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We will continue our discussion on the Schottky barrier diodes their I-V characteristics and
we will go on with how close are they to the ideal characteristics. I have marked this
as 1 because; we will have at least two sessions on this topic because, it is a detailed discussion
on this. Before we get along on to that discussion let us just take a look at some of things
which you have already seen.
We have shown that the Schottky barrier current density J can be expressed as J0 just like
the diode J0 into e to power V by VT minus 1. The only difference is the J0 comes up
by a different mechanism compared to the pn junction. We will have occasion to discuss
today the exact difference. What we have discussed here is that the current transport is through
the transport of carriers over the barrier. All those electrons which have energy acting
in excess of the barrier height will be able to create or give rise to current. Through
that approach we said that, J0 will be given by this expression and the whole thing here
can be put as a constant A star because that depends on the material because the effective
mass comes there and the temperature. The temperature is absorbed here and this A star
is put there which includes the m star and of course it depends on the phiBn. As we have
been telling again and again if the phiBn is higher this is smaller. It is better rectifying
diode if phiBn is smaller it is a very leaky diode which is closer to ohmic contact. Here
what we have seen is A star, I am pulling out this A star alone and you can write it
because all are constants except that mn star is a material dependent quantity. It would
vary from silicon gallium arsenide indium phosphide etcetera. All other constants put
together it becomes mn star by m0. I just discussed this last time, going through it
quickly so that we are focused on our discussion today. This thing is 120 into mn star by m0
amperes centimeter square per Kelvin square. All that is required to evaluate J0 is this
evaluate A star and this phiBn and temperature.
Take the case of gallium arsenide evaluation of J0 for Schottky barrier on gallium arsenide
and we are talking of, of course n type substrate because those are the materials which are
used for making FETs in gallium arsenide. A star is 120 mn star by m0 for gallium arsenide
this ratio is actually equal to 0.067. Effective mass is very small in fact that is one of
the reasons the mobility is very high in the case of gallium arsenide. When you substitute
for that here you get A star is equal to 8.04 that is the Richardson constant the A star
is put to differentiate it from the A. Usually, we call it for area so that you do not confuse
between area and this is just put A star. It also indicates you that star, is related
to the mn star effective mass two stars related to each other. That is the A star and not
related to area. If the phiBn is barrier height is 0.85, which is usually the value that is
obtain for gallium arsenide on n type substrate then you get at temperature T J0 equal to
that A star which is 8.04 multiplied by T square 300 degrees Kelvin square multiplied
by e to power of minus phiBn by kT 25 million volts 850 electron milli electron volts. All
when you evaluate it is about 10 to power of minus 9 amperes per centimeter square.
This is range which we see in the case of gallium arsenide you can see it is excellent
in terms of leakage current etcetera.
Take a look at silicon Schottky barrier there the ratio is about 0.6 mn star by m0. Therefore,
the Richardson constant A star turns out to be 0.6 times 120 that is about 72 amperes
per centimeter square per Kelvin square. If you want to find out J0, we do it at temperature
300. You have got 72 into 300 square into e to power of minus phiBn by kT. That phiBn
I have taken as two thirds of Eg which is value that you get when the Fermi level pinning
is there. You get close to that point. Notice also that this particular band this phiBn
is smaller than that of gallium arsenide. The moral of the story is you get higher phiBn
if the band gap is higher because it is almost related to that Eg. If you want to make Schottky
barrier you are better off with materials which have got higher band gap, silicon it
is border case because band gap is 1.1. If I want to make Schottky barrier germanium
it is a hopeless situation. That is very good point that one should note at higher band
gap materials are better suited for Schottky barrier. Therefore, we can imagine or we can
estipulate that it is better suited for making JFET or MESFET. For making MESFET metal semiconductor
FET’s we can make better with wider band gap semiconductor. That is what I want to
point out here. Now plug in those values here you will get 10 to power minus 7 into 6 which
is almost equal to 10 to power of minus 6. You can see as compared to gallium arsenide
Schottky barrier which is 10 to power minus 9 it is higher than that. The leakage currents
are more reverse bias currents are more per centimeter square.
Comparison of the I-V characteristics of gallium arsenide and silicon Schottky diodes: I am
putting them down again here J0, we just now estimated is 1.24 into 10 to power minus 9
ampere per centimeter square. If you use that J0 you can find out the voltage drop using
the I-V characteristics.
The clarification on what I have written there what we use is J equals J0 e to the power
of V by VT minus 1 that is the forward characteristics. What we are trying to find out is when this
is 1 ampere per centimeter square how much is V? Given that as J0, you should also note
that this is what you are trying to find out just by using this. Also notice we can neglect
that term very comfortably e to power of 4 if you take even if 100 milli volts 100 by
25 e to power of 4 is large compared with 1. When you go to 200 milli volts etcetera
it is totally negligible, we are just taking J is almost equal to J0 e to power V by VT.
We are using that to find voltage for 1 ampere, gallium arsenide; we know that actually J0
is equal to 10 to power minus 9 into 1.24. So, that the number we substitute here find
V there; that is what we are trying to do when you have used this particular term here.
J is equal to J0 into e to power V by VT V is equal to VT logarithm of that, substitute
for all that 10 to power 9 this is 1 this is 1.24 into 10 to power minus 9. That goes
up there V is equal to 0.513 volts. This is actually not bad it is slightly lower than
what you get for pn junction diode is silicon but getting closer to that of the pn junction
characteristics. Let us see what we get in the case of silicon Schottky barrier.
J0, we estimated it to be 6 into 10 to power of minus 7. Same formula V is equal to VT
logarithm of J by J0 substitute here 6 into 10 to power minus 7. 7 gone up there 10 to
power 7 by 6 you get 0.358 volts. You can see the problem with or the differences between
the silicon Schottky and gallium arsenide Schottky.
The sum of all those results in the table, we have seen phiBn is 0.85 you have got higher
band gap, higher phiBn because closer to two thirds here 0.75. That makes the J0 smaller
here compared to this silicon also A star because of smaller effective mass 0.067 compared
to 0.6 in silicon this is 8 and this is 72. All of them make the leakage current here
is large compared to that J0 of gallium arsenide Schottky barrier. When we use that we found
out it is 0.513. There is no sanctity about 0.513 etcetera. You can say about 0.5 0.54
in that range depending upon how much is the barrier height will vary say about 0.5 this
about 0.358 volts. That is the forward drop for one ampere current. This will go through
all the range of the current.
Now see the I-V characteristics of the silicon Schottky barrier this color. This colors here
the silicon like that gallium arsenide smaller J0 larger J0 this one voltage drop is smaller
in silicon compared to voltage drop in gallium arsenide. In other words in a language of
circuit engineering you can say the cutting voltage of silicon Schottky is much smaller
compared to cutting voltage of a gallium arsenide Schottky barrier. The cutting voltage of silicon
Schottky is almost getting closer and closer to that of cutting voltage of germanium pn
junction. Germanium pn junctions all of us know that cutting voltage is about 0.2 volts
because of small band gap barrier are small there. The difference is mainly because of
the barrier and the A star that is the point here. You have got I-V characteristics approaching
to that of pn junction of silicon, we will see that now.
Let us see through the pn junction look to recall what we already know about pn junction
I take p plus n junction. The total formula for this J0 of pn junction diode is this.
These are actually the carriers which are generated in diffusion in one depletion length
diffusion length on either side of depletion layer diffusing into that. When you substitute
this is the minority carrier concentration in the n type region thermal equilibrium.
This is the minority carrier electron concentration is the p type region thermal equilibrium.
You know that product of majority and minority carrier is equal to ni square majority carrier
concentration equal to doping concentration when you do that pn0 is ni square divided
by doping. Similarly, np0 is ni square divide by doping NA. That is all what we done there
this is a very standard formula. Let me not get in to that we have discussed it in many
forums. Between these two terms both will be almost comparable if the doping concentrations
are same. But, when you talk of p plus n diode acceptor doping concentration is large compared
to the donor doping concentration. This term in the denominator is much large compared
to the doping on the ND. This term actually goes off very small four or five order of
magnitude larger here denominator that means this term will be four or but five order magnitude
smaller compared to that. We put that approximately qDP ND LP. I just put this formula that you
can substitute on that for yourself and see I have done that work for you.
p plus n junction in silicon I take doping of 10 to power of 16 centimeters cube per
centimeter cube life time. Life time of holes in n region is about three microseconds typically.
This is only order of magnitude to get an idea. I am putting these numbers. Do not say
this is the number say this is a number which generally will be varying depending upon the
doping and life time characteristics. DP is 12 that is for silicon you say and ni is 1.5
10 to power of 10 per centimeter cube at room temperature. We are calculating J0 at room
temperature. When J is 1 ampere for this case this also has got J equal to J0 e to power
of V by VT. But, J0 we are calculating is a different formula. This is V equal to VT
logarithm of J by J0 J is one amperes J0 is this quantity that is why you get 10 to power
12 divided by 7.2. We are using the same formula.
You are using the same formula: J is equal to J0 e to the power of V by VT this J0 is
something like about 10 to power of minus 12 into 7.2. This is very small compared to
this in the case of Schottky barrier. That is why for a given current 1 ampere you will
get a larger voltage drop. In the case of Schottky barrier the silicon Schottky barrier
we got it as 0.35 volts. Here, when you substitute for J0 and this is 1 this is 1 and this is
7.2 into 10 to power minus 12 amperes per centimeter square. You get a value of V which
actually is much larger than that what you get for the Schottky barrier. What we have
to see here is when you plug in those numbers you get the voltage drop across the diode
for 1 ampere per centimeter square it is 0.64. You are used to here in number cutting voltage
of silicon pn junction is about 0.6 or 0.65. That is the implication of that. About that
current range you get that voltage drop. It does not mean that the diodes starts conducting
at 0.6 it conducts even better. Otherwise if you go to 0.1 amperes per centimeter square
you will get that voltage much smaller than that. It is conductive but when you using
that at particular range you get 0.64. When you say cutting voltage that is the meaning
of that. If you actually spun the I-V characteristics on the lower range you will get small voltage.
Do not be perturbed if somebody comes and tells I get a voltage at a diode is conducting
at 0.4 voltages, it will, but very low current density that is the meaning of that.
Cutting voltage is a concept which is used by circuit engineers for convenience because,
in the normal range of operation that is the voltage across the diode that is about 0.6
volts that is implication. People who are cautious will say 0.6 to 0.7 that is about
the range. Silicon Schottky barrier J0 is that and we have 0.358 volts as the drop that
is the range. What we say now is the voltage drop across the pn junction for a given current
density is more than that of a Schottky barrier diode. That is finally the thing. That is
what we said here V for a given J the voltage V across the pn junction is greater than the
voltage across the Schottky barrier.
Quickly go through the numbers for gallium arsenide Schottky barrier we have seen it
is about 10 to power of minus 9 into 1.29 and 0.513 volts voltage dropped across that
Schottky barrier. p plus n diode used the same formula that we used for silicon it is
about that very low and very lower than that ampere per centimeter per square because ni
is 10 to power 6 as compared to 10 to power of 10 in the case of silicon that low number
makes it very very small. ni square is 6 orders of magnitude smaller that is why it is low.
When we use that for 1 ampere current density same VT logarithm J by J0 you get about 1
volt.
The entire result is shown here in this table, Schottky barrier silicon Schottky barrier
gallium arsenide; we have seen it 0.358 is the voltage dropped cutting voltage about
0.3and 0.5 cutting voltage for gallium arsenide Schottky barrier. Go to silicon PN junction
here on silicon PN junction P plus N junction J0 is much smaller compared to Schottky here
and then correspondingly voltage is you can see 0.358 for Schottky barrier 0.64 for PN
junction gallium arsenide. Gallium arsenide PN junction voltage drop is about 1 volt compared
to Schottky barrier of 0.5volts forward difference.
Take a look at the total I-V characteristics of all the things. I have put the Schottky
barrier of silicon. I have put the Schottky barrier of gallium arsenide. I have also shown
the PN junction of silicon P plus N junction also PN junction of gallium arsenide. Moving
from left-hand side to right hand side, the dotted lines are the junctions the solid lines
are the Schottky. On the forward conduction region here, we can say now minimum voltage
drop 0.35 volts or 0.3volts. I put here as cutting voltage about 0.3 for silicon Schottky
barrier. For gallium arsenide Schottky barrier about 0.5 cutting voltage and you can see
here I have put 0.65 for silicon PN junction all the curves extend to the right and if
I see the gallium arsenide PN junction, 1 volt. Some information related to this. This
property of higher cutting voltage of gallium arsenide makes it useful for solar cells with
the higher open circuit voltage. When you shine light if you take a junction you will
see there is a voltage developed. If the solar spectrum falls on the PN junction and if you
just measure put a multimeter there across that or a voltmeter you will see there is
a voltage developed. If you take silicon we will get something like about it will be less
than the cutting voltage, less than the built-in potential. It will be about 0.55 0.6 if you
get you will hit the roof with joy. You do not get 0.6 after 0.55 volts it becomes tougher
and tougher; whereas, in gallium arsenide pn junction solar cells will get close to
1 volt higher open circuit voltage. We will see if we have occasion to discuss this later
on much later some aspects of these things. This is what I want you to note. Also you
see the current starting from this dotted line higher voltage lower is J0. This is the
curve from here to here and this one silicon junction that dotted line with second dotted
line. Here, you move from this side to this side and here you move from below like that.
Smaller forward voltage larger leakage and if the leakage becomes largest that is the
ohmic contact. If this curve keeps on shifting towards left side this keeps moving down and
you will have ohmic contact close to the ohmic contact. You can see that silicon Schottky
barrier is not so good rectifying. It is not as good gallium arsenide it does the rectifying
job no doubt with a band gap of about 0.75 or so. You do get rectifying contact. It is
not very encouraging like in case of gallium arsenide the gallium arsenide is almost, see
the characteristics close to pn junction of silicon here also.
Let us go further down, benefits of Schottky barrier diode. Before going into that few
things I want to discuss keeping this diagram in mind here, we will discuss one or two things.
One thing is the application of the Schottky barrier diode even in silicon technology the
Schottky barrier diode is used. It is not the proprietary of the gallium arsenide technology
Schottky barrier diode; you see that it is conducting at lower voltages 0.3 volts. You
can see if there is a junction in parallel with this as diode and Schottky barrier diode
and a pn junction in parallel, this will be dominating. The current will pass through
Schottky barrier diode. That property is made use of in fast switching TTL transistor, transistor
logic circuits if you see you have the transistor. In the TTL logic circuits it is slow compared
to emitter coupled logic easier. The reason is in the TTL the device goes into saturation.
I will go through that particular thing now.
I have a simple example: you show this inverter like this, I will put that plus there and
pumping in current I am not sure in the biasing circuit etcetera. If you see when I have the
input actually going like this 0 to 1, what way it goes? In fact, IC versus VCE if you
take, you just recapitulate your memory on these particular switches when the bipolar
transistor is used. I have the characteristics which go like that with different IB. I have
that characteristic now you can draw a load line on that, that RL which is the load line.
We can draw that load line so when the device goes from 0 here it is off and when it is
1 it is totally on that is you are driving base current with 0 that is here you are moving
on to that curve. That means, you are moving from here to that point. The transistor is
driven from the cut off to saturation the transistor goes into saturation here when
the transistor goes into saturation you know that there are charges stored in the region
base region collector region everywhere. If you want to come back from here to turn it
off, the charges stored in transistor various regions will have to be removed that is what
makes is slower. If you want to prevent the transistor to go into saturation, what is
done in the TTL is this I am showing one inverter only of some portion of that TTL where some
inverter comes in we will have and rest are here, is a different story. But I am showing
the principle finally. I do not want to this to go to this thing. What you do is you prevent
the transistor, you know when this goes into this portion. What is the state of voltage
across the emitter base and collector base? When the transistor goes into saturation emitter
base junction is forward biased collector base junction is also forward biased. When
it goes deep into saturation collector base junction voltage will be almost equal to 0.6
volts. If the junction is going to saturation that is conducting may be somewhere here.
I do not want the transistor go into deep saturation when you say deep saturation, lot
of forward voltage across the transistor collector based junction is large almost close to the
emitter base junction voltage is 0.7. This may be 0.7 this may be 0.6 when it goes to
that unless it is less there would not be current flow. Both of them are equal current
will become 0 that never happens. What you do you is put a Schottky barrier diode here.
I think we can put it like this; this is the Schottky barrier diode.
Connect a Schottky barrier diode. In fact you can actually integrate along with the
transistor between the base and the collector. Integrate a Schottky barrier diode or you
can put it separately make it diode connect them within the surface connect internally.
The way they do it I am not getting to that technology right now. The way they do is this
is done by putting a metal over the base region and the collector region you make the Schottky
with the n region collector. You put a metal on n region it becomes the Schottky barrier
and allow that metal to go over to the base region that becomes ohmic. We get an ohmic
contact here p type region any metal on p type becomes ohmic because of low barrier
height and Fermi level pinning. Metal on the collector n region become rectifying very
easily without showing that diagram you can feel that this a single device with just a
metal layer put between the collector and base. That is the technology that is done
there. This is the TTL Schottky TTL STTL. Here, when this goes into this region here,
when the device goes into the saturation this voltage across will never become 0.6. Why?
There is a diode connected across that Schottky barrier which will divert all the current
instead of going through that it is divert diverted through that with a smaller voltage.
There is certain current flowing through that current will flow through this device. The
current instead of going through this it will go through that, that voltage dropped across
that is limited by the voltage drop across the device.
Now, we can see this diagram instead of 0.6 volts here pn junction across when it goes
in saturation you have put a Schottky here this one this dotted and this one first one.
That voltage drop across is 0.3 volts. The voltage drop across the collector base junction
which is forward biased is now 0.3 volts. It has not gone much into deep saturation
now when a diode here in the junction is forward biased you have got carrier concentration
raised as e to power of V by VT. When a forward bias junction is there carrier concentration
is raised as e to power of V by VT.
Now, V is the forward bias you have reduced that V from 0.6 to 0.3. So, e to the power
of 600 by V 25 that is the by V by VT. This is 600 by 25 originally carrier concentration.
If you just recall, I just write that here below. If I take the n region pn boundary
condition is pn0 e to power of V by VT. This is the charge which we have injected into
n region when a forward bias collector base junction is there. V by VT if V is 600 this
600 by 25 this is about 24 e to power of 24 that may be something 10 to power 14 or of
that order very high. Instead of V to power 24 now by connecting this is what you have
done is you have reduced that to e to power of 300 by 25 that is 12. The fact of e to
10 to power of 12, huge reduction in stored charge split up is tremendous that is why
you get split up thing as Schottky barrier diodes. Schottky barrier diodes are very important
device for high speed circuits. That is calling with the theme of our discussions in this
high speed circuits. High speed devices are made in silicon with Schottky barrier connected
across collector base junction integrated like that. May be later on we may have time
to recall but this is the key to the thing. I want to discuss why that difference between
J0 between the pn junction and the metal semiconductor contact. You do not see this explicitly discussed
but, you very closely examine the mechanism of current transport, you can easily figure
out what is happening. If you take metal semiconductor contact, how is the current transport taking
place? Take a look at how the current transport is taking place in the case of p plus n junction.
What is the difference? If you see, why J0 is smaller in pn junction compared to or why
the currents for a given voltage, for a given voltage see for a given current the voltage
is larger in a pn junction compared to the metal semiconductor contact. Or putting in
other way for a given voltage current is smaller in pn junction. Why? We will take a look at
that. Now, this is a very important concept that one should see actually.
Let us draw the energy band diagram which we have been drawing is for Schottky and we
draw the Fermi level there EC EF EV. Recall, we have drawn this sort of diagram in fact
it is a very powerful diagram to visualize what is happening. Those are the electrons
which have energy above that which are able to cross that is the one which gives us J0.
This is the phiBn metal semiconductor. This is the metal or Schottky barrier. People like
to call this more as Schottky barrier than the metal semiconductor. I called it because
when you say metal semiconductor it can be Schottky barrier, rectifier or ohmic general
term.
Now this is thermal equilibrium when you forward bias or here you have got similar distribution
here. I am redrawing the diagram which you have been drawing and this is balanced by
that. These electrons have energy more than this barrier these electrons have more than
the barrier that is balancing. When you forward bias this is forward bias, this is forward
bias situation the conduction band valence band will also go up similarly. By the same
amount equal to V forward bias voltage now this distribution goes up and you have many
more electrons which are able to cross. I am discussing this deliberately because we
have to see the difference between pn junction and this. These extra electrons have energy
above this. We have computed this electron concentration there if you remember what have
we computed. We computed q and whatever electrons are there above that because they are the
electrons which are actually here impinging on the boundary into the mean velocity. That
is all we have done current density. The same thing happens in the pn junction also absolutely
no difference. The difference is what happens beyond these points. Let us take a look at
what happens beyond these points. These electrons whatever have energy above that of course
there some part of it is compensated by this. Let me talk of only total quantity now they
are impinging reaching here now they are here. The quantity ns equal to nn0 e to power V
by VT all those numbers are able to cross that barrier. Now, how much is the current
depends upon what is the fate of this carrier after they cross the barriers. What is the
number that is reaching that metal? How much is the number compared to the total number
available in the metal. That is the key to the whole thing. In the case of Schottky barrier
in the metal you have got 10 to power 22 23 of that order number of electrons per centimeter
cube. What number of electrons which are crossing this barrier here? How much are they? If this
is 10 to power 16 much smaller quantity compared to that nn0 e to power of minus V by VT nn0
into e to power of minus V by VT that quantity. In fact minus VBi minus V if you look back
it is related especially the voltage. You have actually that number much smaller than
equilibrium concentration. If you have 10 to power 16 as the equilibrium concentration
the crossing number is much smaller. May be 10 to power 11 10 to power 10. It is like
putting a drop of water into ocean it disappears immediately. In the sense it is not disturbing
thermal equilibrium here when the carrier concentration is so large from here to the
contact.
If I have metal like this and a semiconductor I am applying a voltage V this is n type what
we are talking of is these electrons which are crossing here. Now, you have got small
even the smallest field that is appearing is there will be able to take away those electrons
because the total electron concentration is very large 10 to power 22 if you have. Even
a very small field will help to remove those charges which have injected give rise to current.
It is again here you calculate it as J ns into V J is equal to q ns into Vx that is
when it is here. Once it goes into metal, what is the current? Same current should flow
if all of them are flowing out. This is across the barrier.
In the metal J will be what q into n in the metal plus whatever ns is injected that is
negligible. Let me write rewrite that in the metal J is equal to q n in the metal plus
ns which has been injected there into velocity in the metal. This is actually small compared
to that nm is very large compared to ns which would mean this is very large compared to
that this is the mean thermal velocity. Whole thing is given by transported because of this
higher velocity because this is hardest of magnitude larger than that. What will be that
quantity? This current continuity is there. Whole thing limitation comes up. How much
can go across the thing? All other will be transported with a very low velocity here;
very low velocity means u is equal to V is equal to mu into T. There is mobility of electrons
is there in a metal. V can be very small that means very small electric field will take
away give rise to this current in the metal. Very small electric field in the metal will
be able to handle this amount of charges which are injected to remove them away because,
it is not this that is flowing through that because the amount that is the voltage across
the metal the current transport to not merely due to that it is due to whole of the things.
We are not dealing with that injected electron alone when you apply voltage total number.
Total number is what existed originally. That is possibly with very low velocity which would
mean that is possible very low field there that means very smallest milli volts drop
across this will be enabling to allow the current flow. What I trying to point out is
in a Schottky barrier diode the current transport the extent of current that is flowing through
the junction is totally controlled by that component. This component is not limiting
it with very small field that will take away all the current because nm is very large there
metal. nm electron concentration in the metal is very large compared to the electron concentration
on the surface of the semiconductor. nm is the electron concentration in the metal that
is the one which is very large therefore, this is not limiting that. They just swept
away whatever is injected total current is limited by this. That is the way, we have
estimated the current. Now, in the case of pn junction also, let us see what happens.
I will just retain this here I will go to this side pn junction. I will say p plus n
junction. I will switch over from there I will retain that there in case, we want that
for discussion. We will focus here now. In the case of the pn junction the diagram is
p plus n. The story is totally different what is energy band diagram? Take a look at this
in the same light as the Schottky barrier diode EF thermal equilibrium. This is p type
material I will put that as a junction and you will have depletion layer there. From
there onwards due to the built in potential and here that is EC. p type material n type
I have not shown the Fermi level not too much difference but if that is p plus it will be
closer to that pn junction or even if you do not put p plus it is okay our discussion
holds good for that. What will be the distribution here? Exactly, we can put same diagram that
we drew for metal semiconductor junction. This goes like that that is the electron distribution.
When the Fermi level has equalized what happened is these have some energy, this is the barrier.
What you saw in the case of Schottky barrier that same type of barrier is there for electrons.
Here electron concentration very small. This is actually like this. Now, look at the fate
of electrons which are injected from here to here when I forward bias that dotted line
becomes like that. I can actually compute the total number of electrons which have energy
to cross this exactly the way you have computed the electron concentration at this surface
here. You can compute electron concentration on surface nn0 e to power VBi minus V by VT
q of course or VT VBi minus V by VT that will be concentration here. That is what you are
doing computing how many have got the energy above that. All of them are capable of crossing
this barrier that extra, previously this was equal to that thermal equilibrium when you
forward bias.
When you forward bias this V, when you do the forward biasing this whole thing gets
lifted up more of the carriers are there. If you give a chance to those electrons to
be removed away once it is injected to the p region the same way that is done in the
case of the metal. The current will be exactly equal to that. Supposing this barrier height
is the same as that phiBn that you have got in the Schottky barrier you would have got
the same current. But, if this barrier height which is the built in potential is the same
0.75 volts, the current that you get will be much smaller than that or same voltage.
Reason is these electrons which are impinging on the surface with the same equation q ns
is actually this quantity q ns into V of x they are striking the surface and are able
to cross the barrier. But, having crossed the barrier they find that they are minority
there. Once they are minority it is actually now the current depends upon which of them
is larger. In the case of Schottky barrier you are impinging the current is transported
across the barrier with q ns into V of x and from that point onwards there is no limiting
because carrier concentrations are very large. q n into V q nm into V very small V or very
small drift field will take away those electrons. Whereas, here n is minority carrier this may
be what you have the electron concentration here may be 10 to power 10 or 10 to power
11. What you have here will be 10 to power 2 10 to power 3. Now, what happens is if you
want to remove them at the same rate you must have a large velocity. For large velocity
there should be electric field coming up. Alternately, the carriers are not removed
as fast as you require because, there is no field available there. They get piled up here
the minority electrons which are injected here be piled up here.
Now you take a look at the boundary condition here pp0 is a hole concentration. I am sorry
this is the pp0 and nn0 is the electron concentration there. You are pumping in electrons at a very
fast rate here governed by that equation, ns into V of x. But, it is not able to take
it away because, there is no field there. Immediately when you inject at a fast rate
what happens you are dumping electrons at a fast rate from the boundary. Therefore their
concentration goes up there and there is no concentration there, from that point onwards
it is carried away by very familiar side. The concentration builds up there because
it is not being taken away fast that is because it is the minority carrier and it will be
very large field to take it away. There is no large field there only small fields are
not sufficient. Carriers get piled up there that is why this builds up. This is slightly
different concept that we usually see but the fact is that. I think we can put it as
np. Hole concentration is p type material hole electron concentration is p type material
and then you have got np0 that goes up. The mechanism by which the current is transported
now through that p layer is by diffusion, because of this concentration gradient, once
it crosses the barrier the transport is by diffusion. The diffusion is actually slow
process this is slow compared to the transport from here to here where equilibrium is hit
so that the rate at which is removed controls the total current. The difference in the Schottky
barrier and the metal semiconductor contact for a given voltage in the Schottky barrier
the current is controlled by rate at which you are able to transport across the barrier.
From there it is taken away fast because large number of electrons is available in the metal.
In the case of pn junction the current is controlled by rate at which we are taking
it away from that. One is the rate at which is coming here other one is rate at which
taken away current is controlled by a slower process which ever is slow is controlling.
In a pn junction this process is slow diffusion driving force is concentration gradient.
This process is faster this is the one, two of them in series. The slower one control
the current flow slower one is the diffusion and that magnitude is much smaller than what
is crossing there. This is smaller compared to what is crossing there and this is the
total current which is actually given by qd and np0 by lp. Whereas, that current would
have been if you have able to take it away completely that will be that A star d square
into power phiBn by VT; whereas, this turns out to be qd and pn0 np0 into e to power of
V by VT. This is small. Going back to sum up what I said about these two currents now
take a look at this particular thing. This portion is able to carry the current fast
so total current is controlled by is the amount coming through this which is much larger compared
to what is been taken away here. This is very small when compared. In both the cases the
rate at which it comes up here for the same barrier height will be same. But, mechanism
by which it is taken away the other region that controls. Here it is taken away by drift
in the presence of large quantity of electrons that why it is fast and in other case it is
taken away by diffusion. Notice you have got this charge storage effect. This is the charge
that is stored here similarly on other side you can talk of whatever holes then same thing
you can talk of here. Holes are hitting the surface at fast rate but they are taken away
by small rate because, hole concentration in n region is very small. These are the two
subtle differences between the two. For the given voltage the current there is smaller
because the current has ultimately decided by diffusion.
That is the reason why physically this current in a pn junction is smaller compared to this
current which has gone up all the way up there. Or for a given current you can drive a particular
current with a smaller voltage Schottky barrier compared to that. That is actually the key
thing for that. In fact, I just went around going through this because this particular
point is not clear this whole thing becomes clear only when you look at it this way. The
electrons having distributed there how many electrons can cross that and once they are
injected beyond that point what happens. I think that is about the main thing that I
wanted to point out to you. One of them is Schottky barrier used in TTL because, lower
voltage other one is the mechanism. Also I have pointed out to you today that this charge
storage here is there because, minority carriers are stored. What about such charge storage
here? Absolutely no charge stored because, whatever injected is very small no charge
storage effect here either here or here. Whenever there is no charge storage effect whenever
it is absent, we can say the device is faster. In high speed circuits it is become very effective
to use a device which has no charge storage that is the Schottky barrier diode. In pn
junction always you will have charge storage effect and that will be slower than the Schottky
barrier. You will see in a logic circuits with gallium arsenide gives you Schottky barrier
diode. With that I will conclude the discussion today. We will continue on some aspects of
this in the next lecture.