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We
have
been discussing the I-V characteristics of schottky barrier diode and when we were doing
that we arrived at the ideal characteristics and we have discussed. I will just summarize
what we said in last lecture. We have seen, or experiments have shown that the I-V characteristics
deviate from the ideal characteristics. For example, the reverse current does not saturate,
it will keep on increasing. Forward I0 e to power of, instead of V by Vt, it is V by nVt.
That is less increases less rapidly with voltage.
So we just also said, we considered only this current that is the thermionic emission current.
B C D is the various terms which were neglected. Out of that, we have seen B and C yesterday,
or in previous lecture. That is a hole injection into neutral region and recombination, that
we said is not going to matter much, we analyzed it making use of law of the pn junction. Similarly
generation recombination current, it is much smaller compared to this particular current,
thermionic emission. So we do not have to worry as it is at least one or two hundreds
of magnitude smaller in gallium arsenide.
Let us see, the other term that we have put it here, which is the quantum mechanical tunneling.
These two terms are negligible Other terms which are there are: image force lowering
of the barrier, quantum mechanical tunneling and insulating interfacial layer. These are
the three things which we are going to discuss today.
Now image force lowering, we have just discussed fairly in detail what it is going to be? It
is due to the negative charge when it comes close to the metal, its mirror image is on
the other side. The charges induced on this plane can be the effect of that on this charge
can be obtained by considering the image of this charge there, its simplification, it
works out very well. Otherwise you have to take… you do not know how the distribution
is, but the entire effect can be accommodated by putting the mirror image here. That means
actually this charge experience the force, which is actually equal to q square by 16
pi epsilon x squared, where x is the distance.
As we go nearer this charge, the energy and the force both of them keep on increasing
negatively. That is why energy is integral of that. I am just going through this quickly
because we have discussed it yesterday. That is the energy integral of the force. q squared
by 4 pi epsilon x, instead of x square. Potential we are talking of the electronic potential
that is actually minus. So what we are just trying to point out is, if I move from here
to here, the electronic potential is that quantity, that is 16 pi epsilon or epsilon0
into x.
As far as this field is concerned plus there and minus there for this, negative charge
and plus charge. So field is in that direction, it moves the energy of the electron lower
negative as you go towards the metal surface. That is why we are actually talking of electronic
energy and electronic potential. The energy level and electron potential will be following
the same path. If energy level is dropping down like that, the electron potential will
also drop down like that. These are lower energy for electron. In fact if you recall
the 0 energy is there and all other energies are negative. So, that is what we have plotted
there. If you look into that, that is the one which
is going down in this fashion. Now in the absence of image force potential, this is
the image force potential psi, we usually plot the energy band diagram like this. In
fact that is the variation of the electronic potential or variation of electron energy.
How to obtain that? You just have the depletion layer there. We call it here as eexternal
because this one is due to its image itself, but this field is due to this depletion layer.
If there is no depletion layer, this would have been only the potential. Because of depletion
layer present, there is additional external force acting on the electron that is actually
given by this. That is electric field eexternal which is given by epeak into 1 minus x minus
by wD. That is what we put here. That is electric field we call it as eexternal. This is regular
depletion field we are talking of. And psi potential is actually integral of that. In
fact in my last lecture I just left out this but this 2 is there. Now, what we do is, what
we trying to find out is, that there are two potentials, that is potential due to this
image psi if and the potential due to this depletion layer, which we called as eexternal.
The total potential is sum of these two. Now, what I have done here is to simplify matters
I just neglected the second term, this term. You can neglect the second term so long as
x by wD is very small compared to 1. We are actually interested in the potential variation
just near the boundary.
Because we are interested in seeing what is the effect of this image force potential and
the external potential, add this and this. The sum of the two is given by, if I approximate
the external potential by this way epeak into x, then, you have got these two terms. You
add up those two, that is, you can call it as delta phi that is the potential with respect
to that point or total phi itself, better call it as total potential. The potential
is sum of the two. Whatever approximation that we have made here will be true if this
xm is small compared to the total depletion layer width. We will see that is really true.
The peak occurs very close to surface in the order of 10 to 15 armstrongs compared to 0.3,
0.4 micron of depletion layer. That is why this approximation is quite good; we are talking
just… In this diagram it does not look so, but it is actually very small. What we want
to find out is what the total variation potential is? More than that, what we are interested
is what the peak value here is? Because after all maybe I can explain here, this potential
is varying like that. That is the external potential due to depletion layer. This potential
is like this. Total potential is this plus that that is this one. In this portion, suppose
you consider this as 0 and this is negative, this follows that and then if you go down
here that is 0 and this total potential is 0 here. That is why the variation is like
that. To find out this value here, what we do is add up the two and we know that it goes
to a maximum differentiate this term with respect to x and when you differentiate with
respect to that what you get?
You get, x is equal to x is xmaximum that is root of that, the simple differentiation.
So, this becomes square 1 by x squared, this x goes off. The location of this particular
peak electric field is actually given by this quantity. Now once you do that, what you want
to find out is, how much is this difference on , top to that point, which we call it as
delta phiBn.
Let me go to the board and show that to you. The actual potential, now instead of being
like this it is like this; It is like that. So, let me when we draw it slightly better
because that is very close by. This entire potential now, instead of being like this,
it will be now like this. Now, actually this quantity is delta phiBn. Why do you call it
delta phiBn? Because, this is the potential barrier which would have been present, but
now, because the barrier does not go all the way up to that point it goes only up to this
point. As a result, there is a reduction in the phiBn. This is called the image force
lowering effect of the barrier height by that amount. How much is that amount? That is same
as this potential. Please remember, the potential we are drawing, taking this as 0 with respect
to that how much is that thing. The psi that we put there is the psi with respect to that
because that was 0 there and that was psi image force. This is, if I take it as 0 that
is the potential. If I find what is the psi value here that straight away gives me delta
phiBn. How do you find psi value their? Add up the two and substitutes that.
Let me just write that down here. We have got psi is equal to minus epeak into x minus
q divided by 16 pi epsilon r epsilon0 into x. We have also seen that… So, psimaximum
will be what, when both of them are equal? See, one is coming like this other one coming
like that, the peak will correspond to the point and both are equal. That is actually
twice epeak into xm. Substitute value of this quantity we have found out the point at which
it is maximum. In fact we can verify that just going through that. That is the value.
Substitute for these quantity from the expression for xm. Let me not right down that, because
you already have wrote that. All that I am doing is we are substituting this xm which
we have determined as… by differentiating we have determined as this quantity.
So, substitute in that equation phimax is equal to phi c into xm. So, you do that, all
that you have to do is multiply this quantity by 2epeak. So, 2 goes inside this square root
that becomes 4. 4 divided by 16 and that is 4 here. You are multiplying this by epeak,
the square root of epeak is their below so, that becomes proportional to epeak. All that
we are doing is that. Hope it is alright. What we are doing is xm that is actually minus
2epeak into square root of q divided by 16 pi epsilonr epsilon0 into epeak. So this becomes
square root of that; this goes inside; this becomes 4. That is how we get that particular
term. The key thing to remember here is that this potential I called it as delta phimaximum
that is this quantity. What you have computed is psi; psi is the delta psimaximum. May be
unnecessarily I have introduced one more term there just differentiate this related to this
barrier. The barrier maximum reduction is delta that; that is delta phib the actual
barrier is this much. So if the actual barrier is this much, the electrons can go from one
end to other end very easily. The current can get affected that what we trying to point
out. Let us just take look at the effect of this particular barrier lowering.
The delta phiBn is a value of phi at xm, we have determined that is proportional to square
root of peak electric field. Now before we go into that, we have derived this by taking
the eexternal is epeak into x. Let us see whether it is valid at x equal to xm. For
that you must find out how much xm is. Let me just give you some numbers their.
How much is the value? Say let us just take this, xm is equal to square root of q divided
by 16 pi epsilonr into epsolin0epeak. All that we have to prove is what is the value
of epeak in the depletion layer? See epeak corresponds to that, this peak field that
will be roughly 10 to power of 5 volts per centimeter or 10 volts per microns. If you
go to 20 volts per microns you may have breakdown. That is why giving that number less than that,
but close to that; 10 volts per micron that is 10 to power 5 volts per centimeter. You
substitute on to this with epsilon taken as 12.8, 8.8 etc., this is approximately about
let us say 20 armstrongs. I can just plug in and see numbers. This is about 20 armstrongs;
2 nanometers. What I am telling is because of that, this peak occurs very close to surface;
this xm is very small compared to WD. So our assumption that, in the region up to this
point, this potential is linear is very much valid. By deriving at this expression for
delta phiBn, we took eexternal is equal to epeak into x. We are talking of that is only
up to x equal to xm. xm is just very small compared to depletion layer, so that is valid.
We are justified in doing that. You are not assuming that the electric field is constant
really but we have actually seen that potential is very linearly up to that point. Now other
thing that we would like to see is how much will be this barrier lowering? We should be
worrying about that. After all we have thrown out other numbers with the hole injection
etc, because they were all not comparable to J0 etc.
Let us take a look at
how much delta phiBn is in order of magnitude? delta phiBn is actually equal to root of q
epeak divided by 4 pi epsilonr epsilon0; this is right here. Delta phiBn is that number.
Now what you do is substitute for this epeak. It will be about 10 to power of 5 volts per
centimeter epeak you have taken that. A depletion layer width is something like 0.3 micron for
about 10 to power of 16 doping. For this situation, this turns out to be substituting all these
value something like 10 to power of 5 volts per centimeter. If I take substitute for epsilonr
is 12.8 for gallium arsenide; 12 for silicon; epsilon0 is 8.854 into 10 to power of minus
14 farads per centimeter. Substitute all that you get in this example about 35 milli electron
volts. That is about 35 milli electron volts. That is not negligible because after all the
J0 value that we talk off is proportional to e to power of minus phiBn by Vt. So if
that were phiBn0 and if it is 700 milli volts, this is 35 milli volts. We will say 35 milli
volts are small compared to 700 milli volts. But if you take e to the power of that quantity,
the currents are proportional to e to power of minus phiBn by Vt. If it is 25 milli volts
reduction, it is e to power 1 which is 2.7. So the current is different by magnitude of
2.7 for 25 milli volts. So even if we are talking about this 10 milli volts change,
that J0 will keep on changing. Because it is exponentially dependent on phiBn, so any
change in phiBn is reflected on J0.
So now whatever changes you see in the reverse bias and forward bias are manifestations of
delta phiBn that is effectively J0 changing. So what is done is let us see how it affects
in the case of forward bias? In the forward bias case what happens to delta phiBn? delta
phiBn depends upon epeak; if delta phiBn increases what about the phiBn? Go back to this diagram
and see, if this increases phiBn falls. In forward bias case what happens to epeak, as
I keep on increasing the forward bias voltage what happens to external state? It reduces.
When you forward bias the depletion layer collapses. So peak electric field falls. When
you forward bias and if it falls, delta phiBn reduces; if epeak falls delta phiBn reduces.
If that happens phiBn increases. So starting from 0 bias, we take 0 bias as some phiBn.
Let us take the J0 corresponding to that phiBn. From there it will either go up or go down.
What we are telling is even at 0 bias the value that you have is not the same thing,
it is different. From that point onwards and it will go up or down depending upon whether
you have forward biasing or reverse biasing. If forward bias electric field falls, therefore
this falls so that goes up and phiBn increases. So if phiBn increases what happens to J0?
J0 reduces. Now, usually we write the expression for the forward characteristics as J is equal
to J0 e to power V by VT minus 1. Now what is done here is to take into account the entire
characteristics. You keep J0 constant. Whatever changes that takes places is J0 you absorb
into that term. When you forward bias, the value of J0 actually what happens? It decreases
because barrier rate increases. That decrease in J0, if you put n is equal to 1 and you
have to keep on changing J0 and that decrease in J0 will reflect in J0 increasing as fast
as e to power of V by VT. So if I say J0 is changing, it is reducing. The implication
is current does not increase as much as it would be.
See for example let me remove that temporarily. If I write J is equal to J0 e to the power
of V by VT minus 1, I get a characteristic which is varying like this and J versus V,
if J0 were constant but if J0 is going on falling as we increase the forward bias because
peak electric field is falling. If J0 is falling slightly, for a given voltage, this current
will be lower. So, you will get actually the current which is slightly lower all through
cases. Strictly it is slightly deviated from this particular curve or slightly from ideal
exponential curve value. So this reduction is taken into account and you assume that
this is constant, absorb that into that factor n , this is sort of fur gee. You are just
cooking up a number n there, which will be greater than 1 and so current will increase
slower than what it would be from the ideal. The entire effect is due to J0 falling. But
you keep J0 constant corresponding to 0 bias and then say that there is a n which is larger
than 1 that is the non-ideality. This term is quite definitely more than 1 because the
delta phiBn from the 0 bias case is always 5 milli volts, 10 milli volts of that order
and that will go on changing. It may vary from 1 milli volt, 2 milli volt, 3 milli volt,
4 milli volts like that, but it affects totally. What about reverse bias? It will show up much
more.
Reverse bias and what about the peak electric field, it increases. The depletion widens,
it increases and if it increases we can see from here delta phiBn actually increases.
If delta phiBn increases, phiBn falls. So as we go on keeping reverse bias, the delta
phiBn goes on increasing and phiBn goes on falling. So this keeps on falling . It is
e to power minus and so if this keeps on falling, JR keeps on increasing. This will be much
more dominant in this case because if you have a 25 milli volts change in the phiBn,
which is quite possible because the peak fields are more increasing quite a bit, where reverse
current can be doubled. That is why we get the reverse saturation current, not saturating,
but keeping on increasing. So both the I-V characteristics in the forward direction and
also in the reverse direction instead of being like this, this curve will keep on increasing
not due to generation recombination but due to image force lowering effect of the barrier.
So barrier height keeps on falling. Now we are in trouble if you have devices made in
certain portion of device, if there are large fields reverse current will dominate quite
a bit. It will leak through those portions because of this image force lowering. Now
let us see what other thing is there on here. I hope this explains as a dominant phenomenon
for deviation from ideality.
This is just a formula which I have just put here. All that we have done here is delta
phiBn is q by this quantity and this whole thing within a square root sign is peak electric
field. This peak electric field is let me just put down here. This is for computing
numbers that is all.
This peak electric field is q ND WD divide by epsilonr epsilon0. Vbi minus V in the forward
bias case, this is negative. If V is reverse bias, this is adding; that is actually equal
to q ND WD square by 2 epsilonr epsilon0. What we do is substitute that from here. Therefore
this is equal to what? So,WD is the standard formula . That is WD. When you put it here,
epeak becomes square root of, substitute from here to there, it becomes equal to root of, this and this cancle there, so q
ND, into Vbi minus V divided by epsilonr epsilon0 into there is a two term. That term that is
2. So all that I have done is whatever I have written on the board there that is a square
root term. Delta phiBn is square root of q by this quantity into epeak that epeak is
this quantity and so the formula looks big but it is actually a simple formula where
we have substituted for peak electric field from the well-known law of junction. Then
when we will do that together it is a square root of 2 to the power 4 and all that comes
because this is square root of to the power of 4.
So delta phiBn is actually a function of V to the power of 1 by 4th and it does not vary
directly as V not as square root of V but it is 1 by 4th total potential. If it is forward
biased Vbi minus V and V is minus VR, Vbi plus VR that keeps on increasing. So, from
here it is very evident that, delta phiBn will keep on increasing if V is minus Vr,
if V is forward biased, this quantity goes on decreasing, delta phiBn goes on decreasing.
Decrease of course will be small in the forward bias case. The impact of will not be too much.
So you will get of ideality factor 1 you may get 1.05, 1.06 that is all we get. We are
making so much fuss about that but you really have to make fuss about it in the reverse
bias, because that current will shoot up. In fact we have seen when you do not take
care of some of these things when you make devises you get a schottky barrier which is
extremely leaky.
So now let us take one more phenomena because these two actually, the image force lowering
effect and the quantum mechanical tunneling effect join hands together to spoil the reverse
characteristics. Now if you recall what is this quantum mechanical tunneling? This is
the fourth phenomena that we are discussing now, the causes for ideality.
In fact by now you must have understood that the non-ideality is more dominating in the
reverse direction than in the forward. Forward, there is a factor n coming up and it is slightly
more than 1. It will not be 1, 2, 3 and all that is just 1.05, 1.06 of that order because
of these effects. Bu it can kill if you do not take care of those peak electric fields.
If there is field crowding somewhere, in the junction region, if there is any crowding
that epeak will go up, the crowding effect. That epeak goes up; the reverse current will
get affected drastically because barrier height will be reduced drastically. After all, barrier
height will get reduced more and more if the field is higher. So any field crowding effect
or concentration of field will reduce the barrier height in the reverse bias direction
and it will increase the leakage current drastically. Similarly this quantum mechanical tunneling,
this is also dominant mostly in the reverse bias. This is due to the electron which is
crossing this barrier, where it is thin. If you take the barrier, if it is thin here goes
on away from the junction, the barrier is becoming thicker and thicker. In this portion
if the electron has energy here, in fact there are a lot of electrons which have energy at
this portion from here to up there are electrons occupying. So, at this energy there is sufficient
number of electrons. Some of them can cross here because of tunneling effect. This is
quantum mechanical tunneling in the sense, there is a probability that good chance that
electrons can cross that. I am not getting into those quantum physics but that is what happens.
Now the width of this barrier depends upon the electric field. If the electric field
is large, this will vary steeply. The slope of this gives the electric field. Between
these two curves, this is at thermal equilibrium or forward bias; whereas, the second curve
is at reverse bias. So when the reverse bias is there, the width of the barrier becomes
smaller. So chance of tunneling is more if there is more reverse bias or the chance of
reverse bias is more, if the peak electric field is more. If the electric field is more
this is steeper. So whatever effect which makes the electric field more will decrease
the barrier width there and if it decreases the barrier width there, you will have reduction
over there. Let us just take a look at that is that. Is the point clear?
Finally, what we are telling here is, let us get down to the practical things now, so
both the effects together will effect if they join hands. Quantum mechanical tunneling is
dominant more and more when the field is high. Image force lowering will become more and
more if the electric field is high. So both of them become more and more when electric
field is high and definitely it is going to affect the reverse bias condition.
Let us look at the junction which is made like that. Particularly, if you take a n type
material and this shaded region here, that is the depletion layer. In this portion where
the metal is put directly below, in that portion there is a depletion layer which is flat.
Now, when you go to the edge, there is a curvature and the depletion layer width on edge is smaller
compared to depletion layer width here. It is much more so if there is an accumulation
layer here. What I put here is some accumulation layer here, which can be present. There can
be accumulation layer that is n becoming n plus there, if there are some positive charges
on the surface. It can be contamination or even if you have some oxide there, it can
give rise to positive charges but whether the positive charge is there or not. That
is, whether the accumulation layer is there or not, the depletion layer crowding effect
is there. Let me just draw that once here just to make it more emphatic.
What I am doing is, I have this n type semiconductor on which I just put this metal; that is a
metal here, which forms a schottky barrier. What we are talking of now is the depletion
layer here will be practically flat, if this width were going like this, then it would
have not been that much of a problem but there will be crowding here. Because field lines
will all is like this in this portion; whereas, from here, there will be need time crowding
there; whether you have an n accumulation layer or not there will be crowding effect
here. So now what we are telling is just like any
power device we usually see that wherever junction curvature is there you will get the
field crowding. Instead of junction, you have an abrupt ending here. As a result, there
will be crowding effect here, field crowding. The moment you have field crowding here what
happens? Epeak is high there. When epeak is high, both the effects come into picture,
the quantum mechanical tunneling due to that extra current and image force lowering is
more due to that J0 becoming high. Both of them join hands together to increase the current
here. So if you make a device like this, just put a metal on silicon and make a schottky
barrier you end up with the characteristics like this. Because of this crowding, you end
up with a characteristic which is very bad.
I-V will be something like that. It is a very highly leaky device, so you would not like
to see it or you would not like to show it to anybody. That is the state of affairs.
In fact I remember when I made my first schottky. I did it like that. After looking into the
theory of this what we realized is, you must cut down the field there. If you cut down
that field somehow near the edge of the depletion layer you can bring this back into this. One
of the methods that are used is shown here.
This is all what I have been explaining that is WDN here in normal direction is smaller
compared to this depletion layer width WDP all that is bringing in the crowding effect.
So, I said here WDN field is higher near the surface causing enhanced quantum mechanical
tunneling and image force lowering effect. In consequence n value in a forward bias mode
is function of voltage and reverse current in the reverse bias mode, JR or IR keeps on
increasing that is the curve we have shown there.
This is off course a typical curve that we get. What I have plotted here on the board
is shown here in typical devices that people have seen. You can see in this portion the
current keeps on increasing, initial to the image force lowering dominant. But when you
go to larger values of the electric field, larger reverse bias and the quantum mechanical
tunneling takes over. Virtually, it looks like a breakdown like in zener diode that
is what happening, it is tunneling. So this portion is quantum mechanical tunneling; this
portion is image force lowering effect. So both together is very bad news, in the sense,
you get a poor diode.
How to overcome that? The ideality factor n is bringing closer to 1. You do not have
to worry so much about the ideality factor, so what if it is just 1.01, 1.02 but what
you will be more concerned will be reverse current. They can be overcome by two typical
methods which are used in practice are: one is actually the guard ring, consisting of
a p plus layer put around that; other one is the field plate. Both of them reduce the
field kept in that portion. Both of them try to reduce the field here because that is the
one which is increasing and it is a cause of increase in the reverse current due to
barrier height lowering and due to quantum mechanical tunneling, which are due to the
thick electric field increase.I will just skip this for the time being. Let us come
back to this afterwards. That is the fifth phenomena which has some effect in the forward
characteristics not in the reverse.
This is the p-type guard ring or p plus guard ring. You can see the plot here shows the
metal here; this is the oxide; this is the metal put over top of that and schottky is
only here. This is actually a schottky which is shorting and the junction is shorted out
here through this hole. Now how does it help? As far as this diode is concerned, I will
just draw that on the board to make a bit clearer.
What you do is that the depletion layer is here, and I want to remove the crowding effect and so just
put this adjacent to that near p plus guard ring and both are connected together. If you
see the diagram, both are shorted together. When I apply voltage here, the depletion layer
is formed here. In the absence of this, it would have crowded down like this. There would
have been crowding effect. Now there is no crowing effect in this portion and depletion
layer moves like that. The depletion layer actually spreads like that. As far as schottky
barrier is concerned, it is being parallel and there is no crowding. All the field lines
are vertical. As far as schottky barrier is concerned there is no change in the peak electric
field, everywhere it is same thing governed by the one dimensional law. You do not have
the increase in the reverse current in that portion, in the schottky diode portion. You
will say there is crowding effect coming up here, but now the crowing effect is in the
pn junction. The crowing effect in the pn junction; the leakage currents in pn junction
are much smaller than that of schottky diode. So you have shifted the crowding effect from
the schottky barrier to the pn junction. In the pn junction, even if the field is higher,
leakage currents are much smaller than that. Here what we have done is, shifted the crowding
effect from the schottky to the pn junction. The current in the pn junction even when its
peak electric field is higher is much lower than that of schottky. When you do that you
get characteristics which are close to ideal flat. We have seen this. This is a real hard truth about this is the leakage
here is what was like that, due to leakage there has shifted up there. Other method which
is popular is… Team is to prevent the crowding taking place whatever be the force.
This is the field plate structure. If I terminate it here, I would have had the crowding here.
Now what I do is, take the metal over the oxide. See you have got a metal here and you
have got the oxide and this is the semiconductor. Now, because it is going over the oxide, this
portion where it would get crowding here, that voltage gets shared between the oxide
and semiconductor. If you see here very carefully, the depletion layer width is wide here and
it becomes narrower here like that. I will just draw that here for clarity further.
You do not do anything here; all that you do is having an oxide over which this metal
will go. So this is N-type region. So that is the metal region. This type of thing people
do with splits in power devices where you cut, where ever there is crowding. If crowding
is there, take the metal over the oxide. Now when I apply voltage here, reverse bias, depletion
layer will be here like this because the total voltage drops across the silicon; whereas,
if you go to this side, metal oxide semiconductor, lot of voltage drop into this depending upon
how much thick it is. So if I add the metal going all the way up to over here it would
have gone like that but because part of the voltage goes into the oxide this is shift
down here. Instead of crowding and coming like this, it is spread out. So depletion
layer actually is spread out like that, so that you do not have the crowding effect.
The moment you do not have the crowding effect there is leakage current here is reduced.
We have seen this also reduces current drastically. In fact this is simple to make. Growing oxide,
open a window, put a metal which is bigger than a window that cuts down leakage drastically.
This in practice you need to make provision for reducing the field crowding, which would
give rise to large current, particularly in the reverse direction due to quantum mechanical
tunneling effect and image force lowering effect.
So one last thing I want to discuss here is the Effect of 5, it is not so important but
it is being talked of sometimes. So out of the different effects which effect the ideality,
we have seen image force lowering and quantum mechanical tunneling these are the two which
really affect particularly in the reverse bias direction. Forward bias direction also
will effect because whatever affects this barrier height, additional current, that will
affect that. Now the other one is actually insulating layer thickness of delta. I have
a thin layer of insulator which is always present due to some native oxide etc that
causes slight problem. We have seen it at the beginning, the voltage whatever is present
gets shared between that insulating layer and the semiconductor, just like this case
. If I apply voltage here, that gets shared. What you are talking of is actually a layer
here; a thin layer present here.
Now let us say, all those effects are overcome by the build plate etc. You put a metal here
and what ever changes in voltage you make particularly in forward direction impact is
much more reverse direction and it is not going to affect because it is going to cut
down leakage current if at all in the region. In forward direction what happens is, when
I apply a voltage here v that gets shared between this and this. The voltage that goes
to this layer is very small, decause it is after all 5 armstrongs or 10 armstrongs. Its
current can tunnel through that and you have got some voltage droping across that. Now
what happens is, when I have a delta V, when I increase the voltage in the forward direction
by delta V, I expect the current to increase by e to power delta V by VT, but it would
not increase by delta e to power delta V by VT because that delta V goes completely to
the depletion layer. You will have e to power delta V by Vt, but part of it goes to 1 milli
volt out of 10 milli volts. The increase in current is not e to power of delta V by VT,
but it is delta V by n VT where n is greater than 1. That is the reason. If you have an
insulating layer there; you get a loose ideality factor to some extent.
In fact in the forward direction apart from the image force is lowering and quantum mechanical
tunneling, quantum mechanical tunneling to less extent that plays role in reverse bias.
More than these two terms, it is this layer which plays role. If you get1.1, 1.5etc, the
ideality factor, then you can say there is a thin layer definitely. If you get 1.01,
1.02, you do not have a thin layer. So this particular quantity V applied voltage is shared
between insulating layer Vi and the depletion layer VD. Delta V will be delta VD plus delta
Vi. If the insulating layer thickness is 0 delta Vi is small and you get ideal factor.
In fact I am not deriving this we can get explicit relations between to get the value
of n from the I-V characteristics. From the value of n, we can actually find out what
is the interface state density. I am not going through that because it involves bit more
derivations. May be for time being let me skip that thing. That is why the factor n
comes here. So, delta V is not proportional current to e to power delta V by Vt, but it
is less than that. That is why that n comes into picture.
Finally, what we have seen the summary of all is these non-ideality factors is recombination
generation current, it may have composite forward characteristics and non-saturation
in IR in some cases. If you take silicon, it can affect to some extent because the J
due to the thermionic emission and this, the difference is less. So it can become more
dominant or it will show up as you go to larger reverse currents. But in gallium arsenide,
you do not have to worry. Image force lowering makes ideality factor greater than 1 not too
much 1; too much more than 1, 1.05, 1.04 in that order and reverse saturation current
will not saturate. It will become worse due to the tunneling, quantum mechanical tunneling.
Because that comes up if there is field crowding is present. The peak electric field goes up
with reverse bias so that is why it is important only in the reverse bias conditions. So, both
the things together are very important in the reverse bias directions. That is what
I am trying to point out.
If you see the insulating layer presence, you have the insulating layer that will actually
affect the forward characteristics because some voltage goes into that. What about hole
injection? You do not have to worry at all because it is the order of magnitude is smaller.
But still some people have made out cases where they say may be if we go to very high
current densities, you may get some hole injection comparable to this one, that is the thermionic
emission. Otherwise you do not have to worry about it. I think with that we have completed
our discussion on schottky barrier diode, we have seen ideality and non-ideality everything.
The main thing that you would require is in the reverse bias operation how does it behave.
You should know how to cut down the reverse leakage current. Those techniques also we
have seen here like providing field plate, oxide and metal running over the oxide you
can reduce that current. So all that we have seen now will take on in the next lecture,
the three terminal device which uses the schottky barrier as the gate that is the MESFET. So
we will see next time on that.