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ECE 874: Physical Electronics

ECE 874: Physical Electronics. Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu. Lecture 04, 11 Sep 12.

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ECE 874: Physical Electronics

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

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

  3. Example problem: the surface of a Si wafer is a (001) plane. Which, if any <110> directions are perpendicular to the Si wafer (001) plane direction? VM Ayres, ECE874, F12

  4. VM Ayres, ECE874, F12

  5. +z a a +x a +y [110] VM Ayres, ECE874, F12

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

  7. Convenient to know two directions: VM Ayres, ECE874, F12

  8. VM Ayres, ECE874, F12

  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 [110] VM Ayres, ECE874, F12

  10. Example problem: VM Ayres, ECE874, F12

  11. Example problem: VM Ayres, ECE874, F12

  12. Example problem: VM Ayres, ECE874, F12

  13. VM Ayres, ECE874, F12

  14. Example problem: pattern is now aligned with [112] direction. [112] VM Ayres, ECE874, F12

  15. VM Ayres, ECE874, F12

  16. [111] • (a) 8 <111> 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. [001] [111] VM Ayres, ECE874, F12

  17. VM Ayres, ECE874, F12

  18. VM Ayres, ECE874, F12

  19. (b) 4 pieces: 4 top cleavage planes split open at the pressure point [001] VM Ayres, ECE874, F12

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

  21. VM Ayres, ECE874, F12

  22. More about non-Si crystal structures:wurtzite and rock salt VM Ayres, ECE874, F12

  23. Motivation to study what crystal structures can do: 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) VM Ayres, ECE874, F12

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

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

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

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

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

  29. 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. VM Ayres, ECE874, F12

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

  31. 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) VM Ayres, ECE874, F12

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

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

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

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