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

2. VM Ayres, ECE874, F12 Lecture 04, 11 Sep 12

3. VM Ayres, ECE874, F12 Example problem: the surface of a Si wafer is a (001) plane. Which, if any directions are perpendicular to the Si wafer (001) plane direction?

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5. +z +x +y a a a [110]

6. VM Ayres, ECE874, F12 Principles of Electronic Devices, Streetman and Bannerjee + Battery - [110]

7. VM Ayres, ECE874, F12 Convenient to know two directions: Crystalline direction lying within the surface plane (transport directions): Primary flat gives a direction Surface orientation: [111] or [001]; also gives n or p type Secondary flat

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9. [110] Primary flat is a reference direction to help you find any in-plane transport direction that you want Note that the directions are different for (1110 and (100) type orientations

10. VM Ayres, ECE874, F12 Example problem:

11. VM Ayres, ECE874, F12 Example problem:

12. VM Ayres, ECE874, F12 Example problem:

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14. [112] Example problem: pattern is now aligned with [112] direction.

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16. [001] (a) 8 type directions to the 8 facets: however: Angles to top facets are all the same values. Angles to bottom facets are all the same values. [111]

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19. (b) 4 pieces: 4 top cleavage planes split open at the pressure point [001]

20. VM Ayres, ECE874, F12 [001] [110] [111] (c) Now locate facet w.r.t in plane direction:

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22. More about non-Si crystal structures: wurtzite and rock salt

23. VM Ayres, ECE874, F12 From The Economist 01-07 Sep 12: Technology Quarterly Very recent recognition: chalcogenides have well-controlled crystalline versus amorphous crystal structures as a function of pulsed thermal energy. Very local ~10 nm 1/0 states and very low energy needed to switch between them. It looks like chalcogenides are going to be the next hot thing for flash drives with a lot of capacity. From Wikipedia: chalcogneides can be: Cadmium telluride Indium sulfide Zinc telluride Sodium selenide Other mixes Crystalline phase: zinc blende, (I’d check wurtzite too) Motivation to study what crystal structures can do:

24. VM Ayres, ECE874, F12 Recall: the diamond crystal structure is formed from two interpenetrating fcc lattices displaced (¼, ¼, ¼):

25. VM Ayres, ECE874, F12 with only 4 of the atoms from the interpenetrating fcc inside a single cubic Unit cell.

26. VM Ayres, ECE874, F12 3 S 7 Cd 7 S 3 Cd 3 S 7 Cd Wurtzite is formed from two interpenetrating hexagonal close packed hcp lattices of different atoms:

27. VM Ayres, ECE874, F12 Wurtzite is formed from two interpenetrating hexagonal close packed hcp lattices of different atoms, inside a hexagonal Unit cell: Note: tetrahedral bonding inside hcp lattice 01

28. VM Ayres, ECE874, F12 Wurtzite is formed from two interpenetrating hexagonal close packed hcp lattices of different atoms, inside a hexagonal Unit cell : Note: tetrahedral bonding inside hcp lattice 02

29. VM Ayres, ECE874, F12 As per in-class demonstration with balls, hcp is a very stable arrangement of atoms. Also: tetrahedral bonds fit well inside the hexagonal Unit cell providing even more stability/balance in very direction.

30. VM Ayres, ECE874, F12 6 from dark contrast atoms (Pr. 1.3) Example problem: what is the Number of equivalent atoms inside the hexagonal Unit cell of height c = ?: 7 S 3 S Also have a contribution from the interpenetrating light contrast atoms:

31. VM Ayres, ECE874, F12 How much of each light contrast atom is inside: all. Total equivalent atoms from triangular arrangement of light contrast atoms in hexagonal Unit cell = 3 x 1 = 3 (do the easy one first)

32. VM Ayres, ECE874, F12 (do the next easiest one second) Also have one inside atom in the middle of the hexagonal layer: 1

33. VM Ayres, ECE874, F12 From the vertex atoms : Hexagonal: In plane: 1/3 inside Therefore: 1 vertex atom = 1/3 X 1 = 1/3 inside 6 atoms x 1/3 each = 2 Hexagonal: Interior plane Top to bottom: all inside: 1

34. VM Ayres, ECE874, F12 Total number of light and dark contrast atoms inside hexagonal Unit cell: 3 + (1 + (6x1/3 = 2) = 6