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Obviously, embedding the
three slices of Si
that form a bipolar transistor into a Si crystal will not do you any
good - we just look at it here to see just how ludicrous this idea would
be: |
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What is the problem with this approach? Many
points
- The transistors would not be insulated. The Si substrate with a
certain kind of doping (either n- or p-type) would simple
short-circuit all transistor parts with the same kind of doping.
- There is not enough place to "put a wire down", i.e. attach the
leads. After all, the base
width should be very small, far less than 1 µm if possible. How do
you attach a wire to that?
- How would you put the sequence of npn or pnp in a piece of
Si crystal? After all, you have to get the right amount of , e.g.
B- and P-atoms at the right places.
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So we have to work with the really small
dimensions in z-direction, into the Si. How about the
following approach? |
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This is much better, but still not too convincing. The
pro arguments are:
- Enough space for leads, because the
lateral dimensions can be as large as you want them to be.
- It is relatively easy to produce the doping: Start with p-type
Si, diffuse some P into the Si where you want the Base to
be. As soon as you overcompensate the B, you will get n-type
behavior. For making the emitter, diffuse lots of B into the crystal and
you will convert it back to p-type.
- The base width can be very small (we see about this later).
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But there is a major shortcoming:
- Still no insulation between the
collectors - in fact the Si crystal is the collector of all transistors
and that is not going to be very useful.
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Easy you say, lets add another layer
of doped Si: |
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This would be fine in terms of insulation, because
now there is always a pn-junction between two terminals of different
transistors which is always blocked in one direction for all possible
polarities. |
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However, you now have to change
an n-doped substrate to p-doping by over-compensating with, e.g.
B, then back to n-type again, and once more back to
p-type. Lets see, how that would look in a diffusion profile diagram: |
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The lg of the concentration of
some doping element as shown in the illustration above is roughly what you must
have - except that the depth scale in modern ICs would be somewhat
smaller. |
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It is obvious that it will be rather difficult to
produce junctions with precisely determined depths. Control of the base width
will not be easy. |
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In addition, it will not be easy to achieve the
required doping by over-compensating the doping already present three times. As
you can see from the diagram, your only way in the resistivity is down. If the substrate, e.g., has a doping of 10
Wcm, the collector can only have a lower
resistivity because the doping concentration must be larger than that of the
substrate, so lets have 5 Wcm. That brings
the base to perhaps 1 Wcm and the emitter to
0,1 Wcm. These are reasonable values, but
your freedom in designing transistors is severely limited |
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And don't forget: It is the relation between the
doping level of the emitter and the base that determines the amplification factor g |
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There must be smarter way to produce integrated bipolar
transistors. There is, of course, but this little exercise served to make clear
that integration is far from obvious and far from being easy. It needs new ideas, new
processes, and new materials -
and that has not changed from the first generation of integrated circuits with
a few 100 transistors to the present state of the art with some
100 million transistors on one chip. |
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And don't be deceived by the low
costs of integrated circuits: Behind each new generation stands a
huge effort of the "best and the brightest" - large scale integration
is still the most ambitious technical undertaking of mankind today. |
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We start with an n-doped wafer
(of course you can start with a p-doped wafer, too; than everything is
reversed) and diffuse the p+ layer into it. We will see what
this is good for right away. |
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On top of this wafer we put an epitaxial layer of p-doped Silicon, an
epi-layer as it is called for short.
Epitaxial means that the crystal is just continued without change in
orientation. The epitaxial layer will always be the collector of the
transistor. |
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Next, we diffuse a closed ring of
n-material around the area which defines the transistor deeply into the
Si. It will insulate the transistors towards its neighbours because no
matter what voltage polarity is used between the collectors of neighbouring
transistors, one of the two pn-junctions is always in reverse; only a
very small leakage current will flow. |
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Then we diffuse the n-base- and
p-emitter region in the epi-layer. |
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Looks complicated because it is complicated. But
there are many advantages to this approach: |
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We only have two "critical"
diffusions, where the precise doping concentration matters. |
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The transistor is in the epitaxial layer which, especially in
the stone age of integration technology (about from 1970 - 1980) had a
much better quality in terms of crystal defects, level and homogeneity of
doping, minority carrier lifetime t, ...)
than the Si substrate. |
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We get one level of wiring for almost
free, the p+ layer below the transistor which can
extend to somewhere else, contacting the collector of another transistor! |
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This leads us to the next big problem in
integration: The "wiring", or how do we connect transistors in the
right way? |
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© H. Föll (Electronic Materials - Script)
