[Slide 1] Welcome back to Unit 3 of our course
and this is fourth lecture.
[Slide 2] Now as you may recall in the second lecture we tried to show
that if you use the diffusion equation and use this new set
of boundary conditions you'll get results
that we have been discussing in this course
that includes the ballistic resistance.
What we did in the last lecture is we introduced this concept
of quasi-Fermi levels the idea that right-moving
and left-moving carriers have different electrochemical
potentials or quasi-Fermi levels
and we just argued what the correct boundary conditions are
for these quasi-Fermi levels and what we want to do,
one of the things in this lecture is
to connect up the two.
What I'll show is this implies that.
From this you can deduce that.
Now in order to get from here to here we need a relationship
which is very important in its own right.
That's this relation that relates the current
to the separation of the quasi-Fermi levels
and this is ballistic conductance in between.
Now often there's some misunderstanding
because when you see the ballistic conductance you think
oh this only applies
to ballistic transport but not really.
This is a general relation that applies to ballistic
or diffusive transport and we'll come back to it later.
And if you take this you could turn it around
and write it in this form.
So this follows algebraically from here and what it says is
that the separation
of quasi-Fermi levels is proportional to the current.
That is, if there's no current the two quasi-Fermi levels will
collapse into one and that of course makes sense
because that tells you how well the right-moving states
are filled.
This tells you how well the left-moving states are filled
and if there's no current then they're all equally filled
and the current flow moves right-moving states are more
filled than the left-moving states and this is this relation
that expresses it quantitatively,
relates the separation of quasi-Fermi levels
to the current that is flowing and this is valid regardless
of whether it's ballistic transport
or diffusive transport.
So we'll try to explain where this comes
from in the next slide but before I go on I just wanted
to show that if you accept this then these quasi-Fermi level
boundary conditions give you the boundary conditions
for mu quite straight forwardly.
It just takes a couple of algebraic steps.
So let me just show that and then we can get
on to discussing where this came from.
So what you do is you first note
that we call this electrochemical potential
that appears here is actually the average
of the two quasi-Fermi levels.
That is mu plus, this is mu minus.
Mu is the average of the two.
Now with a little algebra I can just take this mu plus,
plus mu minus over 2 and write it
as mu plus minus the difference over 2.
So this you can check it out.
It's just a little straightforward algebra.
This can be written like this
and of course once you have this you could take this quasi-Fermi
level difference and express it in terms of the current
that makes user relation we just talked about which we haven't
yet proved but I will come to that.
But if you accept it then you say
that could be written in this the form.
And now you see we immediately have the boundary condition we
are looking for.
Why? Because if you look
at z equals zero the quasi-Fermi level boundary condition tells
us that mu plus at z equals zero must be mu1
so that mu plus becomes mu1 and this mu, that's this mu z equals
to zero and minus qIRB over 2 from there.
So this boundary condition automatically gives you that.
Just one line.
Now the other one, to get the other one what you do is take mu
and instead of writing it as mu plus minus this you write it
as mu minus plus this and that's also again something you can do,
straightforward algebra.
From here you could write it this way
and then this difference
of quasi-Fermi levels you could replace in terms of the current,
again make use of this relation
and once you're here the other boundary conditions
follows automatically.
Why? Because at z equals L mu minus is mu2,
according to the second quasi-Fermi level
boundary condition.
So mu minus is mu2 and then the plus q I RB over 2.
So the main point is what you obtained
in the last lecture gives these boundary conditions
that we discussed in the earlier lecture as long
as you accept this relation.
So what we want to do next is talk
about where this comes from.
[Slide 3] Now this relation in order
to obtain this we first write the current of electrons
that is flowing to the right.
See in general when we look here there's a flow to the right
and then there's a flow to the left
and the actual current is the difference between the two
but let's just consider the flow to the right.
So that's this I plus so how much is the flow to the right?
Well, there's the q which is the charge on an electron so what is
out here tells you the number of electrons per second
and if you multiply it by the charge
on an electron then you get the amount of charge per second.
So what is this quantity?
Well it's got two parts to it.
One is you look at the total number of right-moving electrons
because we're only interested in the right-moving current.
That is we're trying to figure out how many cars are
on the northbound lanes.
You're not worried about the southbound just the right,
northbound lanes.
Well what's the number?
Well first look at how many states are there and then look
at whether they're occupied or empty.
So this f plus is kind of like a Fermi function
but the thing is it just describes the occupation
of the right-moving states
and these are how many states there are which is the density
of states times the energy range of interest.
Why did I divide by two?
Because again I'm only looking at the right-moving states
so if there are a total of eight lanes,
there are only four lanes that are northbound.
The other four are southbound so dE divided by 2 times F.
So that's the number of right-moving electrons
and then I multiply it by the amount of time
that each electron spends on the highway going
from one end to the other.
That time is L divided by the average velocity
of right-moving electrons.
Now why is there an average?
Well, that's what we have discussed
in Unit 1 of the course.
That's because there's actually some electrons going straight,
some electrons going at an angle, some electrons going
at another angle and you need an average
which in the two-dimensional case I think amounts to 2
over pi, some small number but basically,
I mean a number closed to one.
But the point is this is just the average of all the electrons
that are headed to the right.
Of course the ones that are headed
to the left have a negative component of the velocity
and those you are not considering,
only those going to the right.
So that's this L over u-bar here.
Now why is the number times the time?
Why does that give you the current?
That's actually also not obvious unless you have seen
this before.
And here they often use the analogy
that if you had a graduate program
where you have say a hundred students graduate every year.
Then what you can convince yourself is
that if you ask how many students are there
in the graduate program, just steady state?
Every year a hundred come in and hundred leave.
The question is how many students in your program?
Well, the answer is let's say every student say spends five
years in the program then the steady state number
of students is 500.
So if a hundred students per year come in
and leave each student spends five years then the number
of students is 500.
So similarly here you want the flow of electrons.
It's equal to the steady state number of electrons
on the highway, that's like the graduate program,
times the time each electron spends on the highway
which is the length divided by velocity.
So this then with a little algebra we could just rearrange
it a little bit and write it in this form.
Du bar over 2L times f
so I'm just straightforward algebra getting from here
to here and then we note that this quantity is what
in Unit 1 we had defined as the modes as the number of modes,
M. See that's this M so you have q over h times M times f plus
so that then tells you the total current on the states
that are moving to the right.
[Slide 4] Now if you want the net current then what you have
to do is subtract off the current flowing to the left
and that will be exactly the same except
that now the occupation instead of being f plus will be f minus.
So overall the current
in the energy range dE is given by this quantity.
Now how do you get from occupation to quasi-Fermi level?
Well, that's where we use this general Taylor Series Expansion
idea that we have used many times in this course.
That is you assume this Fermi function can be written,
that this occupation can be written in the form
of Fermi function and then use the Taylor series
and so f plus minus f minus becomes this derivative
of f then times mu plus minus mu minus and if this is the current
over an energy dE and if you want the total current you have
to integrate this over energy
and that then gives you this current is q
over h times this part integrated over energy.
That's what you call M zero times mu plus minus mu minus
and the point I want to make is this again is the ballistic
conductance basically and so you then have the net current is the
ballistic conductance times the separation of quasi-Fermi levels
and note that this again holds at any point.
That is if you looked here the current would be that.
If you look somewhere else the current would be the same.
So for example, the continuity equation requires
that the current be the same everywhere.
That immediately tells you that the separation
of quasi-Fermi levels must also be the same everywhere
and as you can see this is a straight line.
That's a straight line.
They're parallel.
There's a difference but the difference is the
same everywhere.
And so the current is the same
and at any point the current is related
to the separation by this relation.
[Slide 5] Now how does this go with the conductance?
You see usually what we write is conductance time mu1 minus
mu2.
And this mu1 minus mu2 is of course the separation
between the two contacts.
Note that the contacts are separated by that much
but the quasi-Fermi levels are separated
by an amount that is less.
So current you can write either
as the conductance times the total separation
or you can write it
as the ballistic conductance times the quasi-Fermi
level separation.
This ballistic conductance is bigger
than the actual conductance
but then the quasi-Fermi level separation is smaller
than this overall separation between the two contacts.
So you could connect the two
and say the quasi-Fermi level separation must be equal
to the separation between the two contacts times G0
over GB and G0 over GB you could write as lambda divided
by L plus lambda where lambda is the mean-free path.
So the point to note is if you have a ballistic conductor
that means L is much less than the mean-free path
so you could drop the L and then the quasi-Fermi levels separation is
exactly equal to the applied voltage,
the separation between the contacts.
On the other hand if you had a diffusive conductor then L is
big and so the quasi-Fermi level separation is much smaller
than the applied potential difference.
In fact that's why in diffusive conductors people tend to forget
that there is a quasi-Fermi level separation.
They tend to say that there's only one mu.
You don't need to worry about mu plus and mu minus.
Forget about that one but as I say in a ballistic conductor
that can be as big as the actual applied potential.
So that then completes our discussion
of this boundary condition and what we'll do next
[Slide 6] in the next lecture let's talk about this Landauer formulas
which were historically very important
in developing this whole point of view and I want
to show you how it connects up to what we have discussed.
So far we have been discussing a channel
with a continous scattering processes everywhere.
That is you have cars that are continually turning
around as they go along the highway.
For if Landauer formula we'll consider a slightly different
situation, you see one where the channel is largely ballistic
but there's a localized scatterer someplace.
You know something like a little construction zone
or a toll booth in the middle of the highway
and that helps clarify certain things.
Anyway so that's what we'll do in the next lecture.
Thank you.