.8 ×1019 ) exp
−
3
− (1.12 − 0.25) 4 = 2.7 ×10 cm 0.0259
(1.04 ×1019 ) exp
N-type Si −
3
Microelectronics I
Change
of Fermi energy causes change of carrier concentration.
no and po equation as function of the change of Fermi energy
− ( E C − E F ) E F − E Fi = ni exp kT kT
no
=
N C exp
po
=
N v exp
−
( E F − E v ) kT
=
ni; intrinsic carrier concentration Efi; intrinsic Fermi energy
ni exp
−
( E F − E Fi ) kT
Microelectronics I
The nopo product
− ( E C − E F ) − ( E F − E v ) exp kT kT − E g = N C N v exp kT no po
=
=
N C N v exp
2
no po
=
2
ni
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
EF
EF donor
Nondegenerate semiconductor
acceptor
Microelectronics I
Large amount of dopant atoms (~effective density of states) 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
EF Ec
e
e Filled states Ec
Ev Ev
Degenerate semiconductor
EF
empty states e
Microelectronics I
Statistic donors and acceptors
Discrete donor level
donor
nd Density of electron occupying the donor level
=
N d
E d − E F 1 + exp 2 kT 1
=
+
N d − N d
Concentration of donors
Concentration of ionized donors
Microelectronics I
Discrete acceptor level
acceptor
pa Concentration of holes in the acceptor states
N a
=
1+
1 g
E F − E a kT
=
N a
−
−
N a
exp
g; degeneracy factor (Si; 4)
Concentration of acceptors
Concentration of ionized acceptor
Microelectronics I
from the probability function, we can calculate the friction of total electrons still in the donor states at T=300 K
nd nd + no
1
=
1+
N C
2 N d
− ( E C − E d ) kT
exp
ionization energy
Consider hos horus do in in Si for T=300K at concentration of 1016 cm-3 (NC=2.8 x1019 cm-3, EC-Ed= 0.045 eV)
nd nd + no =
only
1
=
1+
2.8 ×1019 2 ×1016
− 0.045 0.0259
exp
0.0041 = 0.41%
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 Charge-neutrality condition In thermal equilibrium, semiconductor crystal is electrically neutral “Negative charges = positive charge”
Determined the carrier concentrations as a function of impurity doping
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
Microelectronics I
Charge-neutrality condition −
no + N a
=
po
ega ve c arges
no
+
( N a
−
pa ) = po
+
+
N d
os ve c arges +
( N d − nd )
Microelectronics I
no
+
( N a
−
pa ) = po
+
( N d − nd )
If we assume complete ionization (pa=0, nd=0)
no + N a
=
po
+
N d
From nopo=ni2 2
no + N a
no
=
=
n0
N d − N a
2
+
N d
+
N d − N a 2
2
+
2
ni
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. Electron,
no
=
N d − N a
2 16
=
≈
hole,
po
2
=
no
2 +
2
ni
16 10 2 ( 1 . 5 × 10 ) 2
+
+
1016 cm ni2
+
N d − N a 2
=
−
3
(1.5 × 1010 ) 2 1016
=
2.25 × 10 4 cm
−
3
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
no
=
N d − N a
2
+
N d − N a
2
2
+
2
ni
Very strong function of temperature s empera ure ncreases, ni erm w
om na e.
ows n r ns c c arac er s cs
Freeze-out Partial ionization
0K
Extrinsic no=Nd
Intrinsic no=ni Temperature
Microelectronics I
Hole concentration From charge-neutrality condition and nopo product
no
+
( N a
−
pa ) = po no po
=
N a
=
+
( N d − nd )
2
ni
2
po po
+
=
po + N d
N a − N d
2
+
N a − N d 2
2
+
2
ni
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.
Hole,
po
=
≈
electron,
approximation
=
N a − N d
2
+
1016 − 3 ×1015 2 7 × 1015 cm no
=
ni2 po
=
−
N a − N d 2 +
+
ni2
1016 − 3 × 1015 + (1.5 × 1010 ) 2 2
3
(1.5 ×1010 ) 2
po=Na-Nd
2
7 × 1015
=
3.21× 10 4 cm
−
3
Microelectronics I
Position of Fermi Energy Level As a function of doping concentration and temperature Equations for position of Fermi level (n-type)
E C − E F
=
N C no
kT ln
Compensated semiconductor, n o=Nd-Na
E C − E F
=
N C N d − N a
kT ln
no E F − E Fi = kT ln ni
Microelectronics I
Equations for position of Fermi level (p-type)
E F − E C
=
N v po
kT ln
Compensated semiconductor, p o=Na-Nd
N v E F − E v = kT ln − a
d
po E Fi − E F = kT ln ni
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.
N C N d − N a
E C − E F
=
kT ln
N d − N a
=
N C exp
=
−
C −
kT
− 0.2 16 = 1.24 × 10 cm 0.0259
2.8 ×1019 exp
N d
F
=
1.24 × 1016 cm
3
−
+
N a
=
−
3
2.24 ×1016 cm
3
−
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 re onderance of either electron n-t e or hole -t e is created. Acceptor atoms; Impurity atoms added to a semiconductor to create a ptype 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
n o po
=
ni2
