ECE 474: Principles of Electronic Devices. Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University

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1 ECE 474: Principles of Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University

2 Lecture 07: Quantify physical structures of crystal systems that are important for devices: Lectures: Hexagonal nanosystems: graphene and carbon nanotubes Introduction to graphene and CNTs The Basis Vectors: a 1 and a 2 Nearest neighbor distances The Chiral Vector: C h The CNT diameter d t The Translation Vector: T The Unit Cell of a CNT Number of hexagons N Number of atoms N/2 (π electrons) Areal Density Examples of each

3 Lecture 07: Quantify physical structures of crystal systems that are important for devices: Lectures: Hexagonal nanosystems: graphene and carbon nanotubes Introduction to graphene and CNTs The Basis Vectors: a 1 and a 2 Nearest neighbor distances The Chiral Vector: C h The CNT diameter d t The Translation Vector: T The Unit Cell of a CNT Number of hexagons N Number of atoms N/2 (π electrons) Areal Density Examples of each

Introduction: Graphene: http://www.google.

4 Introduction: Graphene: r.edu/home/baez/graphene.jpg

5 Introduction: Carbon Nanotubes: A single wall Carbon Nanotube is a single graphene sheet wrapped into a cylinder. The endcaps are half-buckyballs. Note: different possible wrappings. Armchair Zigzag Chiral (Ref. lost)

But The particular cylinder wrapping dictates the electronic and mechanical

6 Introduction Buckyball endcaps Many different types of wrapping result in a seamless cylinder. But The particular cylinder wrapping dictates the electronic and mechanical properties. R. Saito, G. Dresselhaus and M.S. Dresselhaus, Physical Properties of Carbon Nanotubes, Imperial College Press, London, 1998, Chapter 3.

7 Introduction Width: C h called the Chiral vector Length: T called the Translation vector

8 Introduction Therefore TX C h = the area of the CNT Unit Cell Note that T and C h are perpendicular.

9 The Basis Vectors a 1 and a 2 : how to get around in a hexagonal system Note that a 1 and a 2 are NOT perpendicular (orthogonal).

10 The Basis Vectors a 1 and a 2 : how to get around in a hexagonal system

11 The Basis Vectors a 1 and a 2 : how to get around in a hexagonal system 120 o 30 o 30 o a 1 a 2 = a cos30 o x + a sin30 o y = a cos30 o x -a sin30 o y where magnitude a = a 1 = a 2

12 The Basis Vectors a 1 and a 2 : how to get around in a hexagonal system 120 o 30 o 30 o a 1 a 2 = 3 a x + 1 a y 2 2 = 3 a x - 1 a y 2 2 where magnitude a = a 1 = a 2

13 The Basis Vectors a 1 and a 2 : how to get around in a hexagonal system Find magnitude a = a 1 = a 2 Length of a side is well known from the Benzene system (very slight value change in CNT to 1.44 angtroms) Length a is related to the length of a side Angstroms

14 The Basis Vectors a 1 and a 2 : how to get around in a hexagonal system a 1 a 2 = 3 a x + 1 a y 2 2 = 3 a x - 1 a y A 1.44 A 120 o Magnitude a = 2 [ (1.44 Angstroms)cos(30) ] a = 2.49 Angstroms

15 The Basis Vectors a 1 and a 2 : how to get around in a hexagonal system 120 o 30 o 30 o a 1 a 2 = 3 a x + 1 a y 2 2 = 3 a x - 1 a y 2 2 where magnitude a = a 1 = a 2 = 2.49 Angstroms

16 The Chiral Vector C h

17 The Chiral Vector C h Basic definition Vector: C h = n a 1 + m a 2 Designation: C h = (n, m) Magnitude of vector: C h = a n 2 + m 2 + mn

18 Example: Designation =?

19 C h = n a 1 + m a 2 = 4 a a 2 This is a (n,m) = (4,2) CNT

20 The Chiral Vector C h Example: Prove that the magnitude of C h is a n 2 + m 2 + mn

21 C hº C h

22 Example: find the magnitude of the chiral vector C h for a (10,10) CNT.

23 Example: find the magnitude of the chiral vector C h for a (10,10) CNT. C h = a n 2 + m 2 + mn = (2.49 Ang) = (2.49 Ang) 10 3 = Ang

24 Magnitude C h defines the CNT tube diameter d t C h π r t d 2 t ( 2π ) C h = arc length = θ r t d t = = ( 2π ) r = ( 2π ) Ch = π a n t 2 + m π 2 dt 2 + mn

25 Magnitude C h defines the CNT tube diameter d t C h = π d t d t = C h /π = (a n 2 + m 2 + mn )/ π

26 Example: find the diameter d t of a (10,10) CNT.

27 C h = a n 2 + m 2 + mn = (1.44 Ang) = (1.44 Ang) 10 3 = Ang d t = Ang/π = Ang 1.4 nm

28 The Translation Vector T: Note that T and C h are perpendicular. Therefore T C h = 0 Let T = t 1 a 1 + t 2 a 2 Take T C h = 0 Solve for t 1 and t 2

29 The Translation Vector T: Vector: T = t 1 a 1 + t 2 a 2 t 1 = 2m + n/ d R t 2 = - (2n + m) /d R Magnitude: T = 3 C h /d R d R = the greatest common divisor of 2m + n and 2n+ m

30 Example: find the magnitude of the Translation vector T for a (10,10) CNT.

32 Example: find the Translation vector T for a (10,10) CNT.

33 for a (10,10) CNT

34 C h = n a 1 + m a 2 = 4 a a 2 This is a (n,m) = (4,2) CNT

35 Example: find the Translation vector T for a (4,2) CNT.

36 Example: find the Translation vector T for a (4,2) CNT.

37 T = n a 1 + m a 2 = 4 a 1-5 a 2 This is a (n,m) = (4,2) CNT

ECE 474: Principles of Electronic Devices. Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University

ECE 474: Principles of Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu Lecture 06: Completed: Chapter 01: quantify physical structures