ece 802-604: nanoelectronics prof. virginia ayres electrical & computer engineering michigan...

ECE 802-604:Nanoelectronics

Prof. Virginia AyresElectrical & Computer EngineeringMichigan State Universityayresv@msu.edu

VM Ayres, ECE802-604, F13

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. DresselhausPhysical Properties of Carbon NanotubesImperial College Press, London, 1998.

CNT Structure

The Basis Vectors: a1 and a2

The Chiral Vector: Ch

The Chiral Angle: cos( The Translation Vector: T The Unit Cell of a CNT

– Headcount of available electrons

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|

1.44 Angstroms is the carbon-to-carbon distance in individual ring

a1 = √3 a x + 1 a y 2 2

a2 = √3 a x - 1 a y 2 2

Magnitude a = 2 [ (1.44 Angstroms)cos(30) ]

= 2.49 Angstroms

1.44 A1.44 A

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

The Basis Vectors

a1 = √3 a x + 1 a y 2 2

a2 = √3 a x - 1 a y 2 2

Example: label the vectors shown in redAnswer: as shown

The Basis Vectors

a1 = √3 a x + 1 a y 2 2

a2 = √3 a x - 1 a y 2 2

Example: what is a1 a2 ?

The Basis Vectorsa1 = √3 a x + 1 a y 2 2

a2 = √3 a x - 1 a y 2 2

Example: what is a1 a2 ?Answer: non-orthogonal in ai system

The Chiral Vector Ch

The Chiral Vector ChBasic definition

Vector:Ch = n a1 + m a2

Ch = (n, m)

Magnitude of vector:

|Ch| = a√n2 + m2 + mn

Lec 17: Introduction

Many different types of wrapping result in a seamless cylinder.

The particular cylinder wrapping dictates the electronic and mechanical properties.

Buckyball endcaps

Example: Prove that the magnitude of |Ch| is a√n2 + m2 + mn

Answer:

Example: Evaluate |Ch| for a (10,10) SWCNT

Example: Evaluate |Ch| for a (10,10) SWCNTAnswer:

The Chiral Vector ChA number associated with the Chiral Vector : the Greatest Common Divisor d(or gcd)

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)

Definition of d:

If Ch = n a1 + m a2

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

Definition of d:

If Ch = n a1 + m a2

The Chiral Vector ChDefine the CNT tube diameter dt(Note that diameter dt is different than greatest common divisor d!)

|Ch | = diameter

dt = |Ch|/ = a√n2 + m2 + mn /

The Chiral Vector ChCNT diameter dt

|Ch | = diameter

dt = |Ch|/ = a√n2 + m2 + mn /

Example: Find the nanotube diameter for a (10, 10) CNT.

The Chiral Vector ChCNT diameter dt

|Ch | = diameter

dt = |Ch|/ = a√n2 + m2 + mn /

Example: Find the nanotube diameter for a (10, 10) CNT.Answer: dt = 136.38 Ang/ = 43.41 Ang

The Chiral Angle

Direction cosine of a1 • Ch:

a1 • Ch = |a1| |Ch| cos

cos = a1 • Ch

|a1| |Ch|

= ( n + m/2) √n2 + m2 + mn

Defines the tilt of Ch with respect to a1

The Chiral Angle

Example: Prove that cos = ( n + m/2) √n2+ m2+ mn

The Chiral Angle

Answer:

The Translation Vector T

Note that T and Ch are perpendicular.

Therefore T•Ch = 0

Let T = t1 a1 + t2 a2

Take T•Ch = 0 Solve for t1 and t2

Vector:

T = t1 a1 + t2 a2

t1 = 2m + n/ dR

t2 = - (2n + m) /dR

Magnitude:

| T | = √3 |Ch |/dR

What is dR?

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

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

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 | | a1 x a2 |

= 2(m2 + n2+nm)/dR

Each primitive cell contains two C atoms.

There is one pz-orbital per each C atom.

Therefore there are 2N pz orbitals available per CNT Unit Cell

Example Problems: CNT Structure

1. Find Ch, |Ch |, cos, , T, |T |, and N for a (10,10) SWCNT

2. What does N count?

3. What type of CNT is this?

5. What type of SWCNT is this?

For use in Example Problems: Ch = n a1 + m a2

|Ch| = a√n2 + m2 + mndt = |Ch|/

cos = a1 • Ch

|a1| |Ch|

T = t1 a1 + t2 a2 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 | | a1 x a2 |

= 2(m2 + n2+nm)/dR

Answer 1:

Answer 1 continued:

| T | = √3 |Ch |/dR = 0.25 nm

Answers 2 and 3:

Answer 4:

Answers 4 and 5:4.

Answer 6:

Answers 6 and 7:6.

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