WEBVTT

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So, today we are going to talk about dopant
redistribution during oxidation. That is you

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know that the silicon is doped either p-type
or n-type, moderately or heavily as the case

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may be and when we carry out the oxidation,
what happens to the impurity? So, that is

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called the dopant redistribution effect. In
this context, first of all we will define

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a term called the segregation coefficient.
Segregation coefficient is something, I think

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once before also we have mentioned. In this
case, we are going to talk about the segregation

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of the dopant impurity, the relative segregation
in silicon and in silicon dioxide. Which way

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the dopant moves? Does it preferentially go
into the oxide or does it preferentially stay

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back in silicon? That is going to vary from
impurity to impurity.

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So, in that context we will define a term
called the segregation coefficient. This will

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be defined as the equilibrium concentration
of dopant in silicon divided by the equilibrium

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concentration of the dopant in silicon dioxide.
That is it is a ratio.

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This

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is defined as the equilibrium concentration
of the dopant, equilibrium concentration of

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the dopant in silicon, it is the ratio of
the equilibrium concentration of the dopant

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in silicon to the equilibrium concentration
of the dopant in silicon dioxide. This ratio

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is called, usually the symbol is K or M. As
this is a ratio, you can understand that its

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value can be either greater than 1 or less
than 1. What do I mean? When K for example

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say is less than 1, that means the equilibrium
concentration of the dopant is less in silicon

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compared to in silicon dioxide. If K is greater
than 1 that means the equilibrium concentration

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of the dopant in silicon is greater than in
case of silicon dioxide.

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Now, depending on these cases whether K is
less than 1 or greater than 1, I may have

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different possibilities. In addition to this
segregation coefficient, there is another

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factor to be considered. That is how fast
the dopant can diffuse through oxide. Remember

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that we said one of the most desirable properties
of silicon dioxide is that it can act as a

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mask against dopants. What do I mean? I mean
that most of the normal dopants that we use

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for silicon, they do not diffuse through silicon
dioxide as fast as they diffuse through silicon.

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That is the silicon dioxide blocks its movement,
silicon dioxide acts as a mask. So, for these

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dopants we can consider that they diffuse
slowly through the oxide and then there are

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some other impurity elements, for example
gallium, which will diffuse very fast through

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silicon dioxide. That is a reason why gallium
in spite of being a Group III element is not

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normally used as a p-type dopant in silicon,
simply because silicon dioxide cannot block

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its movement.

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So, you see now I have two facts to concern,
consider. One is the segregation coefficient

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and the other is whether the dopant diffuses
fast or slow through silicon dioxide. Depending

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on that I can have a variety of situations.

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For example concentrate on figure a. The legend
says K is less than 1. K, you know, is defined

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by the equilibrium concentration of the dopant
in silicon divided by the equilibrium concentration

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of the dopant in silicon dioxide. That is
less than 1 and I have also added rather cryptically

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a term called slow. That means the diffusion
of this impurity through silicon dioxide is

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slow. An example is boron, boron in silicon.
Boron has a segregation coefficient less than

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1 and boron is a slow diffuser through silicon
dioxide. That is why we can use boron as a

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p-type dopant. So, what is the situation?
Since K is less than 1 that means the dopant

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boron will preferentially go into silicon
dioxide, because its equilibrium concentration

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is more in silicon dioxide than in silicon.

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You see, this line is the interface between
silicon dioxide and silicon. At the interface,

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on the silicon side the boron concentration
is less than on the silicon dioxide side.

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Do you see? In other words, at the silicon-silicon
dioxide interface, there is a depletion of

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boron. Near the interface the boron concentration
is less. Do you see that? This is because

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of the fact K is less than 1. Now, the boron
profile inside the silicon dioxide. At the

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interface, this concentration should be higher
on the oxide side that is because K is less

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than 1, equilibrium concentration of boron
in silicon dioxide is high and then it will

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remain more or less constant over quite a
distance inside the oxide that is because

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it is a slow diffuser in oxide, it cannot
move very fast. Only towards the surface it

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starts to fall. Do you understand? This is
when the segregation coefficient is less than

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1 and the diffusion of the dopant through
the oxide is slow.

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Two facts you have to be very careful about.
That is at the interface, on the silicon side

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the concentration is less, signifying K is
less than 1. On the silicon side, the concentration

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of boron is less as shown by the red line.
On the silicon dioxide side it is more and

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this concentration remains more or less constant
over a considerable distance that is because

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the diffusion is slow and then towards the
surface it starts to fall. Let us look at

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a variation of this thing. Let us consider
that the segregation coefficient is still

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less than 1, but the dopant is a fast diffuser.
It diffuses fast through the oxide. A practical

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example is again boron, but this time we will
use a hydrogen ambient. If you use a hydrogen

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ambient, then boron can diffuse fast through
silicon dioxide. That is why normal boron

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diffusion is never done in hydrogen ambient,
but always in oxygen or nitrogen ambient.

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In hydrogen ambient, boron can diffuse fast
through silicon dioxide. That is my case b.

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K is still less than 1, but now instead of
slow, I am considering a fast diffuser through

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oxide. Again you see, if you look at the silicon-silicon
dioxide interface, because K is less than

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1, on the silicon side the boron concentration
is falling near the interface as shown by

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the red line and at the interface on the silicon
dioxide side, I have a higher boron concentration.

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Do you see? This is because segregation coefficient
K is less than 1, but because of the presence

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of hydrogen, now boron can diffuse fast through
the oxide. That is why you see it is, the

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concentration is falling very fast. It is
diffusing out through the oxide. That is the

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difference between these two pictures. In
the first case when the diffusion was slow,

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you have seen that the boron concentration
remains more or less constant in the oxide,

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only towards the surface it starts falling.
Here you see however that it is diffusing

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fast out, right.

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Let us now consider a case where the segregation
coefficient is greater than 1. A case is phosphorus.

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Phosphorus has a segregation coefficient of
K greater than 1 and phosphorus is also a

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slow diffuser through oxide. It stands to
reason because, phosphorus is a very common

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n-type dopant. It can be used because silicon
dioxide can mask phosphorus, its diffusion

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through silicon dioxide is slow. In this case
what will happen?

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Because K is greater than 1, you see on the
silicon side, I see a pile up of phosphorus.

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That is at the interface the phosphorus is
preferentially staying back in silicon. So,

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what you see here is that red line moving
up. In these two figures what did you see?

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The red line was moving down. That is because
K was less than 1. Now, it is moving up because

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K is greater than 1. It is commonly referred
to as a pile up, pile up, phosphorus pile

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up.

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Phosphorus is piling up at the silicon-silicon
dioxide interface. That is why on the silicon

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side I see the phosphorus pile up. On the
silicon dioxide side, you notice that at the

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interface, the concentration of phosphorus
is less on the silicon dioxide side. This

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black line starts at a level lower than the
red line at the interface, right and phosphorus

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is a slow diffuser in the oxide. Therefore
the profile of phosphorus in the oxide is

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also going to be somewhat similar to this.
That is it stays more or less constant over

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a considerable distance and only towards the
surface it starts falling.

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The important difference between a and c is
in the silicon-silicon dioxide interface.

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In one case, the boron concentration at the
interface is low on the silicon side. The

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phosphorus concentration at the silicon-silicon
dioxide interface is high on the silicon side.

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That is because of the difference in the values
of K, whereas the profile inside the silicon

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dioxide is dominated simply by the diffusion
factor and since both of them are slow diffuser,

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look at these two profiles, they look more
or less same inside the silicon dioxide, right.

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I now come to the final variation of the theme.
That is K is greater than 1, but it is a fast

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diffuser. I have already given the example
of gallium, gallium is an example of this.

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Gallium has a segregation coefficient greater
than 1 and gallium is an extremely fast diffuser

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in the oxide.

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This figure however is a bit more confusing.
You see, gallium has a segregation coefficient

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greater than 1. So, what should you have expected?
You should have expected a pile up at the

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silicon-silicon dioxide interface, isn't it?
In the phosphorus case, because K was greater

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than 1, you got a pile up. So, even in the
case of gallium, it stands to reason that

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since K is greater than 1, I would expect
a pile up of gallium at the silicon-silicon

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dioxide interface. Instead of that you see,
if anything, it is falling. That is because,

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here the diffusion factor is going to dominate
the phenomena.

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Gallium is an extremely fast diffuser. There
is no chance for gallium to pile up at the

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interface, it just diffuses out. It just cannot
pile up at the interface, it is forced out

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through the oxide. That is why you see, inside
the oxide it just falls drastically, so sharply

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it is falling and even this effect continues
even at the silicon-silicon dioxide interface.

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It just draws the gallium out from this region,
so that if anything, actually the gallium

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concentration is a bit less at the silicon-silicon
dioxide interface than inside the bulk. It

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does not get a chance to pile up, to build
up. Now, in the light of these four cases,

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now let us see whether we can take a look
once again to the oxide thickness problems

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on a heavily p-type and a heavily n-type silicon.
That is to say, suppose I have a silicon wafer

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which is heavily phosphorus doped, n plus
with phosphorus.

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The situation that you are expecting is case
c, right. This silicon is heavily doped with

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phosphorus. So, at the interface, what do
you expect? At the silicon-silicon dioxide

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interface, you expect a pile up of phosphorus.
Now you remember, thermal oxidation is taking

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place inwards. That is oxide is going to be
formed in this direction. So, it is this heavily

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phosphorus piled up region which is getting
oxidized, agreed. This affects the reaction

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rate. You are modifying the quality of this
region. This is the region where the oxidizing

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species are reacting with the silicon, so
you are affecting the reaction rate. So, so

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long as the oxidation process is governed
by the reaction rate, you can expect a thicker

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oxide growth. That is why you see that when
the silicon wafer is heavily phosphorus doped

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the thicker or the faster oxide growth phenomena
occurs at relatively lower temperature, where

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the reaction rate is dominating the process.

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In contrast, if I have a heavily boron doped
p plus wafer, then the situation I expect

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is case a, right, oxidizing ambient. It is
a slow diffuser. Remember, boron will be a

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fast diffuser only in hydrogen ambient. We
are carrying out the oxidation, so you are

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working in oxidizing ambient. So, if the case
is a, in this case, what do you see? Boron

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has preferentially come inside the oxide.
So, this increased presence of boron in silicon

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dioxide is going to affect the diffusion of
the oxidizing species through the oxide. It

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is going to aid in the diffusion of the oxidizing
species, because now you have a silicon dioxide,

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which is heavily boron doped. That is why
for a p plus silicon wafer, the faster oxide

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growth rate will continue at all temperature,
because you are affecting the diffusion, right.

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This is the two cases for a p plus doped silicon
and for a n plus doped silicon. In both cases,

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the growth rate will be affected, but for
different reasons. When it is phosphorus doped,

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it is affected because the reaction rate is
affected. When it is boron doped, it is affected

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because the diffusion is affected, right.

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So, now that we have talked about the oxidation,
thermal oxidation of silicon in general, let

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us take a look at the practical oxidation
systems. How do we oxidize a silicon wafer?

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As I have already told you, oxidizing a silicon
wafer, thermally oxidizing a silicon wafer

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is a very simple process. What you do is you
put the silicon wafers inside a furnace. Remember,

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contamination is always a key problem in all
VLSI processing. Therefore, to avoid contamination,

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the furnaces, the tubes, the furnace tubes
must be made of quartz. Quartz is silicon

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dioxide, so silicon dioxide will have the
minimum source of contamination and even better

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idea will be to use pure silicon tubes, but
that is very, very expensive. Very few places

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who really are bothered about the stringent
quality of the oxide, they use the silicon

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furnaces, actually silicon furnaces. The furnaces
are also made, furnace tubes are also made

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of silicon. But most commonly the furnace
tubes are made of fused silica, quartz.

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So, you have, you see, two options. One is
you pass pure oxygen through it. Oxidation

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is called the dry oxidation process or you
can pass steam. A very common way of passing

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steam is you have a water bubbler. That is
you have a quartz container containing deionized

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water. Remember, for VLSI processing, we always
use deionized water to minimize contamination.

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So, you put deionized water and this container
is placed in a heater, so that its temperature

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is just below the boiling point of water,
just at or below the boiling point of water

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and you pass a stream of gas. That gas can
be oxygen, it can also be nitrogen, it does

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not matter. That is because, your oxidizing
species are going to be steamed, not oxygen

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and since the steam oxidation rate is much
faster than the dry oxidation rate, it does

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not matter whether you are passing oxygen
or nitrogen, because the oxidation rate will

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be determined by the steam oxidation rate
alone.

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So, you bubble the gas through this container,
so that the outlet is gas saturated with water

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vapour and that is now passed inside the furnace
tube, in a quartz boat. We call it a boat

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that is a carrier which contains the wafers.
So, inside this boat you place the wafers

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and put it inside the furnace where the requisite
temperature is maintained. So, in presence

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of oxygen or in presence of this gas saturated
with water vapour, oxide is formed on silicon.

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Usually, if you are going to grow a thick
oxide, that is for masking purposes, thick

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oxide that is say, 5000 Angstrom or 6000 Angstrom
or thicker than that, then we use a dry wet

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dry sequence. That is first 10 minutes you
carry out dry oxidation, then you carry out

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wet oxidation, for whatever time duration,
then you follow it up again with a dry oxidation.

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The idea is to have the interface of a good
quality. That is why the final dry oxidation.

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Then, you get your requisite thickness and
you take it out. There is another possibility.

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In some systems instead of using the water
bubbler arrangement, you directly introduce

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hydrogen and oxygen in the system where they
form steam.

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This is called pyrogenic oxidation system,
pyrogenic water system or pyrogenic oxidation

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system, where hydrogen and oxygen - a mixture
is introduced and it forms water inside the

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system and this reacts with the wafer to form
the oxide. As I have already mentioned, dry

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oxide has greater density and better quality.
What exactly I mean by better quality, we

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will come to, a little later. Wet oxidation
on the other hand is thicker oxide and is

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used for masking or isolation. Now, there
are a few variations of the theme.

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First of all, let us say I want to grow a
very, very thick oxide. That means I have

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to either increase the temperature or increase
the time of oxidation, agreed. Now, whenever

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you are doing either of these two, normally
the oxidation temperatures are say, from 900

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degree centigrade to 1200 degree centigrade,
depending on what you want. These are the

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normal range of temperature for oxidation.
It can be a little less, it can be 800 also,

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but normal range is 900 to 1200 degree centigrade.
Now, if you want to grow very thick oxide

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that means the wafers have to be subjected
to very high temperature for a prolonged duration.

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This causes a lot of stress on the wafer.
Also, the oxide that is formed, it is amorphous.

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This is what we want, amorphous silicon dioxide.

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On prolonged exposure to high temperature,
this amorphous oxide can get devitrified,

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devitrification or devitrified. Whenever we
want to grow very thick oxide that means we

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have to carry out wet oxidation at very high
temperature, for very long time and then devitrification

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occurs. What is devitrification? Instead of
this amorphous silicon dioxide, we get poor

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quality crystalline silicon dioxide and that
is no good. That will not serve either for

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masking or for isolation. It will just be
poor quality crystalline silicon dioxide with

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a lot of cracks, lot of defects and that will
not serve either as a good mask or for isolation.

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So, what is the way out? How do we grow good,
thick oxide? By thick oxide I mean more than

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1 micron. Normal wet oxidation, the limit
is approximately at 1micron, beyond that it

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tends to devitrify. How do we grow good quality
thicker oxide? That is one of the challenges

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which was posed to the VLSI processing technologists.
I have to ask you to take another look at

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the oxidation rate constants. We have defined
two rate constants, remember. The linear rate

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constant and the parabolic rate constant.

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The parabolic rate constant B was 2 D C star
by N 1 and A was ........, isn't it. So, you

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know, B by A will be C star by N 1 into 1
by K S plus 1 upon h, right. This C star that

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is the equilibrium concentration of the oxidizing
species in the oxide, this is proportional

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to pressure.

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This varies with pressure, gas pressure. So,
if you raise the pressure, if you carry out

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oxidation under high pressure, you are going
to increase both B as well as B by A. Both

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the linear and parabolic rate constants are
going to increase. So, this is something which

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is done. It has a number of interesting fallouts.
First of all, your problem of growing thicker

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oxide is now solved, right. You have increased
both the linear and the parabolic rate constants.

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Therefore, at the same temperature and in
comparable time, compared to the atmospheric

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pressure oxidation, you can now grow a much
thicker oxide without the fear of devitrification,

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because you are not subjecting for any longer
duration or at any higher temperature. That

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is one possibility.

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The other possibility is suppose I do not
have to grow such thick oxide, it will do

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to grow even 5000 Angstrom, I have now one
more parameter to play with. By increasing

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the pressure, I can cut down either on temperature
or on time and grow comparable thicknesses.

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Do you understand? Previously, I had only
two parameters to play with, temperature and

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time. Now, I have a third parameter, pressure.
By increasing the pressure, I can now afford

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to cut down either the temperature or the
time. What are the advantages? If I can cut

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down the time, my throughput increases, right.
For the same oxidation which previously needed

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1 hour, I can probably finish it in half an
hour. That means instead of 1 run, I can now

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have 2 runs, throughput is doubled.

34:36.740 --> 34:43.740
Alternately, let us keep the time same, I
have cut down the temperature; so, my thermal

34:44.730 --> 34:50.030
budget is reduced. It costs money to run the
furnaces, to raise the furnace temperature.

34:50.030 --> 34:57.030
They are all electrically heated. Electricity
costs a lot of money and also your wafers

34:58.740 --> 35:05.740
are now not being subjected to as much thermal
stress. The thermal stress is becoming less

35:06.010 --> 35:13.010
and you take it from me, the thermal stress
problem is getting more and more serious as

35:13.089 --> 35:18.800
we increase the diameter of the wafer, because
you see the wafer thickness is not really

35:18.800 --> 35:25.069
increasing at a proportional rate. For 2 inch
wafers, the thickness used to be, may be 300

35:25.069 --> 35:32.069
odd microns. Now even for a six inch wafer,
it is say, about 500 microns. So, it is not

35:33.510 --> 35:39.920
really increasing at a proportional rate.
That means the wafer now is more prone to

35:39.920 --> 35:41.309
thermal warping.

35:41.309 --> 35:48.309
Warping means it bends, it buckles at high
temperature. So, by cutting down the temperature,

35:51.800 --> 35:58.800
you can let your wafers stay more healthy.
They are not subjected to as much thermal

35:59.010 --> 36:03.510
stress. At the same time, you are cutting
down your thermal budget. So, you are saving

36:03.510 --> 36:10.260
a lot of money and believe me, engineering
is all about saving money. Your product must

36:10.260 --> 36:15.859
be cheaper than others product, only then
you can sell. So finally, that is the bottom

36:15.859 --> 36:18.099
line, the cost is the bottom line.

36:18.099 --> 36:25.099
So, high pressure oxidation in both senses,
in more sense than one, it offers interesting

36:25.260 --> 36:32.260
possibilities. You can grow thicker oxide,
you can also grow normal oxide at a reduced

36:33.950 --> 36:40.950
thermal budget or at a reduced time. So, high
pressure oxidation is a very interesting proposition

36:42.400 --> 36:49.170
today and in our lab, we have one high pressure
oxidation system. It is most probably the

36:49.170 --> 36:55.420
only one in India and this is indigenously
developed. It is developed in our lab, by

36:55.420 --> 37:01.829
our people in our lab and high pressure oxidation
system is available. So, sometime I will take

37:01.829 --> 37:08.829
you to the lab and let you have a look at
this. I said I will talk about the quality

37:15.410 --> 37:22.410
of the oxide. How to, I said that dry oxidation
will give you better quality oxide compared

37:24.290 --> 37:31.290
to wet oxidation. Now, exactly what do I mean
by the quality of oxide?

37:32.410 --> 37:39.410
I have a card here, which outlines some of
the non idealities inside the oxide. See,

37:47.930 --> 37:53.650
the silicon dioxide, ideally it should be
an insulator. There should be no charge in

37:53.650 --> 38:00.650
this, it should be an insulator with a perfect
interface between the silicon and silicon

38:00.670 --> 38:06.839
dioxide. Although silicon dioxide does fulfill
a lot of these requirements, there is still

38:06.839 --> 38:13.839
some non idealities and the most important
way of visualizing what the oxide quality

38:19.280 --> 38:26.000
is, is by identifying the charges that are
present in the oxide, because they represent

38:26.000 --> 38:33.000
the non idealities. So, in this card I have
schematically represented the various types

38:34.520 --> 38:40.099
of charges and where these charges are located.

38:40.099 --> 38:47.099
First of all, note this red line. Red is danger,
red also means very important, right. So,

38:50.170 --> 38:57.170
this is my very important line, demarcation
line that is the line between silicon and

38:57.950 --> 39:04.950
silicon dioxide. It is the interface, silicon
and silicon dioxide interface. Now, this interface

39:07.140 --> 39:14.140
is going to dictate a lot of the things, particularly
in relation to a MOSFET. There are non idealities

39:21.680 --> 39:28.680
present in the interface called the interface
trapped charges. These interface trapped charges

39:32.900 --> 39:38.540
may come about because of various things.
It can be because of some mechanical damages

39:38.540 --> 39:44.910
already present in the wafer, it can be because
of some broken bonds, broken silicon bonds,

39:44.910 --> 39:51.910
it can be because of various things. So, this
is one major source of non idealities and

39:53.890 --> 40:00.410
it adversely affects the MOS device characteristics.

40:00.410 --> 40:07.099
As the name itself suggests, it is present
right at the silicon-silicon dioxide interface

40:07.099 --> 40:14.099
and is generally represented, if you look
at the energy band diagram, this is your silicon

40:19.670 --> 40:26.670
dioxide. It is generally represented as states
present all over the forbidden energy gap.

40:31.309 --> 40:37.260
These are available energy states present
all over the forbidden energy gap. They can

40:37.260 --> 40:44.260
trap and detrap electrons that is it takes
part in the electronic activity. So, this

40:48.390 --> 40:55.390
is one major non ideality. The other major
non ideality is a little distance away from

40:57.380 --> 41:04.380
the interface in the oxide. This distance
is about 100 Angstrom or so, marked with green

41:08.220 --> 41:15.220
and you will notice I have put only positive
signs there. That is because in silicon thermal

41:16.720 --> 41:23.720
oxidation, these charges are mostly always
only positive charges and they come about

41:27.780 --> 41:34.780
because of incomplete oxidation. Again remember,
thermal oxidation is taking place inwards.

41:35.140 --> 41:39.920
That is the interface is the region which
is oxidized the last.

41:39.920 --> 41:46.599
Now think of a practical process. Towards
the end of the oxidation you are going to

41:46.599 --> 41:53.599
switch off the gas supply. So, chances are
that is where the oxidation process is going

41:55.299 --> 42:02.299
to be incomplete. You are reducing the oxidant
species supply. So, chances are this is where

42:04.660 --> 42:11.660
the oxidation is going to be incomplete. That
is why here the composition of the oxide will

42:13.660 --> 42:20.660
not be exactly SiO 2, but it will be SiO x,
x being less than 2. I have less oxygen in

42:24.099 --> 42:31.099
silicon dioxide. This automatically means
that I have more positive charges, isn't it.

42:41.490 --> 42:48.490
That is why this fixed oxide charges are almost
always, they are only positive in nature.

42:56.049 --> 43:03.049
The term fixed oxide charge obviously signifies
that these charges are fixed. They cannot

43:03.099 --> 43:09.750
move about, they cannot be decharged or discharged.

43:09.750 --> 43:16.750
So, this is the second major non ideality
and this fixed oxide charge also affects the

43:17.530 --> 43:24.530
threshold voltage of a MOS device. Because
they are positive, they will push the threshold

43:29.099 --> 43:36.099
voltage of a MOS device into the negative
detection. That is now you first have to apply

43:36.960 --> 43:43.960
a gate voltage in order to neutralize these
positive charges, so as to get back the ideal

43:44.960 --> 43:51.960
condition, so that much extra negative voltage
is added to the actual threshold voltage of

43:52.579 --> 43:58.480
the MOS device. That is why we say that if
the fixed oxide charge is present, it pushes

43:58.480 --> 44:05.480
the threshold voltage to the negative detection
and then, all over the oxide I may have this

44:10.309 --> 44:17.309
orange marks which are called oxide trapped
charges. That is these are the charges trapped

44:18.500 --> 44:20.329
in the oxide.

44:20.329 --> 44:26.930
How did these charges come about? There are
a variety of sources. One source is suppose

44:26.930 --> 44:33.930
your device is exposed to radiation environment,
then hole, electron pairs will be generated.

44:35.510 --> 44:42.510
Holes are relatively slow, therefore they
will get trapped in the oxide. Other possibility

44:43.099 --> 44:50.099
is avalanche injection. Suppose you have a
MOSFET with a very short channel. That means

44:50.849 --> 44:57.380
as the channel length is shorter, the electric
field inside the channel is that much higher.

44:57.380 --> 45:04.380
So now, your electrons moving in the channel
may have acquired sufficient energy to surmount

45:06.450 --> 45:13.450
the oxide silicon barrier and get injected
into the oxide. Incidentally it is also called

45:17.150 --> 45:21.490
hot electron effect. The electrons are hot,
they have acquired a lot of energy. Energy

45:21.490 --> 45:28.309
is equivalent to temperature, right. That
means the electrons are hot, so they can have

45:28.309 --> 45:34.430
enough energy to surmount the barrier and
get injected into the silicon dioxide. So,

45:34.430 --> 45:40.910
there are a variety of reasons, variety of
sources by which these oxide trapped charges

45:40.910 --> 45:47.140
may occur. Most important are these two which
I just outlined. One is if you are exposing

45:47.140 --> 45:52.210
your device to a radiation environment, the
other is if avalanche injection is taking

45:52.210 --> 45:59.210
place and then finally I have what are called
the mobile ionic charges.

46:01.930 --> 46:08.930
I have mentioned the two most important mobile
ions, sodium and potassium and you know what

46:10.670 --> 46:17.670
are the greatest sources of this sodium and
potassium. We humans; you have to handle your

46:19.500 --> 46:26.500
wafer and if you are not careful, from human
body, the sodium and potassium ions can get

46:29.530 --> 46:36.530
into the oxide. Its another major source is
water. Unless your water is properly deionized,

46:37.099 --> 46:42.220
they contain a lot of dissolved sodium and
potassium salts. That means a lot of sodium

46:42.220 --> 46:49.220
and potassium ions and they are mobile. They
play havoc with the system. They can move

46:49.750 --> 46:56.750
all the way from the silicon dioxide surface
to the silicon-silicon dioxide interface,

46:58.260 --> 47:00.430
depending on the detection of the electric
field.

47:00.430 --> 47:06.280
See, they are positively charged, sodium and
potassium. So, suppose on top of this you

47:06.280 --> 47:11.390
have your metal gate that will be a MOS structure,
right, metal oxide and semi conductor. Now

47:11.390 --> 47:17.930
suppose you apply a gate voltage, positive
gate voltage. All these mobile ions will get

47:17.930 --> 47:23.910
pushed to the interface and suppose we apply
a negative gate voltage, all these mobile

47:23.910 --> 47:30.910
ions will move up towards the surface. They
are free to move. This check, whether mobile

47:35.730 --> 47:41.349
ions are present or not, it can be done very
simply by doing a proper CV analysis, capacitance

47:41.349 --> 47:48.349
voltage analysis. It is a routine check in
all fab labs, which make MOS devices.

47:49.450 --> 47:54.640
If you have sodium ions or potassium ions,
might as well close down the lab, get rid

47:54.640 --> 47:59.150
of your sodium and potassium ions, because
if they are present, none of your devices

47:59.150 --> 48:06.150
are going to work. So, it is a routine check.
You must first get rid of sodium and potassium

48:08.299 --> 48:14.549
ions. Well, there are various ways and means
of getting rid of this sodium and potassium

48:14.549 --> 48:21.549
ions. First cardinal rule is your system must
be clean. If there are sodium or potassium

48:23.599 --> 48:30.599
ions, mobile ions present in your system,
it says very badly about the cleanliness of

48:31.700 --> 48:38.700
your system. It says that your system is not
clean enough. So, first of all you have to

48:40.230 --> 48:47.230
tighten the cleanliness of the system and
there are also various other ways of getting

48:48.200 --> 48:55.200
rid of these mobile ions.
One of them is called chlorine oxidation.

48:56.349 --> 49:03.349
In the next class, we will talk about ways
and means of getting rid of these unwanted

49:04.170 --> 49:11.170
oxide charges, mobile ionic charges as well
as the fixed oxide charges and the interface

49:11.710 --> 49:18.039
charges. The oxide trap charges, most often
they are not really related to processing.

49:18.039 --> 49:24.349
They are most often encountered when your
device is fabricated and after that it is

49:24.349 --> 49:29.450
being exposed either to radiation environment
or your MOS device is being operated under

49:29.450 --> 49:36.450
very high channel electric field. So, they
are not really so much related to the processing

49:36.589 --> 49:42.890
conditions, but the other three are - mobile
ionic charges, fixed oxide charges and interface

49:42.890 --> 49:49.030
charges. So, in the next class we will talk
about how to get rid of these charges. That

49:49.030 --> 49:54.420
is very important when you want to talk about
the gate oxide in a MOS device.
