lecture 6 - sjtuhsic.sjtu.edu.cn/.../files/lec6_the_semiconductor_in_equilibrium.pdf · calculate...
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![Page 1: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/1.jpg)
Lecture 6
Semiconductor physics IV The Semiconductor in Equilibrium
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Equilibrium, or thermal equilibrium
No external forces such as voltages, electric fields.
Magnetic fields, or temperature gradients are acting
on the semiconductor.
All properties of the semiconductor will be
independent of time in this case.
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Goal
The concentration of electrons and holes in
the conduction and valence bands with the
Fermi-Dirac probability function and the
density of quantum states.
The properties of an intrinsic semiconductor.
The properties of an semiconductor with
impurities (dopants).
3
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CHARGE CARRIERS IN SEMICONDUCTORS
Two types of charge carrier, the electron and the hole.
The current in a semiconductor is determined largely by the
number of electrons in the conduction band and the number of
holes in the valence hand.
4
The distribution (with respect to energy) of electrons in the
conduction band is
the density of quantum states
in the conduction band
the probability that a state is
occupied by an electron
The total electron concentration per unit volume in the conduction
band is
( ) ( ) ( )c Fn E g E f E
( )n n E dE
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The distribution (with respect to energy) of holes in the valence
band is
the density of allowed quantum
states in the valence band
the probability that a state is not
occupied by an electron
The total hole concentration per unit volume is found by
integrating this function over the entire valance-band energy.
5
( ) ( )[1 ( )]Fvp E g E f E
( )p p E dE
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Fig (b) The splitting of the 3s
and 3p states of silicon into the
allowed and forbidden energy
bands.
At T=0K, valence band is full and the
conduction band is empty in an intrinsic
semiconductor (no impurities and no
lattice damage in crystal).
The location of Fermi energy EF
v F cE E E
At T>0K, the valence electrons gain
energy and a few move to conduction band
and leave empty states.
Electrons and holes are created in pairs by
the thermal energy.
The number of electrons in the conduction
band is equal to the number of holes in the
valence band.
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Fig Density of states functions, Fermi-Dirac
probability function, and areas representing
electron and hole concentrations for the case
when E, is near the midgap energy7
If we assume that the electron and
hole effective masses are equal, then
gc(E) and gv ( E ) are symmetrical
functions about the midgap energy.
The function fF(E) for E > EF is
symmetrical to 1 - fF(E) for E < EF
about the energy E = EF.
The areas representing electron and
hole concentrations are equal
* 3/2
3
4 (2 )( ) n
c c
mg E E E
h
* 3/2
3
4 (2 )( )
p
v v
mg E E E
h
1( )
1 exp( )F
F
f EE E
kT
in the middle of bandgap energyFE
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The n0 and p0 Equations
The thermal-equilibrium concentration of electrons
8
The lower limit of integration is Ec
The upper limit of integration should be the top of the allowed
conduction band energy.
However, since the Fermi probability function rapidly
approaches zero with increasing energy we can take the upper
limit of integration to be infinity.
0 ( ) ( )c Fn g E f E dE
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If (Ec- EF) >> k T , then (E - EF) >> kT
Let
* 3/2
3
4 (2 )( ) n
c cm
g E E Eh
[ ( )]1( ) exp
( )1 exp
FF
F
E Ef E
E E kT
kT
* 3/2
3
4 (2 ) ( )exp[ ]
C
Fno c
E
m E En E E dE
h kT
cE E
kT
* 3/21/2
03
0
4 (2 ) ( )exp[ ] exp( )
c Fnm E En d
h kT
1/2
0
1exp( )
2d
*
3/2
0 2
2 ( )2( ) exp[ ]
c Fnm kT E En
h kT
Boltzmann
approximation
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The thermal-equilibrium electron concentration in the
conduction band
Nc is called the effective density states function in the
conduction band.
The thermal-equilibrium concentration of holes in the valence
band is
11 ( )
( )1 exp
FF
f EE E
kT
* 3/2
3
4 (2 )( )
p
v v
mg E E E
h
0 ( )[1 ( )]v Fp g E f E dE
*3/2
2
22( )n
c
m kTN
h
0
( )exp[ ]c F
c
E En N
kT
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The lower limit of integration is taken as minus infinity instead
of the bottom of the valence band since the probability function
of holes approach zero when energy is minus infinite.
1 ( )1 ( ) exp[ ]
( )1 exp
FF
F
E Ef E
E E kT
kT
* 3/2
0 3
4 (2 ) ( )exp[ ]
vE
p Fv
m E Ep E E dE
h kT
( )F vIf E E kT
* 3/2 0
' 1/2 ' '
0 3
4 (2 ) ( )exp[ ] ( ) exp( )
p F vm kT E E
p dh kT
3/2*
v0 3
2 ( )2 exp
p Fm kT E E
ph kT
' ( )vE E
kT
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The effective density of states function in the valence band is
The thermal-equilibrium concentration of holes in the valence band
The magnitude of Nv is also on the order of 1019 cm-3at T = 300 K
for most semiconductors.
0
( )exp[ ]F v
v
E Ep N
kT
3/2*
2
22
p
v
m kTN
h
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The effective density of states functions, Nc and Nv, are constant
for a given semiconductor material at a fixed temperature.
Table
Effective density of states function and effective mass values
3/2*
2
22
p
v
m kTN
h
3/2*
2
22 n
c
m kTN
h
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Example:Calculate the probability that a state in the conduction band is occupied by an
electron and calculate the thermal equilibrium electron concentration in silicon
at T= 100 K.
Assume the Fermi energy is 0.25 eV below the conduction band. The value of
Nc for silicon at T = 100 K is Nc = 2.8 x 1019 cm-3.
The probability that an energy state at E = Ec is occupied by an
electron is given by
The probability of a state being occupied can be quite small, but the fact that
there are a large number of states means that the electron concentration is a
reasonable value.
Solution:
5( )1 0.25( ) exp[ ] exp( ) 6.43 10
0.02591 exp( )
F
c Fc
c F
E Ef E
E E kT
kT
19 15 3
0
( ) 0.25exp[ ] (2.8 10 )exp( ) 1.8 10
0.0259
c Fc
E En N cm
kT
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Example
Solution
The hole concentration is
Calculate the thermal equilibrium hole concentration in silicon at
T= 400 K.
Assume that the Fermi energy is 0.27 eV above the valence band
energy. The value of Nv for silicon at T = 300 K is Nv = 1.04 ×1019 cm-3.
19 19 3400(1.04 10 )( ) 1.60 10
300vN cm
19
0
15 3
( ) 0.27exp[ ] (1.60 10 )exp( )
0.03453
6.43 10
F vv
E Ep N
kT
cm
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The Intrinsic Carrier ConcentrationFor an intrinsic semiconductor,
The concentration of electrons in the conduction band ni is
equal to the concentration of holes in the valence band pi .
The Fermi energy level is called the intrinsic Fermi energy, or
EF = EFi .
16
Eg is the bandgap energy
0 exp[ ]Fi ci c
E En n N
kT
0 exp[ ]v Fi
i i v
E Ep p n N
kT
2 exp[ ]exp[ ]Fi c v Fii c v
E E E En N N
kT kT
2 exp[ ] exp[ ]gv c
i c v c v
EE En N N N N
kT kT
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Fig The intrinsic carrier concentration of Ge,
Si, and GaAs as a function of temperature.
For Eg = 1.12 eV, ni = 6.95 x 109cm-3 from
the equation for silicon at T = 300 K .
The commonly accepted value of ni for silicon
at T = 300 K is approximately 1.5 × 1010cm-3
This theoretical function does not
agree exactly with experiment.
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The Intrinsic Fermi-Level Position
Since the electron and hole concentrations are equal
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If we take the natural log of both sides of this equation
The midgap energy1
( )2
c v midgapE E E
( ) ( )exp[ ] exp[ ]c Fi Fi v
c v
E E E EN N
kT kT
*
*
1 3( ) ln( )
2 4
p
Fi c v
n
mE E E kT
m
1 1( ) ln( )
2 2
vFi c v
c
NE E E kT
N
3/2*
2
22
p
v
m kTN
h
3/2*
2
22 n
c
m kTN
h
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The intrinsic Fermi level is
below the center of the bandgap.
exactly in the center of the bandgap
above the center of the bandgap.
*
*
3ln( )
4
p
Fi midgap
n
mE E kT
m
* *
p nm m
* *
p nm m
* *
p nm m
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Example
To calculate the position of the intrinsic Fermi level with respect
to the center of the bandgap in silicon at T = 300 K.
The density of states effective carrier masses in silicon are
Solution
The intrinsic Fermi level with respect to the center of the
bandgap is
If we compare 12.8 meV to 560 meV, which is one-half of the
bandgap energy of silicon, we can, in many applications,
simply approximate the intrinsic Fermi level to be in the
center of the bandgap.
* *
0 01.08 0.56n pm m m m
*
*
3 3 0.56ln( ) (0.0259) ln( )
4 4 1.08
0.0128 12.8
p
Fi midgap
n
Fi midgap
mE E kT
m
E E eV meV
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DOPANT ATOMS AND ENERGY LEVELSThe real power of semiconductors is realized by adding small,
controlled amounts of specific dopant, or impurity, atoms.
21
The doped semiconductor, called an extrinsic material
The phosphorus atom without the donor electron is positively
charged. At very low temperatures, the donor electron is bound
to the phosphorus atom.
adding a group V element,
such as phosphorus
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Fig The energy-band diagram showing (a) the discrete donor energy
state and (b) the effect of a donor state being ionized.
The electron in the conduction band can now move through the
crystal generating a current, while the positively charged ion is
fixed in the crystal.
The donor impurity atoms add electrons to the conduction band
without creating holes in the valence band. The resulting
material is referred to as an n-type semiconductor.
The donor
electrons jump to
the conduction
band with thermal
energy
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Adding a group III element, such as boron, as a substitution
impurity purity to silicon.
One covalent bonding
position appears to be
empty
Fig Valence electrons may gain a small amount of
thermal energy and move about in the crystal.
The "empty" position associated with the boron atom becomes occupied, and
other valence electron positions become vacated. These other vacated electron
positions can he thought of as holes in the semiconductor material.
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Fig The energy-band diagram showing (a) the discrete acceptor
energy state and (b) the effect of a acceptor state being ionized.
If an electron were to occupy this "empty" position, its energy
would have to be greater than that of the valence electrons.
The acceptor atom can generate holes in the valence hand
without generating electrons in the conduction band. This type
of semiconductor material is referred to as a p-type material
Acceptor atom
gets electrons
from the
valence band
with thermal
energy.
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Ionization Energy
Energy required to elevate the donor electron into the
conduction band.
Bohr theory
The most probable distance of an electron in a hydrogen atom
from the nucleus from quantum mechanics is the same as
Bohr radius.
The coulomb force of attraction between the electron and ion
equal to the centripetal force of the orbiting electron. This
condition give a steady orbit.
25
2 * 2
24 n n
e m v
r r
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If we assume the angular momentum is also quantized, then we can
write n is a positive integer
The assumption of the angular momentum being quantized
leads to the radius being quantized
The Bohr radius is
normalize the radius of the donor orbital to that of the Bohr radius
the relative dielectric constant of the
semiconductor material
rest mass of an electron
effective mass of the electron
in the semiconductor.
Substitute *
n
nhv
m r into
2 * 2
24 n n
e m v
r r
*
nm r v n
2 2
* 2
4n
nr
m e
2
00 2
0
40.53Aa
m e
2 0
*
0
( )nr
r mn
a m
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If we consider the lowest energy state in which n = 1 , and if we
consider silicon in which = 11.7 and the conductivity effective
mass is m*/m0 = 0.26. then we have thatr
r1 = 23.9 A. This radius corresponds to approximately four lattice
constants silicon. Recall that one unit cell in silicon effectively
contains eight atoms, so the radius of the orbiting donor
electron encompasses many silicon atoms.
=> The donor electron is not tightly bound to the donor atom.
1
0
45r
a
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The total energy of the orbiting electron is given by
The kinetic energy is
Since
The potential energy is
The total energy is
E T V
* 21
2T m v
*
nm r v n2 2
* 2
4n
nr
m e
* 4
2 22( ) (4 )
m eT
n
2 * 4
2 24 ( ) (4 )n
e m eV
r n
* 4
2 22( ) (4 )
m eE T V
n
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Table Impurity ionization energies
in silicon and germanium
In silicon the ionization energy is E = -25.8 meV, much less
than the bandgap energy of silicon. This energy is the
approximate ionization energy of the donor atom.
Germanium and silicon have different relative dielectric constants and
effective masses, resulting in different ionization energy.
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THE EXTRINSIC SEMICONDUCTOR
A material has impurity atoms.
One type of carrier will predominate.
The Fermi energy will change as dopant atoms are added.
n type: the density of electrons is greater than the density of holes
n0>p0 majority carrier: electrons; minority carrier: holes
the Fermi energy is above the intrinsic Fermi energy
p type: the density of holes is greater than the density of electrons
n0<p0 majority carrier: holes; minority carrier: electrons
the Fermi energy is below the intrinsic Fermi energy
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Fig. Density of states functions.
Fermi-Dirac probability function, and
areas representing electron and hole
concentrations for the case when EF
is above the intrinsic Fermi energy.
n type
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Fig Density of states functions,
Fermi-Dirac probability function, and
areas representing electron and hole
concentrations for the case when EF
is below the intrinsic Fermi energy.
p type
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33
The electron concentration in extrinsic semiconductor
The intrinsic carrier concentration
0 exp[ ]F Fii
E En n
kT
0
( )exp[ ]F Fi
i
E Ep n
kT
( )exp[ ]c Fi
i c
E En N
kT
0
( ) ( )exp[ ]c Fi F Fi
c
E E E En N
kT
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The n0 and p0 Product
34
• The product of n0 and p0 is always a constant for a given
semiconductor material at a given temperature.
• The equation is invalid if the Boltzmann approximation is
not valid since it is derived by the Boltzmann approximation
and c F F vE E kT E E kT
2
0 0 in p n
0 0 exp[ ]g
c v
En p N N
kT
0 0
( ) ( )exp[ ]exp[ ]c F F v
c v
E E E En p N N
kT kT
![Page 35: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/35.jpg)
The Fermi-Dirac Integral
35
If the Boltzmann approximation (E-EF>>kT) does not hold.
If we again make a change of variable and let
The integral is defined as
Fermi-Dirac integral𝐹 Τ1 2(𝜂𝐹) =
0
∞
𝜂 Τ1 2𝑑𝜂
)1 + exp(𝜂 − 𝜂𝐹
1/2* 3/2
0 3
( )4(2 )
1 exp( )c
cn
FE
E E dEn m
E Eh
kT
cE E
kT
F c
F
E E
kT
* 1/23/2
0 2
0
24 ( )
1 exp( )
p
F
m kT dn
h
![Page 36: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/36.jpg)
36
Fig The Fermi-Dirac integral as a function
of the Fermi energy
The Fermi-Dirac integral,
is a tabulated function.
The Fermi energy is
actually in the conduction
band.
0Fif
/F C FE E kT
1
2
1/2 ( )0 1 FF
dF
e
![Page 37: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/37.jpg)
37
the Fermi level is in the valence band.
* ' 1/2 '3/2
0 2 ' '
0
2 ( )4 ( )
1 exp( )F
pm kT dp
h
' vE E
kT
' v F
F
E E
kT
' 0Fif
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38
To calculate the electron concentration using the Fermi-Dirac
integral. Let = 2 so that the Fermi energy is above the
conduction band by approximately 52 meV at T = 300 K.
Example
F
Solution
The Femi-Dirac integral has a value of F1/2 (2) = 2.3
With the Boltzmann approximation
no = 2.08 × 1020 cm-30
( )exp( )c F
c
E En N
kT
For silicon at 300K, Nc = 2.8 × 1019 cm-3
19 19 3
0
2(2.8 10 )(2.3) 7.27 10n cm
* 1/23/2
0 0 1/22
0
2 24 ( ) ( )
1 exp( )
nt F
F
m kT dn n N F
h
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STATISTICS OF DONORS AND ACCEPTORS
The probability function of electrons occupying the donor state is
39
nd is the density of electrons occupying the donor level
Ed is the energy of the donor level
The factor 1/2 in this equation is a direct result of the spin factor.
Each donor level has two possible quantum states (spin
orientations). The insertion of an electron into one quantum state,
however, precludes putting an electron into the second quantum
state.
Nd is the donor concentration
11 exp( )
2
dd
d F
Nn
E E
kT
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40
The density of electrons occupying the donor level is equal to
the donor concentration minus the concentration of ionized donors
The concentration of holes in the acceptor states
Na is the concentration of acceptors
Na- is the concentration of ionized acceptors
g is a degeneracy factor(简并因子) g=4 in silicon and gallium
arsenide.
11 exp( )
aa a a
F a
Np N N
E E
g kT
d ddn N N
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Complete Ionization and Freeze-Out
41
The Boltzmann approximation is also valid
The ratio of electron in the donor state to the total number
of electrons in the conduction band plus donor state.
0
( )2 exp( )
( ) ( )2 exp( ) exp( )
d Fd
d
d F c ddd c
E EN
n kTE E E En n
N NkT kT
2 exp( )1
exp( )2
d d Fd d
d F
N E En N
E E kT
kT
0
( )exp( )c F
c
E En N
kT
d FIf E E kT
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42
The factor (Ec - Ed) is just the ionization energy of the
donor electrons.
At room temperature, the donor states are essentially
completely ionized.
=> All donor impurity atoms have donated an electron
to the conduction band.
Small value
At room temperature, there is also essentially complete
ionization of the acceptor atoms.
=>each acceptor atom has accepted an electron from the
valence band.
0
1
( )1 exp( )
d
c c dd
d
n
N E En n
N kT
![Page 43: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/43.jpg)
43
Fig Energy-band diagrams showing complete ionization of
(a) donor states and (b) acceptor states.
Complete Ionization at T=300 K
Partial ionization of donor or acceptor atoms when 0K<T<300K
![Page 44: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/44.jpg)
44
Conduction band Conduction band
Valence band Valence band
Fig Energy-band diagram at T = 0 K for
(a) n-type and (b) p-type semiconductors.
Nd+=0
=>Each donor state must contain an
electron.
=> EF>Ed
Freeze-Out at T=0K
Na- =0
=>Each acceptor state does not
contain electron.
=> EF>Ev
![Page 45: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/45.jpg)
CHARGE NEUTRALITY
In thermal equilibrium, the semiconductor crystal is
electrically neutral. =>The electrons are distributed among the
various energy states, creating negative and positive charges,
but the net charge density is zero.
A compensated semiconductor is one that contains both donor
and acceptor impurity atoms in the same region.
An n-type compensated semiconductor occurs when Nd > Na,
A p-type compensated semiconductor occurs when Na> Nd.
If Nd = Na, it is an intrinsic material.
45
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46
Equilibrium Electron and Hole Concentrations
Fig Energy-band diagram of a
compensated semiconductor
showing ionized and un-ionized
donors and acceptors
The charge neutrality condition is
n0: total electron concentration
= thermal electrons + donor electrons
p0: total hole concentration
= thermal holes + acceptor holes
pa: the hole concentration in acceptor states
nd: the electron concentration in donor states
0 0( ) ( )a a d dn N p p N n
0 0a dn N p N
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47
If we assume complete ionization, nd and pa are both zero
Two factor to impact n0
--- the concentration of impurity atoms
--- the intrinsic carrier concentration
2
0
0
ia d
nn N N
n
0 0a dn N p N
2 2
0 ( )2 2
d a d ai
N N N Nn n
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48
Fig Energy-band diagram showing the redistribution
of electrons when donors are added
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49
Fig Electron concentration versus temperature showing the
three regions: partial ionization, extrinsic, and intrinsic.
As the temperature increases, additional electron-hole pairs are
thermally generated so that the ni term may begin to dominate.
Complete
ionization
( )exp[ ]c Fi
i c
E En N
kT
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50
Similarly
If Na-Nd>>ni , then
0 a dp N N
2 2
0
0 ( )
i i
a d
n nn
p N N
2 2
0 ( )2 2
a d a di
N N N Np n
2
0
0
ia d
nN p N
p
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POSITION OF FERMI ENERGY LEVEL
51
The position of the Fermi energy level is a function of the
doping concentrations and as a function of temperature
For an n-type semiconductor, Nd>>ni then0 dn N
As the donor concentration increases, the Fermi
level moves closer to the conduction band.
where
0
( )exp[ ]c F
c
E En N
kT
0
ln cc F
NE E kT
n
2 2
0 ( )2 2
d a d ai
N N N Nn n
ln cc F
d
NE E kT
N
0lnF Fi
i
nE E kT
n
0 exp[ ]F Fii
E En n
kT
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52
For a p-type semiconductor
0lnFi F
i
pE E kT
n
The difference between the intrinsic Fermi level and the
Fermi energy in terms of the acceptor concentration
As the acceptor concentration increases, the Fermi level
moves closer to the valence band.
ln vF v
a
NE E kT
N
0
ln vF v
NE E kT
p
If we assume that Na>>Ni
![Page 53: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/53.jpg)
53
Fig Position of Fermi level for an (a) n-type and (b) p-type semiconductor.
Fig Position of Fermi level as a
function of donor
concentration (n type) and
acceptor concentration (p type).
![Page 54: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/54.jpg)
54
0lnFi F
i
pE E kT
n
0lnF Fi
i
nE E kT
n
Fig Position of Fermi level as a function of
temperature for various doping concentrations.
As T increases, ni will increase F FiE E
( )exp[ ]c Fi
i c
E En N
kT
![Page 55: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/55.jpg)
Summary
The concentration of electrons and holes
55
• Using the Maxwell-Boltzmann approximation
• The intrinsic carrier concentration is
0 ( ) ( )c Fn g E f E dE 0 ( )[1 ( )]v Fp g E f E dE
2 exp[ ]g
i c v
En N N
kT
0
( )exp[ ]c F
c
E En N
kT
0
( )exp[ ]F v
v
E Ep N
kT
![Page 56: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/56.jpg)
56
• The fundamental relationship of the hole and electron
concentration is 2
0 0 in p n
• The concept of doping the semiconductor with donor atoms and
acceptor atoms to form n-type or p-type material.
• The electron and hole concentrations is a function of impurity
doping concentrations
• The position of the Fermi energy level is a function of impurity
doping concentration
0lnFi F
i
pE E kT
n
0lnF Fi
i
nE E kT
n
2 2
0 ( )2 2
d a d ai
N N N Nn n
2 2
0 ( )2 2
a d a di
N N N Np n
![Page 57: Lecture 6 - SJTUhsic.sjtu.edu.cn/.../files/Lec6_The_Semiconductor_in_Equilibrium.pdf · Calculate the thermal equilibrium hole concentration in silicon at T= 400 K. Assume that the](https://reader033.vdocuments.us/reader033/viewer/2022042005/5e6f60d1f69adb6c0e59b805/html5/thumbnails/57.jpg)
Homework 5Consider germanium with an acceptor concentration of Na =
1015 cm-3 and a donor concentration of Nd = 0. Plot the
position of the Fermi energy with respect to the intrinsic
Fermi level as a function of temperature over the range
200K T 500K