1 Chapter 1: Structure Solids can be divided into 3 categories 1) Crystalline: An ordered structure where each atom can be described by the translation of a small unit cell, repeated over and over throughout the sample. 2) Polycrystalline: Composed of crystalline regions which are randomly oriented with respect to one another. 3) Amorphous: No long range order. In this course we’ll study 4 different crystal structures that are commonly found in electronic materials. 1) Diamond Structure This is the most stable structure for the elemental semiconductors silicon and germanium (and is sometimes taken by carbon). It’s made up of atoms bonded to each other in a tetrahedral (T d ) arrangement forming an extended network. (Imagine how other atoms would attach to this cluster, forming an extended network.)
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Chapter 1: Structure
Solids can be divided into 3 categories1) Crystalline: An ordered structure where each atom can be described by the translation of asmall unit cell, repeated over and over throughout the sample. 2) Polycrystalline: Composed of crystalline regions which are randomly oriented with respect toone another.3) Amorphous: No long range order.
In this course we’ll study 4 different crystal structures that are commonly found in electronicmaterials. 1) Diamond Structure
This is the most stable structure for the elemental semiconductors silicon and germanium (and issometimes taken by carbon). It’s made up of atoms bonded to each other in a tetrahedral (Td)arrangement forming an extended network.
(Imagine how other atoms would attach to this cluster, forming an extended network.)
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To calculate things like density, or the distance between various atoms, we conveniently use thefollowing unit cell.
What I do is to visualize a face-centered cubic (FCC) unit cell. Then I divide it into 8 equalmini-cubes. The center of each mini-cube is a tetrahedrally coordinate site, and if I place an atomin every 2nd one of these sites, then I have the diamond structure. (note: all the atoms areidentical, whether I’m imagining them as being in the “FCC” sites or in the “Td” sites.)
Question: What’s the packing fraction of a diamond structure?Answer: Packing fraction is the ratio of the volume of the atoms in the cell, divided by thevolume of the cell. This cell has (8 · 1/8) corner atoms + (6 · ½ ) face atoms + 4 internal atoms = 8 atoms per cell.So Vatoms = 8·4π/3·r3 where r is the atomic radius.The edge length of any unit-cell is determined by the fact that the nearest neighbor atoms areactually touching each other, ie are separated by 2r. For diamond, the nearest neighbor to each of the “FCC” atoms are “Td” atoms. So, 2r equals
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one-half of the body-diagonal of one of the mini-cubes. (It’s helpful to remember that the face-diagonal of a cube of edge-length x is %2·x and that the body-diagonal is %3·x.)In the diamond case, then, we have 2r = ½ · (%3·a/2) where a is the edge-length of the big unitcell.So Vcell = a3 = (8/%3)3 r3 .
Then the packing fraction is = 323
3
83
3 034π rr
= .
This number is independent of the size of the atom! And 66% of the cell is “empty”!!
Exercise: Look up the atomic radius and atomic mass of silicon, and calculate the density of asilicon crystal, in g/cm3.
2) Sphalerite Structure (also called zinc-blende, or ZnS)
This is the most commonly found structure for the compound semiconductors, such as GaAs andCdSe.
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You can see this is just like the diamond structure, only the “FCC” atoms are one type ofelement, and the “Td” atoms are the other type of element. Keep in mind that we could also havedrawn the cube in this picture such that there were white atoms in the “FCC”positions and yellowatoms in the “Td” positions.
Each atom is tetrahedrally coordinated, and bonded only to atoms of the opposite element.
Exercise: Look up the radius of Ga and As atoms, and calculate the density of GaAs (in this case,we’d have that (rGa + rAs) = ½ the mini-cube body diagonal.)
3) Close-Packed Structures
About two-thirds of the metallic elements have a close-packed structure, ie each atom is bondedto as many neighboring atoms as possible.
Each atom in the structure is identical, and has 12 nearest neighbours. Why 12? Because that’sthe maximum packing density of equivalent spheres that fit together in 3-D space.
Each atom has 6 neighbours in the plane around it, with 3 more sitting in three-fold hollowsabove it and 3 more below it.
Looking at the second layer, you can put the atoms in a choice of two different registries, itdoesn’t matter which. Then in the third layer, you can either put the atoms in the other choice of2nd layer registry, or back into the registry of the 1st layer. This choice creates one of twodifferent close-packed structures, ABCABC or ABAB respectively.
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3a. Hexagonal Close-Packed (HCP) This is the ABAB structure. Looking at the picture above,you should be able to see how it relates to the following unit-cell.
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It’s a hexagonal prism, with 12 atoms at the vertices of the hexagonal faces and 2 atoms in thecenter of each hexagon. There are also 3 atoms half-way up the axis of the prism, sitting in thethree-fold hollow sites of the atoms in the hexagons.
Question: How many atoms are in this cell?Answer: (2·½) atoms in the faces + (12· 1/6) atoms at the corners + 3 internal atoms = 6(the tricky thing is seeing how the corner atoms are shared by 6 cells, three above and threebelow).
It’s a bit of a PITA that the unit-cell isn’t cubic. So instead of just Pythagorus’ theorem we haveto use some trigonometry as well.
Question: What is the ratio of the height of the prism to the atomic radius?
Answer: First consider the projection of a second-layer atom position onto the plane of the 1st
layer. It’s in the center of an equilateral triangle. Each side equals 2r in length (because theatoms in the triangle are nearest neighbors). With some trig you can figure that this point in thecenter of the triangle is r/cos(30E) away from the point in the center of the hexagon face. Second, consider the line from the center of the hexagon face to a second layer atom as thehypotenuse of a right angle triangle, with the base that we’ve found equal to r/cos(30E). Thehypotenuse equals 2r, because these atoms are neighbours, so the height of the second layerabove the first equals ((2r)2-(r/cos(30E))2)1/2 = 1.633r. The third layer is the same distance above again (right??!!) sothe height of the prism equals 3.266 r .
Exercise: Calculate the packing fraction of this unit cell.
3b. Cubic Close-Packed (HCP) This is the ABCABC structure.
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Looking at picture (a) above, a close packed plane is shown (like a rack of billiard balls). Sevenatoms are added to the next layer in picture (b). Note which of the hollow sites in the first layerare covered. Now looking at (b), if we consider a third layer we can either put 3 atoms in registrywith the central three atoms of the first layer (that would be ABAB registry) or we can put oneatom above the center of the second layer (this would be the ABCABC registry). In picture (c)we see that the ABCABC registry leads to a cubic structure. In fact, if you divide (c) up into 8mini-cubes, each one is a face-centered cube. That’s our unit cell! The CCP structure is equal tothe FCC structure.
This is an easy structure to remember. Notice that each corner atom is nearest to a face atom, so2r = ½ of the face-diagonal = ½ · %2·a , where a is the cube edge-length.
It’s not that obvious looking at the FCC cell that each atom is surrounded by 12 nearestneighbours.
In both the HCP and CCP there are 12 nearest neighbours, and they have the same packingdensity. It must be interactions between 1st and 3rd nearest neighbours that make some elementsprefer HCP to CCP (there’s roughly an equal number of each in the periodic table).
