ece 802-604: nanoelectronics
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ECE 802-604: Nanoelectronics. Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University [email protected]. 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 - PowerPoint PPT PresentationTRANSCRIPT
ECE 802-604:Nanoelectronics
Prof. Virginia AyresElectrical & Computer EngineeringMichigan State [email protected]
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.
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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
VM Ayres, ECE802-604, F13
VM Ayres, ECE802-604, F13
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|
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Lec 17: The Basis Vectors
1.44 Angstroms is the carbon-to-carbon distance in individual ring
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Lec 17: The Basis Vectors
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
120o
a
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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
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 redAnswer: as shown
a1
a2
VM Ayres, ECE802-604, F13
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: what is a1 a2 ?
a1
a2
VM Ayres, ECE802-604, F13
The Basis Vectorsa1 = √3 a x + 1 a y 2 2
a2 = √3 a x - 1 a y 2 2
where magnitude a = 2.49 Ang
Example: what is a1 a2 ?Answer: non-orthogonal in ai system
a1
a2
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The Chiral Vector Ch
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The Chiral Vector ChBasic definition
Vector:Ch = n a1 + m a2
Ch = (n, m)
Magnitude of vector:
|Ch| = a√n2 + m2 + mn
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Lec 17: Introduction
Many different types of wrapping result in a seamless cylinder.
But
The particular cylinder wrapping dictates the electronic and mechanical properties.
Buckyball endcaps
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The Chiral Vector ChBasic definition
Example: Prove that the magnitude of |Ch| is a√n2 + m2 + mn
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The Chiral Vector ChBasic definition
Answer:
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The Chiral Vector ChBasic definition
Example: Evaluate |Ch| for a (10,10) SWCNT
VM Ayres, ECE802-604, F13
The Chiral Vector ChBasic definition
Example: Evaluate |Ch| for a (10,10) SWCNTAnswer:
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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.
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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
Then d = the greatest common divisor of n and m.
VM Ayres, ECE802-604, F13
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
Then d = the greatest common divisor of n and m.
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The Chiral Vector Ch
Define the CNT tube diameter dt
(Note that diameter dt is different than greatest common divisor d!)
|Ch | = diameter
= dt
dt = |Ch|/ = a√n2 + m2 + mn /
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The Chiral Vector ChCNT diameter dt
|Ch | = diameter
= dt
dt = |Ch|/ = a√n2 + m2 + mn /
Example: Find the nanotube diameter for a (10, 10) CNT.
VM Ayres, ECE802-604, F13
The Chiral Vector ChCNT diameter dt
|Ch | = diameter
= dt
dt = |Ch|/ = a√n2 + m2 + mn /
Example: Find the nanotube diameter for a (10, 10) CNT.Answer: dt = 136.38 Ang/ = 43.41 Ang
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The Chiral Angle
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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
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The Chiral Angle
Example: Prove that cos = ( n + m/2) √n2+ m2+ mn
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The Chiral Angle
Answer:
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The Translation Vector T
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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
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The Translation Vector T
Vector:
T = t1 a1 + t2 a2
t1 = 2m + n/ dR
t2 = - (2n + m) /dR
Magnitude:
| T | = √3 |Ch |/dR
What is dR?
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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
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The Unit Cell of a CNT (single wall)
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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
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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
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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 pz orbitals available per CNT Unit Cell
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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?
4. Find Ch, |Ch |, cos, , T, |T |, and N for a (9,0) SWCNT
5. What type of SWCNT is this?
6. Find Ch, |Ch |, cos, , T, |T |, and N for a (7,4) SWCNT
7. What type of SWCNT is this?
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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
VM Ayres, ECE802-604, F13
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?
4. Find Ch, |Ch |, cos, , T, |T |, and N for a (9,0) SWCNT
5. What type of SWCNT is this?
6. Find Ch, |Ch |, cos, , T, |T |, and N for a (7,4) SWCNT
7. What type of SWCNT is this?
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Answer 1:
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Answer 1 continued:
-
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| T | = √3 |Ch |/dR = 0.25 nm
Answers 2 and 3:
2.
3.
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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?
4. Find Ch, |Ch |, cos, , T, |T |, and N for a (9,0) SWCNT
5. What type of SWCNT is this?
6. Find Ch, |Ch |, cos, , T, |T |, and N for a (7,4) SWCNT
7. What type of SWCNT is this?
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Answer 4:
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Answers 4 and 5:
4.
5.
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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?
4. Find Ch, |Ch |, cos, , T, |T |, and N for a (9,0) SWCNT
5. What type of SWCNT is this?
6. Find Ch, |Ch |, cos, , T, |T |, and N for a (7,4) SWCNT
7. What type of SWCNT is this?
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Answer 6:
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Answers 6 and 7:6.
7.