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We have been discussing the ohmic contact in our last lecture and towards the end we
touched upon some aspects- one is about the interface state density distribution, why
it is distributed across the bandgap we have discussed; and then we also discussed why
all those donor levels are towards the valence band side in the bandgap and all the acceptor
side levels are towards the conduction band, in the sense they are located above the donor
level. This we have illustrated with an example from gold in silicon because gold can be donor
or acceptor. There those deep donors are .3 electron volts above the valence band and
deep acceptors are mid gap.
With that we understood and we also began some discussion on the effect of interface
state density on a free surface in the sense if no metal is put, what will happen is what
we started on. I have drawn that energy band diagram here. Today we will discuss first
a-part on that then we will go on in the b-part on the I-V characteristics of schottky barrier
diodes, because this particular aspect what we discuss (a) is very important for a field
effect transistors in gallium arsenide, because you can have the devices which are not working
due to these effects.
Now let us take look at this diagram which illustrates the conduction band, valence band,
Fermi level that is n type material and we have the Dit distributed all through the surface
here. I just put this E0 here. Whatever levels are there above E0 they are all acceptors.
Once the Fermi level is in this position, all the levels below the Fermi level will
be occupied. I have just marked the region between EF and E0 with color, just to tell
you that those are the acceptor levels which have been occupied by electrons.
When acceptor level is occupied by electrons the charge is negative. The levels below E0
they are also occupied with electrons but they are donors. Donor level when it is occupied
the charge is 0, it is neutral. We have to consider that this region between EF and E0,
acceptor states below EF and above E0 are occupied by electrons. They are negatively
charged as a result of occupation of electrons. Correspondingly, once there is negative charge
in the surface it has come from somewhere, it has come from the bulk of the semiconductor
and in the bulk it can come from corresponding very close to the point, just near the surface
below the surface. If you look at the top of the surface that is the below the surface,
those charges have come from there. If those electrons have come from there it leaves behind
positive charges. What you mean by that is just below the surface there is a positively
charged layer, in the sense there is a depletion layer.
The n type material when it is depleted it has positive charge. It is depleted because
those electrons have gone and occupied the surface states here. I have marked the total
charge here will be Qit. This particular diagram what I have drawn
is actually the beginning. Just the moment this charges the electrons, there will be
depletion layer and the depletion layer charge will be equal to whatever charge in the space
charge layer. That correct diagram I have drawn next.
You can see if I draw it like this, energy band diagrams are flat which you mean there
is no depletion layer, but the very fact that there is depletion layer energy band diagrams
will bend and will be like this regular schottky barrier diode. You do not have metal there.
Without the metal you have got now a depletion layer formed here up to this point.
The red portion here is the depletion layer, this is the surface state charge Qit and this
is neutral region. Notice one thing, as I go back to the original diagram here entire
charge was here, but as it starts bending down, this point comes down, along with that
the Fermi level in that portion comes down. What happens is as the Fermi level comes down,
the occupants here goes down, because levels below EF and E0 are occupied. And depletion
starts building the Fermi level comes down and less number of charges are occupied.
This process takes place still there is equilibrium between this charge in the interface state
and depletion layer. Till that time that band bending goes on taking place. Like that the
electrons are transferred from the semiconductor to metal or metal to semiconductor. Here the
electrons are transferred from this bulk into the surface till the charge here Qit is exactly
equal to the charge here and charge in the depletion layer is equal to per centimeter
square of course, you are putting q into ND is the donor concentration into the depletion
layer width WD is the width of the depletion layer and that is positive. And that positive
charge is exactly equal to negative charge within this portion that is that region which
is having Qit.
Now Qit is actually equal to EF minus E0 into qDit, qDit is number of states per centimeter
square per electron volt, still we are keeping it number of states per centimeter square
multiplied by EF minus E0 you get that. This is exactly similar to what we did in the case
of schottky barrier. All that has happened is as the equilibrium takes place till the
charge here is equal to charge here. Now you can see for a given doping concentration
what will happen. Here as this Dit goes up for a given q ND WD, EF minus E0 will become
see, as qDit goes up if I am keeping this constant it is not a constant slightly varying,
if this is constant I am just giving as an example if this is constant that is q ND WD
if that is constant when Dit is large, if it is more in the surface EF minus E0 will
actually be smaller, that means actually higher Dit values the Fermi level gets closer and
closer to E0 because a smaller amount of gap between the two is sufficient to supply those
charges. EF minus E0 into Dit, if Dit is large this
gap need not be large. When Dit tends closer and closer to infinity which is not realistic
but when it becomes very very large EF becomes almost matches with the E0 that is what is
known as Fermi level pinning. Fermi level tends to pin to the E0 level, it will not
be in exactly but it will go as close as possible to that, so that the small amount of gap there
between EF and E0 supplies enough some amount of negative charges to the depletion layer.
You can work it out in fact it will be a nice exercise for you to see what will be the potential,
see I have marked here as q phi I have not called it as q Bi. Bi is built in potential
this also is built in potential, but when you talk of free surface you talk of surface
potential. The surface potential is the potential here with respect to the bulk which is same
as the Vbi. Instead of calling it built in potential, we can call it as surface potential.
The potential drop across the depletion layer actually.
Now let us go back to this next diagram. I will just see few things for phi is the surface
potential that is what I have marked here. That phi and this is phi0 is the gap between
the neutral level E0 and EV. Now let us take look at further.
I am rewriting that. I am writing only the magnitudes now. I left out the minus sign;
magnitude of charge in the interface state q Dit into EF minus E0 is equal to q ND into
WD. What is WD? WD is depletion layer width and
that is decided by how much is the potential drop? The usual law phi is equal to q ND WD
square divided by twice of epsilonr epsilon0 very well known formula for depletion approximation.
WD is that. Now you can see the whole thing has hit equilibrium
that location of Fermi level has taken such that EF minus E0 is equal to that quantity
and that quantity is decided by the potential. You can now see, substitutes I have not written
that deliberately so that you can take it work out and see WD when I substitute from
this, what you get is root of q ND twice epsilonr epsilon0 into phi. On the right hand side
you will have a constant into root of phi and within the constant you have got ND.
For a given ND you have got a phi there. What do you have on the left hand side? Left hand
side has got EF minus E0. EF has changed its position such that square root of, when you
are substituting this, square root of q ND twice epsilonr epsilon0 into phi on the right
hand side is equal to that quantity EF minus E0. Can you write EF minus E0 in terms of
phi? But I would like you can actually sit back and workout this. What is EF minus E0?
We want in terms of phi, because ultimately the phi has built up such that EF minus E0
has taken the proper value. You write EF minus E0 in terms of phi.
How do you write that? I have not deliberately written this here so that you can have some
exercise to do with yourself. This quantity now here is the bandgap Eg minus
phi0. Bandgap Eg minus phi0 leaves you with total this thing plus this EF minus E0. Bandgap
Eg minus phi0 minus, this total quantity which is actually equal to q phi and q phi plus
Ec minus EF.
All that you have to do is on the left hand side EF minus E0 you can write from this total
width subtracts whatever is above which is q of plus Ec minus EF and also this phi0.
So what I am trying to point out is you get an equation, transcendental equation on the
left hand side EF minus E0 expressed in terms of phi on the right hand side this quantity
also expressed in terms of ND and phi. So the moment to fix up ND and the moment to
fix up Dit, you can find out what is phi by solving that equation.
I think I would urge that you would look into as an exercise because that is simple matter
to see. All that I am trying to point out from this particular analysis is ultimately
the EF tends to become closer and closer to E0 if Dit is large. What is the maximum value
of phi that you can get? Go back to that diagram. What is the maximum value of phi that you
can get? You get Dit tending to infinity. You can see when Dit tends to infinity EF
gets pinned on to E0. What you get will be actually equal to Eg minus phi0 is that quantity,
Eg minus phi0 minus Ec minus EF. See maximum value is when this merges with
that. This gap goes off. This phi will be equal to Eg minus phi0 minus that quantity.
You can figure it out very easily. It is just what you have taken a look into thing. That
is what I am trying to point out is whatever be the Dit your phi will never become equal
to bandgap it will be two thirds of Eg by q.
In an example where the Dit is infinite or very large the potential there you can easily
immediately be known. The potential there is actually two thirds of Eg by q minus this
quantity Ec minus EF. If EF merges with Ec straight away phi is equal to two thirds of
Eg by q potential. Once you know phi you can actually find out what is the depletion layer
width is because phi is q ND WD squared by twice epsilonr epsilon0 standard formula.
With that I think I will leave this topic of Fermi level pinning or the effect of interface
state density on the barrier height in the schottky and also on the free surface. Please
remember that a free surface with high interface state density can lead to a depleted surface
below which has profound influence on the MESFET. When you go to that we will see between
the gate and the source there can be depleted layer due to free surface. Just let me put
that down so that without leaving you on guessing. For example, if I have
a semi insulating gallium arsenide, all our effort was to realize a MESFET. And then I
have an n type layer gallium arsenide and here what we will do is you will put a schottky
barrier that is a metal gate.
Now you will have ohmic contacts on two sides which will make n plus and metal. You will
have a contact here like this. Notice I am putting it buried inside because you alloy
it n plus into that and on that you have got the metal, so that is the source. Similarly,
I have got drain here and this is the gate, this is the free surface that we are talking
of. In the example that we have considered we
first discussed schottky barrier metal semiconductor contact, we saw the impact of the interface
state density on barrier height. We will see how the barrier height effects the current
or blocking capability soon qualitatively we have discussed we will have quantitative
discussion now. These we have discussed make it n plus to make ohmic contact.
Here this free surface what will be the effect of that? This is n type free surface if you
allow without any passivation if there are infinite number of tangling bonds you will
have depleted layer below that. If you have depleted layer below that you can immediately
see the difficulty. What is the difficulty? Suppose in this, depleted layer comes all
through like this, what happens? Source is disconnected from the channel. That is the
problem that we will face in a MESFET.
This one of the difficulty is that people were facing when making this type of device,
particularly, if this layer become thinner and thinner because after all as we said the
potential here is equal to two thirds of Eg at the most. Correspondingly, how much is
the depletion layer depends upon the doping here, so if I have a thickness only that much.
May be you can say there is still some opening path is there it will lead to increase in
resistance, but if the thickness is this much thin of the layer, entire layer is depleted.
There is no path between source and the channel. The MESFET would not work. It is like just
like the MOSFET without an inversion layer connecting the source and the channel. This
is why depleted layer I will just remove that from there let me put it somewhere else. This
is the depleted layer due to Dit that is all I want to point out right now.
Now we will get back to our discussion on the schottky barrier as I have shown in the
first slide itself, what we now do is, take a look at the conduction mechanisms or the
I-V characteristics of schottky barrier diodes.
This is the metal and this is the semiconductor on this side. As I pointed out earlier we
will be discussing only n type substrate because we are interested in electron transport. Here
this is an n type semiconductor and conduction band C.B valence band V.B all this abbreviated
because later we will say CB and VB and semiconductor SC these are abbreviations I am following
to make it easy to project it here. The schottky junctions we call it also schottky
barrier metal semiconductor contact at the schottky barrier or even schottky junction
with a metal with the semiconductor. You can see here these circles are the electrons.
These open circles are the holes in the valence band. Now the current transport, there are
different mechanisms by which electrons can be transported from the semiconductor to the
metal. From the metal it can be transported if the electrons have energy above that. That
we have already discussed that is called thermionic emission.
Thermionic emission means if you heat up or at room temperature several electrons will
have energy over and above this barrier. If you look into the old vacuum tubes, I do not
know whether I told you the example, if you take the vacuum tube what is the source of
electron there? It is a heated filament. A filament is heated, sufficiently heated red
hot so that it emits electrons. If you take a metal and then if it is in vacuum
there is a barrier that is zero level and what as you have said that is phim.
If I heat at room temperature, if you take a metal no electrons are not coming out of
that. Electrons are staying within that because they have the energy like this, but I have
barrier here like in the case of schottky barrier, these can cross. If I do not have
anything only vacuum these electrons energy have to rise up to that.
If it is raised up to that if I heat up them at metal. It is not the schottky barrier that
I am talking of; it is the vacuum tube actually. You heat up the filament, this entire distribution
will become like this acquire kinetic energy. They can be emitted into the space. In vacuum
tube what they do is you have evacuated tube in which there are two electrodes and just
close by to the cathode we have this filament is heated.
When I apply voltage the electrons which are emitted can be connected by the plus. So you
have the current flow. That is the way the vacuum tube works. In case we do not have
chance to talk of vacuum tube these days though that was the main device in olden days, that
is the transport of electrons from here to here.
From semiconductor there are different mechanisms, one that we have been talking of transport
of electrons across the barrier. I have written that down here, that is the thermionic emission,
the A that I am putting here, that A is the transport of carriers across the barrier.
Transport of electrons from the conduction band CB of the semiconductor over the top
of the barrier into the metal that is called thermionic emission, that mechanism we are
talking about whichever has got energy above the barrier it can cross; second one hole
injection into a neutral region of semiconductor, that is this one B. Holes injected from here,
whatever vacancies are there, it will be injected. They will be transported into the neutral
region and you can see it beyond the depletion layer and in a neutral region it will become
It is just like in case of pn junction when you forward bias, in fact I have shown here
the forward bias device, and it is not thermal equilibrium. In a pn junction when you forward
bias you have hole injection from p to n. Similarly, in a hole injection whatever vacancies
are there, they are injected into neutral region they will recombine.
We are very silent about it so far. We have been talking of only this transport I have
been silent because we will see later on at that. That particular component is very small.
Another component that is present which is negligible in many situations, most of situations
is generation recombination C. Notice here, electrons reaching the depletion layer, holes
reaching the depletion layer recombination. In the forward bias case it will be recombination
if it is a reverse bias case what will be the situation? Electrons will be lifted up
from the valence band to conduction band and separated out. That is generation. That is
why component C is recombination or generation.
Recombination is in forward bias case generation in the reverse bias case of electrons and
holes in the depletion layer. B and C what we have discussed just now, both are same
ultimately, it is a forward bias case. The part B, go back to the diagram, here the recombination
taking place at neutral region, here C recombination taking place in the depletion region. In the
reverse bias case this is absent. This will be generation of current, opposite.There is
one more component here you see, which actually is very close to this top edge electrons which
are having energy not above the top of the barrier but very close to that barrier edge.
Those electrons find a width of that barrier small. As you go from bottom to top, I will
just draw that because it is not very clear effectively there.
Suppose I have a diagram like this and if I have this schottky barrier like this, I
am just drawing it to show it now, let me draw slightly differently; it is not like
heavily doped surface.
Barrier is wide here maybe 200 armstrongs, 300 armstrongs or even .1 micron but when
you go to the top what we are talking of thermionic emission is the injection of electron across
that barrier A. What we are talking of is the electrons which have energy here. They
find that the barrier width is very small. That is quantum mechanical tunneling.I think
the physics people frighten us by say using jargon term but it is actually the fact showing
a request solution gives you all the things. What I am trying to point out is you will
have very easy transfer of electrons from here to here across this thin layer. That
is called tunneling. In fact you can call quantum mechanical tunneling.
In fact tunneling actually simple term, the idea is when the layer is very thin the probability
in the terms of quantum mechanics, the implication is, if I have a thin layer the probability
of finding the electron here and here are the same, because the wave function of this
electron, because electrons is particle and also a wave, the wave function overlaps on
to other side, that is why easily it can be here or here; and if slight amount of force
is there it will find itself on other side.
That is the quantum mechanical tunneling. It is a very simple term. The probability
of finding electrons here and here are same, same in the sense it can easily move on to
that side because of the wave functions spreading into just neighborhood of wherever the electron
is. So to put it in further simple terms, it looks
as if there is no barrier here. That is called tunneling. Sometimes it is also called, as
the top one is thermionic emission, this one is called thermionic field emission TFE. These
electrons have enough energy because of thermal energy here and this is because of the slightest
field, it will push it across that and it can easily occupy the other portion. Because
if the wave function goes on to the other side also. Thermionic field emission there
is another name given to that. Hence you will hear terms like thermionic emission, thermionic
field emission. These are slightly different mechanisms. Right
now what we will see is, among all these three components B, C, D they are not contributing
much. The main contribution to the current will be from here. I just included these terms
just for the sake of completeness and also they will show up under certain conditions.
If the conditions are favorable they will show up. For example you have already seen
if the doping is high that component becomes very high; if the doping is high then the
entire layer becomes thin like that. Tunneling becomes very easy.
There is one example of telling when you provide a proper condition, other terms will also
start bearing its slope. Otherwise normally, for white band semiconductors like gallium
arsenide, silicon carbide, gallium nitrate etcetera and even silicon, your main component
is A, all that we have to see is that if I want to find out the I-V characteristics.
Let us see that. B is the quantum mechanical tunneling through
the barrier. What is that? D. Let me go through that. Thermionic emission generates the recombination
current, hole recombination in neutral region and that one is the quantum mechanical tunneling.
Among all the four, these are not very important and when they are absent, when this thermionic
emission only is present, it shows ideal characteristics; most of the time you get close to ideal characteristics.
That is what I said just now. The I-V characteristics of the schottky barrier diode are mostly determined
by thermionic emission that is process A that we have said.
Processes B, C, D are generally insignificant. When I say generally insignificant that is
generally within quotes, that is if you bring in provides certain circumstances which we
have occasionally see later. One of them I have already pointed out- if a barrier becomes
very thin you will have tunneling. Such junctions show ideal characteristics. I is equal to
I0, e to power of V by VT minus one. Now let us see a very quick analysis of this.
We come back to that diagram which we have been seeing often. Metal, depletion layer,
n type, this is forward biased, this line is Ec under thermal equilibrium conditions.
That is what I have shown here. EF is here.
When you forward biased this goes up, that is the dotted line and the Fermi level you
can call it as a cozy Fermi level EFN that also goes up. Telling you that you can apply
for finding out the total carrier concentration here we can use thermal equilibrium situation
and that will be practically equal to the dopant concentration. Entire area under this
curve there gives you total concentration n n zero which is virtually equal to ionized
dopants. When you forward bias it, this is lifted up and this particular region where
I have marked shaded there, this region, that region is having energy above that barrier;
all these electrons have energy above that barrier. Before thermal equilibrium that was
whole thing was down and now more number of electrons has the barrier energy more.
I am repeating some of the things of course but just get into that, write down equations,
that is the one if you have to find out how many electrons are able to cross the barrier
what we have to do is find out what are these quantities. You can see that is less than
the total quantity area under the curve. It is less than the thermal equilibrium electron
concentration. Over here, this region, this shaded portion there, those are the electrons
which can cross the barrier, which can give rise to current JMS in this direction due
to electron injection from here to here, JMS from here. That is this one.
Whenever I forward biased or reverse biased this quantity is not changing because this
phiBN is not changing. All that happens is here. This will give rise to JMS which is
J0 mutually you call that, whereas JSM is due to these electrons. How do you compute
the current? It is different from the way you compute the current in the simple pn junction.
That is the real difference between the two, why it is we will also see. Now let us see,
all that you do is supposing I can find out how many electrons are there above this barrier.
That is if I take a move from here to here, how many electrons are there? Here there are
the ones which are having energy above that.
As I move from the bulk to this surface, what happens to electron concentration? Keeps on
falling and the total concentration here is equal to the one which is above that. If there
is electron concentration NS at the surface, if they have a velocity component Vx in this
direction, due to their kinetic energy they are above the valence band. Due to their kinetic
energy, that is due to the temperature they have got certain velocity Vx. Suppose see
N is a concentration, V is the velocity, what is the current, N into V into Q, as simple
as that. All that you have to find out, if I want to
find out this JMS, that is the current due to the electrons injected from the semiconductor
to the metal, all that I have to find out is what is that electron concentration on
surface and multiplied by the velocity? This is simple ohms law that we use or simple laws
that we use, not merely ohms law. This is all that I have said now.
JMS the current component due to electrons crossing from the metal to semiconductor,
JSM is current component due to electrons crossing from the semiconductor to the metal.
Reverse bias condition, total current is actually equal to go back reverse bias condition the
whole thing will be pressed down, so none of these carriers will have energy to cross,
we have discussed this earlier. Only these electrons will have the energy
to cross that is JMS, see please remember JMS is in that direction due to injection
of electrons from here to here, metal to semiconductor. JMS is actually equal to J0. In reverse bias
condition you will have current due to the excess electrons which have energy from the
metal to the semiconductor that is J0. In forward bias condition JMS remains same thing
that is J0. J is actually J0 which is from which is flowing from semiconductor to metal
due to transport of electrons from metal to semiconductor.
And JMS is due to, I think this is other way I think there is a mistake here; just this
actually is JSM minus J0. Is it right?
That is JSM minus J0 because JMS is the one which is changing, what we have to find out
if I have to find out forward bias condition current is, find out JSM. Make V equal to
0 in the expression what will I get? J0. I do not have to compute J0 separately find
out JSM as a function of forward bias voltage find its value when V equal to 0 that gives
me J0. So that is the thing, JSM equal to JMS equal to J0 at thermal equilibrium. Let
us calculate that. I can quickly run through this now, because all the formulae are known
to you.
The only formula that you do not know is or you were not applying is JSM current due to
the electrons injected from the semiconductor to the metal is number of electrons or electron
density near the surface at x equal to zero. We can go back once. x equal to zero is on
surface, metal semiconductor junction at x equal to zero what is electron concentration
there? Multiply it by velocity that we are computing. Multiply it by the velocity component
that is Vx bar is actually the mean velocity component in the X direction. There is a standard
formula for any gas particle from the statistics that comes up.
ns is the electron concentration is that is per centimeter cube, number again one more
error there, I will just correct it right here.
These are all per centimeter cube how much it is same. At the very close to that surface
in the bulk the carrier concentration is that much having energy higher than the barrier
height on semiconductor side. That is what we are trying to see here. Current
is ns Vx and ns is per centimeter cube, density, having an energy higher than the barrier height
on the n side. What is the barrier height here?
When I apply voltage V thermal equilibrium it is Vbi, when I forward bias it is Vbi minus
V, supposing that is barrier height potential difference what is the ns?
This is the nn0 is the carrier concentration in the bulk. From the bulk to the surface
if there is a potential difference phi, the carrier concentration surface will be, whatever
here is present on the bulk, multiplied by e to power of phi by VT, q phi by kT that
is phi by VT, this is Boltzmann approximation. Carrier concentrations vary as exponentially
as the potential divided by VT. That is a well known law so I have just used that.
What is phi? Phi is that quantity, minus q Vbi minus V by kT. I am using minus V, minus
because of the potential, go back from here to here when I go, you have potential drop
is negative. If the carrier concentration is nn0 here, the potential carrier concentration
here is equal to nn0 into e to power of minus q Vbi minus V by kT because the potential
difference is Vbi minus V here. That is what we have done.
Very simple actually and once you have written this equation, if you accept that, that is
because it is a boltzmann’s law which you are have used, carrier concentration between
two points can be related by ratio of the carrier concentration is e to power phi by
VT, that is what I have used. The ratio of the carrier concentration here
and at the bulk is equal to e to power that. You have to use the sign properly. As you
go towards the negative potential the electron concentration low, that tells you that should
be minus. Even if you do not understand more than that, that is clear enough.
Now what is nn0? It is thermal equilibrium carrier concentration. That is related to
EC minus EF and density of states at the conduction band. This again is a standard formula with
semiconductors, nn0 thermal equilibrium value equal to Nc e to power of minus EC minus EF
by kT. What I do is I substitute those three into that and then ns becomes equal to that
quantity I have rewritten this, instead of nn0 I have put Nc e to power of minus EC minus
EF and this I have put within that bracket that becomes plus q Vbi because minus is there
and this becomes minus minus plus. What is this quantity now? If you take a look at this
quantity EC minus EF plus q Vbi let me just put that diagram.
Thermal equilibrium situation if you take, this is thermal equilibrium situation and
this is EC and this is EF. What is this quantity? q Vbi, and what is q Vbi plus EC minus EF?
phiBN that is what we have done.
What we have done is, we have substituted for q Vbi plus EC minus EF is equal to phiBN;
this is in electron volts; this is electron volts, electron volts; this also in electron
volts Vbi in volts. We have written that. That is what we have done. I have taken this
out all these three quantities together put it as phiBN, so e to power of minus phiBN
by kT and this minus and this minus together will become e to power of V by VT.
Now you can see the whole thing has become simple. These are the electrons which are
capable of crossing the barrier, so they are present at the boundary there and they will
be hitting the boundary at the velocity, hitting, impinging of the surface with a velocity V
of x in that component.
Current now is equal to q into V of x and current is this quantity multiplied by V of
x into q. That is what I have written there. Now what you do is, you have to just see substitute
for Nc, substitute for V of x, you get the equation for current. Nc is this quantity,
effective density of states is related to temperature and effective mass, twice 2pi
mn star kT by h square to the power 3 by 2, V of x is the mean velocity component. From
the assembly of particles statistics there from the energy considerations people have
arrived at the mean velocity component in that direction as proportional to root kT.
That is the quantity so I substitute this quantity and this quantity on to these two.
Root kT by 2pi mn star is mean velocity component, substitute that you get that.
That is all put together, 4pi mn star q k square T square by h cube e to power of minus
phiBN divided by kT e to power V by VT. Now the whole thing inside that is J0. Now you
see you have got the very famous equation J0 e to power V by VT. The whole effort has
been to find out what is this J0 quantitatively. You can see that.
Now let me just go one step further here. If I have put V equal to 0 what do I get?
That is JSM at V equal to 0 is thermal equilibrium value. That is exactly balanced by JMS. That
is JMS is same J0, see in thermal equilibrium V equal to 0 and JSM is J0 and JMS what comes
from metal semiconductor must be balanced by that. Total current is actually JSM minus
JMS. I have just subtracted J0 from here because one is in that direction other is in the other
direction. These are diode equations where J0 is given by this expression. Let us take
a look at that.
The J0 is this quantity whatever I have said just now. Now this is further simplified to
make life easy for us. You do not have to remember that big formula. The entire thing
here leaving the T that is a constant independent of temperature, you call it as the A star.
A star is actually Richardson constant, the person who called it by its name, A star Richardson
constant. The idea is this A star you still could have
used it as constant but you can see that there is mn star coming, that means A star depends
upon the semiconductor. For silicon it will be higher because mn star is higher, for gallium
arsenide it will be lower because mn star is smaller. That into T square into that quantity.
You can see that J0 depends upon barrier height and temperature plus this constant.
Let us now go further down because you do not have to remember even this. What are these
quantities? A star is actually this quantity 4pi mn star q k square by h cube. What I do
is you do not like to have that mn star there. You know always mn star by m0 because you
know for gallium arsenide it is .067 or so. That is why we multiply this by m0 divided
by m0. You get a constant independent of the semiconductor multiplied by mn star by m0
and this quantity turns out to be 120. I think if you remember that, that is sufficient.
A star is 120 into mn star by m0 ampere per centimeter square per k square, degree Kelvin.
Now you can compute A star.
Coming back to this, therefore J0 is A star T square into e to power minus phiBN by kT.
This is the one which decides what is the voltage dropped across this device. A star
is 120 mn star by m0, you have just now derived, in that quantity. For gallium arsenide we
are estimating now evaluation of J0, how much is J0 in gallium arsenide?
This quantity is mn star by m0 is .067; A star turns out to be 8.04. Once you put that
as 8.04 A star, T at room temperature 300 degree Kelvin 300 square, when you do that
you get 9 into 10 to the power 4 into e to power minus phiBN kT. I take phiBN as .85
electron volts close to that EF merging with E0, you get 1.24 into 10 to power minus 9
ampere per centimeter square. Let us see just remember this number it is
about 10 to power minus 9 amperes per centimeter square.
Silicon if you take mn star by m0 is .6; gallium arsenide it was .067, this is not very close
to 1, A star is 72. What was it in gallium arsenide? 8.04 here, it is 72.
Now let us see phiBN, two thirds of Eg I will take, we cannot have better than that. It
is about .75 then J0 is for this case is 72 is A star T square this quantity and e to
power this quantity is 6 into 10 to power of minus 7. What you say now? J0 in gallium
arsenide was 10 to power minus 9 because of better phiBN is one; number two A star is
smaller. It is A star T square e to power minus phiBN by VT. A star is smaller so J0
is smaller mainly because of the smaller effective mass, so you get that.
Comparison of characteristics of gallium arsenide you get schottky diodes to do.
We have just now found out J0 is 10 to power minus 9 in gallium arsenide, silicon it is
10 to power minus 7, now J0 is this quantity. Supposing I say J equal to 1 ampere per centimeter
square, corresponding voltage across the junction will be VT long J by J0 where J is 1 ampere
J0 is that quantity for gallium arsenide it is .513 volts, for silicon it will be, for
silicon J0 is higher therefore V will be smaller .358.
What you are telling is when you take a silicon schottky barrier and gallium arsenide schottky
barrier diode the voltage drop across the gallium arsenide schottky barrier diode for
a given current will be larger. In this example of 1 ampere per centimeter square it was .513
this case it is .358. Sum up here.
Comparison of gallium arsenide schottky and silicon schottky phiBN is .85 and .75that
itself increases this current; A star is 8.04 and 72; J0 minus 9, minus 7; forward voltage
drop at 1 ampere is that quantity. As a result we will get, I will just go one step ahead,
only compare these two. These two curves are the I-V characteristics of the schottky barrier.
Schottky barrier blue that is gallium arsenide about .5 volts for a current of 1 ampere per
centimeter square and silicon the forward voltage drop is smaller. Corresponding reverse
currents I have shown here, gallium arsenide reverse current is smaller compared to silicon
schottky, so these two, we will do the comparisons between the pn junction of gallium arsenide
with metal semiconductor contact of gallium arsenide and pn junction of silicon metal
semiconductor contact in silicon. Those things we will compare in the next discussion
because right now we have good idea about schottky barrier in gallium arsenide use a
cutting voltage of about .5 volts which is close to the cutting voltage of silicon pn
junction. It is almost like that. If I take a semiconductor which has larger
bandgap it will get even larger forward voltage drop. So with metal semiconductor contact
itself you can get quite close performance as good as pn junction.
With that I think I will close down today, we will continue on this discussion some aspects
of this and the non-idealities etc., in the next lecture.