Physics PY4118 Physics of Semiconductor Devices€¦ · Physics PY4118 Physics of Semiconductor...

Physics PY4118Physics of Semiconductor Devices

Hybrid Bonds

Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland 2.1

ROINN NA FISICE

Department of Physics

PY4118 Physics of Semiconductor Devices

Why?

Orbitals?� They explain the subsequent crystal structureCrystal Structure?� This is important in generating band structure� The crystal also has interesting symmetrySymmetry & Band Structure?� Leads to physical properties


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Hybrid Orbitals� One might expect the number of bonds formed by an

atom would equal its unpaired electrons

� Chlorine, for example, generally forms one bond as it has one unpaired electron - 1s22s22p5

� Oxygen, with two unpaired electrons, usually forms two bonds - 1s22s22p4

� However, Carbon, with only two unpaired electrons, generally forms four (4) bonds

C (1s22s22p2) [He] 2s22p2

The four bonds come from the 2 (2s) paired electrons and the 2 (2p) unpaired electrons


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Hybrid Orbitals� Linus Pauling proposed that the valence atomic orbitals in

a molecule are different from those of the isolated atoms forming the molecule

� Quantum mechanical computations show that if specific combinations of orbitals are mixed mathematically, “new” atomic orbitals are obtained

� The spatial orientation of these new orbitals lead to more “stable” bonds and are consistent with observed molecular shapes

� These new orbitals are called: “Hybrid Orbitals”


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Hybrid Orbitals� Types of Hybrid Orbitals

� Each type has a unique geometric arrangement

� The hybrid type is derived from the number of s, p, d atomic orbitals used to form the Hybrid

Hybrid Orbitals

(Hybridization)

GeometricArrangements

Number of Hybrid Orbitals

Formed byCentral Atom

Example

sp Linear 2 Be in BeF2

sp2 Trigonal planar 3 B in BF3

sp3 Tetrahedral 4 C in CH4


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sp Hybrid Orbitals� SP Hybridization

� 2 electron groups surround central atom

� Linear shape, 180o apart

� VB theory proposes the mixing of two nonequivalent orbitals, one “s” and one “p”, to form two equivalent “sp” hybrid orbitals

� Orientation of hybrid orbitals extend electron density in the bonding direction

� Minimizes repulsions between electrons

� Both shape and orientation maximize overlap between the atoms


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“sp” Hybrid Orbitals

hybrid orbitalsEx: BeCl2

The Be-Cl bonds in BeCl2 are neither

spherical (s orbitals) nor dumbbell (p

orbitals)

The Be-Cl bonds have a hybrid shape

In the Beryllium atom the 2s orbital

and one of the 2p orbitals mix to form

2 sp hybrid orbitals

Each Be Hybrid sp orbital overlaps a

Chlorine 3p orbital in BeCl2

orbital box diagrams

Beryllium Hybrid Orbital Diagram



“sp2” Hybridization� sp2 - Trigonal Planar geometry

� (Central atom bonded to three ligands)

� The three bonds have equivalent hybridized shapes

� The sp2 hybridized orbitals are formed from:

1 “s” orbital and 2 “p” orbitals

Note: Of the 4 orbitals available (1 s & 3 p) only the s orbital and 2 of the p orbitals are used to form hybrid orbitals

Note: Unlike electron configuration notation, hybrid orbital notation uses superscripts for the number of atomic orbitals of a given type that are mixed, NOT for the number of electrons in the orbital, thus,

sp2 (3 orbitals), sp3 (4 orbitals)


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“sp2” Hybridization

BF3

Boron (B) 1s22p1

Forms 3 sp2 hybrid orbitals

The 3 B-F bonds are neither

spherical nor dumbell shaped

They are all of identical shape

In Boron, the “2s” orbital and two of

the “2p” orbitals mix to form 3 sp2

hybrid orbitals, each containing one

of the 3 total valence electrons

Each of the Boron hybrid sp2 orbitals

overlaps with a 2p orbital of a

Fluorine atom

Hybrid Orbital DiagramBF3


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sp3 Hybrid Orbitals� sp3 (4 bonds, thus, Tetrahedral geometry)

� The sp3 hybridized orbitals are formed from:

1 “s” orbital and 3 “p” orbitals

� Example”

� Carbon is the basis for “Organic Chemistry”

� Carbon is in group 4 of the Periodic Chart and has 4 valence electrons – 2s22p2

� The hybridization of these 4 electrons is critical in the formation of the many millions of organic compounds and as the basis of life as we know it


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C atom (ground state)

2s

2p

En

ergy

2s

2p

1s 1s

C atom (promoted)

sp3 Hybrid Orbitals

This structure implies different shapes and energies for the “s” and “p” bonds in carbon compounds.

Observations indicate that all fours bonds are equivalent


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Hybridization of Carbon in CH4

C atom

(ground state)

En

ergy

1s

2p

2s

sp3

1s

sp3

1s

C atom

(hybridized state)

C atom

(in CH4)

C-H bonds

4 sp3 orbitals formed


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Oxygen Atom Bonding in H2O

O

Central Atom

(ground state)

En

ergy

1s

O atom

(hybridized state)

O atom

(in H2O)

sp3

O-H bonds

lonepairs

1s

sp3

1s

2p

2s Tetrahedral

4 sp3 Hybridized Orbitals


Spatial Arrangement ofsp3 Hybrid Orbitals Shape of sp3 hybrid orbital different than either s or p


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sp3 orbitals - details� 1. sp3 = ½ s - ½ px - ½ py + ½ pz

� 2. sp3 = ½ s - ½ px + ½ py - ½ pz

� 3. sp3 = ½ s + ½ px - ½ py - ½ pz

� 4. sp3 = ½ s + ½ px + ½ py + ½ pz

Linear Combination of Atomic Orbitals

Scalar product:

(n.sp3; m.sp3) = 0


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Hybridization visualisation

� http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/hybrv18.swf


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Semiconductor Crystals


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Crystals

The language used for crystals is required: � To describe semiconductors� To understand details of band structure� To design certain types of devices

� Will revisit during discussion on symmetry


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Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland

Valence e’s for “main group” elements


Group 4 Semiconductors

Si atom

(ground state)

En

ergy

2s

3p

3s

sp3

2s

sp3

2s

Si atom

(hybridized state)

Si crystal

Si-Si bonds


3s23p2


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III-V Semiconductors

Ga atom

(hybridized state)

En

ergy

sp3

3s

sp3

3s

As atom

(hybridized state)

GaAs crystal

Ga-As bonds


sp3

3s


4s24p1 4s24p3


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Zincblende (diamond structure)

Si, Ge, GaAs, InP, etc.Diamond: The 2 atoms are the same. Zincblende: The 2 atoms are different.


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Wurtzite (hexagonal structure)

GaN, AlN etc. (semiconductors for blue lasers)


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Survey of Solid State Physics

Taken from:� EP364 Solid State Physics

Prof.Dr. Beşire GÖNÜL

� A link is provided, please look over for a description of the crystal lattice

� Now a very short summary…


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25

CRYSTAL LATTICE

What is crystal (space) lattice?

In crystallography, only the geometrical properties of the crystalare of interest, therefore one replaces each atom by ageometrical point located at the equilibrium position of that atom.

Platinum Platinum surface Crystal lattice and structure of Platinum

26

Crystal Structure

� Crystal structure can be obtained by attaching atoms, groupsof atoms or molecules which are called basis (motif) to thelattice sides of the lattice point.

Crystal Structure = Crystal Lattice + Basis

A two-dimensional Bravais lattice with

different choices for the basis

27

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Types Of Crystal Lattices

1) Bravais lattice is an infinite array of discrete points with anarrangement and orientation that appears exactly the same,from whichever of the points the array is viewed. Lattice isinvariant under a translation.

Nb film

29

Translational Lattice Vectors – 2D

A space lattice is a set of points such thata translation from any point in the latticeby a vector;

Rn = n1 a + n2 b

locates an exactly equivalent point, i.e. apoint with the same environment as P .This is translational symmetry. Thevectors a, b are known as lattice vectorsand (n1, n2) is a pair of integers whosevalues depend on the lattice point.

P

Point D(n1, n2) = (0,2)

Point F (n1, n2) = (0,-1)Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland

ROINN NA FISICE Department of Physics

30

� The two vectors a and bform a set of lattice vectorsfor the lattice.

� The choice of lattice

vectors is not unique.Thus one could equally welltake the vectors a and b’ asa lattice vectors.

Lattice Vectors – 2D

31

Lattice Vectors – 3D

An ideal three dimensional crystal is described by 3fundamental translation vectors a, b and c. If there is a latticepoint represented by the position vector r, there is then also alattice point represented by the position vector where u, v and ware arbitrary integers.

r’ = r + u a + v b + w c

32

Unit Cell in 2D

� The smallest component of the crystal (group of atoms, ionsor molecules), which when stacked together with puretranslational repetition reproduces the whole crystal.

S

a

b

S

S

S

S

S

S

S

S

S

S

S

S

S

S

33

Unit Cell in 2D

� The smallest component of the crystal (group of atoms, ionsor molecules), which when stacked together with puretranslational repetition reproduces the whole crystal.

S

S

The choice of unit cell

is not unique.

a

Sb

S

34

2D Unit Cell example -(NaCl)

We define lattice points ; these are points with identical

environments

35

Choice of origin is arbitrary - lattice points need not be atoms - but unit cell size should always be the same.

36

This is also a unit cell -it doesn’t matter if you start from Na or Cl

37

- or if you don’t start from an atom

38

This is NOT a unit cell even though they are all the same - empty space is not allowed!

39

Unit Cell in 3D

40

Three common Unit Cell in 3D

41

UNIT CELL

Primitive Conventional & Non-primitive

� Single lattice point per cell� Smallest area in 2D, or�Smallest volume in 3D

� More than one lattice point per cell� Integral multiples of the area of

primitive cell

Body centered cubic(bcc)

Conventional ≠ Primitive cell

Simple cubic(sc)Conventional = Primitive cell

42

The Conventional Unit Cell

� A unit cell just fills space whentranslated through a subset ofBravais lattice vectors.

� The conventional unit cell ischosen to be larger than theprimitive cell, but with the fullsymmetry of the Bravais lattice.

� The size of the conventional cell isgiven by the lattice constant a.

43

Primitive and conventional cells of FCC

44

� A primitive unit cell is made of primitivetranslation vectors a1 ,a2, and a3 such thatthere is no cell of smaller volume that canbe used as a building block for crystalstructures.

� A primitive unit cell will fill space by repetitionof suitable crystal translation vectors. Thisdefined by the parallelpiped a1, a2 and a3.The volume of a primitive unit cell can befound by

� V = a1.(a2 x a3) (vector products) Cubic cell volume = a3

Primitive Unit Cell and vectors

45

� The primitive unit cell must have only one lattice point.� There can be different choices for lattice vectors , but the

volumes of these primitive cells are all the same.

P = Primitive Unit CellNP = Non-Primitive Unit Cell

Primitive Unit Cell

1a

Wigner-Seitz Method

A simply way to find the primitivecell which is called Wigner-Seitzcell can be done as follows;

1. Choose a lattice point.2. Draw lines to connect these

lattice point to its neighbours.3. At the mid-point and normal to

these lines draw new lines.

The volume enclosed is called as a Wigner-Seitz cell.

46

Wigner-Seitz Method

47

Wigner-Seitz Cell - 3D

48

Crystal Directions

Fig. Shows [111] direction

� We choose one lattice point on the line asan origin, say the point O. Choice of originis completely arbitrary, since every latticepoint is identical.

� Then we choose the lattice vector joining Oto any point on the line, say point T. Thisvector can be written as;

R = n1 a + n2 b + n3c

� To distinguish a lattice direction from alattice point, the triple is enclosed in squarebrackets [ ...] is used.[n1n2n3]

� [n1n2n3] is the smallest integer of the samerelative ratios.

49

210

X = 1 , Y = ½ , Z = 0[1 ½ 0] [2 1 0]

X = ½ , Y = ½ , Z = 1[½ ½ 1] [1 1 2]

Examples

50

X = -1 , Y = -1 , Z = 0 [110]

Examples of crystal directions

X = 1 , Y = 0 , Z = 0 [1 0 0]

_

1Negative One = 51

Miller Indices

Miller Indices are a symbolic vector representation for the orientation of anatomic plane in a crystal lattice and are defined as the reciprocals of thefractional intercepts which the plane makes with the crystallographic axes.

To determine Miller indices of a plane, take the following steps;

1) Determine the intercepts of the plane along each of the three crystallographic directions

2) Take the reciprocals of the intercepts

3) If fractions result, multiply each by the denominator of the smallest fraction

52

Axis X Y Z

Intercept points 1 ∞ ∞

Reciprocals 1/1 1/ ∞ 1/ ∞Smallest

Ratio 1 0 0

Miller İndices (100)

Example-1

(1,0,0)

53

Axis X Y Z

Intercept points 1 1 ∞

Reciprocals 1/1 1/ 1 1/ ∞Smallest

Ratio 1 1 0


Example-2

(1,0,0)

(0,1,0)

54

Axis X Y Z

Intercept points 1 1 1

Reciprocals 1/1 1/ 1 1/ 1Smallest

Ratio 1 1 1

Miller İndices (111)(1,0,0)

(0,1,0)

(0,0,1)

Example-3

55

Axis X Y Z

Intercept points 1/2 1 ∞

Reciprocals 1/(½) 1/ 1 1/ ∞Smallest

Ratio 2 1 0

Miller İndices (210)(1/2, 0, 0)

(0,1,0)

Example-4

56

Axis a b c

Intercept points 1 ∞ ½

Reciprocals 1/1 1/ ∞ 1/(½)

Smallest Ratio 1 0 2


Example-5

57

Axis a b c

Intercept points -1 ∞ ½

Reciprocals 1/-1 1/ ∞ 1/(½)

Smallest Ratio -1 0 2


Example-6

58

Miller Indices

Reciprocal numbers are: 2

1 ,

2

1 ,

3

1

Plane intercepts axes at cba 2 ,2 ,3

Indices of the plane (Miller): (2,3,3)

(100)

(200)

(110)(111)

(100)

Indices of the direction: [2,3,3]a

3

2

2

bc

[2,3,3]

59

� There are only seven different shapes of unit cell which can bestacked together to completely fill all space (in 3 dimensions)without overlapping. This gives the seven crystal systems, inwhich all crystal structures can be classified.

� Cubic Crystal System (SC, BCC, FCC)� Hexagonal Crystal System (S)� Triclinic Crystal System (S)� Monoclinic Crystal System (S, Base-C)� Orthorhombic Crystal System (S, Base-C, BC, FC)� Tetragonal Crystal System (S, BC)� Trigonal (Rhombohedral) Crystal System (S)

3D 3D 3D 3D –––– 14 BRAVAIS LATTICES AND THE SEVEN CRYSTAL SYSTEM14 BRAVAIS LATTICES AND THE SEVEN CRYSTAL SYSTEM14 BRAVAIS LATTICES AND THE SEVEN CRYSTAL SYSTEM14 BRAVAIS LATTICES AND THE SEVEN CRYSTAL SYSTEM

TYPICAL CRYSTAL STRUCTURESTYPICAL CRYSTAL STRUCTURESTYPICAL CRYSTAL STRUCTURESTYPICAL CRYSTAL STRUCTURES

61

The only semiconductor crystals we are interested in

62

Coordinatıon Number

� Coordinatıon Number (CN) : The Bravais lattice points closestto a given point are the nearest neighbours.

� Because the Bravais lattice is periodic, all points have thesame number of nearest neighbours or coordination number.It is a property of the lattice.

� A simple cubic has coordination number 6; a body-centeredcubic lattice, 8; and a face-centered cubic lattice,12.

63

Atomic Packing Factor

� Atomic Packing Factor (APF) is defined as thevolume of atoms within the unit cell divided bythe volume of the unit cell.

APF =Volume of Atoms in the Unit Cell

Volume of the Unit Cell

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1-CUBIC CRYSTAL SYSTEM

� Simple Cubic has one lattice point so its primitive cell.� In the unit cell on the left, the atoms at the corners are cut

because only a portion (in this case 1/8) belongs to that cell. Therest of the atom belongs to neighboring cells.

� Coordinatination number of simple cubic is 6.

a- Simple Cubic (SC)

a

b c

65

Atomic Packing Factor of SC

66

b-Body Centered Cubic (BCC)

� BCC has two lattice points so BCC isa non-primitive cell.

� BCC has eight nearest neighbors.Each atom is in contact with itsneighbors only along the body-diagonal directions.

� Many metals (Fe,Li,Na..etc), includingthe alkalis and several transitionelements choose the BCC structure.

a

b c

67

0.68 = V

V = APF

cell unit

atomsBCC

Atomic Packing Factor of BCC

Can you work this out?

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c- Face Centered Cubic (FCC)

� There are atoms at the corners of the unit cell and at the center ofeach face.

� Face centered cubic has 4 atoms so its non primitive cell.� Many of common metals (Cu,Ni,Pb..etc) crystallize in FCC

structure.

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0.68 = V

V = APF

cell unit

atomsBCCFCC 0.74

Atomic Packing Factor of FCC

Can you work this out?

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Atoms Shared Between: Each atom counts:corner 8 cells 1/8face centre 2 cells 1/2body centre 1 cell 1

lattice type cell contentsP 1 [=8 x 1/8]I 2 [=(8 x 1/8) + (1 x 1)]F 4 [=(8 x 1/8) + (6 x 1/2)]

Unit cell contents

Counting the number of atoms within the unit cell

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2 - HEXAGONAL SYSTEM

� A crystal system in which three equal coplanar axesintersect at an angle of 60 , and a perpendicular to theothers, is of a different length.

73

Bravais Lattice : Hexagonal LatticeHe, Be, Mg, Hf, Re (Group II elements)ABABAB Type of Stacking

Hexagonal Close-packed Structure

a=b a=120, c=1.633a,basis : (0,0,0) (2/3a ,1/3a,1/2c)

74

A A

AA

AA

A

AAA

AA

AAA

AAA

Sequence ABABAB..-hexagonal close pack

Sequence ABCABCAB..-face centered cubic close pack

Close packed

B

AA

AA

A

A

A

A A

B

B B

Sequence AAAA…- simple cubic

Sequence ABAB…- body centered cubic

Packing

B B

B

B

B B

B

B

B

BB

C C C

CC

C

C

C C C

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Hexagonal Closest Packing

76

Cubic Closest Packing

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4 - Diamond Structure

� The diamond lattice is consist of two interpenetrating facecentered bravais lattices.

� There are eight atom in the structure of diamond.

� Each atom bonds covalently to 4 others equally spread about atom in 3d.

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4 - Diamond Structure

� The coordination number of diamond structure is 4.

� The diamond lattice is not a Bravais lattice.� Si, Ge and C crystallizes in diamond

structure.

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5- Zinc Blende

� Zincblende has equal numbers of zinc and sulfurions distributed on a diamond lattice so thateach has four of the opposite kind as nearestneighbors. This structure is an example of alattice with a basis, which must so describedboth because of the geometrical position of theions and because two types of ions occur.

� AgI,GaAs,GaSb,InAs,

80

Zincblende (ZnS) Lattice

Zincblende LatticeThe Cubic Unit Cell.

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Crystal Symmetry


2.82ROINN NA FISICE Department of Physics

Symmetry?

This is actually really important for some semiconductor devices, especially:

Inversion Symmetry:

This is (not) required for:� Second harmonic generation� The electro-optic effect� Piezo-electric effect� etc.Coláiste na hOllscoile Corcaigh,

Éire University College Cork, Ireland

3.83ROINN NA FISICE Department of Physics

� Each of the unit cells of the 14 Bravais lattices has one ormore types of symmetry properties, such as inversion,reflection or rotation,etc.

SYMMETRY

INVERSION REFLECTION ROTATION

ELEMENTS OF SYMMETRYELEMENTS OF SYMMETRYELEMENTS OF SYMMETRYELEMENTS OF SYMMETRY

84

Lattice goes into itself through Symmetry without translation

Operation Element

Inversion Point

Reflection Plane

Rotation Axis

Rotoinversion Axes85

Reflection Plane

� A plane in a cell such that, when a mirror reflection in thisplane is performed, the cell remains invariant.

86

Rotation Axis

� This is an axis such that, if the cell is rotated around itthrough some angles, the cell remains invariant.

� The axis is called n-fold if the angle of rotation is 2π/n.

90°

120° 180°

87


Symmetry

� The geometry of the crystal lattice can be described by its symmetry

� The symmetry is key in understanding certain physical properties of material

� Let’s see how and why


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Department of Physics 2.88


Crystal Optics - susceptibility


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For a real material, the dielectric susceptibility is

not necessarily the same in all directions, thus:

=

z

y

x

z

y

x

E

E

E

P

P

P

333231

232221

131211

0

χχχ

χχχ

χχχ

ε

The dielectric susceptibility is a type 2 tensor


Type 2 Tensor


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A thermodynamic argument shows that the tensor

is symmetric:

=→=

332313

232212

131211

χχχ

χχχ

χχχ

χχχ jiij

Let’s now consider symmetry operations…

Link 1 Link 2Here is the proof:


Symmetry Operations


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A vector can be rotated, mirrored, inverted etc., using a

matrix operator. For example

−

−

−

=

100

010

001

iThe inversion matrix is:

Thus:

−

−

−

=

−

−

−

z

y

x

z

y

x

100

010

001

Link to a list of more symmetry operators


Symmetry Operations


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−

=

100

010

001

zσReflection in x-y plane:

( )

−

=

100

02

cos2

sin

02

sin2

cos

nn

nn

zCn

ππ

ππ

Rotation around z:


Symmetry Operations


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So a 4-fold rotation symmetry, will be:

( )

−=

−

=

100

001

010

100

04

2cos

4

2sin

04

2sin

4

2cos

4

ππ

ππ

zC

Thus:

−=

−

z

x

y

z

y

x

100

001

010


Symmetry Operations


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The symmetry operators can also be used to rotate the matrixT

MTMT =′

So a 4-fold rotation of our tensor is:

−

−

100

001

010

100

001

010

332313

232212

131211

χχχ

χχχ

χχχ

−

−−

−

=

−

−

−

−=

331323

131112

231222

331323

231222

131112

100

001

010

χχχ

χχχ

χχχ

χχχ

χχχ

χχχ

NOTE


Symmetry Operations


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Thus, we can then equate the rotated and non-rotated

tensors:

−

−−

−

=

331323

131112

231222

332313

232212

131211

χχχ

χχχ

χχχ

χχχ

χχχ

χχχ

Thus: 2211χχ =

01212

=−= χχ

0132313

=−== χχχ

=→

33

11

11

00

00

00

χ

χ

χ

χ

If cubic, then there will be 4-fold rotation along other axis

as well, in which case 332211χχχ ==

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