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
T>0K
e
Conduction
band
e
Valence
band
Ec
Ev
Particles that can freely move and contribute to the current flow (conduction)
carrier
1. Electron in conduction band
2. Hole in valence band
Microelectronics I
How to count number of carriers,n?
Assumption; Pauli exclusion principle
If we know
1. No. of energy states
2. Occupied energy states
Density of states (DOS)
The probability that energy states is
occupied
“Fermi-Dirac distribution function”
n = DOS x “Fermi-Dirac distribution function”
Microelectronics I
Density of state
E
No of states (seats) above EC for electron
e
Conduction
band
4π (2m*)3 / 2
g (E) =
h3
E − EC
Ec
Ev
e
Valence
band
No of states (seats) below Ev for hole
4π (2m*)3 / 2
g (E) =
h3
g (E)
Ev − 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
Ec
1
 E − EF 
1 + exp

kT


Fermi energy, EF
Ev
Hole having
energy below
Ev
hole (red line)
e
0.5
1−
1
f (E)
1
 E − EF 
1 + exp

kT


EF; the energy below which all states are filled with 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
e
1
g(E) x f (E)
free hole
e
Microelectronics I
Thermal equilibrium concentration of electron, no
∞
∫ g ( E ) f ( E )dE
no =
EC
4π (2m*)3 / 2
g (E) =
h3
f F (E) =
E − EC
1
 − ( E − EF ) 
≈ exp

kT
 E − EF 


1 + exp

 kT 
3/ 2
Boltzmann approximation
 2πm kT 
 − ( EC − E F ) 


no = 2
exp 


h
kT




 − ( EC − E F ) 
NC; effective density of states
= N C exp 

kT
function in conduction band


*
n
2
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
 − ( EC − EF ) 
no = N C exp 

kT


 − ( EC − EC + 0.25) 
19
= 2.8 ×10 ⋅ exp 

0.0259


= 1.8 ×1015 cm −3
Microelectronics I
Thermal equilibrium concentration of hole, po
Ev
po = ∫ g ( E )[1 − f ( E )]dE
∞
4π (2m*)3 / 2
g (E) =
h3
1 − f F (E) =
Ev − E
1
 − ( EF − E ) 
≈ exp

kT
 EF − E 


1 + exp

 kT 
3/ 2
Boltzmann approximation
 2πm kT 
 − ( E F − Ev ) 


exp 
po = 2

 h

kT




 − ( E F − Ev ) 
Nv; effective density of states
= N v exp 

function in valence band
kT


*
p
2
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
Ev
0.27 eV
 − ( E F − Ev ) 
po = N v exp 

kT


 − ( Ev + 0.27 − EV ) 
19
= 1.04 ×10 ⋅ exp 

0.0259


= 3.09 ×1014 cm −3
Microelectronics I
 2πmn* kT 

no = 2
2
 h

 2πm kT 

po = 2


 h

*
p
2
3/ 2
3/ 2
 − ( EC − EF ) 
 − ( EC − EF ) 
N
exp 
=
exp
C



kT
kT




 − ( E F − Ev ) 
 − ( E F − Ev ) 
exp 
=
N
exp
v



kT
kT




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
EF
0.25 eV
Eg-0.25 eV
Ev
 − ( EC − EF ) 
no = N C exp 

kT


 − ( EC − EC + 0.25) 
19
= 2.8 ×10 ⋅ exp 

0.0259


= 1.8 ×1015 cm −3
Hole concentration
Eg=1.12 eV
 − ( E F − Ev ) 
po = N v exp 

kT


 − (1.12 − 0.25) 
19
= 1.04 ×10 ⋅ exp 

0.0259

= 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
 − ( EC − EFi ) 
 − ( EFi − Ev ) 
exp
ni = pi = N C exp 
=
N
v



kT
kT




 − Eg 
 − ( EC − Ev ) 
ni = N C N v exp 
= N c N v exp 


kT
kT




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.
 − 1.42 
12
n = (4.7 × 10 )(7.0 ×10 ) exp 
=
5
.
09
×
10
 0.0259 
ni = 2.26 × 106 cm −3
2
i
17
18
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
 − ( EC − E Fi ) 
 − ( E Fi − Ev ) 
N C exp 
= N v exp 


kT
kT




Ec
 m*p 
3
E Fi = Emidgap + kT ln * 
m 
Emidgap
4
 n
Ev
Mp = mn
Mp ≠ mn
EFi = Emidgap
EFi shifts slightly from Emidgap
Microelectronics I
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
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;
Majority carrier;
Donor impurity atom
electron
p-type
Acceptor impurity atom
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;
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<EFi
Microelectronics I
 − ( EC − E F ) 
no = N C exp 

kT


 − ( E F − Ev ) 
po = N v exp 

kT


Thermal equilibrium concentration of electron
Thermal equilibrium concentration of hole
Ex. 4
Ec
0.25 eV
EF
1.12 eV
Band diagram of Si. At T= 300 K,
Nc=2.8x1019cm-3 and Nv=1.04x1019cm-3.
Calculate no and po.
Ev
 − 0.25 
15
−3
no = (2.8 × 1019 ) exp
 = 1.8 ×10 cm
 0.0259 
 − (1.12 − 0.25) 
4
−3
po = (1.04 ×1019 ) exp
 = 2.7 ×10 cm
0.0259


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
 − ( EC − E F ) 
 E F − E Fi 
no = N C exp 
=
n
exp
i

 kT 
kT




 − ( E F − Ev ) 
 − ( E F − EFi ) 
po = N v exp 
=
n
exp
i



kT
kT




ni; intrinsic carrier concentration
Efi; intrinsic Fermi energy
Microelectronics I
The nopo product
 − ( EC − E F ) 
 − ( E F − Ev ) 
no po = N C N v exp 
exp



kT
kT




 − Eg 
= N C N v exp 

kT


= ni2
no po = ni2
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
nd =
= N d − N d+
1
 Ed − E F 

Density of electron 1 + exp
2
 kT 
occupying the
donor level
Concentration of
donors
Concentration of
ionized donors
Microelectronics I
Discrete acceptor level
acceptor
Na
pa =
= N a − N a−
1
 E − Ea 
1 + exp F

Concentration of
g
 kT 
holes in the
acceptor states
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
N
 − ( EC − Ed ) 
1 + C exp 

2Nd
kT

ionization energy
Consider phosphorus doping 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
1
2.8 × 1019
 − 0.045 
1+
exp 
16
2 ×10
 0.0259 
= 0.0041 = 0.41%
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
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
concentration
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 + N d+
Negative charges
Positive charges
no + ( N a − pa ) = po + ( 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
ni2
no + N a =
+ Nd
n0
2
Nd − Na
 Nd − Na 
2
no =
+ 
+
n

i
2
2


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.
2
Electron,
no =
Nd − Na
 N − Na 
2
+  d
 + ni
2
2


 1016 
1016
 + (1.5 × 1010 ) 2
=
+ 
2
 2 
≈ 1016 cm −3
hole,
ni2 (1.5 × 1010 ) 2
4
−3
po =
=
=
2
.
25
×
10
cm
no
1016
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
Nd − Na
 Nd − Na 
2
no =
+ 
 + ni
2
2


Very strong function of temperature
As temperature increases, ni2 term will dominate. Shows intrinsic characteristics
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 + ( N d − nd )
no po = ni2
ni2
+ N a = po + N d
po
2
Na − Nd
 Na − Nd 
2
po =
+ 
+
n

i
2
2


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
Hole,
po =
Na − Nd
 N − Nd 
2
+  a
 + ni
2
2


 1016 − 3 × 1015 
1016 − 3 × 1015
 + (1.5 × 1010 ) 2
=
+ 
2
2


≈ 7 × 1015 cm −3
ni2 (1.5 × 1010 ) 2
4
−3
=
=
3
.
21
×
10
cm
electron, no =
po
7 × 1015
approximation
po=Na-Nd
Microelectronics I
Position of Fermi Energy Level
As a function of doping concentration and temperature
Equations for position of Fermi level (n-type)
 NC
EC − E F = kT ln
 no



Compensated semiconductor, no=Nd-Na
 NC 

EC − EF = kT ln
 Nd − Na 
n 
E F − E Fi = kT ln o 
 ni 
Microelectronics I
Equations for position of Fermi level (p-type)
 Nv 

E F − EC = kT ln
 po 
Compensated semiconductor, po=Na-Nd
 Nv
E F − Ev = kT ln
 Na − Nd
 po 
E Fi − EF = 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.
 NC 

EC − EF = kT ln
 Nd − Na 
 − ( EC − E F ) 
N d − N a = N C exp 

kT


 − 0.2 
19
16
−3
= 2.8 ×10 exp 
=
1
.
24
×
10
cm
 0.0259 
N d = 1.24 × 1016 cm −3 + N a = 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
preponderance of either electron (n-type) or hole (p-type) 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
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
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)