ECE 874: Physical Electronics

1
ECE 874Physical Electronics

• Prof. Virginia Ayres
• Electrical Computer Engineering
• Michigan State University
• ayresv_at_msu.edu

2
Lecture 02, 04 Sep 12
3
Silicon crystallizes in the diamond crystal
structureFig. 1.5 (a)
Notice real covalent bonds versus definition of
imaginary cubic unit cell
4
(No Transcript)
5
Corners 8 x 1/8 1 Faces 6 x ½ 3 Inside 4
x 1 4 Atoms in the Unit cell 8
6
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)
7
(No Transcript)
8
1
Which pair are nearest neighbors? O-1 O-2 O-3
2
3
O
9
1
Which pair are nearest neighbors? O-1 O-2 O-3
2
a/2
a/2
3
O
z
y
x
10
(1/4, ¼, ¼) is given To get this from drawing
would need scale marks on the axes
11
1
Which pair are nearest neighbors? O-1 O-2 O-3
2
a/2
a/2
3
O
z
y
x
12
Directions for fcc (like Pr. 1.1 (b))8 atoms
positioned one at each corner.6 atoms centered
in the middle of each face.
13
The inside atoms of the diamond structure are all
the ( ¼, ¼, ¼) locations that stay inside the box
14
All the ( ¼, ¼, ¼) locations are
15
You can match all pink atoms with a blue fcc
lattice, copied from slide 12
16
You can match all bright and light yellow atoms
with a green fcc lattice copied from slide 12
17
Therefore the diamond crystal structure is
formed from two interpenetrating fcc lattices
18
But only 4 of the (¼, ¼, ¼) atoms are inside the
cubic Unit cell.
19
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)
20
Note that there are 4 complete molecules of GaAs
in the cubic Unit cell.
21
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.
22
(No Transcript)
23
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
24
Example problem verify that the hexagonal Unit
cell volume is the value given on page
13.
25
Hexagonal volume 2 triangular volumes plus 1
rectangular volume
a
a
120o
a
c
26
(No Transcript)
27
(No Transcript)
28
Hexagonal volume 2 triangular volumes plus 1
rectangular volume
a
a
120o
a
c
29
Work out the parallel planes for CdS
3 S
7 Cd
7 S
3 Cd
3 S
7 Cd
Note tetrahedral bonding inside
30
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) )
3 S
7 S
3 S
31
How much of each S atom is inside
Hexagonal In plane 1/3 inside
Hexagonal Interior plane Top to bottom all
inside 1
Therefore vertex atoms 1/3 X 1 1/3 inside 6
atoms x 1/3 each 2
Also have one inside atom in the middle of the
hexagonal layer 1
Total atoms from the hexagonal arrangement 3
32
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
33
How many equivalent S atoms are inside the
hexagonal Unit cell?Total equivalent S atoms
inside hexagonal Unit cell 6
3 S
3/2
7 S
3
3 S
3/2
34
Pr. 1.3 plus an extra requirement will be
assigned for HW
You are required to indentify N as the Cd atoms
in Fig. 1.6 (a)
35
So youll count the numbers of atoms for the red
layers for HW, with c the distance between
Cd-Cd layers as shown.
Cd ? N
7 Cd
3 Cd
7 Cd