[L3.2 Slide 1] Well let's continue our discussion,
in the previous lecture we talked about the Fermi function
the reason we need to understand
the Fermi function is because it is used
to compute carrier densities in semi-conductors.
When we do that computation we'll encounter
an integral known as the Fermi-Dirac integral.
And that integral is what I would like
to discuss in this lecture.
[Slide 2] Just a reminder, the Fermi function gives us the probability
that a state is occupied in equilibrium,
if that state exists.
It's given by this simple expression.
The two key parameters in this expression
are the Fermi level E sub F and the temperature T.
[Slide 3] So our goal is to understand what the carrier density
is in the conduction band, and what the hole density
is in the valence band given a particular location
of this Fermi energy, okay?
We argued that in the nondegenerate approximation
that the probability that a state at the bottom
of the conduction band is occupied is
e to the EF minus EC over kT.
So we expect n to be proportional to that.
We argue that the probability that a state near the top
of the valence band is empty is e to the EV minus EF over kT
so the whole concentration should be proportional to that.
This applies to a nondegenerate semiconductor.
But, what is the constant of proportionality?
And what if the semiconductor is not nondegenerate?
Then what do we do?
That's the subject of today's lecture.
[Slide 4] Well, there are two options in doing this calculation,
our goal is to relate the carrier density
to the Fermi energy, and to the other properties
of the semiconductor, and we can do the calculation
in k-space or we can do it in energy space.
I'll explain how we do it in k-space
but I won't actually do it,
and then we'll discuss how to do it in energy space
using the density of states.
[Slide 5] So consider again a chunk of semiconductor,
a cube carved out of an infinite slab
of three dimensional volume of a semiconductor.
This slab that we're doing the calculation
in has a volume omega.
Length in the x direction, y direction,
and the length in the z direction.
Now if I wanted to calculate the number of electrons
in this volume, we saw earlier that when we applied
periodic boundary conditions we had a finite number
of k-states that were very closely spaced.
I would just add up the probability
that each one of those states is occupied
in equilibrium, that would get me the total number
of electrons in this volume.
Now the volume itself is arbitrary so I'm really
more interested in what is the carrier density
per unit volume, so I would divide by volume.
Okay now if I were doing this calculation,
I would realize that these states in any kind of,
you know significant sized volume are very finely spaced
so I could convert this sum to an integral.
As long as I have the correct number of states
in each element of k-space, the correct number
will be given by the density of states in k-space.
And remember we computed the density of states
in k-space, it was 2 for spin times volume
over 8 pi cubed.
So we can put this all together,
convert the sum to an integral, perform the integral
and get the answer that we're after.
[Slide 6] Okay, alright I'm not going to do that calculation
it's one that you might attempt and see if you can get
the answers we're going to get in a minute here
though the energy space argument.
But it's a calculation that proceeds straight forwardly
using the density of states in k-space
and using this prescription for converting the sum
over these finely spaced states in this volume omega
into an integral but accounting for the fact
that we have the proper number of states there.
We have to be careful when doing this that we include
the valley degeneracy also, remember if silicon had
six equivalent valleys along the six directions,
plus x, minus x, plus y, minus y, plus z, minus z.
[Slide 7] Alright what I would like to do now,
is to do the calculation in energy space
and to actually work through it.
To do the calculation in energy space
we're going to need the density of states.
Now we've seen how it behaves in 1D,
we've seen how it behaves in 2D,
we've seen how it behaves in 3D.
Let's do the calculation in 3D where the density
of states goes to zero at the bottom of the conduction band
and then increases as the square root of energy,
as we go above the bottom of the conduction band.
[Slide 8] So first of all let's take a look at some energy
in the conduction band, and ask ourselves,
how many electrons are there
in this shaded energy range here?
Well the number of states in that region,
is the density of states at that energy
times the width of that region in energy dE, okay.
Now if I want to know how many electrons there are
in that region, I take the number of states
in that region and I multiply them by the probability
that those states are occupied,
that's given by the Fermi function.
Here's my Fermi function, I can make my non-degenerate
approximation because the states are way above
the Fermi energy, and I'll remember that the density
of states goes as the square root of energy
and then we conclude that the number of states
in that region is proportional to the square root of energy
times this exponential factor.
[Slide 9] So if I go ahead and plot that function,
I'll get a expression that looks like this,
that tells me where the electrons are located
in the conduction band.
As I increase the energy the square root starts
to increase and the number goes up.
But as I continue to increase the energy
the exponential gets large and damps everything out
and it turns around.
So there is a peak in this, most of the electrons are located,
the highest probability of finding
an electron is right there at that location.
How do we find that location?
Well we differentiate this quantity
with respect to energy and set it equal to zero.
If you do that, you'll be able to show that this peak
occurs at an energy that is 1/2 kT above
the bottom of the conduction band.
That's a relatively small energy,
I mean, if kT at room temperature is 0.026 electron Volts.
So that peak occurs at 0.013 electron Volts
above the bottom of the conduction band,
the conduction band is 1.11 electron Volts wide.
So this is, you know I have mentioned from time to time
that most of the electrons in the conduction band
are near the bottom of the conduction band.
This calculation shows that they actually are
quite close to the bottom of the conduction band.
Similarly the holes are located within about 1/2 kT
from the top of the valence band.
So electrons and holes are located
very near the band edges.
[Slide 10] So now let's go head and do the calculation,
let's compute the electron density by integrating
the number of electrons in an energy range
over the entire conduction band.
I'm going to integrate from zero to infinity.
Now I have a problem here you might worry about,
my density of states, the square root of energy
only describes the density of states near the bottom
of the conduction band, so it's not going to be a good
approximation as I go to very high energies.
But the Fermi function is going to help me out there.
There will be virtually zero probability that those
states are occupied so it doesn't matter
that I'm making an error in the density of states.
So if I put in the number of states,
times the probability that the states are occupied,
integrate from the bottom of the conduction band
to infinity, I know what my Fermi function is,
I know what my density of states is.
It's simply a matter of doing some math,
I can see that the result is going to depend
on Fermi level and temperature.
Because those two parameters are in the Fermi function.
The result is also going to depend on the effective mass
of the semiconductor and the number of equivalent valleys
because those quantities are in the density of states.
[Slide 11] Okay so let's do the math, we have this expression
to evaluate, we know the Fermi function
and the density of states.
If we plug those quantities in, we get an integral
that we have to evaluate.
Now let me make a couple of definitions,
I'll define a normalized Fermi level with respect
to the bottom of the conduction band.
Fermi level minus bottom of the conduction band
in units of kT, that's what eta F is,
and I'll define a normalized energy,
energy minus the bottom of the conduction band
in units of kT.
If I define those two normalized energies
I can rewrite the integral this way,
and now it's simply a matter of evaluating that integral
and then I have my answer.
Alright there's only one problem,
that integral cannot be done analytically.
It has to be done numerically
with Simpson's rule or whatever.
But we'll give that integral a name.
That integral is called a Fermi-Dirac integral
of order 1/2, okay.
The constants that we have out front here,
we'll wrap them all together and those constants depend
on the effective mass and the temperature.
Just as I expected that they would.
So that particular constant N sub C is called
the effective density of states.
It results from integrating the density of states
over this energy range.
So we have the expression that we're after,
and let's talk a little bit more about this
Fermi-Dirac integral because,
I glossed over that pretty quickly.
[Slide 12] In general, we can define a Fermi-Dirac integral
of order j this way.
It's an integral of eta to the jth power,
d eta over one plus e to the eta minus eta F.
And there's a normalization factor out front
which involves a gamma function.
You know remember that this gamma function
behaves like a factorial, the gamma function
of an integer n is n minus one factorial.
But the gamma function can also have you know,
non-integer values, gamma of 1/2 is the square root
of pi and there's a recursion relation
that if need to know, if I know gamma at some argument p
and I need gamma at an argument p plus one
I can use this recursion relation to do that.
The integral that we have been, that we encountered
when we worked out the carrier density had a value
of j equals 1/2, but it's simply an example
of a more general function.
Now this integral has a very nice property,
if the normalized Fermi energy is much less than one
then it says that the Fermi energy is much below
the conduction band then this integral simplifies
and becomes an exponential and that makes life easy.
That's a nondegenerate semiconductor.
In a nondegenerate semiconductor this complicated
Fermi-Dirac integral reduces to the familiar exponential.
There's also a nice property related to the derivative.
So we are going to avoid the use of this,
throughout the remainder of the course,
I want you to understand how to do the calculations
and use Fermi functions because you might encounter
situations where you need to do that.
These notes that I can refer to you really contain
all you need to know about using Fermi-Dirac integrals
in semiconductor problems.
One thing I want to point out is that if you're reading
papers or books you have to be very careful.
People sometimes define a Fermi-Dirac integral
without this normalization factor out front.
Usually when they do that if they're being careful
they use a roman F when they're defining it that way.
Whereas we're using a script F here.
But people aren't always even careful about that,
so you have to be very careful when reading
books and papers that you understand
which version of the Fermi-Dirac integral
that they are talking about.
[Slide 13] The one that we have encountered is the Fermi-Dirac integral
of order 1/2, J is equal to 1/2
and that Fermi-Dirac integral is defined this way,
the quantity eta F is the normalized Fermi level.
If I plot that function as a function of eta F,
when eta F is much less than zero, the Fermi level
is way below the conduction band,
and the semiconductor is nondegenerate.
When eta F is above zero, the Fermi level is above
the bottom of the conduction band
and the semiconductor is degenerate.
You can see for nondegenerate semiconductors,
the Fermi function is equal to the exponential
and that simplifies matters greatly.
The Fermi function is always equal to or less than
the exponential for a given argument eta F.
So in this nondegenerate assumption, when the Fermi level
is way below the bottom of the conduction band,
we have a very simple relation between
the carrier density and the Fermi level.
It's just an exponential relation, that's the one
we're going to use throughout the bulk of this course.
[Slide 14] I'll remind you that a nondegenerate semiconductor
is just one where the Fermi level stays
within this shaded region and doesn't get too close
to the bottom of the conduction band or too close
to the top of the valence band.
[Slide 15] Okay so we have succeeded in relating the electron density
to Fermi energy, in general that relation involves
Fermi-Dirac integrals, that are a little bit complicated
because we need to do them numerically.
For a nondegenerate semiconductor the relation
becomes much simpler, a nondegenerate semiconductor,
n-type semiconductor has the Fermi level well below
the bottom of the conduction band,
the Fermi-Dirac integral reduces to an exponential
and the carrier density becomes the effective density
of states times e to this quantity,
this normalized Fermi energy.
[Slide 16] Now we could go through the same calculation for holes,
we would be interested in the probability that the state
is empty, if we were to do that calculation
we would get answers that look much the same.
We would get an effective density of states
for the valence band times a Fermi-Dirac integral
that depends on the normalized Fermi energy.
Which in this case, is how far the Fermi energy
is below the top of the valence band in units of kT.
The effective density of states for the valence band,
would involve the effective mass for the valence band
in a nondegenerate p-type semiconductor
the Fermi-Dirac integral would reduce to an exponential
and we would have a simple exponential relation
between the hole density.
Equilibrium hole density and the Fermi level,
in a p-type semiconductor.
[Slide 17] Okay so, to summarize, we have discussed how one uses
the Fermi function, and the density of states
to determine the concentration of electrons
in the conduction band, and holes in the valence band
if we know the location of the Fermi energy.
In general these relations involve this complicated
Fermi-Dirac integral of order 1/2,
for parabolic bands, if the semiconductor
is nondegenerate the Fermi-Dirac integrals simplify
to exponentials and we are left with these
two simple expressions that I'd like you to remember.
Because, we're going to use them over
and over and over again.
The electron density is equal to the effective density
of states for the conduction band,
which involves the conduction band of effective mass
times e to the Fermi level minus model
in the conduction band over kT.
So it's easy to remember the order of the Fermi level
and the bottom of the bottom of the conduction band
because the higher the Fermi level is the more probable
the states on the conduction bands are occupied.
The density of holes in the valence band is the effective
density of states for the valence band.
Which involves the valence band effective mass
and it depends on how far below the top
of the valence band the Fermi level is.
So these are expressions that are valid
for nondegenerate semiconductors and they're the expressions
that we will be making use of for the remainder
of the course, we will continue in the next lecture
to discuss in a little more detail,
how the carrier concentrations vary
as a function of the Fermi level.