ece 874: physical electronics
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ECE 874:Physical Electronics
Prof. Virginia AyresElectrical & Computer EngineeringMichigan State Universityayresv@msu.edu
VM Ayres, ECE874, F12
Lecture 02, 04 Sep 12
VM Ayres, ECE874, F12
Silicon crystallizes in the diamond crystal structureFig. 1.5 (a):
Notice real covalent bonds versus definition of imaginary cubic unit cell
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VM Ayres, ECE874, F12
Corners: 8 x 1/8 = 1
Faces: 6 x ½ = 3
Inside: 4 x 1 = 4
Atoms in the Unit cell: 8
VM Ayres, ECE874, F12
What do you need to know to answer (b)?
Size of a cubic Unit cell is a3, where a is the lattice constant
(Wurtzite crystal structure is not cubic)
VM Ayres, ECE874, F12
VM Ayres, ECE874, F12
Which pair are nearest neighbors?
O-1
O-2
O-3
O
1
2
3
VM Ayres, ECE874, F12
Which pair are nearest neighbors?
O-1
O-2
O-3
O
1
2
3
a/2
a/2
x
z
y
VM Ayres, ECE874, F12
(1/4, ¼, ¼) is givenTo get this from drawing would need scale marks on the axes
VM Ayres, ECE874, F12
Which pair are nearest neighbors?
O-1
O-2
O-3
O
1
2
3
a/2
a/2
x
z
y
VM Ayres, ECE874, F12
Directions for fcc (like Pr. 1.1 (b)):8 atoms positioned one at each corner.6 atoms centered in the middle of each face.
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The inside atoms of the diamond structure are all the ( ¼, ¼, ¼) locations that stay inside the box:
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All the ( ¼, ¼, ¼) locations are:
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You can match all pink atoms with a blue fcc lattice, copied from slide 12:
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You can match all bright and light yellow atoms with a green fcc lattice copied from slide 12:
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Therefore: the diamond crystal structure is formed from two interpenetrating fcc lattices
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But only 4 of the (¼, ¼, ¼) atoms are inside the cubic Unit cell.
VM Ayres, ECE874, F12
Now let the pink atoms be gallium (Ga) and the yellow atoms be arsenic (As). This is a zinc blende lattice.
Compare with Fig. 1.5 (b)
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Note that there are 4 complete molecules of GaAs in the cubic Unit cell.
VM Ayres, ECE874, F12
Example problem: find the molecular density of GaAs and verify that it is: 2.21 x 1022 molecules/cm3 = the value given on page 13.
VM Ayres, ECE874, F12
VM Ayres, ECE874, F12
Increasing numbers of important compounds crystallize in a wurtzite crystal lattice:
Hexagonal symmetry Fig. 1.3 forms a basic hexagonal Unit cell
Cadmium sulfide (CdS) crystallizes in a wurtzite lattice Fig. 1.6 inside a hexagonal Unit cell
Wurtzite: all sides = a
VM Ayres, ECE874, F12
Example problem: verify that the hexagonal Unit cell volume is:
= the value given on page 13.
ca2
2
33
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a
a
c
Hexagonal volume = 2 triangular volumes plus 1 rectangular volume
a
120o
VM Ayres, ECE874, F12
VM Ayres, ECE874, F12
VM Ayres, ECE874, F12
a
a
c
Hexagonal volume = 2 triangular volumes plus 1 rectangular volume
a
120o
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Work out the parallel planes for CdS:
3 S
7 Cd
7 S
3 Cd
3 S
7 Cd
Note: tetrahedral bonding inside
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3 S
7 S
3 S
How many equivalent S atoms are inside the hexagonal Unit cell?Define the c distance as between S-S (see Cd-Cd measure for c in Fig. 1.6 (a) )
VM Ayres, ECE874, F12
How much of each S atom is inside:
Hexagonal: In plane: 1/3 inside
Therefore:vertex atoms = 1/3 X 1 = 1/3 inside6 atoms x 1/3 each = 2
Hexagonal: Interior planeTop to bottom: all inside: 1
Also have one inside atom in the middle of the hexagonal layer: 1
Total atoms from the hexagonal arrangement = 3
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How much of each atom is inside:
Note: the atoms on the triangular arrangement never hit the walls of the hexagonal Unit cell so no 1/3 stuff.
But: they are chopped by the top and bottom faces of the hexagonal Unit cell: = ½ atoms.
Therefore have:3 S atoms ½ inside on each 3-atom layer layers: = 3/2 each layer
Total atoms from triangular arrangements = 2 x 3/2 = 3
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3 S
7 S
3 S
How many equivalent S atoms are inside the hexagonal Unit cell?
Total equivalent S atoms inside hexagonal Unit cell = 6
3/2
3
3/2
VM Ayres, ECE874, F12
You are required to indentify N as the Cd atoms in Fig. 1.6 (a)
Pr. 1.3 plus an extra requirement will be assigned for HW:
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Cd N:
7 Cd
3 Cd
7 Cd
So you’ll count the numbers of atoms for the red layers for HW, with c = the distance between Cd-Cd layers as shown.
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