Class 19_BB

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November 11, 2010
Class 19
Intrinsic Carrier Concentration
3
m k T  2
n  2 e B2  e
 2 
  Ec
3
m k T  2
p  2 h B2  e
 2 
k BT
Ev  
k BT
Multiplying n by p
Eg
3

3
k T 
np  4 B 2  me mh  2 e k B T
 2 
which is a magnitude independent of the Fermi level and only depends on the temperature. From
material to material, Eg will be different and also the effective mass of electrons and holes will be
different, but given a material, np only depend on T.
For instance, at 300K for Si, Ge, and GaAs, np=2.10x1019 cm-6, 2.89x1026 cm-6, and 6.55x1012
cm-6
In an intrinsic semiconductor, the number of electrons is equal to the number of holes, since each
electron leaves a hole behind, thus
3
Eg

3
k T  2
ni  pi  np  4 B 2  me mh  4 e 2 k B T
 2 
Also using this equality, we can find a value for the Fermi level. We make n and p equal and
found, from the top of the valence band Ev=0 (what implies Ec=Eg)
e
2
k BT
3
Eg
m  2
  h  e k B T
 me 
or  
m 
1
3
Eg  k BT ln  h 
2
4
 me 
Thus, if the effective mass for electrons and holes is the same, the Fermi level is right at the
middle of the gap.
Mobility
Mobility is defined as the drift velocity (absolute value) divided by the electric field. The symbol
for mobility is the same as for chemical potential , to distinguish when we refer to the mobility
a subindex will be used (e for electrons and h for holes). In terms of the mobility (and this can
be used as a definition for mobility) conductivity is given by
  nee  neh 
1
And by using the definition for conductivity  
ne 2
the mobility in terms of microscopic
m
parameters is given by
e h 
e e h
me h
for electrons and holes respectively.
Extrinsic Semiconductors-Impurity Conductivity
The conductivity of a semiconductor can also be increased by modifying the lattice from its
perfect conformation. For instance, the addition of 1 boron atom for every 105 atoms of silicon
increases the conductivity of silicon in three orders of magnitude.
To fix the concepts let’s consider tetravalent semiconductors such us Si or Ge. They form a
diamond-like structure where each atom is covalently bond to four other atoms. Lets supposed
one of these atoms is replaced by a pentavalent atom. Then there will be 1 extra electron that can
be given up to the lattice for conduction. This extraneous atom is known as donor impurity. If
instead a trivalent atom is substituted for one of the atoms in the lattice, then there will be a
missing electron that can attract one of the electrons in the lattice, thus generating a hole. This is
known as an acceptor impurity.
Donor states
Let’s assume a pentavalent atom is added to the crystal (for instance As), studies reveal that
indeed this impurity enters as a substitution (replaces existing atoms) and not as an interstitial (in
between atoms already in the lattice).
The crystal is neutral because the electron remains in the crystal, but it remains near the ion of
the inserted atom since all other atoms are neutral. The only effect at this point is that the
coulombic attraction between As+ and the electron is partially decreased by the polarization these
charges produce in the medium, thus the coulombic potential experienced by the electron is
given by 1/4εo e/εr, where ε is the dielectric constant in the crytal.
We will consider this as a hydrogen-like atom where the coulombic interaction is shielded by a
dielectric medium and the mass of the electron is the effective mass. The binding energy of the
e4m
electron in a hydrogen atom is given by Eh  
thus, the binding energy of the
2
24o   2
electron to the impurity ion is:
e 4 me
13.6 m
Ed  
 2 e
2 2
 m
24o  
The radius of the orbit of these electrons is given by (doing the same correction)


 ao  
ad  
 where ao si the bohr radius 0.529Å.
 me 
 m
2
Considering as an example me/m=0.1 for Ge and 0.2 for Si, and using the dielectric constant from
table 4, the binding energies are found to be 5meV and 20meV. Compare those values with the
band gaps of 1.11eV for Si and 0.66 eV for Ge at room temperature. The important information
here is that those electrons are very weakly bounded and can go into the conduction band very
easily. Precise calculations predict ionization energies of 9.05 meV for Ge and 29.8 meV for Si.
(This is how much into the gap these donor states are from the bottom of the conduction band)
The radius of the orbit for this electrons is around 80 Å and 30 Å for Ge and Si respectively, thus
at moderate doping level, the electron orbit overlaps forming a band of impurity that lies right
below the conduction band. Conduction in the impurity band can occur when the charges hop
from one ion to the next one. This conduction is increased if some acceptors are present so some
of the ions remains unoccupied thus making the hop more effective.
Acceptor states
The problem is what happens when an impurity with one less electron than the host atoms is
present. In that case, we can say that a hole is bounded to the impurity that can be filled up if one
electron from the valence band gets attached to the impurity completing the four bonds; this
process leaves a hole in the valence band. This process is equivalent to the process described
above where a donor is ionized by losing the extra electron that then goes to the conduction band.
The mathematical solution is similar to that of donor impurities; however the effective mass of
the electrons in the valence band depends on what valence band they are in.
Ionization of acceptor species are also in the order of a few meV, thus a band of acceptor level is
positioned very close to the valence band making it easy for electrons in the valence band to
occupy those levels.
In a particular material, if the number of donor species is larger than the acceptor species, then
conduction is dominated by electron conduction and the materials is known as n-type
semiconductor (n for negative), if instead acceptor species are more than donor species,
conduction is dominated by holes and the material is known as p-type semiconductor (p for
positive).
Thermal ionization of Donors and Acceptors
By applying chemical equilibrium, the number of electrons due to donors, assuming that there
are no acceptors is given by
1
  m k T  3 2  2  Ed
nd  2 e B2  N d  e 2 k B T
  2 

For the number of holes, we can use that np must be constant np=ni2 where ni is the intrinsic
carrier concentration. If we assume that all donors have been ionized, then in the case of
extrinsic semiconductors n=(ni+nd)~ nd for a donor concentration larger that the intrinsic carrier
concentration. In that case p= Ni2/nd. In any case, the addition of donor impurities, increase the
number of negative carriers over that of intrinsic semiconductors leading to a decrease on holes
so that np remains constant. It can be shown than in any case, n+p is larger than for the intrinsic
semiconductor.
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A similar equation is found for acceptors when no donors are present.
1
  m k T  32  2  Ea
p  2 h B2  N a  e 2 k B T
  2 

The number of holes is larger than for the intrinsic case while the number of electrons is smaller
such that np is the same as for the intrinsic case and n+p results to be larger than for the intrinsic
case.
When both are present the problem is very complicated.
Paramagnetic and Diamagnetic material
In simple terms both paramagnetic and diamagnetic materials are materials that respond to an
external magnetic field with a magnetization on their own. For paramagnetic materials, the
magnetization M (defined as the total magnetic moment per unit of volume) is parallel to the
applied electric field while for diamagnetic materials M is in the opposite direction.
There are three source of magnetization in a material, the electron spin, the electron orbital
angular momentum, and changes in the orbital momentum due to an external magnetic field.
The first two, intrinsic electron spin and orbital momentum both are “permanent” magnetic
moments and magnetization occurs when these align with the external field. The last source of
magnetic moment leads to diamagnetism.
The relationship between an applied field and the magnetization is the magnetic susceptibility.
M
 o
where o is the magnetic permeability of vacuum.
B
Paramagnetic materials are characterized by positive susceptibility while diamagnetic materials
are characterized by negative susceptibility.
Diamagnetism
Consider a material with closed-shell atoms, in each atom, there exist pair of electrons that spin
around the nucleus at similar speed but in opposite direction, their magnetic moments thus cancel.
The centripetal force on each of these electrons is giving by
v2
e2
Fc  m  k 2 Where e is the positive of the electron charge and the minus in the equation
r
r
indicate the force is towards the center of the orbit (attractive electron-nucleus interaction)
When an external magnetic field is applied, both electrons experience an extra force due to that
field.
Fm=q vB
Consider two electrons spinning in opposite directions with the same speed with the plane of the
orbit parallel to the surface of the paper; each electron will have a magnetic moment L (the
angular moment) in opposite directions. The electron spinning clockwise will have a magnetic
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moment out of the page while the electron spinning counterclockwise will have a magnetic
moment pointing into the page.
Consider that the applied magnetic field is into the page. The new centripetal force will be
v2
e2
Fc  m  k 2  ev  B
r
r
Where vB will be radial, the minus implies a magnetic force pointing towards the center of the
orbit and the plus when pointing outwards
For the electron spinning in the clockwise direction, the magnetic force is towards the
nucleus and thus the centripetal force increases. The electron increases its speed to compensate
for the extra centripetal force and thus its magnetic moment, which points out of the page
increases. The opposite will happen to the electron spinning counterclockwise, it responds
decreasing its momentum which is into the page. The overall effect is a net (weak) magnetic
moment pointing out of the page, opposite to the applied field.
Diamagnetism is present in all materials, however if net magnetic moments exist,
diamagnetism so weak that is not appreciated.
Diamagnetism is usually treated in terms of the Larmor Theorem that says that in a
magnetic field the electron’s motion is that of the electron when the field is not present plus a
precession with frequency

eB
This is half of the cyclotron frequency of an electron in a magnetic field.
2m
A classical treatment of the electron precession is as follow. Consider an electron in an orbit with
a frequency equal to the Larmor frequency. The current associated with this electron is given by
 1 eB 
I  ef  e

 2 2m 
The magnetic moment associated with a current in a loop is given by current x area, thus the
momentum associated with the precession motion of an electron is given by
e2 B 2

 where  is the mean square of the radius of the projection of the electron orbit
4m
over a plane perpendicular to B <2>=<x2>+<y2> For an spherically symmetric distribution,
<r2>=<x2>+<y2>+<z2>, thus <r2>= 3/2 <2> and

e2 B 2
r .
6m
The above is the magnetic moment per electron, if there are Z electrons in each atom and N
atoms per unit of volume, then the diamagnetic susceptibility is given by
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
o M
B

o NZ
B

o NZe2
6m
r2
These results hold even if the electron is treated within the quantum mechanics formalism.
Paramagnetism
We can imagine a number of individual atoms with permanent magnetic moments pointing in all
directions due to the random thermal effect. Then a magnetic field is applied, the potential
energy of a magnetic moment  in a magnetic field B, is given by U=-.B where the minimum
energy is obtained when  and B are parallel to each other. There is now a competition between
the temperature to misalign the moment and the magnetic field to align them. The collective
contribution of a number of individual magnetic moments leads to a mild (in most materials)
magnetic field parallel to the external applied field.
According to quantum theory, the magnetic moment of an atom or ion is given by =ħJ where
ħJ is the total angular momentum which is the sum of the orbital plus the spin angular
momentum.  is known as the gyromagnetic or magnetogyric ratio.
g B
e
where  B 
is the Bohr magneton, which is closely equal to the spin

2m
magnetic moment of a free electron. g is known as the g factor and for an electron spin is ~2.
For a free atom, it is given by the Landé equation
J J  1  S S  1  LL  1
(notice that for a just a spin, L=0, J=S, and g=2
g 1
2 J J  1
Also,   
The magnetic energy is given by U=-.B=mjgBB where mj is the angular momentum
component in the direction of the magnetic field with values between J and –J.
As example consider a state with no orbital momentum (s levels for instance), in this case
mj=½. U=½ 2.0 BB=BB (for this case with L=0, =B). In the absence of an external
magnetic field, these two levels will be degenerate (they will have the same energy), however,
when in a magnetic field, they split in two level EoBB where Eo is the energy of the level
before the magnetic field is applied.
In general, the relative population of these levels depends on the temperature as:
B
N1

N
e
B
e
k BT
k BT
e
 B
k BT
N2

N
e
 B
B
e
k BT
k BT
e
 B
k BT
with N1 and N2 are the population of the lower and upper level. (Notice that for very low
temperature N1  N and N2  0 while for very high temperature both then to ½N. The
interpretation is that at low temperature the sample is completely magnetized, thermal effect is
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too small and all the magnetic moments get to align with the external field. For high temperature
instead, magnetization is completely lost since half of the spins are up and half of the spins are
down. The total magnetization is given by
B
M  N1  N 2   N
k BT
e
B
e
k BT
e
e
 B
 B
k BT
k BT
 B 

 N tanh
 k BT 
At high temperature (BB<<kBT), the argument of tanh is very small and tanh(x)~x thus
M ~ N B
B B
k BT
The temperature dependence of M/B is general, even when L is not zero and it is known as the
curie law
M C

B T
C is the Curie constant. Thus

o M
B

 oC
T
Ferromagnetic materials
Ferromagnetic materials have atoms grouped together is domains. Inside each domain atoms
have their magnetic moment aligned. Let’s consider that domains are misaligned with other
domain and thus the total magnetic moment is zero. If an external magnetic field is applied, the
domains will try to rotate to align to the magnetic field. Since the entire domain needs to be
rotated, there is some resistance for that to happen and the magnetization curve is not linear. As
the applied field increases, more and more domains align until
saturation is reached (all domains are align to each other).
The thermal effect is a lot less important than for
paramagnetic materials.
Thermal energy is able to rotate
individual atoms, but not entire domains, thus the magnetic
susceptibility (depends on the field) is in the order of thousands.
If after the material is completely magnetized the magnetic
field is decreased, domains do not completely rotate back to their random orientation but due to
mechanical resistance to this rotation, the demagnetization is not complete even when the applied
field is brought back to zero. The material is now permanently magnetized. Only if a magnetic
field in the opposite direction is applied, the magnetization can be brought back to zero.
However if further increased in the opposite direction, magnetization in the opposite direction
occurs. The cycle continues forming what is called and hysteresis curve.
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