[L2.5 Slide 1] Well now that we're comfortable
with some key concepts of quantum mechanics,
we can talk about a really important topic
for semiconductors.
The topic of this lecture is density of states.
[Slide 2] So let me remind you that we have these valance bands
that is mostly filled.
There's a top of the valence band
and a bottom of the valance band and that is so far down
that we don't have to worry about it.
We have a conduction band that consists of a set of states
that are mostly empty.
There's a bottom of the conduction band
there's a top of the conduction band
that we generally don't need to worry about
because it's so high in energy.
It's going to be important for us to know now
how the states are distributed within these bands.
They're not uniformly distributed within energy.
Now, we know that most of the empty states
in the valence bands,
the holes are near the top of the valance band.
Most of the electrons in the conduction band
are near the bottom of the conduction band.
So we only really need to be worried
about how the states are distributed in energy
near the edges of the band.
That's what our focus will be in this lecture.
[Slide 3] The key quantity that we're going to be discussing
is called the density of states
and it's simply a way to count the number of states
between a certain energy and an energy plus dE.
The number of states in that range is the density of states
times the width of that energy range.
So this density of states
is the quantity that we will see in detail,
we will make use of in detain in the next unit.
We want to learn how to appreciate what it is
and how to calculate it for semiconductors in this lecture.
We're going to take a simple description of energy
versus crystal momentum and these simple descriptions
work near the top of the valence band
and near the bottom of the conduction band,
and that's going to be adequate for most semiconductor work.
[Slide 4] So let's begin by thinking about a large chunk
of semiconductor crystal with a volume omega
with the length Lx in the x direction,
Ly in the y direction, Lz in the z direction.
So this is, let's think of this as a large
but finite volume that we have carved out
of an infinite bulk volume.
Well, there's going to be a finite number of states
because the states come from atoms
and there are a finite number of atoms in this volume.
We want to count the states.
There are different ways of doing this.
They end up all giving you the same answer.
I'm going to take one particular way
that makes a lot of physical sense.
Since this is a piece of a volume from an infinite solid
I'm going to assume that at this end of the solid
and at the other end of the solid
the wave function is the same.
We'll assume periodic boundary conditions
so that everything just repeats itself
throughout all of space in the x direction,
in the y direction, and in the z direction.
[Slide 5] So let's just look in the x direction.
This is x equals 0.
This is x equals Lx, which is, we're going to measure things
in units of the lattice spacing
so capital N is the number of atoms
and a is the lattice spacing.
Well, we know what the solutions to the wave equation are
in this region.
They're these bloch functions.
So periodic part times a wave solution.
So at x equals 0 we have to have the same wave function
at x equals L and we know that the periodic part will be the same
because that repeats itself at every lattice spacing.
So this condition imposing the periodic boundary conditions
leads us to conclude that e to the ikx times Lx
must equal 1
in order to satisfy these periodic boundary conditions.
Well, we know that some integer number of 2 pis,
either the i 2 pi or some integer number of 2 pis
will give us 1.
So, we have to require that kx times Lx
equals an integer number of two pis.
So the integer is one, two, three,
it just keeps repeating itself.
Solving for kx we see that kx is two pi over the length
that we're doing all of our counting of states in
times this integer j.
If we plot that we'll get a set of states.
So whenever we apply boundary conditions to a wave function
we get a set of discrete states.
We now have a set of discrete k states.
The difference is that we're thinking about this,
it's not a particle in a box problem.
This is a very large volume.
These are very finely spaced states.
Now, we're going to ask the question.
In some range dk, how many states are there?
Well, we know that each of the states
are spaced by 2 pi over Lx.
So the number of states in that region
is the width of the region dk
divided by the spacing between states,
then we'll multiply by 2.
Because each state can hold an electron with spin up
or with spin down and this quantity we will say
is the density of states in k-space
times the width of the region in k-space.
So we conclude that the density of states in k-space
is the length that we're doing the counting in
divided by pi.
This is in 1 dimension, okay?
And remember, we always have to remind ourselves
to account for the factor of two for spin
for our electrons.
Now, we can also look at this result a little more closely.
The length is some integer number of lattice constants.
That means I can insert N times a here for Lx
and then I have kx is two pi over a.
The integer j divided by the number of atoms
in this region N, capital N.
Now, as I let j get bigger and bigger,
eventually it will hit N.
When it hits N then the phase of this exponential
has gone from zero to 2 pi.
If I continue to let j increase,
I'm just going around the circle again
and getting another set of results
that are completely equivalent.
So, the unique solution to this equation
are between k equals 0 and k equals 2 pi over a.
That means the maximum k that can occur is 2 pi over a.
This is similar to the Brillouin zone
that we talked about in the previous lecture.
There is a region in k-space of which 2 pi over a
in which the unique solutions are found.
We can shift this region.
Sometimes it's usually convenient to shift it
so that we go from minus pi over a to plus pi over a,
but the width of that region,
over which we have a unique set of answers,
is 2 pi over a.
This is the Brillouin zone.
[Slide 6] Okay, alright so we've gone through the counting in 1D
and we've seen that the density of states in k-space
is two for spin, L over 2 pi.
So the result is L over pi.
If we were to do this in 2D, if we had a 2D material,
a quantum well that we've made
with semiconductor heterostructure or whatever,
the density of states would be two for spin
but now they're spaced L over 2 pi in the x direction
and L over 2 pi in the y direction.
So we get L squared, which is the area,
over 2 pi squared, which is 4 pi squared.
So when we reduce this we get A over 2 pi squared.
In 3 dimensions we have the factor of two for spin.
Now we have states spaced in x, y, and z.
Lx times Ly times Lz is the volume of this material
that we're doing the counting of our states in.
This is 2 pi cubed.
So we get omega over 4 pi cubed.
In 1D the width of the region in k-space is dk.
In 2D it's dkx, dky.
In 3D it's dkx, dky, dkz.
It's a volume in k-space.
Alright, now the important thing to point out
is this applies to any type of band structure,
any type of E of k, it's independent of E of k.
The states are, the density of states in k-space
is the same no matter what the material is.
[Slide 7] Now we will more commonly ask ourselves
a different question.
We will ask ourselves the question
how are the states distributed in energy?
Then it gets a little bit different.
So let's assume we have a parabolic band,
energy is h-bar squared k squared over 2m.
This would be the bottom of the conduction band here.
We've seen that the states are spaced uniformly in k.
At any particular k I have a corresponding energy.
If you simply map these states
to their corresponding energies you can see now
that they are not distributed uniformly in energy.
How they are distributed will depend
on the shape of the E versus k curve.
So what we're going to want to do now
is to figure out how these states are distributed in energy
and find a density of states in energy space.
What we will do is to count the states in k-space,
map them into a corresponding region in energy,
and we will relate
the number of states in this region in k space,
the same number of states have to occur in this region
in energy space, and that will allow us
to define a density of states in energy.
I'm going to write this as D prime of E.
dE is the number of states in that region,
and that's in this region right here.
And that's equal to N(k) dk, the density of states in k-space
times the width of the region in k-space.
That's the number of states in this region of k-space
and they are the same.
So by using this relation we will be able to deduce
from the known density of states in k-space,
the density of states energy space.
So let's do that.
[Slide 8] So first example
we will do it for a 1 dimensional nanowire.
These used to be academic exercises years ago.
Today they're real exercises because we can produce
semiconducting nanowires
and we often need to know what the density of states is.
We'll assume just a single subband.
And recall that there are different subbands.
If we assume a single subband with an energy E sub n,
then the kinetic energy
is associated with the motion in the x direction,
it's h-bar squared kx squared
over 2 times the effective mass.
[Slide 9] Alright, so here's our problem.
Here's our k in one dimension, here's our region dk.
We're going to map the states in that region
because there's corresponding regions in energy space.
Well, we know that the number of states in this region
in k-space, we've worked that out.
It's N of k dk,
is L over pi times the width of that region in k-space.
That has to be the same total number of states
in the energy, in the corresponding energy range.
That's D prime of E dE.
Now, I'm going to drop the prime.
Typically when we talk about density of states
we are interested in the density of states per unit length.
We're not interested in the actual number of states
because it's sort of arbitrary.
If I take double the length
I'll get double the number of states.
So, D, 1D of E,
the density of states that we're going to compute,
is the density of states per unit length.
So I have to divide by L.
The units then are going to be the number of states,
not really a unit, per unit energy, per unit meter
for the length, the units of density of states.
[Slide 10] Alright, so now we are prepared
to calculate the 1D density of states.
This is the number of states in energy space.
This is the number of states in the corresponding k-space.
We have our relation
between the number of states in k-space.
We have our dispersion.
We can put it all together.
This comes from the density of states in k-space.
We can now use the dispersion to compute dE
from this relation.
We can also compute k from the dispersion,
and we can then compute dE in terms of energy,
plug it all into this relation,
and we find the density of states in energy
is one over pi h-bar square root of effective mass
over 2 times the energy
with respect to the bottom of the conduction band.
This applies of course only when the energy
is greater than the bottom of the subband
because there are no states below that.
So this is the density of states
and we can pick the first subband, second subband,
third subband, or whatever.
[Slide 11] Now, I should just point out
we've done the counting over here for plus kx
and we found the number of states
that are in that corresponding energy range.
The band structure is symmetric.
There's an equal number of states over here at minus kx
that map into the same energy range.
So I need to multiply by 2.
And this is my final answer.
This is the density of states, the number of states
between E and E plus dE as a function of energy,
and we've derived the result that we were after.
Remember, we've assumed parabolic bands.
So the answer could be different
if we had a much different band structure.
[Slide 12] So in general we'll have multiple subbands,
we'll have a density of states for each one of those.
Note that the density of states goes to infinity
at the bottom of the subband when E is equal to E sub n.
And we'll have a series of those.
If this is the first quantized level
of the particle-in-a-box type solution,
we'll have a density of states associated with electrons
that are in that subband.
If this is the bottom of the second subband quantized state
then we'll have a density of states
associated with that as well, and on and on.
[Slide 13] Alright, that's 1D.
Let's see if we can do it for 2 dimensional electrons.
We can produce 2 dimensional electrons
by artificial semiconductor structures.
There are also 2 dimensional materials these days
that people are quite interested in.
So, consider a thin semiconducting slab of thickness t.
The electrons are confined in the z direction.
So I'll have particle-in-a-box subbands
associated with that.
The electrons are free to move in the x-y plane
and we would like to count the states
within this area A.
We're going to assume again that the bands are parabolic
so the energy is the bottom of the subband
plus h-bar squared times k in the x-y plane squared
divided by 2 times the effective mass.
[Slide 14] So let's have a look at that.
We're going to approach it the same way.
We'll take the density of states
times the width and energy range.
We'll map that to the corresponding number of states
in the k-space range.
And the way we'll do this is we will assume
a 2 dimensional k-space.
So there is a parabolic band.
So energy is h-bar squared k squared over 2 m,
that's this circle, and we'd like to know
how many states are there between k and k plus dk.
The red ring around here is width dk.
The area of that region
that we're asking to count the states in
is the circumference of the circle, 2 pi k
times the thickness of that little ring, which is dk,
and we're asking how many states are there
in that red area.
Well, each state has an area of 4 pi squared over A,
as we've seen
when we apply the periodic boundary conditions.
And then we multiply by two for spin, okay.
So, when we do this calculation
we have A over 2 pi squared.
We're dividing by A because we're going to compute
the density of states per unit area.
And then we have the volume of k-space, 2 pi dk.
We use our dispersion to compute 2 pi k dK
in terms of energy.
We do a little bit of math and we end up
with this simple expression
for the 2 dimensional density of states.
Note that the 2 dimensional density of states
is proportional to the mass divided by pi h-bar squared.
Of course these states only exist
if we're above the bottom of the subband.
[Slide 15] Okay, so the density of states in 2D then
is quite different from the density of states in 1D.
Let's do 3D, and most of the course
we will be concerned with 3 dimensional semiconductors.
So, we will assume a simple dispersion.
Parabolic bands, the constant energy surfaces are spheres.
We will equate the number of states in the energy range
to the number of states in the corresponding k range.
We now have a volume of k-space, d cubed k.
We have computed the density of states in k-space,
so we know that.
We will equate that, the volume, the volume of k-space
that we're going to do the counting in then
is the surface area of the sphere, 4 pi k squared
times the thickness of a ring around that sphere dk.
So that's the volume of k-space
that we are counting the states in.
Well, now we'll make use of the dispersion
and we will express k squared dk in terms of energy.
When we do that algebra we end up finding
that the density of states is proportional
to the square root of the effective mass
times the effective mass
and it varies as the square root of energy.
So we've calculated the density of states now
for 1 dimensional electrons, 2 dimensional electrons,
and 3 dimensional electrons all under the assumption
that the bands are parabolic.
[Slide 16] Now, things can get a little more complicated.
I'll just mention this;
although, we won't go into the calculation.
Recall that the band structure of silicon
is the bands are parabolic
but they have different curvatures in different directions
and there are six bands at the same energy.
So the constant energy surfaces,
the minima of the conduction bands are located
along a [1 0 0] direction, or these equivalent
[0 1 0] direction, [0 0 1] direction.
There are six of these valleys along these six directions
and they have different effective masses
in the x and the y direction.
So that band structure is a little more complicated.
We've been assuming a simple spherical band
centered at k equals zero.
We could compute the density of states for silicon as well.
It would be a little more complicated calculation
but it's something that you go through
in slightly more advanced courses.
I'll simply give the answer.
You can wrap everything together
and you can make it look like our simple result.
If we express, if we define a quantity
that we will call the density of states effective mass,
this is not a real effective mass.
The real effective masses are related to the curvature
of E versus k.
But this is a definition such that if we use it
in our simple spherical band approximation,
we'll get the correct answers
and the correct densities of states.
So, when we are computing the density of states
in a semiconductor, we have to be careful in this formula
to use the appropriate density of states effective mass.
If it's a simple spherical semiconductor
centered at k equals zero, it's just the effective mass
associated with the curvature of the band.
If it's a more complicated material like silicon
it is a more complicated quantity.
So you should remember that.
[Slide 17] Alright, so some things that we should take away.
This is the expression for the density of states.
It's worth remembering that the density of states varies
as the three halves power of the effective mass.
So if I'm dealing with a semiconductor
that has a light effective mass,
I know it's going to have a small density of states.
It's also worth understanding that it varies
as the square root of energy.
So as I increase in energy I will get more and more states.
So, for example if we look at silicon
it has a much higher density of states effective mass
than gallium arsenide.
We would expect then the density of states of silicon
to be significantly higher than gallium arsenide.
[Slide 18] So just to summarize,
we've derived the density of states in 1D, 2D, and 3D
assuming parabolic bands.
If we plot them we can see that the density of states in 1D
goes as one over the square root of energy
when the electrons get down near the bottom of the subband
the density of states goes to infinity.
In 2D there are states above the bottom of the first subband
and the density of states is independent of energy
above the bottom of that subband.
In 3D the density of states is zero
at the bottom of the conduction band
and then it increases as the square root of energy.
So, these are quantities
that we will make significant use of
in the next unit of this course.
[Slide 19] As an exercise, if you'd like to test yourself
and see if you understand the basic concepts here
compute the density of states for graphene.
Graphene does not have a parabolic band structure.
It is a 2D material but it will have a density of states
that is significantly different
than the 2D density of states for parabolic bands
that we just worked out.
The answer will be something like this.
The density of states will increase linearly with energy
as we go above E equals zero
and as we go below E equals zero.
So you might see if you can derive that result.
If you can do that then you understand the process
of deriving a density of states for any band structure.
[Slide 20] Alright, we've covered a lot of ground
but we've introduced this important quantity
that we're going to need to use again and again
throughout the course.
They quantity is the density of states
and it's one that we're going to frequently use.
The important points to remember
are that the density of states
depends on the dimension of the semiconductor.
We have different density of states in 1D, 2D, and 3D.
The density of states in k-space is constant
and does not depend on band structure
but the density of states in energy space
does depend on band structure and can be quite different
for different band structures.
For 3 dimensional bulk semiconductors
with parabolic bands it's worth remembering
that the density of states is proportional
to the three-halves power of effective mass
and to the square root of the energy
above the bottom of the conduction band.
So, these are some important concepts.
The quantum mechanical background that we've had
helps us understand where these come from.
We will make significant use of them,
especially the 3D density of states
in the next unit. Thank you.
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