EE 362 Electric and Magnetic Properties of Materials

1
EE 362 Electric and Magnetic Properties of
Materials

• Dr. Brian T. Hemmelman
• Chapter 4 Slides

2
The Semiconductor in Equilibrium (2)
The number of electrons in the conduction band at
a given energy, E, is
The number of holes in the valence band at a
given energy, E, is
To find the total number of electrons in the
entire conduction band, we will have to integrate
n(E) over the conduction band energies. Similarly
we will have to integrate p(E) over the valence
band energies to find the total number of holes
in the valence band. We can get an idea of what
is happening though by graphically integrating
n(E) and p(E).
3
The Semiconductor in Equilibrium (3)
4
The Number of Conduction Electrons (4)
Graphs are great but we need to be able to
compute specific numbers, so we must integrate
n(E) over the conduction band energies to find
n0, the total number of electrons in the
conduction band at zero bias.
We will make a change of variables and let
. Then we also have
5
The Number of Conduction Electrons (5)
Making these substitutions we can rewrite the
formula for the number of electrons in the
conduction band as
Thus,
All the terms out front are constants so let us
just call that some new constant, NC, the
effective density of states in the conduction
band.
6
The Number of Conduction Electrons (6)
We finally arrive at a compact version of the
formula for the number of the electrons in the
conduction band of a crystal under zero bias.
Similarly, we can derive a formula for the number
of holes in the valence band of a crystal under
zero bias to be
7
Intrinsic Carrier Concentrations (7)
In an intrinsic semiconductor (a perfect crystal,
i.e. no defects and no impurities) the number of
electrons in the conduction band must equal the
number of holes in the valence band. Hopefully
this is obvious since the only way for an
electron in a pure crystal to become a conduction
electron is if it breaks out of its atomic bond,
and therefore leaves behind an empty bond (that
is, a hole). The intrinsic carrier concentrations
in a crystal at zero bias have the special
notation, ni and pi. Thus, in a pure crystal ni
pi.
8
Intrinsic Carrier Concentrations (8)
Multiplying these together we get
Conclusion For a given temperature ni is a
constant.
9
Intrinsic Carrier Concentrations (9)
10
Fermi Level Position In An Intrinsic
Semiconductor (10)
We have derived a formula to calculate what the
carrier concentrations are in a pure crystal.
These concentrations must be related somehow to
the Fermi energy what correlates to probabilities
there are electrons and holes in specific quantum
states. Therefore it is logical to want to
understand where the Fermi energy level is in an
intrinsic crystal. Our starting point for this
derivation is again that in a pure crystal the
number of electrons and holes must be equal, so
ni pi.
11
Fermi Level Position In An Intrinsic
Semiconductor (11)
All terms here are the same except
If
then
12
Dopants (12)
While there are some free electrons in a pure
crystal at room temperature (ni for silicon at
300 K is 1.51010 cm-3), there are not enough
available to conduct useful amounts of current
for most applications. The reason semiconductors
are so prevalent in electronics is because their
properties can be intentionally changed and
controlled. One way to do this is to add
impurities to the crystal in very precise
quantities. These intentional impurities are
called dopants. In silicon crystals for instance
(Group IV of the periodic table) we can add
elements from Group V (e.g. phosphorous) that
have one extra electron that is not needed for
atomic bonding or elements from Group III (e.g.
boron) that have one too few electrons for atomic
bonding (one too many holes). Impurities that add
extra electrons are called donors. Impurities
that add extra holes are called acceptors.
13
Dopants Donors (13)
14
Dopants Acceptors (14)
15
Ionization Energy (15)
If we add donors (with extra electrons) we can
look at how tightly bound those extra electrons
are to the donor atom. Our hope is that those
extra electrons can be broken loose from the
atoms and are therefore available to serve as
current. The amount of energy required to free up
those electrons is called the ionization energy.
We can calculate what this should be as well as
what the orbiting radius of the extra electrons
are to the donor atoms. This gives us a good feel
for how easy or hard it will be to create extra
conduction electrons from the donor dopants. We
can begin by setting the coulombic force of
attraction between the extra electron and the
donor nucleus to the centripetal force of the
orbiting electron. If the extra electron is still
bound to the donor atom, this will allow us to
compute the radius at which it is orbiting.
16
Ionization Energy (16)
(coulombic force centripetal force)
where ? is the permittivity of silicon, v is the
velocity of the electron, and rn is the radius of
the nth orbit. As angular momentum will also be
quantized we also have
and thus
To get a feel for the size of rn we can compare
it to what is called the Bohr radius, a0, which
is a standard measure in atomic phenomenon.
Å
Å
The electron will orbit over MANY silicon atoms.
17
Ionization Energy (17)
The total energy of one of these orbiting
electrons would be the sum of its kinetic and
potential energies, E T V.
Experimentally measured values of dopant
ionization energies in silicon are 0.045 eV for
phosphorous, 0.05 eV for arsenic, 0.06 eV for
aluminum. Compared to the bandgap energy of
silicon (1.12 eV) we see that it takes relatively
minimal thermal energy to ionize the dopant
atoms. We would thus expect that most of the
dopant atoms will create additional free charge
carriers available for conduction.
18
Equilibrium Distribution of Electrons and Holes
(18)
When we add dopants to a semiconductor crystal we
know we will be changing the number of free
electrons or holes in the material.
19
Equilibrium Distribution of Electrons and Holes
(19)
In an intrinsic semiconductor we have
Thus, in general we can rewrite
Similarly, we can derive analogous expressions
for holes
20
n0p0 Product (20)
Let us take a look at what happens when we
multiply electrons and holes
This is the same result we had for intrinsic
semiconductor, so we can conclude
Since ni is a constant for a given semiconductor
material at a fixed temperature we can see that
if electron concentration goes up, the
corresponding hole concentrations must go down
(and vice versa, if p0 increases then n0
decreases).
21
Statistics of Dopants (21)
If we have added dopants, for example donors,
then the number of electrons, nd, actually
occupying the donor energy level Ed if the doping
concentration was Nd is given as
where Nd is the number of ionized dopant atoms.
Under the Boltzmann Approximation we have
22
Statistics of Dopants (22)
We can then make a relative comparison of how
many electrons are in the donor state energies
versus the total number of electrons in both the
donor states and the conduction band states.
This ratio is
where EC-Ed is the ionization energy. For
phosphorous dopants in silicon at T 300 K with
Nd 1016 cm-3 we find in Example 4.7 that this
ratio is
So at room temperature we see that virtually all
common dopants will ionize.
23
Charge Neutrality (23)
When a semiconductor is in thermal equilibrium
the net charge density must be zero. This should
be obvious since for every thermally generated
electron from the semiconductor there is a hole
created. For the dopants, every dopant that
ionizes will generate an additional electron (or
hole) but left behind will be a corresponding
positive (or negative) ion. Of course, nothing
prevents us from using both donors and acceptors
in the same material. This type of material is
called a compensated semiconductor. In general
then, charge neutrality requires that
Assuming there is complete ionization (a safe
assumption at room temperature)
24
Charge Neutrality (24)
Using the n0p0 product we can rewrite this as
There are two roots to this quadratic equation.
If we are assuming the material is n-type then we
know Nd gt Na and we can select the larger root to
find
for n-type
To find the hole concentration in this n-type
material we use the n0p0 product.
25
Charge Neutrality (25)
In Example 4.9 we have n-type silicon at T 300
K with doping concentrations of Nd 1016 cm-3,
Na 0, and ni 1.51010 cm-3. Then
By a similar derivation we can find the hole
concentration in p-type material to be given as
for p-type
26
Positions of Fermi Level Energy (26)
These formula derivations are really nothing new,
but merely take formulas we already have are
rearrange them. We can start with n-type
material where
From one of our other n0 formulas we could also
derive
These were derived for n-type material.
27
Positions of Fermi Level Energy (27)
For p-type material the corresponding formulae are