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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 02, 04 Sep 12. Silicon crystallizes in the diamond crystal structure Fig. 1.5 (a):.

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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 02, 04 Sep 12 VM Ayres, ECE874, F12

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

  4. VM Ayres, ECE874, F12

  5. Corners: 8 x 1/8 = 1 Faces: 6 x ½ = 3 Inside: 4 x 1 = 4 Atoms in the Unit cell: 8 VM Ayres, ECE874, F12

  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) VM Ayres, ECE874, F12

  7. VM Ayres, ECE874, F12

  8. 1 Which pair are nearest neighbors? O-1 O-2 O-3 2 3 O VM Ayres, ECE874, F12

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

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

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

  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. VM Ayres, ECE874, F12

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

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

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

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

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

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

  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) VM Ayres, ECE874, F12

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

  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. VM Ayres, ECE874, F12

  22. VM Ayres, ECE874, F12

  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 VM Ayres, ECE874, F12

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

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

  26. VM Ayres, ECE874, F12

  27. VM Ayres, ECE874, F12

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

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

  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 VM Ayres, ECE874, F12

  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 VM Ayres, ECE874, F12

  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 VM Ayres, ECE874, F12

  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 VM Ayres, ECE874, F12

  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) VM Ayres, ECE874, F12

  35. So you’ll 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 VM Ayres, ECE874, F12

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