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ECE 802-604: Nanoelectronics

ECE 802-604: Nanoelectronics. Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu. Lecture 18, 29 Oct 13. Carbon Nanotubes and Graphene Carbon nanotube/Graphene physical structure Carbon bond hybridization is versatile : sp 1 , sp 2 , and sp 3

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ECE 802-604: Nanoelectronics

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  1. ECE 802-604:Nanoelectronics Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu

  2. Lecture 18, 29 Oct 13 Carbon Nanotubes and Graphene Carbon nanotube/Graphene physical structure Carbon bond hybridization is versatile : sp1, sp2, and sp3 sp2: origin of mechanical and electronic structures Carbon nanotube/Graphene electronic ‘structure’ R. Saito, G. Dresselhaus and M.S. Dresselhaus Physical Properties of Carbon Nanotubes Imperial College Press, London, 1998. VM Ayres, ECE802-604, F13

  3. CNT Structure • The Basis Vectors: a1 and a2 • The Chiral Vector: Ch • The Chiral Angle: cos(q) • The Translation Vector: T • The Unit Cell of a CNT • Headcount of available p electrons VM Ayres, ECE802-604, F13

  4. VM Ayres, ECE802-604, F13

  5. Lec 17: The Basis Vectors a1 = √3 a x + 1 a y 2 2 a2 = √3 a x - 1 a y 2 2 where magnitude a = |a1| = |a2| VM Ayres, ECE802-604, F13

  6. Lec 17: The Basis Vectors 1.44 Angstroms is the carbon-to-carbon distance in individual ring VM Ayres, ECE802-604, F13

  7. Lec 17: The Basis Vectors a1 = √3 a x + 1 a y 2 2 a2 = √3 a x - 1 a y 2 2 1.44 A 1.44 A 120o a Magnitude a = 2 [ (1.44 Angstroms)cos(30) ] = 2.49 Angstroms VM Ayres, ECE802-604, F13

  8. The Basis Vectors a1 = √3 a x + 1 a y 2 2 a2 = √3 a x - 1 a y 2 2 where magnitude a = 2.49 Ang Example: label the vectors shown in red VM Ayres, ECE802-604, F13

  9. The Basis Vectors a1 = √3 a x + 1 a y 2 2 a2 = √3 a x - 1 a y 2 2 a1 where magnitude a = 2.49 Ang a2 Example: label the vectors shown in red Answer: as shown VM Ayres, ECE802-604, F13

  10. The Basis Vectors a1 = √3 a x + 1 a y 2 2 a2 = √3 a x - 1 a y 2 2 a1 where magnitude a = 2.49 Ang a2 Example: what is a1 a2 ? VM Ayres, ECE802-604, F13

  11. The Basis Vectors a1 = √3 a x + 1 a y 2 2 a2 = √3 a x - 1 a y 2 2 a1 where magnitude a = 2.49 Ang a2 Example: what is a1 a2 ? Answer: non-orthogonal in ai system VM Ayres, ECE802-604, F13

  12. The Chiral Vector Ch VM Ayres, ECE802-604, F13

  13. Vector: Ch = n a1 + m a2 Ch = (n, m) Magnitude of vector: |Ch| = a√n2 + m2 + mn The Chiral Vector ChBasic definition VM Ayres, ECE802-604, F13

  14. Lec 17: Introduction Buckyball endcaps Many different types of wrapping result in a seamless cylinder. But The particular cylinder wrapping dictates the electronic and mechanical properties. VM Ayres, ECE802-604, F13

  15. The Chiral Vector ChBasic definition Example: Prove that the magnitude of |Ch| is a√n2 + m2 + mn VM Ayres, ECE802-604, F13

  16. The Chiral Vector ChBasic definition Answer: VM Ayres, ECE802-604, F13

  17. The Chiral Vector ChBasic definition Example: Evaluate |Ch| for a (10,10) SWCNT VM Ayres, ECE802-604, F13

  18. The Chiral Vector ChBasic definition Example: Evaluate |Ch| for a (10,10) SWCNT Answer: VM Ayres, ECE802-604, F13

  19. Definition of d: If Ch = n a1 + m a2 Then d = the greatest common divisor of n and m. The Chiral Vector ChA number associated with the Chiral Vector : the Greatest Common Divisor d(or gcd) VM Ayres, ECE802-604, F13

  20. Definition of d: If Ch = n a1 + m a2 Then d = the greatest common divisor of n and m. The Chiral Vector ChThe Greatest Common Divisor (d, or gcd) Example: Find d for the following SWCNTs: (10,10), (9,9), (9,0), (7,4), and (8, 6) VM Ayres, ECE802-604, F13

  21. Definition of d: If Ch = n a1 + m a2 Then d = the greatest common divisor of n and m. The Chiral Vector ChThe Greatest Common Divisor (d, or gcd) Example: Find d for the following SWCNTs: (10,10), (9,9), (9,0), (7,4), and (8, 6) Answer: (10,10): d=10; (9,9): d = 9; (9,0): d=9; (7,4): d=1; and (8, 6): d=2 VM Ayres, ECE802-604, F13

  22. |Ch | = p diameter = p dt dt = |Ch|/p = a√n2 + m2 + mn / p The Chiral Vector ChDefine theCNT tube diameter dt(Note that diameter dt is different than greatest common divisor d!) VM Ayres, ECE802-604, F13

  23. |Ch| = p diameter = p dt dt = |Ch|/p = a√n2 + m2 + mn / p The Chiral Vector ChCNT diameter dt Example: Find the nanotube diameter for a (10, 10) CNT. VM Ayres, ECE802-604, F13

  24. |Ch | = p diameter = p dt dt = |Ch|/p = a√n2 + m2 + mn / p The Chiral Vector ChCNT diameter dt Example: Find the nanotube diameter for a (10, 10) CNT. Answer: dt = 136.38 Ang/p = 43.41 Ang VM Ayres, ECE802-604, F13

  25. The Chiral Angle q VM Ayres, ECE802-604, F13

  26. The Chiral Angle q Defines the tilt of Ch with respect to a1 Direction cosine of a1 • Ch: a1• Ch = |a1| |Ch| cos q cos q = a1• Ch |a1| |Ch| = ( n + m/2) √n2 + m2 + mn VM Ayres, ECE802-604, F13

  27. The Chiral Angle q Example: Prove that cosq = ( n + m/2) √n2+ m2+ mn VM Ayres, ECE802-604, F13

  28. The Chiral Angle q Answer: VM Ayres, ECE802-604, F13

  29. The Translation Vector T VM Ayres, ECE802-604, F13

  30. The Translation Vector T Note that T and Ch are perpendicular. Therefore T•Ch = 0 Let T = t1a1 + t2a2 Take T•Ch = 0 Solve for t1 and t2 VM Ayres, ECE802-604, F13

  31. The Translation Vector T Vector: T = t1a1 + t2a2 t1 = 2m + n/ dR t2 = - (2n + m) /dR Magnitude: | T | = √3 |Ch|/dR What is dR? VM Ayres, ECE802-604, F13

  32. The Translation Vector TdR is a NEW greatest common divisor: dR = the greatest common divisor of 2m + n and 2n+ m dR = d if n-m is not a multiple of 3d VM Ayres, ECE802-604, F13

  33. The Unit Cell of a CNT (single wall) VM Ayres, ECE802-604, F13

  34. The Unit Cell of a CNT (single wall) Note that T and Ch are perpendicular. Therefore T X Ch = the area of the CNT Unit Cell VM Ayres, ECE802-604, F13

  35. The Primitive Cell of a CNT (single wall) Also a1 x a2 is the area of a single basic or primitive cell. Therefore: the number of hexagons N per CNT Unit Cell is: N = | T X Ch | | a1xa2 | = 2(m2 + n2+nm)/dR VM Ayres, ECE802-604, F13

  36. The Unit Cell of a CNT (single wall) Each primitive cell contains two C atoms. There is one pz-orbital per each C atom. Therefore there are 2N pzorbitals available per CNT Unit Cell VM Ayres, ECE802-604, F13

  37. Example Problems: CNT Structure • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (10,10) SWCNT • What does N count? • What type of CNT is this? • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (9,0) SWCNT • What type of SWCNT is this? • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (7,4) SWCNT • What type of SWCNT is this? VM Ayres, ECE802-604, F13

  38. Ch = n a1 + m a2 |Ch| = a√n2 + m2 + mn dt = |Ch|/p cos q = a1 • Ch |a1| |Ch| T = t1a1 + t2a2 t1 = (2m + n)/ dR t2 = - (2n + m) /dR dR = the greatest common divisor of 2m + n and 2n+ m |T| = √ 3 |Ch| / dR N = | T X Ch | | a1xa2 | = 2(m2 + n2+nm)/dR For use in Example Problems: VM Ayres, ECE802-604, F13

  39. Find Ch, |Ch |, cosq, q, T, |T |, and N for a (10,10) SWCNT • What does N count? • What type of CNT is this? • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (9,0) SWCNT • What type of SWCNT is this? • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (7,4) SWCNT • What type of SWCNT is this? VM Ayres, ECE802-604, F13

  40. Answer 1: VM Ayres, ECE802-604, F13

  41. - Answer 1 continued: VM Ayres, ECE802-604, F13

  42. Answers 2 and 3: 2. 3. | T | = √3 |Ch|/dR = 0.25 nm VM Ayres, ECE802-604, F13

  43. Find Ch, |Ch |, cosq, q, T, |T |, and N for a (10,10) SWCNT • What does N count? • What type of CNT is this? • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (9,0) SWCNT • What type of SWCNT is this? • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (7,4) SWCNT • What type of SWCNT is this? VM Ayres, ECE802-604, F13

  44. Answer 4: VM Ayres, ECE802-604, F13

  45. Answers 4 and 5: 4. 5. VM Ayres, ECE802-604, F13

  46. Find Ch, |Ch |, cosq, q, T, |T |, and N for a (10,10) SWCNT • What does N count? • What type of CNT is this? • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (9,0) SWCNT • What type of SWCNT is this? • Find Ch, |Ch |, cosq, q, T, |T |, and N for a (7,4) SWCNT • What type of SWCNT is this? VM Ayres, ECE802-604, F13

  47. Answer 6: VM Ayres, ECE802-604, F13

  48. Answers 6 and 7: 6. 7. VM Ayres, ECE802-604, F13

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