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Welcome back. This is now the 17th lecture of the condensed matter, of course.
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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.
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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.
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So let's start by drawing the bottom of a conduction band.
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So we have some E and some K and there's going to be some quadratic minimum of a conduction band.
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Let's call this point here came in where the bottom occurs and this point here, we'll call it E,
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not the energy of the bottom of the conduction band very generally already over here, very generally,
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we can we can write for an electron near the bottom of the band that E is approximately e
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nought plus which is the Taylor expansion D squared e d squared times k minus came in squared,
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maybe other terms which will ignore. And given this, we will define an effective mass.
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An electron effective mass. Electron effective.
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Mass effective. Mass.
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I'm a star. I'm stars of electron via one over am star electron.
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Equals one over a squared d squared.
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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.
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Now this looks very much like a free electron, except the momentum, the the wave vector should be measured with respect to the wherever.
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The minimum in the momentum is is the usual free electron parabola.
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Now given this, we can also calculate actually given any form of the energy of the of the electron in this,
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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,
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since we can replace omega by one over h bar times in times in energy d decay.
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This is the general relationship for the group velocity of an electron in a band.
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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.
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So it means that if you have this picture over here,
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if you if you move the electron to the right of the minimum, then the electron is moving to the right.
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If you move the electron to the left of the minimum, then the electron is moving to the left.
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This is true independent of where in the bronze down the minimum occurs.
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So really the issue is where your position with respect to the minimum,
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if you want to figure out which direction your group velocity is now, there are a couple of caveats here.
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Caveat, one are generally effective masses, right?
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This masses may be anisotropic, masses can be anisotropic, meaning that the energy is it may be parabolic in all directions,
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but depending on which crystal direction you choose, the curvature may be different.
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This is sort of natural. If you have something like a tetrahedral crystal where the axes are different lengths in different directions,
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you might expect that the curvature of the energy in the different displays in K in the different directions might be different.
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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.
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We're going to ignore this generally, so ignore this complication, but more generally you should be aware that it does occur.
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The second complication that you should be aware and it occurs is you can have situations if you plot E versus K,
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you can have a situation where there are two minima that occur and exactly the same energy.
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This E so that the minimum of the conduction band can occur at two places in the burn zone, K1 and K2.
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These are known as valleys. Valleys for obvious reason.
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This may look like it is something that is very unusual to have happen,
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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.
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But in fact it's actually quite common to have this be due to the symmetries of the crystal.
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So generally this will happen. If the crystal is particularly symmetric,
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you'll have to minimum in the conduction band at two different points in the bronze
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zone or even four different points or even six different points of of the barren zone.
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So you have to keep track of whether the electron is in this minimum or this minimum.
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But the general rule applies that if you displace yourself in the minimum to the right, you're moving to the right.
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If you replace yourself from to the minimum are to the left of the minimum, then you're moving to the left.
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Okay. Now, the final point. And generally, we're actually we're going to ignore this complication and also ignore this as well.
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Just be aware that it sometimes occurs. The final complication is perhaps a much more important complication.
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If you have a band structure, here's a broken zone.
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If you have a band structure like the tight binding band structures we drew before, it looks like this.
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The generally the effective mass is given by the second derivative of the energy with respect to K.
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So that would mean that the star of the electron here is greater than zero.
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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.
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That actually makes quite a bit of sense. And really we should think of the mass of the electron as being negative.
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The general definition that now so the electron one over h by square squared e.g. case where will always hold.
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And that is what we should always think of as the effect of mass.
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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.
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So the reason it gets hard is because minus signs will come back and bite you one way or the other.
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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.
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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.
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You have the even number of minus signs. At the end of the day, you got the right answer.
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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.
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And I think most books agree with me on this, but not all.
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So be warned about that. So this is the situation with holes at the top of valence bands.
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Top of bands. So so that is means up here where we found this odd result that the mass of the electron is
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negative when we're thinking about the top of a band like over there where we found this result,
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we're usually thinking when we're at the top of the band about a nearly filled band and nearly filled valence band.
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So we draw again E versus K, and we'll imagine that we have the top of some down,
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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.
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We imagine if we put an electron up at the top of this band somewhere, its energy is going to be okay.
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So for the case of Paul's top of band, top of band, the energy of an electron, e electron.
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It's really good to keep keeping your head track of whether you're talking about an electron energy or a whole energy,
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because you'll see that they're going to get confused in a second.
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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,
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e d k squared times k minus k max squared exactly the same way as it was before.
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But now this quantity here is negative. So this is less than zero.
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And this is why we should think of the mass of the electron here as being a negative quantity,
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because the energy actually goes down as you go away from the maximum.
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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
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of whole to be h bar squared over m hall r star is minus d squared e for the electron decay squared.
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So this thing here is greater than zero.
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At top of that top of band. And you'll notice from comparing this definition to the above definition,
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that the star of the whole is actually equal to minus the star of the electron.
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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.
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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?
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And why should it be the mass of the hall? Be positive at the top of the Valence Band.
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So let's think about this for a second. So here we have the same top of the Valence band.
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We imagine the whole Valence band is filled except for one hall, which we're putting at the top of the band here.
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So the energy of the hall e hall be some constant in our Twitter and then we want to find
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out how does the energy of this hole change as we displace it from the top of the band?
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So let's imagine this point here, just so we don't have to carry around these K axes and K means anymore.
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Let us decide this point. Here is cable zero as we displace K from.
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This is just for simplicity. It doesn't have to be a capable zero.
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But let's imagine is let us ask what happens to the energy as you displace the hall from the top of the band?
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So we're going to move the hall from here to here.
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Well, when you do that, the energy actually has gone up from this place in the hall from the top of the band.
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Why is that as squared to star hall times?
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I guess k squared here we're displacing it from Capel zero.
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The reason that the energy has gone up when we displace the hall from the top of
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the band is because the electrons are trying to go to the lowest energy possible.
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The holes try to bubble up to the highest energy possible.
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It's like is like an air pocket or a bubble or a like pushing a balloon underwater.
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If you take this balloon here and you push it underwater, that costs you energy.
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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.
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That clear, that is the intuition of why it is that you should think of the mass of the hall as being positive,
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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.
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So far, so good. Okay, good. All right.
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So that's the easy sign to keep track of, but the signs get more complicated.
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Let us ask about the momentum of a whole. What is the momentum of a hall?
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Momentum of a whole. And a good way to think about this is to realise that a filled band, filled band has no momentum.
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Has no crystal momentum. No crystal momentum.
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Carrying no current inert has no momentum at all.
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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.
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We fill the entire band like this, except for one state which is missing out of a state like this.
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We imagine taking this hole, this missing particle from the band.
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If we add it to a band, it's empty. Except for that one electron placed in.
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Then this plus this equals a fill down equals filled van filled with file filled band.
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So momentum. So total momentum equals zero. Total h party equals zero.
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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.
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Sorry, the filled band with the absence of this electron plus the empty band with the presence of this electron.
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When you put them together, their total momentum is going to be zero.
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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.
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So far so good. So that means that h pa k for a hall must be minus h bar k for the missing electron.
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Okay. So that is a good way to keep track of what the momentum of a horse should be.
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There is one of these confusing minus signs.
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Now let's think about a given. This what the velocity is of a hole at the top of are.
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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.
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Keeping track of it for an electron at the bottom of the band, everything seems very natural,
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but hole at the top of the band, everything starts getting confused.
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So it is important to keep track of these correctly. So let's think about velocity of hall at top of band.
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Top of band. So again, we start with the energy for the electron at the top of the band.
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So that is E not minus H bar squared, k squared K electron squared over two and Mr. Hall.
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Okay. Now why did I draw it this way? So Mr. Hall is greater than zero.
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The energy for the electron decreases as you increase the momentum of the electron away from zero.
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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.
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So this is group velocity for the electron.
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The electron decay electron is then minus h for k electron over Mr. Hall or since Mr. Hall is minus Mr. of the electron,
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we can write this is h bar k e over m star electron.
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But Mr. Electron is is less than zero.
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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,
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the velocity of this electron is actually going this way.
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The group is going this way, right.
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The MC because you can think of that as the electron mass is going negative is negative because of the negative curvature.
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So the group velocity is actually in the opposite direction as the wave vector.
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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.
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Let's compare that to the energy situation if we think about it in terms of holes.
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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.
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So the energy of the hole increases as you increase its momentum away from the top of the band.
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So the group velocity, the group hole equals one over h bar.
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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.
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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.
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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.
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This is an extremely important principle that will come back many times or at least several times, that the velocity of a state,
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the velocity of an orbital of an orbital orbital is independent of whether that orbital is filled.
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Whether the orbital is filled.
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So you can think of it as you have a bunch of orbitals in your system and the orbitals have some velocity,
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you can put an electron in that orbital and the electron will move that out with that velocity.
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If you remove the electron from that orbital, it is still the orbital is still moving with the same velocity it is.
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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.
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Even when we think about the we think about the hydrogen atom or something, we think about these orbitals being there,
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whether or not that there's an electron in the orbital and the orbitals can be moving with some velocity and
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they're still there and moving independent of whether we put the electron in or we didn't put the electron in.
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All right. So let us now actually try to take what we've learned and calculate some dynamics.
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Dynamics of electrons and holes. Dynamics of electrons and holes, channels and halls.
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And the simplest way to handle dynamics is what we started the term with, which is to think about the theory.
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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
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equations with a drag term p over tau and we can write that out for electrons.
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For electrons we will have the mass of the electron.
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I'm going to put a star here. Times V Dot.
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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.
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So it's the affective mass, the curvature at the bottom of the band that matters,
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not the bear electron mass, which is not an important quantity in a semiconductor or an insulator.
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We will have the force, the usual, the usual Lorentz force, which is always with you as a another joke.
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I never mind the Tao for the electron, the scattering time for the electron.
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So this should look fairly familiar. This is the Duda transport equation that we had before.
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There is some acceleration and mass times acceleration.
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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.
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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.
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However, near the top of the band, the sign of the mass gets to be negative, and that seems a little bit weird,
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but if we think about it, near top of band, near top of a band, we can consider the equation for holes.
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So we would have an m star whole times, the velocity of the whole same equation.
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But the charge on the whole is now plus e right?
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This again equals plus e, the charge on the whole is plus E.
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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.
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So we have a equation for the holes at the top of the band as well.
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Now remember that the we have two things here.
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We have two equations, one for the electrons, one for the holes.
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And the difference in the two equations is that the sign of the charge is different for the electrons in the holes.
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But also Mr. Hall was defined to be minus the star of the electron.
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So in fact these two equations are exactly the same if this equation has been multiplied by an overall minus one
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and Star Hall is minus M star of the electron and so is actually just repeating exactly the same JUDA equation.
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So this is actually quite important. It's exactly the same statement that I just wrote over here.
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The dynamics of the state are exactly the same.
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Whether you're talking about an electron moving the state or a whole moving the state, the absence of an electron,
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the orbital still moves with the same dynamics, whether or not the orbital is filled with an electron gun.
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So the thing that is funny about the equation in the semiconductors is if you're up at the top of the band,
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you have a choice of two things that both seem a little bit weird.
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One possibility is that you have to think about a negative mass for the electrons.
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If you're thinking about electrons in the state,
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the other possibility is you think about positive mass holes, but now these are positively charged particles.
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Now this resolves one of our all puzzles, all puzzle from the second day or the third day of term of the puzzle,
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which was why is sign of hall coefficient hall coefficient.
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Sometimes positive.
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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.
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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,
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then you have the theory for these electrons moving around,
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just like we thought about in Sommerfeld Theory or in Judith Theory at the beginning of the term.
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And you'll get a whole coefficient which is negative.
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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,
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then either way you look at it, what you think about it as holes running around with a positive charge,
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or you think about it as electrons running around, but now they're massive electrons negative.
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What you'll get is you'll get the opposite sign because now the things carrying the charge have the have the wrong sign.
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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,
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for some reason, the charge carrier had the wrong sign.
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And indeed that is what's going on in some materials. The charge is carried by holes rather than electrons.
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So if you have a semiconductor and you doped with electrons, you always get this sign of the hall coefficient.
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If you do it with holes, you'll get the opposite sign of the hall coefficient.
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Now for metals, real, real metals like calcium or beryllium or manganese or something like that,
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which are not just lightly doped where they, you know, we have some complicated band structure.
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It is less clear what the sign of this is supposed to come out as,
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because you can have a situation like we discussed with calcium in the last
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lecture where you have a partially filled lowest band in a partially filled,
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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.
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So the holes are trying to give you a positive hall coefficient.
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The electrons in the second band are trying to give you a negative coefficient of the two of them compete
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with each other and then you have to start worrying about which one has a greater scattering time.
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Okay, so it gets a little bit more complicated for real, for a genuine real metals like beryllium or or manganese or calcium.
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But in the case of a doped semiconductor, it comes out very clearly.
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If you have basically electron transport, you get this sign of the hall coefficient.
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If you have basically hole transport, you have the opposite side of the hall coefficient.
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So far so good. So we resolve this one puzzle.
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There is something else that I mentioned when we talked about due to theory all the way back at the beginning of the term,
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we remember that due to theory, works well, works well for semiconductors.
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For semiconductors. Why is that? Besides the fact that it can get the sign intelligently and the reason why the theory works
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particularly well for semiconductors is because usually when you're talking about semiconductors,
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you're talking about a very low density of charge carriers.
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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.
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Whether we're talking about electrons in the conduction band or holes in the valence band, the density is fairly low.
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And why does that help you?
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The reason that the theory failed back when we were studying it for the first time is because we neglected Fermi statistics.
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We neglected the fact that the electrons were fermions and that made us mess up all of these things, like the the heat capacity.
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We got that completely wrong because we neglected Fermi statistics.
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But if the density is really low, you're actually entitled to neglect Fermi statistics.
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Why is that?
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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,
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then you can just treat these as classical particles.
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Maximum Boltzmann particles can, because it's very unlikely they're going to sit in the same orbital in the first place.
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A certain better defined way of saying that is the Fermi energy.
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Given this density, the Fermi energy increases with density.
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If the density is low, the Fermi energy is low, if the Fermi energy is much less than the temperature, which would be true.
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If the density is low, then you don't have to worry about Fermi statistics.
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You can treat the particles as being classical. So.
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So particles are classical. Our classical and you can consider just regular classical due to theory classical Boltzmann
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dynamics maximal Boltzmann distributions without worrying about Fermi distributions.
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So this is yes, it is similar based.
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The whole yes world will come to that in a moment. But also you obey for me, statistics.
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You can only put if you think about it for a second. You can only put one hall in a particular case state.
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If you start with the Valence band filled, you can take it out once, but you can't take it out again.
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Okay. Yes. Holes of the Fermi statistics, they are fermions.
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All right. So this is actually bringing us to the to the realm of statistical mechanics or thermodynamics of our particles in in semiconductors.
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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?
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How did they get there and what are they doing once they're there? What is the effect of temperature?
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So one thing that I told you from the in the last lecture that I should be a little bit more honest about.
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We're going to go back to think about doping for a second and how doping is affected by temperature, doping and temperature.
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Okay. So let us recall that when we took a and opened and opened like, like a phosphorus box in silicon and they opened,
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we realised that phosphorus equals a silicon plus one proton plus one electron.
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And I told you the electron goes into the conduction band, into conduction band band.
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And I told you to ignore the fact that there's an extra proton ignore.
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Now this issue of being able to ignore the extra proton that we put in, in addition,
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that is something that is actually we have to be a little bit more careful about and we have
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to worry about temperature and things like that to ask whether it's it's okay to ignore it.
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So think about this a little bit more carefully. We put in a proton and we put in an electron.
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So what is an electron and a proton do when you put them together?
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Which is really going to happen is you are going to form form a hydrogen atom or a hydrogen atom.
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Hydrogen atom. We put the electron in the conduction man, but we also put a positive charge behind.
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And the electron is going to bind to that positive charge is attracted to that positive charge.
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By Coulomb interactions, you form a hydrogen atom, but it is not a regular hydrogen atom.
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It is slightly different from a hydrogen atom because one m star not m not the mass of the electron.
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It is the effective mass of the electron, not the massive electron.
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The dynamics of the electron running around in the conduction man is given by M Star, not the bear mass of the electron.
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So that is the difference. One. The second difference is something you learned in A&M in the Coulomb interaction.
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The attraction back to the proton is is squared over for pi epsilon, not epsilon relative times r r epsilon,
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a relative dielectric constant relative dielectric constant, not the usual e squared over four pi epsilon not R.
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Okay, so the, the Coulomb interaction is, is reduced by the relative dielectric constant because you're living in a semiconductor,
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not living in the vacuum of outer space. So if you recall, we are regular hydrogen atom, regular hydrogen, regular hydrogen atom.
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There's a binding energy binding which is well, okay, you probably well,
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maybe do remember this formula is eight epsilon not squared Planck's constant squared,
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but you probably remember the magic number 13.6 EV, which is a red berg.
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So this is the binding energy of a, of a regular hydrogen atom.
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But in a semiconductor the binding energy e binding of the electron back to the proton.
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The additional charge that I put in with the doped is now m star electron.
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It is a fourth over eight epsilon relative epsilon not squared over h squared which I can then write as r m
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star of the electron divided by the mass of the electron times one over epsilon relative squared times 13.65.
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Now this expression can be a whole lot different from 13.6, and generically it's going to be a lot smaller.
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Typically effect of masses of electrons and semiconductors can be a third of the
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mass of an electronic can be even a 10th of the mass with electron or even smaller.
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The dielectric constant in semiconductors can be numbers like ten.
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So that means that this number,
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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.
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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.
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Now, why is that important? Okay. Oh, look at this.
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I need. That can be. With this in here, I'm going to draw.
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So here is the bottom of the conduction band. Here's the top of a valence band.
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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.
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Binding. A binding below the bottom of the conduction band.
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So you gain a tiny bit of energy by binding. Instead of just sitting with cake with zero in the conduction band,
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you gain a little bit of energy by binding to the positive nucleus that you came from.
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However, this energy here is very, very small and at room temperature, at room temperature, write this down.
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A tea room. The electron breaks free.
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Breaks free. And goes off into the conduction band.
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Breaks free and goes into conduction band.
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Is thermally excited into the conduction band.
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And this is why this is why it is okay to think about when we add our phosphorus doping to silicon.
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We really should just think a room temperature or higher temperature.
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We should think about this as just directly adding electrons into the conduction band.
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If you cool it down, if you cool this, maybe I'll write that down.
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So doping adds electrons adds directly to two conduction band.
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Or similarly, if you do the same thing with holes, when you add in the hole,
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it would form a a bound state to the to the negative charge on the nucleus that we left behind.
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We are putting in a boron or an aluminium or something.
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And at a finite temperature, that hole breaks free of its nucleus and runs off into the valence band.
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However, at low temperature, at low temperature, the electron becomes becomes bound, becomes bound again.
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Bound to the nucleus. To the open nucleus.
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Nucleus again. So if you are measuring the density of electrons in the conduction band, if you cool the system down a lot,
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all of a sudden you would discover that the density of electrons in conduction band was dropping
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because all of these electrons were getting frozen back onto their the nucleus that they came from.
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This is known as carrier freeze out. Carrier freeze out.
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Okay. All right. So, so far, so good.
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Everyone happy with this? All right, so now it's time to do some serious statistical mechanics thinking about how many electrons
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we should have in the conduction band and how many holes we should have in the Valence Band.
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This is a rather important calculation that actually shows up on exams very frequently.
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So it's a good thing to go through several times.
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All right. So let's start with this generic picture of the bottom of a conduction band, the top of a valence band.
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They do not have to have the same curvature. In general, they could have a different effect of masses of John.
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The effect of masses being slightly different. They don't have to be a line.
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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.
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We'll call this energy conduction easy for in conduction.
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We call this energy even for a valence.
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This energy gap is a D and e gap equals e conduction minus e valence.
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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.
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All right.
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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.
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So a couple of things we're going to need. One is what the spectra are.
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So for energy greater than energy conduction,
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we have energy as a function of K is energy conduction plus H bar squared minus K mean squared over to Mr. Electron.
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And then for energy less than E valence,
349
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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.
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And Mr. Hall, remember, is positive. So we have a negative curvature of the top of the band.
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And then we have to think about counting states. So counting states.
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And we'll think back to when we did Sommerfeld theory for free electrons.
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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.
354
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So the density of states per unit volume while g heavy was given by this expression which we calculated to mass.
355
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So three halves are over two pi squared h bar cubed epsilon to the one half.
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We calculated this, I think, on the second or third day of the term.
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This is the density of states. Of states per unit volume.
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States per volume. So we'd like to use exactly the same calculation, except for the conduction band and the Valence band here.
359
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So without actually doing the work, it's pretty easy to see what the result is going to come out.
360
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As the density of states in the conduction band for E greater than E conduction is going to be too massive.
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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,
362
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but it's e minus e conduction band. So you're measuring the energy with respect to the bottom of the conduction band.
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The parabola starts at the bottom of the conduction band, but otherwise it's exactly the same calculation.
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People happy with that OC again. So G Valence may be a little bit more complicated, but not too much more complicated.
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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.
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And we're measuring going in the downward direction while we need this thing to the one half to be a positive number.
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So it has to be e valence minus E we measure going down.
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We get an increasing number of states as we go down to lower energy.
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Good. All right. So then we can write in conduction band.
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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,
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that's an n equals the integral from the bottom of the conduction band up to energy infinity,
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the energy density of states as a function of energy and the Fermi factor theta energy minus mu.
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Okay, so the integral of the density of states is basically a sum over all eigen states.
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And the Fermi factor is the probability that that eigen state is going to be filled.
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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.
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So now we're going to. Well, let's write out the Fermi function here.
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The Fermi function is one over e to the beta energy minus mu plus one.
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Now if, if we're at a fairly high temperature or if the density is low or strictly speaking,
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what we need is beta energy minus mu is much greater than one.
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This means that NU is much below, much below, below ac meaning that are below by at least the amount of the temperature.
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So, you know, fairly well below the bottom of the conduction band.
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And that is pretty easy because the gaps between the conduction band and the Valence band,
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there can be a couple of electron volts, much bigger than typical temperatures.
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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.
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This is just Boltzmann. Maxwell Boltzmann. Maxwell Boltzmann.
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So let us put that in and we have and of t then is going to be integral to energy zero.
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Very easy from the bottom of the conduction band up to infinity of g of energy.
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And then the maximal Boltzmann factor here, the minus beta energy minus mu making that approximation and this temperature.
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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?
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We're going to write out what this is.
391
00:43:53,370 --> 00:44:03,150
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
392
00:44:03,870 --> 00:44:11,459
the energy and conduction band to infinity times e minus e conduction to the one half and then E to the minus.
393
00:44:11,460 --> 00:44:19,160
Beta energy is mu. So this is the one half that we have up there and we have to do this integral.
394
00:44:19,160 --> 00:44:29,700
And the way we do this integral is by shifting out a factor of E to the minus beta c minus mu out frontal.
395
00:44:29,700 --> 00:44:40,170
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.
396
00:44:41,580 --> 00:44:44,920
Which is nice. We can then write if we want to.
397
00:44:45,300 --> 00:44:56,460
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.
398
00:44:57,360 --> 00:45:02,050
Then we can do that integral by defining y squared equals x.
399
00:45:02,070 --> 00:45:06,120
This becomes a simple Gaussian integral, and we get better than minus three.
400
00:45:06,120 --> 00:45:09,420
Half times square root of pi over two.
401
00:45:10,730 --> 00:45:22,160
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
402
00:45:22,160 --> 00:45:39,590
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.
403
00:45:39,590 --> 00:45:43,310
The conduction band minus mu. So this is our expression.
404
00:45:43,490 --> 00:45:49,160
We now know how many electrons there are in the conduction band is a function of temperature and is a function of MU.
405
00:45:49,400 --> 00:45:53,420
The important thing to take away. There are some pretty factors here which aren't that crucial.
406
00:45:53,630 --> 00:46:03,650
The important thing here to take away is this basically activated that you have to activate the energy up from the chemical
407
00:46:03,650 --> 00:46:09,500
potential into the conduction band in order to get an electron into the conduction band as you drop the temperature.
408
00:46:09,740 --> 00:46:12,920
The number of electrons in the conduction band is going to drop exponentially.
409
00:46:13,560 --> 00:46:17,970
Okay. All right. So far, so good. Now we have to do the same thing for holes.
410
00:46:18,000 --> 00:46:22,790
I think we can still do this in time. So the number of holes in the valence band.
411
00:46:24,020 --> 00:46:28,850
Well, okay, let's write that. That's going to be integral from minus infinity to the top of the valence band.
412
00:46:29,270 --> 00:46:33,290
The energy, the density of states in the Valence Band.
413
00:46:33,710 --> 00:46:39,920
And then I'll write one minus the Fermi factor in the Valence Band.
414
00:46:39,920 --> 00:46:43,130
Beta Epsilon minus new, minus me. Why is that?
415
00:46:43,160 --> 00:46:49,370
Well, if the probability of an electron being a state is the Fermi function, then the probability of a hole being the state.
416
00:46:49,370 --> 00:46:52,460
The absence of electron is one minus the Fermi function.
417
00:46:54,880 --> 00:47:00,540
All right. So this factor here, we're going to basically go through the same manipulations.
418
00:47:00,540 --> 00:47:11,530
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?
419
00:47:12,460 --> 00:47:22,420
All right, well, this thing becomes, ah, this whole thing here becomes one over one plus either the minus beta,
420
00:47:23,050 --> 00:47:28,390
a minus mu that maybe took two steps, but but you can check algebraically.
421
00:47:28,390 --> 00:47:36,010
That's true. And again, we are going to assume something analogous to well, do I have more space somewhere?
422
00:47:36,080 --> 00:47:42,400
Now I maybe I don't. So we're going to assume something analogous to what we assumed.
423
00:47:42,670 --> 00:47:54,129
But the electrons will assume that that chemical potential is not too close to close to close,
424
00:47:54,130 --> 00:48:00,600
meaning within the within the temperature, within the range of cavity of the top of the valence band E.V.
425
00:48:01,270 --> 00:48:11,889
In which case this thing can be approximated as this whole thing here can be approximately equal to the Boltzmann factor.
426
00:48:11,890 --> 00:48:15,400
Either the beta epsilon minus mu, much less than one.
427
00:48:17,360 --> 00:48:22,099
Okay. So it is, again, more or less the same, the same story.
428
00:48:22,100 --> 00:48:27,170
But here you have to realise that energy is less than you and you're activating a whole
429
00:48:27,380 --> 00:48:32,570
downwards into the valence band rather than a electron up into the conduction band.
430
00:48:33,290 --> 00:48:39,950
So plugging in the expression for the density of states in the valence band,
431
00:48:40,280 --> 00:48:49,399
we get to our whole to the three halves to pi squared h bar cubed integral the energy
432
00:48:49,400 --> 00:48:58,880
minus infinity to valence e valence minus E the one half in the beta epsilon minus mu.
433
00:49:00,140 --> 00:49:08,150
And exactly the same manipulations are with maybe a couple of minus signs put in various places.
434
00:49:09,170 --> 00:49:13,160
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.
435
00:49:13,790 --> 00:49:25,340
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
436
00:49:26,450 --> 00:49:36,470
over pi h by a squared all to three halves e to the minus beta mu minus e valence.
437
00:49:36,920 --> 00:49:41,260
And again. It is the same story. You have some free factors here,
438
00:49:41,560 --> 00:49:46,810
but the important part is that you have to activate a whole down from the chemical potential
439
00:49:47,080 --> 00:49:52,899
into the valence band because it takes energy to push and hold down to the Valence band.
440
00:49:52,900 --> 00:49:58,270
The holes want to go up as much as possible. So if we knew the chemical potential and we knew the temperature,
441
00:49:58,420 --> 00:50:04,960
we can figure out the density of electrons in the conduction man and the density of holes in the valence band.
442
00:50:05,230 --> 00:50:12,400
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.
443
00:50:13,480 --> 00:50:24,280
We get a bunch of factors. So one half out front kb t over a pi squared cubed.
444
00:50:24,740 --> 00:50:30,190
Then there's a mass of an electron star, mass of whole star to three halves.
445
00:50:30,430 --> 00:50:32,410
And then if you look at the exponential factors,
446
00:50:32,620 --> 00:50:38,620
you'll notice that the exponential factors in the minus beta news and the the plus beta new will cancel.
447
00:50:38,830 --> 00:50:46,630
So the only thing that's left over is even the minus beta e conduction minus e valence e conduction minus valence is egad.
448
00:50:48,430 --> 00:50:51,340
So if you know the gap energy and you know the temperature,
449
00:50:51,490 --> 00:50:58,149
you know the product of the number of electrons in the conduction band and the number of holes in the valence band.
450
00:50:58,150 --> 00:51:04,480
And this, this law, very important law is known as that.
451
00:51:04,780 --> 00:51:07,780
That is known as the law of mass action.
452
00:51:08,440 --> 00:51:15,490
Law of mass action. If you talk to chemists, they know this.
453
00:51:16,090 --> 00:51:20,200
They have a law of mass action. And the nomenclature comes from this, from chemistry.
454
00:51:20,200 --> 00:51:25,750
Actually, I won't maybe go through why it is it's called the same thing is a law of mass action in chemistry.
455
00:51:25,960 --> 00:51:30,250
But the important thing here to realise is that the product is independent of the
456
00:51:30,250 --> 00:51:34,390
chemical potential and that means it is independent of how we have doped the system.
457
00:51:34,660 --> 00:51:39,550
So that is a couple more things. If we have an intrinsic, oh, did I use that word?
458
00:51:39,970 --> 00:51:46,510
I never used that word before. So intrinsic means undocked.
459
00:51:48,740 --> 00:51:54,080
So if I have an intrinsic semiconductor, I should have probably used that word in the last lecture.
460
00:51:54,380 --> 00:52:01,190
For an intrinsic semiconductor, the density of electrons in the conduction band equals the density of holes in the valence band.
461
00:52:01,190 --> 00:52:03,769
Because each time you take an electron out of the valence manning,
462
00:52:03,770 --> 00:52:07,400
you put it up in the conduction band, you created one hole and you created one electron,
463
00:52:07,670 --> 00:52:14,660
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.
464
00:52:14,720 --> 00:52:21,620
So you are good there. But for a doped system, doped then is more complicated.
465
00:52:22,580 --> 00:52:27,320
Then we add doping and are.
466
00:52:28,850 --> 00:52:33,079
We have the number of electrons in the conduction band,
467
00:52:33,080 --> 00:52:45,500
minus the number of holes in the valence band equals the number of donor impurities impurities minus the number of acceptor impurities.
468
00:52:51,180 --> 00:52:59,940
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.
469
00:53:00,240 --> 00:53:05,460
I recently had an acceptor. You add a hole into the Valence band without adding an electron into the conduction band.
470
00:53:05,740 --> 00:53:14,550
And this is true. As long as as long as T is greater than T frees out.
471
00:53:17,590 --> 00:53:27,310
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.
472
00:53:27,580 --> 00:53:35,110
So the key thing to realise here is that the product of LMP is fixed independent of the doping.
473
00:53:35,110 --> 00:53:45,040
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.
474
00:53:45,340 --> 00:53:51,190
So you get more electrons and fewer holes. But the product of the number of electrons in the hole stays fixed.
475
00:53:51,400 --> 00:53:57,219
As you add more acceptors, the chemical potential goes down, you get more holes than electrons.
476
00:53:57,220 --> 00:54:05,980
But the product of the Tuesdays suspects maybe even right that before calling it quits, adding donors.
477
00:54:08,140 --> 00:54:19,540
New goes up. New goes up where adding acceptors, acceptors mew goes down.
478
00:54:20,320 --> 00:54:25,570
But because of the law of mass action, the product of those two is always going to stay fixed.
479
00:54:25,930 --> 00:54:27,309
All right, so next time,
480
00:54:27,310 --> 00:54:32,920
we'll pick up and actually talk about some real semiconductor devices and the devices that change the world, like the transistor.
481
00:54:33,190 --> 00:54:36,520
And I'll see you I think Wednesday. Wednesday. I think it's Wednesday.
482
00:54:36,720 --> 00:54:37,480
Okay. Have a good weekend.