Welcome back. This is now the 17th lecture of the condensed matter, of course.
When we left off last time, we were talking about band structure and we're talking about electrons in conduction bands and holes in valence bands.
And what we're going to do today is we're going to start by thinking about how electrons and conduction bands and holes in valence bands move around.
So let's start by drawing the bottom of a conduction band.
So we have some E and some K and there's going to be some quadratic minimum of a conduction band.
Let's call this point here came in where the bottom occurs and this point here, we'll call it E,
not the energy of the bottom of the conduction band very generally already over here, very generally,
we can we can write for an electron near the bottom of the band that E is approximately e
nought plus which is the Taylor expansion D squared e d squared times k minus came in squared,
maybe other terms which will ignore. And given this, we will define an effective mass.
An electron effective mass. Electron effective.
Mass effective. Mass.
I'm a star. I'm stars of electron via one over am star electron.
Equals one over a squared d squared.
E d k squared. Such that the energy is not plus the usual h bar squared k minus came in a squared over two an electron star.
Now this looks very much like a free electron, except the momentum, the the wave vector should be measured with respect to the wherever.
The minimum in the momentum is is the usual free electron parabola.
Now given this, we can also calculate actually given any form of the energy of the of the electron in this,
in this band we can write a group velocity V group very generally is d omega decay if it's a general expression for a group velocity,
since we can replace omega by one over h bar times in times in energy d decay.
This is the general relationship for the group velocity of an electron in a band.
In our case it will be using this dispersion up here it will be h bar k minus came in over m star of the electron.
So it means that if you have this picture over here,
if you if you move the electron to the right of the minimum, then the electron is moving to the right.
If you move the electron to the left of the minimum, then the electron is moving to the left.
This is true independent of where in the bronze down the minimum occurs.
So really the issue is where your position with respect to the minimum,
if you want to figure out which direction your group velocity is now, there are a couple of caveats here.
Caveat, one are generally effective masses, right?
This masses may be anisotropic, masses can be anisotropic, meaning that the energy is it may be parabolic in all directions,
but depending on which crystal direction you choose, the curvature may be different.
This is sort of natural. If you have something like a tetrahedral crystal where the axes are different lengths in different directions,
you might expect that the curvature of the energy in the different displays in K in the different directions might be different.
And indeed it often is. So one has to worry about the fact that depending on which way you're going, your effect of mass may seem different.
We're going to ignore this generally, so ignore this complication, but more generally you should be aware that it does occur.
The second complication that you should be aware and it occurs is you can have situations if you plot E versus K,
you can have a situation where there are two minima that occur and exactly the same energy.
This E so that the minimum of the conduction band can occur at two places in the burn zone, K1 and K2.
These are known as valleys. Valleys for obvious reason.
This may look like it is something that is very unusual to have happen,
that you have just such a coincidence that the minima that the same energy occurs at the minimum in two places in the bronze zone.
But in fact it's actually quite common to have this be due to the symmetries of the crystal.
So generally this will happen. If the crystal is particularly symmetric,
you'll have to minimum in the conduction band at two different points in the bronze
zone or even four different points or even six different points of of the barren zone.
So you have to keep track of whether the electron is in this minimum or this minimum.
But the general rule applies that if you displace yourself in the minimum to the right, you're moving to the right.
If you replace yourself from to the minimum are to the left of the minimum, then you're moving to the left.
Okay. Now, the final point. And generally, we're actually we're going to ignore this complication and also ignore this as well.
Just be aware that it sometimes occurs. The final complication is perhaps a much more important complication.
If you have a band structure, here's a broken zone.
If you have a band structure like the tight binding band structures we drew before, it looks like this.
The generally the effective mass is given by the second derivative of the energy with respect to K.
So that would mean that the star of the electron here is greater than zero.
But up here m star of the electron would be less than zero, which seems rather odd to have a negative mass for a particle, but we'll see in a moment.
That actually makes quite a bit of sense. And really we should think of the mass of the electron as being negative.
The general definition that now so the electron one over h by square squared e.g. case where will always hold.
And that is what we should always think of as the effect of mass.
So far so good. This was the easy part. The hard part is when we start thinking about holes at the top of of a valence band.
So the reason it gets hard is because minus signs will come back and bite you one way or the other.
So it is really hard to keep track of the minus signs. And to make matters worse, some books put the minus signs in different places.
So this is one of these key situations where as long as you know what you're doing, you can move the minus sign around.
You have the even number of minus signs. At the end of the day, you got the right answer.
But what I am going to tell you is what I think is the most natural way to think of where you put the minus signs.
And I think most books agree with me on this, but not all.
So be warned about that. So this is the situation with holes at the top of valence bands.
Top of bands. So so that is means up here where we found this odd result that the mass of the electron is
negative when we're thinking about the top of a band like over there where we found this result,
we're usually thinking when we're at the top of the band about a nearly filled band and nearly filled valence band.
So we draw again E versus K, and we'll imagine that we have the top of some down,
which is mostly filled and we'll think about, okay, we'll put this at K max and okay, and we will call this E e not also here.
We imagine if we put an electron up at the top of this band somewhere, its energy is going to be okay.
So for the case of Paul's top of band, top of band, the energy of an electron, e electron.
It's really good to keep keeping your head track of whether you're talking about an electron energy or a whole energy,
because you'll see that they're going to get confused in a second.
So the electron energy, if you put an electron near the top of the band, that will be some E0 plus one half D squared,
e d k squared times k minus k max squared exactly the same way as it was before.
But now this quantity here is negative. So this is less than zero.
And this is why we should think of the mass of the electron here as being a negative quantity,
because the energy actually goes down as you go away from the maximum.
So here, what we're going to do to make our life a little bit simpler is to define a mass of a whole define mass
of whole to be h bar squared over m hall r star is minus d squared e for the electron decay squared.
So this thing here is greater than zero.
At top of that top of band. And you'll notice from comparing this definition to the above definition,
that the star of the whole is actually equal to minus the star of the electron.
Okay, so up here where the mass of the electron appears to be negative, we've defined the mass of the hall to be minus of that.
Now, why is it that this makes sense to do? Why is it we should define the mass of a hall to be negative, the mass of electron?
And why should it be the mass of the hall? Be positive at the top of the Valence Band.
So let's think about this for a second. So here we have the same top of the Valence band.
We imagine the whole Valence band is filled except for one hall, which we're putting at the top of the band here.
So the energy of the hall e hall be some constant in our Twitter and then we want to find
out how does the energy of this hole change as we displace it from the top of the band?
So let's imagine this point here, just so we don't have to carry around these K axes and K means anymore.
Let us decide this point. Here is cable zero as we displace K from.
This is just for simplicity. It doesn't have to be a capable zero.
But let's imagine is let us ask what happens to the energy as you displace the hall from the top of the band?
So we're going to move the hall from here to here.
Well, when you do that, the energy actually has gone up from this place in the hall from the top of the band.
Why is that as squared to star hall times?
I guess k squared here we're displacing it from Capel zero.
The reason that the energy has gone up when we displace the hall from the top of
the band is because the electrons are trying to go to the lowest energy possible.
The holes try to bubble up to the highest energy possible.
It's like is like an air pocket or a bubble or a like pushing a balloon underwater.
If you take this balloon here and you push it underwater, that costs you energy.
It costs you positive energy to take this hall and push it underneath the electrons because the electrons want to go to the lowest energy possible.
That clear, that is the intuition of why it is that you should think of the mass of the hall as being positive,
because when you displace it from the top of the band, it cost you positive energy to move away from the top of the band.
So far, so good. Okay, good. All right.
So that's the easy sign to keep track of, but the signs get more complicated.
Let us ask about the momentum of a whole. What is the momentum of a hall?
Momentum of a whole. And a good way to think about this is to realise that a filled band, filled band has no momentum.
Has no crystal momentum. No crystal momentum.
Carrying no current inert has no momentum at all.
So if we want to know what the momentum is of a hole in the top of the band, let's imagine, here's our band.
We fill the entire band like this, except for one state which is missing out of a state like this.
We imagine taking this hole, this missing particle from the band.
If we add it to a band, it's empty. Except for that one electron placed in.
Then this plus this equals a fill down equals filled van filled with file filled band.
So momentum. So total momentum equals zero. Total h party equals zero.
Okay, so the sum of these two put together the fill band with the absence of the whole, plus the empty band with this one electron.
Sorry, the filled band with the absence of this electron plus the empty band with the presence of this electron.
When you put them together, their total momentum is going to be zero.
So what we have then is h bar K for the hall over here plus h bar K for the electron should which is over here should give us a total of zero.
So far so good. So that means that h pa k for a hall must be minus h bar k for the missing electron.
Okay. So that is a good way to keep track of what the momentum of a horse should be.
There is one of these confusing minus signs.
Now let's think about a given. This what the velocity is of a hole at the top of are.
These things are actually going to be end up being important because you need to know which way the electrons or the holes are moving.
Keeping track of it for an electron at the bottom of the band, everything seems very natural,
but hole at the top of the band, everything starts getting confused.
So it is important to keep track of these correctly. So let's think about velocity of hall at top of band.
Top of band. So again, we start with the energy for the electron at the top of the band.
So that is E not minus H bar squared, k squared K electron squared over two and Mr. Hall.
Okay. Now why did I draw it this way? So Mr. Hall is greater than zero.
The energy for the electron decreases as you increase the momentum of the electron away from zero.
The band is curving down, so the energy for the electron is decreasing the group velocity then which is always one over h bar D.
So this is group velocity for the electron.
The electron decay electron is then minus h for k electron over Mr. Hall or since Mr. Hall is minus Mr. of the electron,
we can write this is h bar k e over m star electron.
But Mr. Electron is is less than zero.
So what this means is if I have a, if I had an empty band like this and I had an electron here near the top of the band,
the velocity of this electron is actually going this way.
The group is going this way, right.
The MC because you can think of that as the electron mass is going negative is negative because of the negative curvature.
So the group velocity is actually in the opposite direction as the wave vector.
Okay, I know it's I know it's confusing, but but if you just keep track of the science carefully, everything comes out correct.
Let's compare that to the energy situation if we think about it in terms of holes.
So the energy of a hole near the top of the band is some constant plus h bar squared K hole squared over two and a star hole.
So the energy of the hole increases as you increase its momentum away from the top of the band.
So the group velocity, the group hole equals one over h bar.
The hall decay hall equals h bar k hall over Mr. Hall and K Hall, we just decided is the same thing as minus K for the electron.
So this is the same thing as minus k, h, bar h parquet electron over Mr. HALL Which is in fact the same thing we just calculated up here.
So the point here is that if we have a filled band like this and one empty state here, the group velocity of that is going in this direction as well.
This is an extremely important principle that will come back many times or at least several times, that the velocity of a state,
the velocity of an orbital of an orbital orbital is independent of whether that orbital is filled.
Whether the orbital is filled.
So you can think of it as you have a bunch of orbitals in your system and the orbitals have some velocity,
you can put an electron in that orbital and the electron will move that out with that velocity.
If you remove the electron from that orbital, it is still the orbital is still moving with the same velocity it is.
There is no electron in it. Okay. So this is kind of an important principle which sort of actually makes a very good amount of sense.
Even when we think about the we think about the hydrogen atom or something, we think about these orbitals being there,
whether or not that there's an electron in the orbital and the orbitals can be moving with some velocity and
they're still there and moving independent of whether we put the electron in or we didn't put the electron in.
All right. So let us now actually try to take what we've learned and calculate some dynamics.
Dynamics of electrons and holes. Dynamics of electrons and holes, channels and halls.
And the simplest way to handle dynamics is what we started the term with, which is to think about the theory.
So we're back to Judith Theory after many weeks. And so let's recall that the equation that we derived some p dot equals F just Newton's
equations with a drag term p over tau and we can write that out for electrons.
For electrons we will have the mass of the electron.
I'm going to put a star here. Times V Dot.
So the only thing different from what we did at the beginning of the term is I'm now putting a star next to the mass.
So it's the affective mass, the curvature at the bottom of the band that matters,
not the bear electron mass, which is not an important quantity in a semiconductor or an insulator.
We will have the force, the usual, the usual Lorentz force, which is always with you as a another joke.
I never mind the Tao for the electron, the scattering time for the electron.
So this should look fairly familiar. This is the Duda transport equation that we had before.
There is some acceleration and mass times acceleration.
There's some force, there's a charge of minus E on the electron, there's Lorentz force and there's a drag force here.
This equation holds true whether we're talking about an electron near the bottom, a band, or an electron near the top of the band.
However, near the top of the band, the sign of the mass gets to be negative, and that seems a little bit weird,
but if we think about it, near top of band, near top of a band, we can consider the equation for holes.
So we would have an m star whole times, the velocity of the whole same equation.
But the charge on the whole is now plus e right?
This again equals plus e, the charge on the whole is plus E.
Then we have E plus the Crosby Ren's force and then minus m star hole velocity of the hole over some scattering time for the hall as well.
So we have a equation for the holes at the top of the band as well.
Now remember that the we have two things here.
We have two equations, one for the electrons, one for the holes.
And the difference in the two equations is that the sign of the charge is different for the electrons in the holes.
But also Mr. Hall was defined to be minus the star of the electron.
So in fact these two equations are exactly the same if this equation has been multiplied by an overall minus one
and Star Hall is minus M star of the electron and so is actually just repeating exactly the same JUDA equation.
So this is actually quite important. It's exactly the same statement that I just wrote over here.
The dynamics of the state are exactly the same.
Whether you're talking about an electron moving the state or a whole moving the state, the absence of an electron,
the orbital still moves with the same dynamics, whether or not the orbital is filled with an electron gun.
So the thing that is funny about the equation in the semiconductors is if you're up at the top of the band,
you have a choice of two things that both seem a little bit weird.
One possibility is that you have to think about a negative mass for the electrons.
If you're thinking about electrons in the state,
the other possibility is you think about positive mass holes, but now these are positively charged particles.
Now this resolves one of our all puzzles, all puzzle from the second day or the third day of term of the puzzle,
which was why is sign of hall coefficient hall coefficient.
Sometimes positive.
So you'll recall that we derived due to theory that the whole coefficient is one over n times the charge on the particle moving around.
This is the charge on the electron. So if you have a band which is mostly empty and there are just a couple of electrons moving around,
then you have the theory for these electrons moving around,
just like we thought about in Sommerfeld Theory or in Judith Theory at the beginning of the term.
And you'll get a whole coefficient which is negative.
But if you have a a band which is mostly filled and you just have a couple of holes in the top of the band running around,
then either way you look at it, what you think about it as holes running around with a positive charge,
or you think about it as electrons running around, but now they're massive electrons negative.
What you'll get is you'll get the opposite sign because now the things carrying the charge have the have the wrong sign.
They have positive sign instead of negative sign. So this is what you might have to do it way back in 1901 that it appeared that for some materials,
for some reason, the charge carrier had the wrong sign.
And indeed that is what's going on in some materials. The charge is carried by holes rather than electrons.
So if you have a semiconductor and you doped with electrons, you always get this sign of the hall coefficient.
If you do it with holes, you'll get the opposite sign of the hall coefficient.
Now for metals, real, real metals like calcium or beryllium or manganese or something like that,
which are not just lightly doped where they, you know, we have some complicated band structure.
It is less clear what the sign of this is supposed to come out as,
because you can have a situation like we discussed with calcium in the last
lecture where you have a partially filled lowest band in a partially filled,
highest band. So you can think of it as having some holes in the first bronze and some holes in the first band, but some electrons in the second band.
So the holes are trying to give you a positive hall coefficient.
The electrons in the second band are trying to give you a negative coefficient of the two of them compete
with each other and then you have to start worrying about which one has a greater scattering time.
Okay, so it gets a little bit more complicated for real, for a genuine real metals like beryllium or or manganese or calcium.
But in the case of a doped semiconductor, it comes out very clearly.
If you have basically electron transport, you get this sign of the hall coefficient.
If you have basically hole transport, you have the opposite side of the hall coefficient.
So far so good. So we resolve this one puzzle.
There is something else that I mentioned when we talked about due to theory all the way back at the beginning of the term,
we remember that due to theory, works well, works well for semiconductors.
For semiconductors. Why is that? Besides the fact that it can get the sign intelligently and the reason why the theory works
particularly well for semiconductors is because usually when you're talking about semiconductors,
you're talking about a very low density of charge carriers.
You're talking about just a few electrons in the conduction band or just a few holes in the valence band because because density is low is low.
Whether we're talking about electrons in the conduction band or holes in the valence band, the density is fairly low.
And why does that help you?
The reason that the theory failed back when we were studying it for the first time is because we neglected Fermi statistics.
We neglected the fact that the electrons were fermions and that made us mess up all of these things, like the the heat capacity.
We got that completely wrong because we neglected Fermi statistics.
But if the density is really low, you're actually entitled to neglect Fermi statistics.
Why is that?
Well, if the probability of having two electrons ever sit in the same orbital is extremely low because the density of the electrons is very low,
then you can just treat these as classical particles.
Maximum Boltzmann particles can, because it's very unlikely they're going to sit in the same orbital in the first place.
A certain better defined way of saying that is the Fermi energy.
Given this density, the Fermi energy increases with density.
If the density is low, the Fermi energy is low, if the Fermi energy is much less than the temperature, which would be true.
If the density is low, then you don't have to worry about Fermi statistics.
You can treat the particles as being classical. So.
So particles are classical. Our classical and you can consider just regular classical due to theory classical Boltzmann
dynamics maximal Boltzmann distributions without worrying about Fermi distributions.
So this is yes, it is similar based.
The whole yes world will come to that in a moment. But also you obey for me, statistics.
You can only put if you think about it for a second. You can only put one hall in a particular case state.
If you start with the Valence band filled, you can take it out once, but you can't take it out again.
Okay. Yes. Holes of the Fermi statistics, they are fermions.
All right. So this is actually bringing us to the to the realm of statistical mechanics or thermodynamics of our particles in in semiconductors.
So we have to start thinking about things like how many are how many electrons get excited from the Valence band to the conduction band?
How did they get there and what are they doing once they're there? What is the effect of temperature?
So one thing that I told you from the in the last lecture that I should be a little bit more honest about.
We're going to go back to think about doping for a second and how doping is affected by temperature, doping and temperature.
Okay. So let us recall that when we took a and opened and opened like, like a phosphorus box in silicon and they opened,
we realised that phosphorus equals a silicon plus one proton plus one electron.
And I told you the electron goes into the conduction band, into conduction band band.
And I told you to ignore the fact that there's an extra proton ignore.
Now this issue of being able to ignore the extra proton that we put in, in addition,
that is something that is actually we have to be a little bit more careful about and we have
to worry about temperature and things like that to ask whether it's it's okay to ignore it.
So think about this a little bit more carefully. We put in a proton and we put in an electron.
So what is an electron and a proton do when you put them together?
Which is really going to happen is you are going to form form a hydrogen atom or a hydrogen atom.
Hydrogen atom. We put the electron in the conduction man, but we also put a positive charge behind.
And the electron is going to bind to that positive charge is attracted to that positive charge.
By Coulomb interactions, you form a hydrogen atom, but it is not a regular hydrogen atom.
It is slightly different from a hydrogen atom because one m star not m not the mass of the electron.
It is the effective mass of the electron, not the massive electron.
The dynamics of the electron running around in the conduction man is given by M Star, not the bear mass of the electron.
So that is the difference. One. The second difference is something you learned in A&M in the Coulomb interaction.
The attraction back to the proton is is squared over for pi epsilon, not epsilon relative times r r epsilon,
a relative dielectric constant relative dielectric constant, not the usual e squared over four pi epsilon not R.
Okay, so the, the Coulomb interaction is, is reduced by the relative dielectric constant because you're living in a semiconductor,
not living in the vacuum of outer space. So if you recall, we are regular hydrogen atom, regular hydrogen, regular hydrogen atom.
There's a binding energy binding which is well, okay, you probably well,
maybe do remember this formula is eight epsilon not squared Planck's constant squared,
but you probably remember the magic number 13.6 EV, which is a red berg.
So this is the binding energy of a, of a regular hydrogen atom.
But in a semiconductor the binding energy e binding of the electron back to the proton.
The additional charge that I put in with the doped is now m star electron.
It is a fourth over eight epsilon relative epsilon not squared over h squared which I can then write as r m
star of the electron divided by the mass of the electron times one over epsilon relative squared times 13.65.
Now this expression can be a whole lot different from 13.6, and generically it's going to be a lot smaller.
Typically effect of masses of electrons and semiconductors can be a third of the
mass of an electronic can be even a 10th of the mass with electron or even smaller.
The dielectric constant in semiconductors can be numbers like ten.
So that means that this number,
the actual binding energy of the electron to the proton in the semiconductor can be as much as a factor of a thousand less than this.
For example, in silicon in silicon e binding of the electron back to its doped nucleus is 0.04 for electron volts, a much, much, much smaller number.
Now, why is that important? Okay. Oh, look at this.
I need. That can be. With this in here, I'm going to draw.
So here is the bottom of the conduction band. Here's the top of a valence band.
So when you put a dope in, when you put that phosphorus opened in, you form a bound state, which is a very small distance.
Binding. A binding below the bottom of the conduction band.
So you gain a tiny bit of energy by binding. Instead of just sitting with cake with zero in the conduction band,
you gain a little bit of energy by binding to the positive nucleus that you came from.
However, this energy here is very, very small and at room temperature, at room temperature, write this down.
A tea room. The electron breaks free.
Breaks free. And goes off into the conduction band.
Breaks free and goes into conduction band.
Is thermally excited into the conduction band.
And this is why this is why it is okay to think about when we add our phosphorus doping to silicon.
We really should just think a room temperature or higher temperature.
We should think about this as just directly adding electrons into the conduction band.
If you cool it down, if you cool this, maybe I'll write that down.
So doping adds electrons adds directly to two conduction band.
Or similarly, if you do the same thing with holes, when you add in the hole,
it would form a a bound state to the to the negative charge on the nucleus that we left behind.
We are putting in a boron or an aluminium or something.
And at a finite temperature, that hole breaks free of its nucleus and runs off into the valence band.
However, at low temperature, at low temperature, the electron becomes becomes bound, becomes bound again.
Bound to the nucleus. To the open nucleus.
Nucleus again. So if you are measuring the density of electrons in the conduction band, if you cool the system down a lot,
all of a sudden you would discover that the density of electrons in conduction band was dropping
because all of these electrons were getting frozen back onto their the nucleus that they came from.
This is known as carrier freeze out. Carrier freeze out.
Okay. All right. So, so far, so good.
Everyone happy with this? All right, so now it's time to do some serious statistical mechanics thinking about how many electrons
we should have in the conduction band and how many holes we should have in the Valence Band.
This is a rather important calculation that actually shows up on exams very frequently.
So it's a good thing to go through several times.
All right. So let's start with this generic picture of the bottom of a conduction band, the top of a valence band.
They do not have to have the same curvature. In general, they could have a different effect of masses of John.
The effect of masses being slightly different. They don't have to be a line.
It could be an indirect gap at the top of the Valence band. Could be in different place than the bottom of the conduction band.
We'll call this energy conduction easy for in conduction.
We call this energy even for a valence.
This energy gap is a D and e gap equals e conduction minus e valence.
And we'll imagine that the chemical potential is somewhere in the middle of the band and we'll try to figure out where it is.
All right.
And so the we start with the Valence band being completely filled and we're going to try to figure out what happens when we add temperature.
So a couple of things we're going to need. One is what the spectra are.
So for energy greater than energy conduction,
we have energy as a function of K is energy conduction plus H bar squared minus K mean squared over to Mr. Electron.
And then for energy less than E valence,
we have the upside down parabola e k equals E valence minus h bar squared k minus k max squared over two m star whole.
And Mr. Hall, remember, is positive. So we have a negative curvature of the top of the band.
And then we have to think about counting states. So counting states.
And we'll think back to when we did Sommerfeld theory for free electrons.
For free electrons. We had our well, our energy was e equals H per squared K squared over two M and the density of states.
So the density of states per unit volume while g heavy was given by this expression which we calculated to mass.
So three halves are over two pi squared h bar cubed epsilon to the one half.
We calculated this, I think, on the second or third day of the term.
This is the density of states. Of states per unit volume.
States per volume. So we'd like to use exactly the same calculation, except for the conduction band and the Valence band here.
So without actually doing the work, it's pretty easy to see what the result is going to come out.
As the density of states in the conduction band for E greater than E conduction is going to be too massive.
The electrons star to the three halves divided by the same two pi squared h bar cubed and it is the same e to the two one half,
but it's e minus e conduction band. So you're measuring the energy with respect to the bottom of the conduction band.
The parabola starts at the bottom of the conduction band, but otherwise it's exactly the same calculation.
People happy with that OC again. So G Valence may be a little bit more complicated, but not too much more complicated.
R is two star hall r three has over two pi squared h bar cubed and here we measure are e minus e valence minus E to the one half.
And we're measuring going in the downward direction while we need this thing to the one half to be a positive number.
So it has to be e valence minus E we measure going down.
We get an increasing number of states as we go down to lower energy.
Good. All right. So then we can write in conduction band.
We just use our statistical mechanics. The density of electrons, density N of electrons of E's is given by N as a function of T,
that's an n equals the integral from the bottom of the conduction band up to energy infinity,
the energy density of states as a function of energy and the Fermi factor theta energy minus mu.
Okay, so the integral of the density of states is basically a sum over all eigen states.
And the Fermi factor is the probability that that eigen state is going to be filled.
So this is just counting up all the states that are filled at a given temperature, but we need to know what the chemical potential is.
So now we're going to. Well, let's write out the Fermi function here.
The Fermi function is one over e to the beta energy minus mu plus one.
Now if, if we're at a fairly high temperature or if the density is low or strictly speaking,
what we need is beta energy minus mu is much greater than one.
This means that NU is much below, much below, below ac meaning that are below by at least the amount of the temperature.
So, you know, fairly well below the bottom of the conduction band.
And that is pretty easy because the gaps between the conduction band and the Valence band,
there can be a couple of electron volts, much bigger than typical temperatures.
That means that an F can be approximated by well, you basically you can just drop the plus one and you get the just the regular Boltzmann factor.
This is just Boltzmann. Maxwell Boltzmann. Maxwell Boltzmann.
So let us put that in and we have and of t then is going to be integral to energy zero.
Very easy from the bottom of the conduction band up to infinity of g of energy.
And then the maximal Boltzmann factor here, the minus beta energy minus mu making that approximation and this temperature.
Then how do we do this integral? Well, first thing we're going to do is we are going to well, what are we going to do?
We're going to write out what this is.
So this has a to m star of the electron to the three halves and it has a two pi squared h bar cubed and is integral
the energy and conduction band to infinity times e minus e conduction to the one half and then E to the minus.
Beta energy is mu. So this is the one half that we have up there and we have to do this integral.
And the way we do this integral is by shifting out a factor of E to the minus beta c minus mu out frontal.
Then we're left with an integral D energy from AC to infinity D minus easy to the one half, and then the minus beta E minus sc.
Which is nice. We can then write if we want to.
Well, x equals you want us to see if we want. So this becomes integral of the X from zero to infinity x to the one half, even minus beta x.
Then we can do that integral by defining y squared equals x.
This becomes a simple Gaussian integral, and we get better than minus three.
Half times square root of pi over two.
Okay. So we put that all together and we end up getting the density of electrons in the conduction band is a function of t is then putting together
all the factors one quarter two mass of the electron star kb t over pi h bar squared r to the three half e to the minus beta energy.
The conduction band minus mu. So this is our expression.
We now know how many electrons there are in the conduction band is a function of temperature and is a function of MU.
The important thing to take away. There are some pretty factors here which aren't that crucial.
The important thing here to take away is this basically activated that you have to activate the energy up from the chemical
potential into the conduction band in order to get an electron into the conduction band as you drop the temperature.
The number of electrons in the conduction band is going to drop exponentially.
Okay. All right. So far, so good. Now we have to do the same thing for holes.
I think we can still do this in time. So the number of holes in the valence band.
Well, okay, let's write that. That's going to be integral from minus infinity to the top of the valence band.
The energy, the density of states in the Valence Band.
And then I'll write one minus the Fermi factor in the Valence Band.
Beta Epsilon minus new, minus me. Why is that?
Well, if the probability of an electron being a state is the Fermi function, then the probability of a hole being the state.
The absence of electron is one minus the Fermi function.
All right. So this factor here, we're going to basically go through the same manipulations.
So this is we can write this as a one minus one over E to the beta, minus new plus one that becomes, Oh, I'm running out of space, aren't I?
All right, well, this thing becomes, ah, this whole thing here becomes one over one plus either the minus beta,
a minus mu that maybe took two steps, but but you can check algebraically.
That's true. And again, we are going to assume something analogous to well, do I have more space somewhere?
Now I maybe I don't. So we're going to assume something analogous to what we assumed.
But the electrons will assume that that chemical potential is not too close to close to close,
meaning within the within the temperature, within the range of cavity of the top of the valence band E.V.
In which case this thing can be approximated as this whole thing here can be approximately equal to the Boltzmann factor.
Either the beta epsilon minus mu, much less than one.
Okay. So it is, again, more or less the same, the same story.
But here you have to realise that energy is less than you and you're activating a whole
downwards into the valence band rather than a electron up into the conduction band.
So plugging in the expression for the density of states in the valence band,
we get to our whole to the three halves to pi squared h bar cubed integral the energy
minus infinity to valence e valence minus E the one half in the beta epsilon minus mu.
And exactly the same manipulations are with maybe a couple of minus signs put in various places.
I think you're supposed to do it for homework, so maybe I won't go through it. And also I'm running out of time.
So this whole thing becomes, at the end of the day, a very analogous expression to what we found above is one quarter to Mr. Hall capped
over pi h by a squared all to three halves e to the minus beta mu minus e valence.
And again. It is the same story. You have some free factors here,
but the important part is that you have to activate a whole down from the chemical potential
into the valence band because it takes energy to push and hold down to the Valence band.
The holes want to go up as much as possible. So if we knew the chemical potential and we knew the temperature,
we can figure out the density of electrons in the conduction man and the density of holes in the valence band.
But now there's a trick. And the trick is to look at the product of energy, times p of t let us do that.
We get a bunch of factors. So one half out front kb t over a pi squared cubed.
Then there's a mass of an electron star, mass of whole star to three halves.
And then if you look at the exponential factors,
you'll notice that the exponential factors in the minus beta news and the the plus beta new will cancel.
So the only thing that's left over is even the minus beta e conduction minus e valence e conduction minus valence is egad.
So if you know the gap energy and you know the temperature,
you know the product of the number of electrons in the conduction band and the number of holes in the valence band.
And this, this law, very important law is known as that.
That is known as the law of mass action.
Law of mass action. If you talk to chemists, they know this.
They have a law of mass action. And the nomenclature comes from this, from chemistry.
Actually, I won't maybe go through why it is it's called the same thing is a law of mass action in chemistry.
But the important thing here to realise is that the product is independent of the
chemical potential and that means it is independent of how we have doped the system.
So that is a couple more things. If we have an intrinsic, oh, did I use that word?
I never used that word before. So intrinsic means undocked.
So if I have an intrinsic semiconductor, I should have probably used that word in the last lecture.
For an intrinsic semiconductor, the density of electrons in the conduction band equals the density of holes in the valence band.
Because each time you take an electron out of the valence manning,
you put it up in the conduction band, you created one hole and you created one electron,
in which case you can just take the square root of the law of mass action and you can tell what and in PR at any temperature.
So you are good there. But for a doped system, doped then is more complicated.
Then we add doping and are.
We have the number of electrons in the conduction band,
minus the number of holes in the valence band equals the number of donor impurities impurities minus the number of acceptor impurities.
Because basically each time you add a donor impurity, you add one electron into the conduction band without adding a hole into the valence band.
I recently had an acceptor. You add a hole into the Valence band without adding an electron into the conduction band.
And this is true. As long as as long as T is greater than T frees out.
Where the electrons in the conduction man get get down to their donors and the holes in the in the valence band get bound to their acceptors.
So the key thing to realise here is that the product of LMP is fixed independent of the doping.
The difference of N.A. depends on the doping. This means that as you add electrons, as you add donor impurities, the chemical potential goes up.
So you get more electrons and fewer holes. But the product of the number of electrons in the hole stays fixed.
As you add more acceptors, the chemical potential goes down, you get more holes than electrons.
But the product of the Tuesdays suspects maybe even right that before calling it quits, adding donors.
New goes up. New goes up where adding acceptors, acceptors mew goes down.
But because of the law of mass action, the product of those two is always going to stay fixed.
All right, so next time,
we'll pick up and actually talk about some real semiconductor devices and the devices that change the world, like the transistor.
And I'll see you I think Wednesday. Wednesday. I think it's Wednesday.
Okay. Have a good weekend.