 First we will take a close look at some small
angle grain boundaries in Silicon. Whereas they are the most simple boundaries imaginable,
they are still complex enough to merit some attention. They are also suitable to demonstrate
a few more essential properties of grain boundary dislocations. 
  Lets first look at a pure tilt boundary as
outlined in the preceding paragraph. Below is shown how edge dislocations can accommodate
the misfit relative to the S = 1 orientation (for a boundary
plane that contains the dislocations lines). 

 
 
  The distance d between the dislocations is for small tilt angles
a as before given
by 






 This
is a simple version of a general relation betwen Burgers vectors and misorientation in small
angle grain boundaries called
Franks formula
(more correctly FrankBilby formula).



  In real life it
looks slightly more complicated  but not much: 
  


 
This is an early HRTEM of a small angle tilt boundary in Si. The red lines mark
the edge dislocations, the blue lines indicate the tilt angle. 

This picture nicely illustrates that we have indeed a S
1 relationship in the area between the dislocations, i.e. a perfect crystal. The dislocations
are not all in a row, but that does not really matter. 

 
 Next, we look at twist boundaries. 
  These and some of the other
boundaries were artificially produced to study the structure. Two specimens
of Si with a desired orientation relationship were placed on top of each other and "sintered"
or "welded" together at high temperatures. This process, first called "sintering"
is now known as "waferbonding" and used for technical
applications. 
  The
result for a slight twist between {100} planes is shown in the next
picture: 
  

This is a remarkable picture. As ascertained
by contrast analysis, it shows a square network of pure screw dislocations
. The picture is remarkable not only because it shows a rather perfect square network
of screw dislocations, but because it is obviously a bright field TEM micrograph, however with a resolution akin to weakbeam conditions.

  Pictures like this one are
obtained by orienting the specimen close to a {100} (or, in other cases, {111})
orientation thus exciting many reflections weakly. All dislocations are then imaged, but the
detailed contrast mechanism causing the superb resolution is not too clear. 

Lets first find out why a network like this can produce the required
twist. We do this in reverse order, i.e. we will construct a screw dislocation network in
a perfect crystal and see what it does. 
  We start by looking at {100} lattice planes below and above
the (future) grain boundary plane. They are exactly on top of each other, we obtain a (trivial)
picture of a formal lowangle twist boundary with twist angle a
= 0^{o}. 
 



Now we introduce two screw dislocations running from left to
right. Referring to the same
kind of picture in chapter 5, we see that the lattice planes below the screw dislocations
are bent to the right (blue lines), above the screw dislocations to the left: 
 

  

 With many dislocations, the
average orientation of the lattice planes below the small angle grain boundary will rotate
to the right, above to the left. The combined effect is shown below. 
  
 If we want to
rotate not just one set of lattice planes, but all of the top part of the crystal, we need
at least a second set of screw dislocations. This produces a screw dislocation network of
the kind shown in the TEM micrograph above. 
  The relation between the twist angle a
and the dislocation spacing d is again a simple version of the general case given
in Franks formula: 
  
  A detailed
drawing of this dislocation network structure can be viewed in the link. 
  With luck, it is possible to image the lattice plane in
a HRTEM micrograph. The link shows examples  the only HRTEM image of screw dislocations obtained so far. 

The exact geometry of the network for the same twist angle a
in an arbitrary twist boundary depends on many factors: 
  The
twist angle a which determines the spacing between the
dislocations. For S = 1 boundaries it simply the twist angle
itself, for arbitrary boundaries it is the twist angle needed to bring the boundary to the
closest low S orientation 


The Burgers vectors of the
dislocations. Even in smallangle grain boundaries they could be perfect, or split into partials.
In arbitrary boundaries they must be grain boundary dislocations with a Burgers vector of
the proper DSC lattice. Note that a network of grain boundary
screw dislocations simply superimposes some twist to whatever orientation the boundary has
without those dislocations. 
  The
type of the dislocations. For an arbitrary twist plane,
the Burgers vectors of the possible dislocations are not necessarily contained in the grain
boundary plane; the required pure screw dislocations do not exist. In this case mixed dislocations
must be used with a component of the Burgers vector in the grain boundary plane. 
  The symmetries
of the two crystal planes in contact at the grain boundary  even lowangle twist boundaries
with a twist around the <100> axis can be joined on planes other than {100}. 
  The
complications that may arise because the (perfect) dislocations split
into partials. Obviously, that has not happened in the case shown above. The reason
most likely is that the splitting would have to be on two different {111} planes inclined
to the boundary plane (look at your Thompson tetrahedron!) which leads to very unfavorable
knot configurations. Since the distance between dislocations is of the same order of magnitude
as the typical distance between partials, we do not observe splitting into partials or dissociation
of the knots. 
 We now can understand the very regular square network shown
in the picture above  it is really about the most simple structure
imaginable. 
  But we still need to explain the interruptions in the network; the lines along
which the net is shifted. In fact, to see a very regular network like this you must be pretty
lucky; more often than not often (artificially) made twist grain boundaries look like this: 
  
  Both pictures show the result
of the attempt to make a pure twist boundary. Whereas the left one still looks like the picture above  just with more interruption of the network  the
right one does not convey the impression of a square network at all. 
  The answer is that these grain boundaries must accommodate
more that just a pure twist: There is also a tilt component and the plane of the grain boundary
is not exactly {100}. We will pick up this subject again in the next paragraph; more information about the righthand side picture can be
found in the link. 