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We will have yet another session on HBT. We have had three sessions focusing mainly on
gallium arsenide-based heterojunction bipolar transistors. The reason for switching over
to this gallium arsenide-based HBT is to reduce the base width and still accommodate high
doping in the base region. This is possible because in the emitter region, you can reduce
the doping when the band gap is large, without hurting the emitter efficiency ñ you can
still have very high emitter efficiency.
We also saw how the surface effects control the performance of this device. We saw that
it is important to passivate the surface of gallium arsenide if we want to get reasonable
beta, reasonable performance. Otherwise, the entire performance will be masked by the surface
recombination effects. In fact, the first result that was done by Bell Labs, New Jersey
was disastrous in the beginning ñ they expected beta of about 2000 but they got a beta of
about 30; the whole reason was the surface effect. Once they passivated the surface,
they found that the beta goes up by a factor of 60 right up to 1800, but there were problems
with stability and that was sorted out over several years by switching over from sodium
sulphide passivation to ammonium suphide ñ ammonium suphide with a coating of PECVD nitride
or polymide. Things like that have been going on over the past decade now. I have brought
that scenario almost up to date with gallium arsenide-based HBT.
The other types of HBTs are gallium indium arsenic phosphide ñ that is quaternary, it
is gallium indium arsenide on a ternary and on a binary compoundÖ. Here, all these are
lattice patch, that is, gallium indium arsenide matches with gallium indium phosphide ñ you
add a little bit of phosphorus to that and the band gap goes closer to indium phosphide.
In fact, I just wanted to put that diagram.
For example, we haveÖ I am just drawing in the diagram indium arsenide, gallium arsenide.
I am not putting the axis, etc. In the axis, you have got the band gap and here you have
got the lattice constants. This is indium arsenide. Somewhere here, you have got gallium
indium arsenide where one of them is 0.47 and the other one is 0.53. Now you add and
here, you have got indium phosphide. The band gap of that is higher than gallium indium
arsenide and the gallium indium arsenide band gap is lower than that of gallium arsenide.
Now to this ternary compound, you add phosphorus ñ you can call instead of galliumx, indiumx,
etc., you can add phosphorus and arsenicy. You can call it like this: galliumx indium1-x
arsenicy phosphorous1-y. We are matching all those things, so if I add phosphorus to that,
you move along this vertical line ñ keeping the same lattice constant x, you move up and
so this will be actually somewhere here gallium indium arsenic phosphide. Why I do this diagram
is, you know that this lattice matches with gallium indium arsenide and indium phosphide,
because after all, adding phosphorus does not change the lattice constant with respect
to that, because in the same phosphorus atom, lattice. Now why should you go for this?
This type of structure has a benefit. One is gallium indium arsenide as you know has
much better mobility than gallium arsenide, because indium arsenide has got higher mobility.
When you mix with gallium, gallium indium arsenide has got better mobility ñ but in
between indium arsenide and gallium arsenide. The band gap is also different ñ lower than
that of gallium arsenide, but we get a fairly decent band gap there. The main reason is
that it gives better mobility and of course, you put this material that has got a wider
band gap region on top of that.
Here, the advantage of gallium indium arsenide over gallium arsenide is not merely higher
mobility but it also has lower surface state densities ñ that is what I have seen. Gallium
indium arsenide is easier to passivate compared to gallium arsenide. This is one of the key
reasons why people are trying to look into that. Of course, the added attraction for
those who are working on indium phosphide substrates, it is the same material ñ that
is of course on a lighter sense. Now there is another HBT. I am not going into details
of that ñ it is only the change in the material but performance-wise, it gives slightly better
performance compared to gallium arsenide. Of course, it has the disadvantage of lower
band gap ñ you pay for that in terms of lower band gap and associated effect on the breakdown
voltages, etc. Now you can have a double heterojunction. In fact, when starting with the gallium arsenide
HBT, I pointed out that the structureÖ. Let me just put it on the board here.
The structure of aluminum gallium arsenide. That is n and in fact lightly doped, then
p-type gallium arsenide. This is the emitter, this is the base and the collectorÖ in a
heterojunction transistor, we have one heterojunction. In a double heterojunction, we have this also
ñ AlGaAs. On that, you can have.Ö This may be n minus and then you have got n-plus layer.
The n-plus layer can be gallium arsenide, but the main thing is that both this junction
and this junction are heterojunctions. What is the merit of this? The merit of that is
whatever you gain here, you gain here also.
When the transistor is in the active region, not a big role for this, because this layer
will be depleted. But if the transistor goes into saturation as in switching circuits,
if you are not using ECL, then in TTL and structures like that, the transistor goes
into saturation, collectorñbase junction gets forward-biased and if the collectorñbase
junction is forward-biased, what happens? You have the carrier concentration like that,
stored charge there, this is n and of course here, you have got hole injection. Since this
is a wide band gap material, we saw that the hole injection into this region is negligibly
small: ni square by doping; ni square is very small, so the carrier concentration here is
very low and that is why you get higher beta ñ emitter efficiency is very high in spite
of lower doping; here also, it is lower doping, but we have got wider band gap. So if it is
forward-biased, the device is in the saturation region, if this were the same gallium arsenide,
you would have got the hole concentration reached like that. Of course, I just draw
a small region there to signify the transition region. You would have got high concentration
hole injection like that if it is a narrow band gap material.
If it is a wide band gap material, what happens here? Same thing. The hole injection from
the np region to the n region when it is getting forward-biased is very very small, because
the boundary value here is pn0 ñ p in the n region; pn0 is related to ni square; ni
square is small, because it is a wide band gap material, so you will have very small
injection there. That is what we mean by saying that the transistor will go to saturation.
Still, you will not have the stored charge effect; in double heterojunction or switching
transistors, this will be ideal. You do not have to use lifetime control techniques, etc.
All that you have to worry about will be charge in that region. No stored charge here, no
stored charge here, because this is a wide band gap material. We can minimize the stored
charge here by shrinking the base width, so you are in business for high-speed circuits.
That is one of the merits of double heterojunction transistors .
That is why I just thought I will point this out to you because this is something that
people have been exploring, but if you are working on any emitter-coupled logic circuit,
you do not have to worry about that, because there is no forward bias on the collectorñbase
junction and there is no saturation ñ it is only active. Still, you can make use of
that. Without using ECL, you can use it for high-speed devices. So you can see that bipolar
is a potential candidate for high-speed devices ñ HBTs. The other one that has gained popularity
over the past decade is the silicon germanium HBT.
What is the merit of silicon germanium? The merit is apparently or evidently, you have
got silicon germanium in the base, silicon in the emitter, because the band gap of germanium
is 0.7, the band gap of silicon is 1.1 and silicon germanium will be somewhere in between,
depending upon how much germanium is there. You have a narrow band gap material and a
wide band gap material and if you can grow one over the other and still get a defect-free
layer, then you can make heterojunctions with silicon. This is where the joy of the silicon
technology person is, because silicon technology is so much advanced today ñ if we can bring
in HBT in that, nothing like that. People are working on that and they came up with
the silicon germanium transistor.
Compatibility with silicon technology: I can have a silicon germanium transistor HBT and
I can have CMOS of silicon, I can have BiCMOS ñ bipolar and CMOS combination, which will
give real high speed; in fact, IBM has been working on it and it has succeeded in doing
this ñ combination of silicon germanium transistors and BiCMOS. Let us go through that and see
why and how this gives better performance or comparable to that of compound semiconductors
like gallium arsenide-based devices.
That is the structure: silicon emitter, silicon germanium p type, silicon n minus, silicon
n plus. You will be wondering why we are putting a lower band gap material, but that is the
whole key here. Let us go through this quickly and see what benefits we get ñ other benefits.
The band gap of silicon germanium, which is this base, depends upon x ñ the mole fraction
of germanium; x = 0 is silicon and x = 1 is germanium. Now the lattice constant of germanium
is different from that of silicon; germanium would match with gallium arsenide, 5.63 Angstroms
ñ that is; silicon is lower than that ñ 5.43 or so; you may say it is just 4 percent,
but 4 percent is quite large when you grow one layer over the other, because for each
25 atoms, there will be one mismatch. If you have to grow a layer over that, how do you
do that?
This diagram is a very popular diagram ñ silicon germanium people always flash this.
What we have on the y-axis is the critical thickness of germanium on silicon or silicon
germanium on silicon substrate. What is the maximum? I have discussed this earlier when
I talked of chain layers. You can grow a lattice match layer, epitaxial layer on a substrate
up to a certain thickness. I am not flashing the diagram again, but if you recall, if the
lattice is wider, the top layer gets compressed like that and elongated that way so that there
is no defect, there is no misfit, no dislocation, you get a single crystal but a strained layer
ñ that is better than a defective layer.
Now if you keep on growing, the stress keeps on building up and at a particular point,
the bonds will break and you get a defect. We discussed all this earlier ñ just recall
that. There is a maximum thickness layer that you can tolerate, which does not have a defect
but it is strained. You can see here the strained layer maximum thickness. This curve shows
you one of the theoretical calculations that has made strain equilibrium ñ the maximum
layer when the strain is in equilibrium. For example, I can grow germanium on silicon but
the thickness of layer that you can grow is about 10 Angstroms. It may be just two atomic
layers, but you keep on reducing x, go towards silicon. You can see that I can go even to
1000 Angstroms; x is 0.2 ñ 20 percent germanium, I can grow a layer about 100 Angstroms that
is defect-free. This was the initial theoretical value, but if you do this growth by ultra-high
vacuum techniques like MBE, etc., you can get thicker layers that are defect-free in
the sense that as you grow, you do not provide the high temperature required for them to
break up the bond and so it grows without breaking the bond for thicker layers ñ that
is what is shown here in the dotted line.
The dotted line is actually the one that is called metastable ñ you grow it at lower
temperature; though it has a tendency to create defective layers at this point, if you grow
at lower temperatures, you can get thicker layers; not too thick ñ you are not pitching
up on very thick layers, you are talking of base width microns and you are happy with
that; if we are able to get 1000 Angstroms, that is quite sufficient. I do not know whether
you are able to see but there are points one , two, three, four, five ñ those are the
experimental points. People have grown silicon germanium by MBE and found that strained layers
that are defect-free can be grown with that thickness ñ that matches with this curve,
dotted line. What we are telling is I can have 0.2 x and still get a strained defect-free
layer that is about 2000 Angstroms. Now that gives us confidence in this technology and
people have gone ahead with making these devices.
Let us take a look at the band gap. There are a number of curves we have. All that you
have to understand is if you just have a silicon germanium alloy, you will get a band gap some
particular band ñ no strain, just make an alloy of silicon germanium, single crystal
and grow a thick layer, the intermediate layer will be defective, but the top layer will
be silicon germanium with no defect ñ thick layer; that is not suitable for making HBT,
but if you do that, that will be the band gap, the top curve.Germanium is 0.72, silicon
is slightly more than 1.1 and in fact, we will talk of 1.111 or 1.115 ñ that range
we will talk of. This curve shows it is varying from about 1.15 toÖ it all depends also on
what temperature you are measuring that at ñ 2.72, but what we are talking of is silicon
germanium that is strained ñ the top layer is unstrained silicon germanium. When the
layer is strained, this lattice is under stress, the energy band diagrams also undergo changes
and because of that, you get different band gap ñ in fact, you get lower band gap and
you can see that here and in fact, that suits you, you can mix very small quantities of
germanium and still get a band gap that is lower than that of silicon.
The top curve here is unstrained silicon germanium. This particular curve is actually strained
silicon germanium on unstrained silicon. If you grow thick silicon substrate n type, grow
silicon germanium, that is a strained layer, then this is the band gap that they have estimated.
You get a band gap that follows this , which is much smaller than that of silicon germanium
unstrained. You can see that for about 0.2, you will have about 1 electron volt; instead
of 1.15 here, you have got 1 electron volt ; you have got about 0.15, which is about
0.2 ñ in that ballpark; you have got a material that is strained but with a band gap definitely
smaller than that of silicon by about 0.1 electron volts; that is good enough ñ to
get that e to the power of delta EgB by kT, e to the power of 4 ñ 50 times more. What
is the benefit of using that? Now let me go back to this diagram. You can
put a layer of 1000 Angstroms very comfortably ñ with a base width 1000 to 2000 Angstroms,
with a band gap that is smaller than that of the emitter. Compared to a silicon NPN
transistor ñ silicon germanium base with changeover, what benefit do you get? We have
seen already what benefit you get ñ you can dope that region heavily.
Apart from that even in gallium arsenide-based devices, that holds good. The collector current
Ic is larger for a given VBE. How do you say so? We have been writing this expression on
and off for the collector current. That depends upon q DnB ni square by total integrated doping
into e to the power of VBE by VT ñ this is a very familiar formula for you. You know
that you can increase the doping for a given current or you can actually.Ö When I reduce
this band gap, ni square goes up. If ni square goes up for everything same, Ic goes up exponentially,
ni square is proportional to e to the power of minus EgB by kT, so the collector current
is larger for the same VBE. What I am comparing is a silicon bipolar junction transistor and
a silicon germanium base HBT. A silicon germanium base HBT will give higher collector current
for the same VBE. In other words, you can go to lower VBE to get the same current.
We can go to the active region. How is that? That is the way the carrier concentration
is. Now the band gap is smaller and you get a larger current. How do you do that in this
diagram? This is fixed, that goes up and it is like that for the same voltage because
after all, np(0) is equal to np0 e to the power of VBE by VT and np0 is actually the
minority carrier concentration, thermal equilibrium. For a given doping, when ni square is larger,
this will be larger and that is why that is larger. np(0), the minority carrier concentration,
is larger if the band gap is here, ni square divided by pp0. This is ni square divided
by doping. If that increases, that increases and if that increases, np(0) increases, current
decreases. It is a simple way of understanding from the formula we have put in the slide.
So for a given voltage, the current is more; this current does not change ñ what you have
injected to the metal, total current increases.
What is the impact of that? The impact of that is that beta improves ñ collector current
increases means beta improves, the base current does not change much because the base width
we are talking of is very small, volume recombination we saw is negligible in HBT ñ it all depends
upon the emitter efficiency. Then, the increase in Ic decreases the emitter charging time
or any charging time for a given VBE for a given change in collector current; the collector
current change is more for a given change in VBE in this case compared to BJT and so,
all the charging times are smaller; particularly, the emitter charging time is smaller and this
improves the switching speed.
It is a well-known fact that if the currents are larger for a given change in voltage,
the switching are large; even an integrated circuit has a device, because after all, the
cut-off frequency will depend upon the transit time in the base and the charging time in
the emitter ñ emitter base capacitor charging time. The emitter base charging time depends
upon how much the current is ñ that is larger here for VBE. You have to charge the smaller
voltage to get a particular current or for reaching a particular voltage, you have a
larger current available, so it charges faster. That is why the charging time of the emitter
base capacitor is smaller, so .
Another very important advantage that is there for all HBTs but is mainly exploited in the
case of HBT silicon germanium transistors is this factor ñ germanium content (x) in
silicon germanium can be controllably graded across the base. I do not know whether it
rings a bell. As I move from here to here, I can vary the germanium content, I can make
the germanium content more here and I can keep on changing that; that means I can vary
the band gap across the base ñ narrower band gap, wider band gap here. You can do that
by just growing, varying the mole fraction of germanium by adjusting the flow rates or
by adjusting the germanium content in MBE ñ you can vary the band gap. What do you
get out of that? You get of course the band gap change; if you change the band gap, it
introduces an electric field.
Do you remember? When I draw a diagram like this, the carrier concentration is linear
varying ñ injected carrier concentration. What you are assuming in this carrier transport
is through diffusion ñ there is no electric field here and the transit time is w square
by 2Dn. You can increase the speed or reduce the transit time by not only reducing w but
if I can bring in an electric field, the transport will be faster. The velocity by diffusion
is smaller compared to velocity by drift. If I introduce drift by some means, I am in
this. I can increase the speed and the transit time can be reduced by a factor of 5 to 10.
From 10 GHz, you may go to 100 GHz ñ people go there; 100 GHz is not a joke ñ it is a
reality, it is not a myth; reported 100 GHz silicon germanium transistors, HBT. One of
the reasons is this ñ you can introduce that. I will just quickly go through the analysis
of this. Now I have put this as a statement here: you have an electric field that aids
the electron transport across the base, thereby decreasing the base transit time.
When I say an electric field is present here aiding the transport of electrons, the field
direction is plus here, minus there ñ electrons are moving in that direction and it attracts
towards positive, so there is a field in this direction, a built-in field comes in and in
the next diagram, you will see it clearly.
The tautb is of course reduced, fT is better due to smaller transit time.That diagram tells
us the whole story of grading the band gap across the base. I just put of course both
the conduction band and valance band. It could be that the valance band is not changing,
only the conduction band , but I have shown both. Now what we are interested in seeing
is the impact of this change in the conduction band and the valence band ñ you can see smaller
band gap Eg at WB, larger band gap Eg at 0. I hope you would understand that the x that
I am using here is for the distance ñ I tried to change it but I am stuck with the thing,
so I am just keeping that; this x is the distance ñ it is not the mole fraction that we are
talking of here now when you write the equations.
x silicon germanium, there we can put x is the mole fraction, but this is actually the
distance. So you can see that if I have an energy band diagram like this, conduction
band like this, what does it mean? Whenever there is a tilt in the band gap, there is
an electric field. A tilt in the band gap like this would mean that this is positive
here, minus here. The electrons that are injected here find themselves in an electric field
that pushes them in that direction towards the collector ñ drift velocity.
You do not worry about what is happening to the hole here, because we do not care about
the holes. In fact, in this type of thing, the electric field, the nice thing about this
added thing is if you have a band gap like this, the field for electrons will be in this
direction, the field for the holes will be in the other direction. The holes are being
prevented from moving into the emitter and the emitter efficiency further improves. This
is a classic thing ñ the nice thing about this type of structure band gap varying continuously
across the base region. Eg is a function of x. In fact, what you can also understand isÖ.
We will come back to this if we have time to see the physical meaning of this, but right
now from this diagram itself, you can see that there is an electric field like this.
Whenever there is a band gap changing like that, the field is plus on the lower energy
side and minus on that side.
Now let us go back and see what is the magnitude of this field ñ we will rush through that
very quickly, very simple equation. Instead of writing the current as due to diffusion
alone, since you know that if there is electric field, you have to write diffusion current
and drift current, it is q Dn dn by d of x. I am writing this equation at thermal equilibrium;
no bias ñ we are not applying any bias, so it is q Dn dn by d of x. I am putting np0
instead of n ñ thermal equilibrium and it is a function of x and q np0 , E is the electric
field and that should be equal to 0. No electron current under thermal equilibrium conditions.
np0 is actually the thermal equilibrium value of np ñ electron concentration in the p region.
If I remove the 0, it will be actually injected carrier concentration and it is a function
of distance x from the emitter to the base, from the emitterñbase junction. Why? Because
np0 is ni square by pp and pp is the doping; doping may also be function of x; even if
it is constant, you will have the np varying; even if it is constant, if the band gap is
varied, ni is varying, ni is actually increasing as you move from the emitter to the other
end and so np actually increases. The thermal equilibrium concentration value is more in
that region as x goes on. Now to find out the field, you have to equate this current
to 0, because it is thermal equilibrium. There will be a concentration gradient existing,
but if the current flow has to be 0 when the concentration gradient is present, the only
way possible is presence of electric field and that comes because of this charge redistribution.
So that electric field is the one we are going to find out. Equate this diffusion current
to drift current and plug in this np0 equal to ni square by NA(x).
I have rewritten that equation. Instead of np0, now I am putting ni square by NA, so
q Dn d by d of x np0, this is ni square by NA, q np is ni square by NA. That is 0 in
thermal equilibrium condition. That is what we mean by equation 1 and 2. Now, all that
you do is differentiate this. When I differentiate that, the first term, I am retaining the second
term retaining as it is here ñ q ni square by NA mun E is retained as it is; remember
that NA may be a function of x, so we will take that also into account and E is a function
In the first term, if I take NA also as a function of doping, then I get q Dn, I pull
out 1 over NA, qdV formula: first function into differential of second function ñ it
is 1 over NA, differential of second function is d by dx of ni square I am not differentiating
it right now but keeping it intact plus second function is ni square into differential of
1 by NA, which is minus 1 by NA square into dNA by d of x, because NA is a function of
x; if NA is independent of x, that term goes off. This term is written here. Now, E is
given by equating to 0. What I do is I find E equal to.Ö I take all these to the other
side, divide by q ni square by NA. Let us take this term and divide by q ni square by
NA.
ni square by NAÖ when you do that, you get that. Just rearranging the things, what you
get is actually that the electric field is equal to Dn by mun. We can just see it here.
You get Dn by mun when you divide the whole thing. Dn by mun is there for both the terms
and in this term , ni square cancels and you get 1 by NA and since it goes to other side,
it is plus, so you get Dn by mun into 1 by NA dNA by d of x. Similarly, the second term
becomes Dn by mun into 1 by ni square d by dx ofÖ.
You can just see for yourself by rearranging those ñ this is what you get. Now, the electric
field. Dn by mun is actually equal to VT kT by q, so VT by NA into dNA by d of x ñ this
is a very familiar formula for some of you who have taken more courses. If there is a
doping gradation, there is an electric field due to the built-in field due to that. Wherever
there is doping, gradation will be there, that is the first term. You get now additional
terms due to the band gap variation. That is VT by ni square d by dx of ni square. Now
let us take a look at that. Now this ni square is actually proportional to E to the power
of minus Eg by kT. I differentiate this and divide by ni square.
So d by dx ni square is actuallyÖ When I differentiate that, you get minus 1 by kT
ni square, is the exponential term. When you differentiate this with respect to x, you
get ni square itself, because it is exponential e to the power of minus k ñ minus Eg by kT,
but 1 by kT into differential of dE by dx, df by dx is a function you are differentiating
once and then.Ö Simple differential equations, differential
formula, so you get that because when you differentiate, you get whole term on the numerator,
by kT, differential of this, minus is there because of that. I plug in this quantity into
the previous equation, which actually has VT by ni square into dEg ñ we will just go
back to that. See here, the first term is VT by NA into dNA by d of x ñ that I am retaining
as it is; second term, I am substituting for d by dx ni square, which is nothing but minus
1 by kT into dEg by dx into ni square; ni square cancels, so I get that.
What you get now is VT by NA into dNA by d of x plus VT by kT into dEg by d of x. It
tells you totally now that the electric field is decided by doping gradation plus gradation
in the mole fraction ñ that is gradation in the band gap. Now let us take a look at
that. That was missing there. This is the formula that we have got now, which tells
us what is the electric field thermal equilibrium. Now when you forward bias the emitterñbase
junction and inject carrier electrons, that electron experiences this built-in field.
What is the direction of this field? Dn by d of x. If the doping is changing from here
to here, more doping and less doping there , if there is doping naturally and if you
do a diffusion, it will have doping here, n0 e to the power of minus x by L, something
like that ñ exponentially it is falling. Then, the field actually will beÖ. Dn by
dx is negative in that direction. What about the second term ñ plus dEg by d of x? If
the band gap is going on falling here, Eg is going on falling and that also in that
direction. Both are in this direction . Both the electric fields actually support the faster
movement of electrons from the emitter to the collector ñ this is the added benefit
that you get for silicon germanium HBT compared to BJT with silicon alone. You can also put
this term particularly asÖ sometimes, you will see this put as VT is kT by q, so that
becomes 1 by q.
It is 1 by q dEg by d of x. The first term is due to doping gradation and it gives the
electric field component aiding the electron flow, provided doping gradation is in that
direction from the emitter to the base collector, which is usually the case when you do diffusion.
Take a substrate, diffuse from top, doping is higher on the top and so as you move down
towards the collector, doping is reduced ñ that is aiding the electron flow. What about
the second term? The second term is due to the band gap gradation and it aids electron
flow if dEg by dx is negative. It means actually that the band gap at the collector edge, at
WB should be less than the band gap at the emitterñbase junction .
Now let us get an idea of what sort of fields are there in the transistor. You see here
this 1 by q dEg by d of x ñ that is the second term; we are seeing what is the magnitude
of the field due to the band gap gradation ñ that is that. For a linear gradation, you
have got the linearly varying band gap ñ delta Eg varies as a function of x linearly,
then you have got delta Eg by d of x, the negative sign is there, that minus that by
WB. Your dEg by d of x is negative, that is why it is put it there. Magnitude is that
quantity ñ it is delta Eg by WB. Now if delta Eg is 0.15 electron volts ñ
vary the band gap from the collector to the emitter by 0.15, then for a 0.1 micron base
width, we can still get a strained layer that has band gap and has no defects. The electric
field is 0.15 by 10 power of ó5, 15 KV per cm. This is the benefit that you get. Current
by drift and diffusion. Now, there are a couple of things I want to point out here. If you
take the electric field in the case of doping gradation, what sort of electric field do
you get?
Let me put that. If I take the NA(x) equal to NA0 into e to the power of minus x by L,
what is the electric field? No band gap gradation. You get this all the time in the case of BJT
ñ all modern BJTs have an electric field built in the base because of this gradation
that you saw just now. What is that actually? VT by NA of x into dNA by dx. We just now
saw this; go back and see that term ñ that is due to doping gradation. In fact, you can
derive that equation right from the fundamentals when ni is constant ñ we will get only that
one term. The way we started, we have taken ni varying and nA also as varying. When you
take ni constant, you will get the second term dropped out, if delta Eg by dx is 0.
If that is the thing, what is the electric field? L is a constant, the decay constant
for doping. What is the electric field? VT by L. dNA by d of x is this into divided by
L and that is the quantity. You will get minus ñ that is minus. From here, the minus sign
comes and this tells you that the electric field is in that direction ñ the minus sign
is right and it must be there; otherwise, whatever we have been telling is not correct.
The minus sign comes from there and in fact, it tells that dNA by dx is negative and it
is a constant electric field. What will be the magnitude of this quantity VT by L? Some
idea we must have. That depends upon L.
How much will L be? I go back down here. NA(0) is NA0. NA at WB is NA0 into e to the power
of minus WB by L. What I am getting at is this: that field depends upon the base width
and L. Therefore, NA(0), the doping at the emitter base , divided by NA at WB is actually
equal to e to the power of WB by L. This WB by L therefore depends on logarithm of that
. Supposing WBÖ. You will get the benefit of this only if the doping here is smaller
compared to doping there ñ gradation. When we have doping concentration like this ñ
0 and x, some variation like that, NA0 and that is nA(WB), that is what you are talking
of ñ exponential doping concentration gradation, you will get the doping variation like this.
Now WB Ö.
Supposing this ratio is 100 or let us say this ratio is 54 ñ doping at the emitter
and doping at the collector edge, that ratio is 54, which is reasonable to assume, what
will be this factor? 4, e to the power of 4 is 54; if this ratio is 54, you get this
as e to the power of 4, so WB by L is equal to 4. If WB by L is 4, if WB is let us say
0.4 micron, I am just giving some example, L will be equal to 0.1 micron ñ I am just
computing some number to see what is the order of the field here. If this ratio is 54, this
is e to the power of 4; that means WB by L is 4 and if WB is 0.4, L is 0.1.
If L is equal to 0.1, I am just computing the . What is the order of this L? It may
be 0.1, 0.2, whatever. If it is 0.1, what is the VT by L field now? 25 millivolts divided
by 0.1. So for this case, this is actually minus VT by L, which is actually equal to
25 millivolts into 10 to the power of ó3 divided by 10 to the power of ó5. How much
is that? This is 2.5 KV per cm. I just took this example to illustrate that you will have
in the conventional transistor itself a built-in field of this order. In fact, the factor WB
by L is called the field factor, which actually decides what is the potential change there
due to that electric field. You have this much field and you also saw that in this case,
I have taken an example where the base width is 0.4. If I take the same example with base
width is 0.1, one-fourth of that will be 4 times that ñ 10 KV per cm. I took a wrong
example there of L equal to 0.1, taking WB as 0.4, but if you have to compare with WB
equal to 0.1 microns and if this is 4, L will be 0.1 by 4 and this will be 10 KV per cm.
It just depends upon the base width and the doping gradation there. In other words, it
depends upon the WB by L ratio and that L depends upon the doping concentrations.
Now, let us go further down. This is the structure that.Ö Now, we can run through the results
that people have reported. They have made devices that can go right up to 100 GHz. You
go to the Web and say ësilicon germanium HBTí, you will see IBM results put across
ñ they have made BiCMOS, they have made ECS and all sorts of things using the silicon
germanium technology. It is a big contender for gallium arsenide technology ñ silicon
germanium. This is the answer, challenge thrown by silicon technology towards gallium arsenide-based
HBT, but still the high electron mobility transistor stays there and there is no contender
here ñ that is one thing that you have to remember. But there are other issues: when
you want to go to harsh environments like higher temperatures, you cannot use base materials
that have smaller band gap. Silicon germanium has a smaller band gap compared to gallium
arsenide, etc., and so for those harsh environments, you still have to look at gallium arsenide-based
devices.
This is the type of structure that you see. In a bipolar technology, you have the integrated
circuit, you have the p minus layer substrate and n plus collector contact from the top,
then n-type layer, collector and you can see the narrow p plus epitaxial base made of silicon
germanium and contact is taken laterally there. Of course, I am not going through the technology
of this ñ this is a self-aligned technology where you can grow layer by layer without
too much difficulty in aligning single alignment defined already, you can realize this. This
is that. This is the emitter, base, collector; the emitter is polysilicon, n-plus polysilicon
ñ this is another thing that is actually very attractive for self-aligned techniques;
you also put n plus contact ñ it is slightly diffused into that region; then you make a
single crystal ñ a very shallow n plus junction or n-type junction and then a contact is taken
on the n plus layer. This is the structure here. Of course, people use silicon technology
for trench isolation, etc., where you put a trench, oxidize it, put polysilicon... all
those techniques are there for isolating devices integrated circuits. IBM and the layers, the
device structures are microns.
Doping profile. I will just quickly run through this because there is no more explanation
needed. This particular structure that is used here does not show doping concentration
changes; arsenic emitter doping , base doping gradation and germanium; we have not shown
that gradation that much in this case, but you can see from here to here that the gradation
is there; germanium gradation is there, that is, the band gap is varying from larger to
smaller value other way and there is a built-in field coming up; the percentage of germanium
goes right up to about 10 percent, 0.1, so band gap is narrower compared to that.
As I said right at the beginning, the collector current is larger in the case of silicon germanium
BJT compared to the.Ö There are two curves: collector current in the silicon germanium
HBT , collector current in the case of silicon BJT ñ that is about a factor of 4.5 larger;
one is due to the field and the other one is heterojunction effect ñ both. The base
current is the same and the device size as we can see is 0.5 by 2.5 micron ñ very small
strip for the emitter, which reduces all the capacitance.
Then, cut-off frequencies. Once you get actually larger current for a smaller voltage, the
charging times are all smaller and you get much higher cut-off frequencies. For this
particular device that has been reported quite some years back, the cut-off frequency depends
upon the collector current, because that gives better charging capability for the emitterñbase
junction and that is why as you go to higher currents, you get higher cut-off frequency,
but what you have to do is compare with the single BJT ñ much higher. For 20 or 25, we
get about 50. This is an older result that I am putting here, but I have seen 100 GHz;
these all are GHz, 50 GHz. In the emitter-coupled logic circuit they have made, there BJT, if
we get 30 picoseconds, the gate delay ñ unloaded of course, of 1, fan-in by fan-out 1, that
is 30 in the case of BJT if you have minimum and here you get 20 for HBT . In fact, they
have gone down further below that picosecond range ñ much lower than that when you go
to GHz. These are the merits that we get in the case of silicon germanium. In fact, I
think I do not need to discuss more about these things. In summary, silicon germanium
is playing a key role today and a lot of people are putting in efforts, because we can realize
silicon germanium transistors bipolar along with CMOS integrated circuits. With that note,
I will close today. We will take on the circuit aspects in the next lecture.