ECE 474: Principles of Electronic Devices · ECE 474: Principles of Electronic Devices Prof....
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ECE 474:Principles of Electronic Devices
Prof. Virginia AyresElectrical & Computer EngineeringMichigan State [email protected]
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V.M. Ayres, ECE474, Spring 2011
Example for VA Pr. 01 1.13: The {111} family of planes with the <111> family of directions. There are X = 8 of these.
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V.M. Ayres, ECE474, Spring 2011
Va Pr. 01 is: do likewise for the {110} family. You need only draw four out of the X possibilities.Hint: X is NOT equal to 8.
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V.M. Ayres, ECE474, Spring 2011
Lecture 08:
Lectures: Hexagonal nanosystems: graphene and carbon nanotubes
Introduction to graphene and CNTsThe Basis Vectors: a1 and a2Nearest neighbor distancesThe Chiral Vector: ChThe CNT diameter dtThe Translation Vector: TThe Unit Cell of a CNTNumber of hexagons NNumber of carbon atoms 2N (π electrons)Areal Density
Examples of each
Quantify physical structures of crystal systems that are important for devices:
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V.M. Ayres, ECE474, Spring 2011
Lecture 08:
Lectures: Hexagonal nanosystems: graphene and carbon nanotubes
Introduction to graphene and CNTsThe Basis Vectors: a1 and a2Nearest neighbor distancesThe Chiral Vector: ChThe CNT diameter dtThe Translation Vector: TThe Unit Cell of a CNTNumber of hexagons NNumber of carbon atoms 2N (π electrons)Areal Density
Examples of each
Quantify physical structures of crystal systems that are important for devices:
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V.M. Ayres, ECE474, Spring 2011
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
= | T || Ch |sin(90o)
= | T || Ch |
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V.M. Ayres, ECE474, Spring 2011
Example: find the area of the Unit cell for a (10,10) CNT.
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V.M. Ayres, ECE474, Spring 2011
Example: find the area of the Unit cell for a (10,10) CNT.
From Lec 07:
|Ch| = 43.13 Ang
| T | = 2.49 Ang
Area of the Unit cell = | T || Ch | = 107.39 Ang2
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V.M. Ayres, ECE474, Spring 2011
Example: find the formula for the area of the Unit cell in terms of n, m and dR.
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V.M. Ayres, ECE474, Spring 2011
Example: find the formula for the area of the Unit cell in terms of n, m and dR.
Area of the Unit cell = | T || Ch |
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V.M. Ayres, ECE474, Spring 2011
Number of hexagons N in the Unit Cell:
Basic unit is the hexagon.
The number of hexagons in the Unit cell is called N
Therefore:the number of hexagons N per CNT Unit Cell is:
N = area of the Unit cellarea of one hexagon
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V.M. Ayres, ECE474, Spring 2011
Called: dividing by the Primitive Cell of a CNT:
a1 x a2 is the area of a single basic hexagon. This is also called the primitive cell.
Note: the area of the dotted rhombus is equal to the area of a single hexagon. The magnitude is:
Note the two inequivalent carbon atoms A and B inside the rhombus
23 2a
C C
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V.M. Ayres, ECE474, Spring 2011
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V.M. Ayres, ECE474, Spring 2011
The number of hexagons N per CNT Unit Cell is:
N = | T X Ch || a1 x a2 |
= 2(m2 + n2+nm)/dR
How many carbon atoms per hexagon?
Number of hexagons N versus the number of carbon atoms in the Unit Cell:
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V.M. Ayres, ECE474, Spring 2011
120o
120o
120o
Each C atom is 1/3 inside the hexagon.
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V.M. Ayres, ECE474, Spring 2011
120o
120o
120o
Each C atom is 1/3 inside the hexagon.
6 X (1/3) = 2 atoms per hexagon
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V.M. Ayres, ECE474, Spring 2011
The number of hexagons N per CNT Unit Cell is:
N = | T X Ch || a1 x a2 |
= 2(m2 + n2+nm)/dR
How many carbon atoms per hexagon?
Answer: 2
Number of hexagons N versus the number of carbon atoms in the Unit Cell:
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V.M. Ayres, ECE474, Spring 2011
How many carbon atoms in the Unit cell?
Answer:First solve for the number of hexagons N.
The number of carbon atoms in the Unit cell = 2N.
Number of hexagons N versus the number of carbon atoms in the Unit Cell:
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V.M. Ayres, ECE474, Spring 2011
Each primitive cell contains two carbon atoms.
There is one pz-orbital per each carbon atom.
Therefore there are 2N pzorbitals available per CNT (or graphene) Unit Cell = cloud of electrons = greatconductivity.
The shared electrons from the pz orbitals are called πelectrons.
π electrons and electrical properties:
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V.M. Ayres, ECE474, Spring 2011
a) Example:
Find the number of hexagons N in the unit cell of a (10, 10) CNT.Find the number of carbon atoms in the unit cell of a (10, 10) CNT.
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V.M. Ayres, ECE474, Spring 2011
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V.M. Ayres, ECE474, Spring 2011
Example: find the areal density of the unit cell of a (10, 10) CNT.
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V.M. Ayres, ECE474, Spring 2011
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V.M. Ayres, ECE474, Spring 2011
Example: find the general areal density formula and evaluate it.
This is the conclusion of the discussion we started on the board. The answer is obvious when you think about it!
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V.M. Ayres, ECE474, Spring 2011
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V.M. Ayres, ECE474, Spring 2011
= the same number as for the (10,10) CNT. Of course! The density of a material doesn’t change when you cut it.The areal density on different faces within the cubic system changes but graphene has only one face.
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V.M. Ayres, ECE474, Spring 2011
Interesting fact: there are Endcap restrictions on CNT diameter dt:
CNTs have closed endcapswhich are ½ a buckyball.
Armchair
Zigzag
Chiral
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V.M. Ayres, ECE474, Spring 2011
C60
XZ Ke, et al, Physics Letters A 255 (1999) 294-300
C36
C60 mean ball diameter 6.83 Å
C60 ball outer diameter 10.18 Å
C60 ball inner diameter 3.48 Å
http://www.sesres.com/PhysicalProperties.asp
Buckyballs are ‘round CNTs’. They have some pentagons amongst their hexagons for curvature. C60 forms quite easily, C36 is a more strained structure.
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V.M. Ayres, ECE474, Spring 2011
The smallest possible buckyball endcap is half of the dodecahedral C20, a shape consisting of 12 pentagonal faces and no hexagonal faces. So the smallest CNT diameter dt is about 4 Angstroms.
C20
~4 angstroms