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[Slide 1] Okay good morning now in lecture
two, we're going to continue our discussion of
the Landover approach. Mostly what I'm going to
be discussing in this lecture is this concept
of modes and what it means physically, but
we're also going to discuss some applications of
these formulas and then connections to experiments to
convince ourselves that this really does work in
practice and describes experimental situations. [Slide 2] So
just by way of a quick summary, we
have this simple model for a nano device.
The device itself is described by some distribution
of channels for current flow in energy, that's
M of E, and by some transmission, some
probability between zero and one that an electron
that comes in from contact one goes out
to contact two. Two large thermodynamic equilibrium contacts
but possibly with two different Fermi levels if
we have different voltages applied to those contacts.
We developed a simple expression for the current,
current is driven by differences between the Fermi
functions of the two contacts, and we also
simplify that expression for small voltages, linear response.
And in that case we got an expression
for the conductance that was given by this
last expression. [Slide 3] Now we talked a
little bit about transmission and modes, and we'll
get more comfortable with that as the course
goes, if the course progresses. I'll just remind
you that I postulated this simple expression for
transmission, you can see that this is a
number between zero and one, it's one in
the ballistic limit. It's very small, the ratio
of the main free path to the length
of the device in the diffusive limit, but
often these days devices operate somewhere between those
two events in what we would call a
quasiballistic limit, quasiballistic regime. And M is number
of channels and is given by an expression
which when you look at it, it's hard
to interpret physically, but if I'm computing the
number of channels from some given ban structure,
this is the way I compute it. It's
Planck's constant divided by four, times the average
velocity in the direction of transport, times the
density of states per unit energy per unit
area, times the width of the conductor. One
of our goals in this lecture is to
get comfortable with that formula and to understand
what it really means physically. [Slide 4] But
first let's take a look at just using
it. We'll assume that this equation is right,
let's look under low current, this might be
a lowvoltage between the drain and the source
of the mosfet, and let's calculate the conductance.
So here's our Fermi function, it's one way
below the Fermi level, it's zero way above
the Fermi level, it makes the transition right
around the Fermi level. What's important is the
derivative of that function, and notice that the
derivative is peaked and has a sharply negative
value around the Fermi energy. So minus the
F naught dE, the term that's important here,
is peaked near the Fermi energy and it's
something that I'll call the Fermi window, so
it basically picks out which of the energy
channels are important to current flow, the energy
channel are those that are inside this Fermi
window and that Fermi window has a width
of a few KT, so it's relatively broad
it at room temperature, very narrow at lower
temperatures. Now if I look at lower temperatures,
then I can make a very simple approximation,
this looks like a delta function, in fact
this has an area of one so it
has all the properties of the delta function.
That makes this integral easy to do, it's
just a delta function peaked at the Fermi
energy, so it tells us that the conductance
is 2Q squared over H, times the transmission
at the Fermi energy, times the number of
channels available for conduction at the Fermi energy.
So we get a very nice, simple expression
for the conductance. Valid at lower temperatures when
the width of this derivative is very narrow
and it acts like a delta function. [Slide
5] Now let's look at experiments, this really
does happen in practice. The physics was sorted
out in the 1980s and early 1990s, and
here's one example of an experiment. This is
a twodimensional resistor, here are my large Landauer
contacts, then its neck down to a region
where the width gets very thin. Turns out
that this width can be controlled electrically because
if I look vertically at the structure, there's
a layer of gallium arsenide where the electrons
flow. There's the layer of aluminum gallium arsenide
which is actually dope and the electrons spilled
from the aluminum gallium arsenide into the undoped
gallium arsenide, that's a technique that people call
modulation doping and it's a trick to get
very high mobility to very long mean free
paths. But around this and not shown in
the two sides are two shocky barriers, reverse
bias junctions that can complete the semiconductor, and
that's how these experimentalists electrically controlled the width
of the conductor. So if we measure the
conductance as a function of the width of
this constriction, we would expect that the wider
is the higher the conductance, the narrower the
smaller of the conductance, that's the way resisters
work. But when you do the experiment, you
find something very interesting. The conductance does increase
as the conductor gets wider and wider, but
it doesn't increase smoothly. It increases in discrete
steps and this conductance is plotted in units
of 2Q squared over H. So it starts
at some level, if you make it a
little bit wider it increases by 2Q squared
over H, another 2Q squared over H. So
the important point is that the conductance is
quantized in the steps of 2Q squared over
H. To understand this experiment, we would take
our expression because the experiment was on the
low temperatures so we'll take our low temperature
expression. The mobility in mean free paths is
a very long, so we'll assume the transmission
is one. So what must be happening is
that we must be changing the number of
conducting channels discretely. To understand this experiment we
need to understand M of E in a
little more detail. [Slide 6] So let's take
a look at that, that's our main goal
for this lecture is to get a physical
feel for what M is. So here's a
2D ballistic channel. You know electrons are free
to move in the xyplane, there's some magnitude
of velocity dV, moving at some angle, theta,
to the xyplane. Now if I ask, what
is the average velocity in the X direction
in the direction of transport? It's got to
be less than the magnitude of the V
because of a lot of them are pointing
at an angle, and if you just do
that angle average, it's easy to see that
the average velocity is two over pi, times
the magnitude of the velocity. So that's my
average velocity in the direction of transport. Okay
let's use that and see if we can
evaluate this quantity M, and figure out what
it is. [Slide 7] So we have the
average velocity in the direction of transport. If
this is a material with a simple band
structure, then one half MV squared is equal
to the kinetic energy, E with respect to
the bottom of the conduction band. So I
know what the magnitude of the velocity is
at that energy. If it's a twodimensional shape
with a simple parabolic band structure, I know
what the density of states per unit area,
per unit energy is. So I can simply
take these quantities, plug them into my formula
for M, and this is what I'll get.
It's beginning to look simpler, but we're still
not completely cleared as to what it is.
Well let's also remember that in a simple
band structure, then the kinetic energy E minus
the bottom of the conduction band, is H
bar squared K squared over 2M. Remember we
think of H bar K as crystal momentum,
so this is like momentum squared over 2M.
We get this simple expression for parabolic bands.
So there's a relation between E and K.
And remember that K is the wave vector.
So the electron wave is propagating as E
to the minus I, K, X. K is
the wave vector 2 pi over the wavelength,
I'll call that lambda B, the DeBroglie wavelength
of the electron. So if I solve this
expression for K, then I see that what
I'm solving for is the electron wavelength. So
let me take this expression, plug it back
into my expression for M, and reexpress M
in terms of the electron wavelength. Now we
get a really simple looking expression. The number
of channels is the width of the channel
divided by the half wavelength of electrons at
that particular energy. Okay that's beginning to look
like something we should be able to interpret
physically. So this is what the formula says,
it's the number of half wavelengths that fit
into the width of the conductor. [Slide 8]
So let's see how that works, this is
also why people refer to the channels as
modes, because they're like modes in a wave
guide. So if I think of my conductor
like a 2D waveguide, there's an electron wave
propagating along this 2D waveguide and the electron
wave is confined to be inside the conductor,
so the wave function has to go to
zero at the two edges of the conductor.
So we take our expression and I see
that there is one channel when the half
wavelength is equal to the width of the
conductor, and what does that look like? That
means that what I have here is an
electron wave where half of the wavelength fits
into the width so the wave function goes
to zero at both ends of the conductor
and satisfies the boundary condition. So this is
something that will happen at a certain value
of KY and it will give me an
expression that looks like this. [Slide 9] Now
let's take another look, let's look at our
expression. We can see that we have two
channels when the wavelength itself is equal to
W. Then when I plug it into this
expression I get M is equal to two.
When the wavelength is equal to W, I
fit an entire wavelength in and that also
satisfies the boundary conditions. That occurs at some
value of KY, but a smaller KY means
that there's a smaller wavelength that could also
fit in. So in this case I have
two channels at this particular energy. Depending on
the value or the direction that is propagating
to the xyplane, and depending on the value
of KY, I'll have a wavelength in the
Y direction that will have two possible conditions
that will satisfy the boundary conditions and most
give me two possible modes. [Slide 10] We
can see that a little clearer if we
look down on the top in the xyplane
and if I draw a circle of constant
energy for this parabolic energy bands. If I
have an electron that's propagating at some angle
to the X direction, there will be some
wave vector KY in the transverse direction. So
that will give me a wavelength in that
direction that is two pi over lambda one,
because I'm talking about the first KY. To
satisfy the boundary condition, if a half wavelength
is equal to W, I will have satisfied
that boundary condition. But staying on this constant
energy surface, an electron with the same energy
but with a higher KY, a higher KY
implies a smaller wavelength, it has the same
energy but it has a smaller wavelength. In
fact, if it has a wavelength of W
then the entire wavelength will fit into the
width, I still satisfy the boundary conditions. So
for this particular case I have two different
values of K in the transverse direction, both
of which give me wavelengths that satisfy the
boundary condition, so at the particular energy given
by this circle, I have two possible electron
channels that will satisfy the boundary conditions and
those of are two channels. [Slide 11] So
that gives us a simple physical understanding of
what's going on here, as you increase the
width, you increase the number of wavelengths that
can fit into the channel, each time we
do that we have one more conduction mode,
each time we get one more conduction mode
we get one more channel. And we can
only have a finite number of those because
there's a discrete number of KYs that give
me wavelengths that fit the boundary conditions. So
that's what leads to quantized conductance. [Slide 12]
So you're all familiar with density of states.
We will use the density of states to
compute the number of electrons in the device.
But now we've introduced another quantity, the distribution
of channels in energy, and we have an
expression for that. So we use M to
compute the current. We use density at states
to compute the electron density. And most of
you are familiar with what the density of
states versus energy looks like in a twodimensional
semiconductor, it's just constant effective mass over pi
H bar squared, but the distribution of channels
for the same case goes as the square
root of energy. So depending on whether I'm
working in 1D, 2D or 3D, I'll get
a distribution of a density of states and
a distribution channels that will have different shapes.
I'll point out here a very important assumption,
we're making an assumption here that the energy
bands are parabolic. If I have a different
band structure I'll get different results. [Slide 13]
So to recap we've been talking about the
number of channels, we have a formula that
given any ban structure any dimension, 1D, 2D,
3D conduction, we can compute this quantity, so
it depends on both band structure and dimensionality,
but it basically has a simple physical interpretation,
it's the number of half wavelengths to fit
into this 2D resistor or more generally into
a 3D resistor. And we have transmission, and
we have given you an expression that looks
plausible although we haven't talked about it in
a lot of detail, when we talk more
about scattering we will discuss more about transmission.
It depends on band structure because that's what
gives us this velocity, but also depends on
the physics of scattering. So that's something that
I'll ask you to accept for the time
being and we'll get more comfortable with it
later on in the course. [Slide 14] Okay
now just to conclude here, let's apply this
to a 2D resistor that might be something
like a channel of a mosfet till we
get warmed up for where we're going with
mosfets. So looking down on the top eye
of a contact with zero voltage ground one
where I apply a voltage, I have a
2D conductor. I have an expression for the
conductance, or one over the resistance, that's given
by our land hour expression in the small
voltage regime. [Slide 15] So at 0k the
ballistic conductance when the transmission is one, is
just these fundamental constants times the number of
channels at the Fermi energy EF. If I
want the ballistic resistance, I take one over
the conductance is just one of a number
of channels times those fundamental constants. Those fundamental
constants turn out to have a value of
12.8 kilo ohms. So fundamentally the resistance of
anything is 12.8 kilo ohms divided by the
number of channels at the Fermi energy. Okay
so we use this to explain quantized conductance
but in the bigger conductor, the number of
channels might be so large that it's hard
to count but still the same principles apply.
Important to note that the resistance is independent
of channel length, because the transmission is one.
Doesn't matter how long the resistor is, an
electron that comes in contact one goes out
contact two. [Slide 16] Now let's look at
T greater than zero. Still ballistics so the
transmission is still one. Now we have to
work out this integral. That involves a little
more math, and I won't get into that
today, but you can show the integral I
could write the result as 2q squared over
h, times the number of channels inside that
Fermi window. And when you do that math,
the expression for the number of channels inside
that Fermi integral or window involves a Fermidirect
interval order minus one half, so it's something
that we can compute. If I want the
resistance then I just take one over the
ballistic conductance, and I get the same sort
of thing, 12.8 kilo ohms divided now not
by the amount of channels at the Fermi
energy, but by the number of channels in
the Fermi window, same kind of concept. Still
the resistance is independent of link because we're
talking about ballistic conditions [Slide 17] Now if
we remove the assumption that we're ballistic but
begin with a T equals zero case, then
it I can replace this derivative again with
the Fermi function, what will be important is
the transmission at the Fermi energy and I'll
get an expression for the conductance that will
just be the transmission at the Fermi energy
times the ballistic conductance that we got before.
Or if I flip it upside down and
look at the resistance in the non ballistic
case it's one plus the length of the
conductor, divided by mean free path times the
ballistic resistance. So if I plot resistance as
a function of the length of the of
the channel, I'll get something that I'll find
out that the resistance doesn't go to zero
as a channel length goes to zero, the
resistance goes to this fundamental lower limit, this
ballistic channel resistance which is related to fundamental
parameters and numbers of channels. [Slide 18] We
could do the same exercise T greater than
zero, it would involve Fermidirect integrals, I'll just
leave that as an exercise, there's no new
physics involved in that. [Slide 19] So to
wrap up, these two lectures have tried to
get as comfortable with this Landauer approach. Current
is related to transmission, number of channels and
is driven by differences in Fermi functions of
the two contacts. Under low bias, current is
proportional to voltage and we have an expression
for the conductance. We have simple expressions for
the transmission as a function of energy and
for the number of channels as a function
of energy. When the number of channels gets
very small we can just count the number
of channels, and when it's larger we compute
them with this expression. Okay so we now
have the mathematical formalism to understand diffusive, ballistic,
and quasiballistic transport. We're not ready to apply
this to mosfets and that's what we'll do
in the next lecture. Thank you.