elec lecture 2

1Slide #8/20/2006 Dr. F. Nkansah Physical Electronics

Crystal = Lattice + basis

Infinite number of possible crystals but finite number of possible crystal types, so they are combined together as lattice and basis


Cubic LatticesSimple Cubic Body-centred cubic Face-centered cubic

Cubic lattice types are one of the most common ones. There are only three cubic bravais lattices: they are the simple cubic lattice, the body centered cubic lattice and the face centered cubic lattice. All other cubic crystal structures (for instance the diamond lattice) can be formed by adding an appropriate base at each lattice point to one of those three lattices.

Si and GaAs, two most common type of semiconductors have face centered cubic lattice (actually two interpenetrating FCC lattice or two bases)Both Si and GaAs have two bases at (0,0,0) and (, , ), but in GaAs two different atoms are at the two bases, while in Si they are the same


Diamond Lattice (Si)

Diamond Lattice, can be treated as two interpenetrating FCC sub-lattices

Top View (along any direction) of an extended diamond lattice.

Face-centered cubic plus a basis 1/4, 1/4, 1/4 from each atom

Ref. [1]

Ref. [1]


Unit cell and primitive cell of Diamond Lattice

Primitive cell

Unit cell

Ref. [1]

Note that each atom of Ga or As is bonded to 4 atoms of the opposite type in a tetrahedral arrangement


Diamond Lattice (Si) and ZincblendeLattice (GaAs): A comparison

Diamond Lattice Zincblende Lattice


Example # 1

Find the fraction of the FCC unit cell volume filled with hard spheres in the figure below:

FCC Unit Cell

Atoms per cell = 1 (corners) + 3 (faces) = 4

Nearest neighbor distance =

Radius of each sphere = 241 a

221 a

Volume of each sphere = 3

241

34

aMaximum fraction of cell filled =

No. of spheres X vol. of each sphere / total vol. of each cell = 74 %


Example # 2Calculate the density of GaN (zincblende-type lattice; this is diamondtype lattice but Ga and N atoms are arranged on alternating sites)from the lattice constant (a = 4.5 ), atomic weight (Ga = 69.7 gmol-1,N = 14.0 gmol-1) and Avogadros number (NA = 6.021023particles/mole). Solution:

.1.61002.6)105.4(

)147.69(44313

1

233833 cmg

molcmgmol

Naanm

Vm

A

GaNGaN =+====


Example # 3The X-Ray analysis shows that Ge has cubic lattice structure. Prove thatit is diamond type lattice. Use the following data: the lattice constant is5.65 , density is 5.32 g/cm3, atomic weight is 72.6 g/mole. Solution: Different cubic cells have different atoms per cell (sc has 1 atom/cell, bcc has 2 atoms/cell, fcc has 4 atoms/cell, diamond type lattice has 8 atoms/cell). Lets calculate the number of atoms nper cell in Ge.

.3anm

Vm Ge==

Mass of one Ge atom mGe is: A

GeGe N

m= (NA is Avogadros number).

Thus,

.86.72

1002.6)1065.5(/35.51

12338333

====

gmolemolecmcmgNa

man

Ge

A

Ge

So, Ge has diamond-type lattice.


Diamond Lattice (Si) and ZincblendeLattice (GaAs): A comparison

Diamond Lattice Zincblende Lattice


Example # 1

Find the fraction of the FCC unit cell volume filled with hard spheres in the figure below:

FCC Unit Cell

Atoms per cell = 1 (corners) + 3 (faces) = 4

Nearest neighbor distance =

Radius of each sphere = 241 a

221 a

Volume of each sphere = 3

241

34

aMaximum fraction of cell filled =

No. of spheres X vol. of each sphere / total vol. of each cell = 74 %


Example # 2

Calculate the density of GaN (zincblende-type lattice; this is diamondtype lattice but Ga and N atoms are arranged on alternating sites)from the lattice constant (a = 4.5 ), atomic weight (Ga = 69.7 gmol-1,N = 14.0 gmol-1) and Avogadros number (NA = 6.021023particles/mole). Solution:

.1.61002.6)105.4(

)147.69(44313

1

233833 cmg

molcmgmol

Naanm

Vm

A

GaNGaN =+====


Example # 3The X-Ray analysis shows that Ge has cubic lattice structure. Prove thatit is diamond type lattice. Use the following data: the lattice constant is5.65 , density is 5.32 g/cm3, atomic weight is 72.6 g/mole. Solution: Different cubic cells have different atoms per cell (sc has 1 atom/cell, bcc has 2 atoms/cell, fcc has 4 atoms/cell, diamond type lattice has 8 atoms/cell). Lets calculate the number of atoms nper cell in Ge.

.3anm

Vm Ge==

Mass of one Ge atom mGe is: A

GeGe N

m= (NA is Avogadros number).

Thus,

.86.72

1002.6)1065.5(/35.51

12338333

====

gmolemolecmcmgNa

man

Ge

A

Ge

So, Ge has diamond-type lattice. Note: SC has 1 atom, BCC has 2 atoms, FCC has 4 atoms inan unit cell.

elec lecture 2

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