ECE 474:
Principles of Electronic Devices
Prof. Virginia Ayres
Electrical & Computer Engineering
Michigan State University
[email protected]
Example for VA Pr. 01 Ù 1.13: The {111} family of planes with
the <111> family of directions. There are X = 8 of these.
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.
V.M. Ayres, ECE474, Spring 2011
Lecture 08:
Quantify physical structures of crystal systems that are
important for devices:
‰ Lectures: Hexagonal nanosystems: graphene and
carbon nanotubes
¾Introduction to graphene and CNTs
¾The Basis Vectors: a1 and a2
¾Nearest neighbor distances
¾The Chiral Vector: Ch
¾The CNT diameter dt
¾The Translation Vector: T
¾The Unit Cell of a CNT
¾Number of hexagons N
¾Number of carbon atoms 2N (π electrons)
¾Areal Density
Examples of each
V.M. Ayres, ECE474, Spring 2011
Lecture 08:
Quantify physical structures of crystal systems that are
important for devices:
‰ Lectures: Hexagonal nanosystems: graphene and
carbon nanotubes
¾Introduction to graphene and CNTs
¾The Basis Vectors: a1 and a2
¾Nearest neighbor distances
¾The Chiral Vector: Ch
¾The CNT diameter dt
¾The Translation Vector: T
¾The Unit Cell of a CNT
¾Number of hexagons N
¾Number of carbon atoms 2N (π electrons)
¾Areal Density
Examples of each
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 |
V.M. Ayres, ECE474, Spring 2011
Example: find the area of the Unit cell for a (10,10) CNT.
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
V.M. Ayres, ECE474, Spring 2011
Example: find the formula for the area of the Unit cell in
terms of n, m and dR.
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 |
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 cell
area of one hexagon
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: 3a 2
2
C
V.M. Ayres, ECE474, Spring 2011
C
Note the two inequivalent
carbon atoms A and B
inside the rhombus
V.M. Ayres, ECE474, Spring 2011
Number of hexagons N versus the number of
carbon atoms in the Unit Cell:
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?
V.M. Ayres, ECE474, Spring 2011
Each C atom is 1/3
inside the hexagon.
120o
120o
120o
V.M. Ayres, ECE474, Spring 2011
Each C atom is 1/3
inside the hexagon.
120o
120o
120o
V.M. Ayres, ECE474, Spring 2011
6 X (1/3) = 2 atoms
per hexagon
Number of hexagons N versus the number of
carbon atoms in the Unit Cell:
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
V.M. Ayres, ECE474, Spring 2011
Number of hexagons N versus the number of
carbon atoms in the Unit Cell:
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.
V.M. Ayres, ECE474, Spring 2011
π electrons and electrical properties:
Each primitive cell contains
two carbon atoms.
There is one pz-orbital per
each carbon atom.
Therefore there are 2N pz
orbitals available per CNT (or
graphene) Unit Cell = cloud
of electrons = great
conductivity.
The shared electrons from
the pz orbitals are called π
electrons.
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.
V.M. Ayres, ECE474, Spring 2011
V.M. Ayres, ECE474, Spring 2011
Example: find the areal density of the unit cell of a (10, 10)
CNT.
V.M. Ayres, ECE474, Spring 2011
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!
V.M. Ayres, ECE474, Spring 2011
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.
V.M. Ayres, ECE474, Spring 2011
Interesting fact: there are Endcap restrictions on
CNT diameter dt:
CNTs have
closed endcaps
which are ½ a
buckyball.
Armchair
Zigzag
Chiral
V.M. Ayres, ECE474, Spring 2011
Buckyballs are ‘round CNTs’. They have some pentagons amongst their
hexagons for curvature. C60 forms quite easily, C36 is a more strained
structure.
C60
C36
C60 mean ball
diameter 6.83 Å
C60 ball outer
diameter 10.18 Å
XZ Ke, et al, Physics Letters A 255 (1999) 294-300
C60 ball inner
diameter 3.48 Å
V.M. Ayres, ECE474, Spring 2011
http://www.sesres.com/Phys
icalProperties.asp
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
V.M. Ayres, ECE474, Spring 2011