ECE 874: Physical Electronics

ECE 874:Physical Electronics

Prof. Virginia AyresElectrical & Computer EngineeringMichigan State [email protected]

VM Ayres, ECE874, F12

Lecture 02, 04 Sep 12


Silicon crystallizes in the diamond crystal structureFig. 1.5 (a):

Notice real covalent bonds versus definition of imaginary cubic unit cell


Corners: 8 x 1/8 = 1

Faces: 6 x ½ = 3

Inside: 4 x 1 = 4

Atoms in the Unit cell: 8


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)


Which pair are nearest neighbors?

O-1

O-2

O-3

O

1

2

3



O-1

O-2

O-3

O

1

2

3

a/2

a/2

x

z

y


(1/4, ¼, ¼) is givenTo get this from drawing would need scale marks on the axes



O-1

O-2

O-3

O

1

2

3

a/2

a/2

x

z

y


Directions for fcc (like Pr. 1.1 (b)):8 atoms positioned one at each corner.6 atoms centered in the middle of each face.


The inside atoms of the diamond structure are all the ( ¼, ¼, ¼) locations that stay inside the box:


All the ( ¼, ¼, ¼) locations are:


You can match all pink atoms with a blue fcc lattice, copied from slide 12:


You can match all bright and light yellow atoms with a green fcc lattice copied from slide 12:


Therefore: the diamond crystal structure is formed from two interpenetrating fcc lattices


But only 4 of the (¼, ¼, ¼) atoms are inside the cubic Unit cell.


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)


Note that there are 4 complete molecules of GaAs in the cubic Unit cell.


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.


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


Example problem: verify that the hexagonal Unit cell volume is:

= the value given on page 13.

ca2

2

33


a

a

c

Hexagonal volume = 2 triangular volumes plus 1 rectangular volume

a

120o


Work out the parallel planes for CdS:

3 S

7 Cd

7 S

3 Cd

3 S

7 Cd

Note: tetrahedral bonding inside


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) )


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


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


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


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:


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.

ECE 874: Physical Electronics

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