1. ECE 874: Physical Electronics Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu

2. VM Ayres, ECE874, F12 Lecture 02, 04 Sep 12

3. VM Ayres, ECE874, F12 Silicon crystallizes in the diamond crystal structure Fig. 1.5 (a): Notice real covalent bonds versus definition of imaginary cubic unit cell

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5. Corners: 8 x 1/8 = 1 Faces: 6 x ½ =3 Inside: 4 x 1 =4 Atoms in the Unit cell: 8

6. VM Ayres, ECE874, F12 What do you need to know to answer (b)? Size of a cubic Unit cell is a 3, where a is the lattice constant (Wurtzite crystal structure is not cubic)

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8. Which pair are nearest neighbors? O-1 O-2 O-3 O 1 2 3

9. VM Ayres, ECE874, F12 Which pair are nearest neighbors? O-1 O-2 O-3 O 1 2 3 a/2 x z y

10. VM Ayres, ECE874, F12 (1/4, ¼, ¼) is given To get this from drawing would need scale marks on the axes

11. VM Ayres, ECE874, F12 Which pair are nearest neighbors? O-1 O-2 O-3 O 1 2 3 a/2 x z y

12. 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.

13. VM Ayres, ECE874, F12 The inside atoms of the diamond structure are all the ( ¼, ¼, ¼) locations that stay inside the box:

14. VM Ayres, ECE874, F12 All the ( ¼, ¼, ¼) locations are:

15. VM Ayres, ECE874, F12 You can match all pink atoms with a blue fcc lattice, copied from slide 12:

16. VM Ayres, ECE874, F12 You can match all bright and light yellow atoms with a green fcc lattice copied from slide 12:

17. VM Ayres, ECE874, F12 Therefore: the diamond crystal structure is formed from two interpenetrating fcc lattices

18. VM Ayres, ECE874, F12 But only 4 of the (¼, ¼, ¼) atoms are inside the cubic Unit cell.

19. 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)

20. VM Ayres, ECE874, F12 Note that there are 4 complete molecules of GaAs in the cubic Unit cell.

21. VM Ayres, ECE874, F12 Example problem: find the molecular density of GaAs and verify that it is: 2.21 x 10 22 molecules/cm 3 = the value given on page 13.

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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. VM Ayres, ECE874, F12 Example problem: verify that the hexagonal Unit cell volume is: = the value given on page 13.

25. VM Ayres, ECE874, F12 a a c Hexagonal volume = 2 triangular volumes plus 1 rectangular volume a 120 o

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28. a a c Hexagonal volume = 2 triangular volumes plus 1 rectangular volume a 120 o

29. VM Ayres, ECE874, F12 Work out the parallel planes for CdS: 3 S 7 Cd 7 S 3 Cd 3 S 7 Cd Note: tetrahedral bonding inside

30. VM Ayres, ECE874, F12 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) )

31. 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 inside 6 atoms x 1/3 each = 2 Hexagonal: Interior plane Top 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

32. VM Ayres, ECE874, F12 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. VM Ayres, ECE874, F12 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

34. 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:

35. VM Ayres, ECE874, F12 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.