solid state physics 1 - · pdf file1.1. stress and strain tensors ssp1 1.1 stress and strain...

Solid State Physics 1

Vincent Casey

Autumn 2017

Contents

1 Crystal Mechanics 11.1 Stress and Strain Tensors . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Physical Meaning . . . . . . . . . . . . . . . . . . . . . . 61.1.2 Simplification of the ’Matrix’ . . . . . . . . . . . . . . . 91.1.3 Deformation Modes of Cubic Crystals . . . . . . . . . . . 11

2 Elastic Waves in Solids 152.1 Elastic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Waves in Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2.1 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2.2 Static and Dynamic Models . . . . . . . . . . . . . . . . 22

2.3 Linear Lattice Models . . . . . . . . . . . . . . . . . . . . . . . . 232.3.1 Linear 1D Chain Model . . . . . . . . . . . . . . . . . . 232.3.2 Diatomic Linear Chain Model . . . . . . . . . . . . . . . 27

3 Specific Heat Capacity 293.1 Specific Heat Capacities . . . . . . . . . . . . . . . . . . . . . . 30

3.1.1 Historical . . . . . . . . . . . . . . . . . . . . . . . . . . 303.1.2 Specific Heat: Classical Model . . . . . . . . . . . . . . . 303.1.3 Einstein Model . . . . . . . . . . . . . . . . . . . . . . . 30

3.2 Debye Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2.1 Key Features . . . . . . . . . . . . . . . . . . . . . . . . 353.2.2 Debye Model . . . . . . . . . . . . . . . . . . . . . . . . 35

3.3 Phonons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.3.1 Actual Phonon Density of States . . . . . . . . . . . . . . 403.3.2 Phonon Interactions . . . . . . . . . . . . . . . . . . . . 403.3.3 Anharmonicity and Thermal Expansion . . . . . . . . . . 423.3.4 Normal and Umklapp Phonon Interactions . . . . . . . . . 433.3.5 Neutron Scattering by a Crystal . . . . . . . . . . . . . . 433.3.6 Inelastic Neutron Spectroscopy . . . . . . . . . . . . . . 46

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CONTENTS SSP1

4 Drude Free Electron Theory 494.1 Classical Theory and Conduction in Solids . . . . . . . . . . . . . 50

4.1.1 Free Electron Theory (Drude’s Free Electron Theory) . . . 504.1.2 Case I: No electric field . . . . . . . . . . . . . . . . . . . 524.1.3 Case II: Constant uniform electric field . . . . . . . . . . 524.1.4 Case III: Time dependent sinusoidal electric field . . . . . 534.1.5 Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.2 Drude Model and Metal Reflectivity . . . . . . . . . . . . . . . . 604.2.1 Drude model and plasma frequency of metals . . . . . . . 614.2.2 Plasma oscillations in metals . . . . . . . . . . . . . . . . 614.2.3 Plasma oscillations in metals with scattering . . . . . . . . 62

5 Band Theory of Solids 655.1 Band Theory of Solids . . . . . . . . . . . . . . . . . . . . . . . 66

5.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 665.1.2 Bloch’s Theorem . . . . . . . . . . . . . . . . . . . . . . 675.1.3 The Nature of the Crystal Potential Field . . . . . . . . . 685.1.4 Solving Schrodinger’s equation for the K-P 1D model . . 685.1.5 Dispersion Relation for the Kronig-Penney Model . . . . 73

5.2 Some consequences of the band theory of solids . . . . . . . . . . 755.2.1 Band Filling . . . . . . . . . . . . . . . . . . . . . . . . 755.2.2 Band Shapes of Real Semiconductors . . . . . . . . . . . 755.2.3 Effective Mass and Holes . . . . . . . . . . . . . . . . . . 775.2.4 Filling of the energy bands with electrons. . . . . . . . . . 795.2.5 Occupation of allowed states. . . . . . . . . . . . . . . . 815.2.6 Properties of Semiconductors . . . . . . . . . . . . . . . 835.2.7 Carrier Density . . . . . . . . . . . . . . . . . . . . . . . 835.2.8 Fermi Level in Semiconductors . . . . . . . . . . . . . . 855.2.9 Law of Mass Action . . . . . . . . . . . . . . . . . . . . 855.2.10 Temperature Dependence of Carrier Density . . . . . . . 88

5.3 Current Transport and Current Density Equations . . . . . . . . . 88

6 Unipolar Devices 916.1 High Field Effects . . . . . . . . . . . . . . . . . . . . . . . . . . 926.2 Explanation of NDR based on band structure . . . . . . . . . . . 936.3 Negative Differential Resistance . . . . . . . . . . . . . . . . . . 93

6.3.1 NDR and Instability . . . . . . . . . . . . . . . . . . . . 956.4 Transfer Electron Devices TEDs . . . . . . . . . . . . . . . . . . 95

6.4.1 Modes of Operation . . . . . . . . . . . . . . . . . . . . 956.4.2 Accumulation . . . . . . . . . . . . . . . . . . . . . . . . 976.4.3 Domain Mode . . . . . . . . . . . . . . . . . . . . . . . 98

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SSP1 CONTENTS

7 pn Junction Diode 1017.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1027.2 Fabrication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1027.3 Idealized Junction Formation . . . . . . . . . . . . . . . . . . . . 103

7.3.1 Energy band diagrams . . . . . . . . . . . . . . . . . . . 1047.3.2 Pn Junction Equilibrium Band Diagram . . . . . . . . . . 1057.3.3 Currents flowing at equilibrium. . . . . . . . . . . . . . . 106

7.4 Barrier potential VB. . . . . . . . . . . . . . . . . . . . . . . . . 1077.5 Device Equations at equilibrium . . . . . . . . . . . . . . . . . . 108

7.5.1 The Depletion Approximation. . . . . . . . . . . . . . . . 1087.5.2 Depletion Region Width . . . . . . . . . . . . . . . . . . 1117.5.3 Maximum Electric Field . . . . . . . . . . . . . . . . . . 111

7.6 Junction Capacitance . . . . . . . . . . . . . . . . . . . . . . . . 1127.6.1 Diode Information from CJ measurements. . . . . . . . . 113

7.7 Pn Junctions under external bias. . . . . . . . . . . . . . . . . . . 1157.7.1 Reverse Bias . . . . . . . . . . . . . . . . . . . . . . . . 1157.7.2 Forward Bias . . . . . . . . . . . . . . . . . . . . . . . . 1157.7.3 The diode or rectifier equation. . . . . . . . . . . . . . . . 118

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Chapter 1

Crystal Mechanics

1

1.1. STRESS AND STRAIN TENSORS SSP1

1.1 Stress and Strain TensorsThis section is drawn largely from:

The Physics and Chemistry of Solids, Stephen Elliot, Wiley 1998A solid is stressed by applying external forces such that both the net force and

the net torque are zero, i.e. static equilibrium prevails.

Figure 1.1: (a) Zero net force and (b) zero net torque, applied to a cubic volumeelement of face area A. The first suffix indicates the direction in which the stress(force) acts and the second suffix indicates the direction of the normal to the faceupon which the stress acts.

Zero Net Force achieved by exerting forces equal in magnitude, but in oppositedirections, perpendicularly to opposite faces, tending to compress/elongatethe sample, 1.1(a).

Zero Net Torque achieved by applying equal and opposite forces in a parallelfashion to opposite faces, tending to shear the sample, 1.1(b).

Therefore, the cube does not accelerate or rotate. Two kinds of stress are identi-fied:

Normal, σi j with i = j;Shear, σi j with i 6= j.

A force Fx applied in the x-direction to a plane with area Ax whose normal is alsoin the x-direction produces a stress component:

σxx =Fx

Ax

Likewise, the same force applied to a plane whose normal is in the y-directionproduces the stress component:

σxy =Fx

Ay

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SSP1 1.1. STRESS AND STRAIN TENSORS

Since stresses on opposite faces of the cube are equal and opposite we only need toconsider three faces. Thus, stress in general is described by a second-rank tensorσ with nine components σi j(i, j = x,y,z), conveniently written as a 3×3 array:

σ =

σxx σxy σxzσyx σyy σyzσzx σzy σzz

However, only six of these stress components are independent because of the con-straint of zero torque, which implies that:

σxy = σyx;σyz = σzy;σzx = σxz

Special cases of stress have most of the six independent stress components equalto zero. For instance for uniaxial stress in the x-direction, the only non-zerocomponent is σxx = σ.For hydrostatic stress, the non-zero components are the diagonal components

σxx = σyy = σzz =−P

For pure shear, in which the shape but not the volume of the solid is changed, isrepresented by say

σxy = σyx = σ

being the only non-zero components.A particular orientation of the spatial coordinates can always be found such

that the general stress matrix has diagonal components only, i.e.

σ′=

σx 0 00 σy 00 0 σz

the six independent variables defining the stress system now being the three prin-cipal stresses σi(i = x,y,z) acting along the principle axes, and the three variablesneeded to determine the orientation of the principle axes with respect to the origi-nal coordinate system.

A general stress matrix, transformed to the principal axis system can alwaysbe written as

σx 0 00 σy 00 0 σz

=

σ0 0 00 σ0 00 0 σ0

+ (σx−σ0) 0 0

0 (σy−σ0) 00 0 (σz−σ0)

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Figure 1.2: Uniform hydrostatic compressive stress and pure shear deformation(at constant volume).

where the new stress component is given by

σ0 = (σx +σy +σz)/3

The first term of the transformed stress matrix represents a purely hydrostaticterm, with

σ0 =−P = (σx +σy +σz)/3,

which causes a change in volume(dilatoric), but not of shape, of an elasticallyisotropic solid.The second term of the transformed stress matrix represents a pure shear (devia-toric), causing a change of shape, but not of volume, of a solid, since the sum ofthe diagonal components equals zero, i.e. ∑x,y,z σii = 0

The application of an external force to a solid causes a deformation becausedifferent points in the material are displaced by different amounts. Consider apoint P initially at r and a neighbouring point Q initially at r+u, displaced underthe action of stress to P

′at (r+∆r) and Q

′at (r+∆r+u+∆u) respectively.

For small relative displacements (|∆u| << |∆r|), the components of the relativedisplacement are given by

∆ui =∂ui

∂x∆x+

∂ui

∂y∆y+

∂ui

∂z∆z

for i = x,y,z. The strain components ei j ≡ εi j are then defined in terms of thedimensionless displacement gradients as

eii =∂ui

∂i(i = x,y,z)

and

ei j = e ji =

(∂ui

∂ j+

∂u j

∂i

)(i = x,y,z)

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Figure 1.3: 2D representation of infinitesimal homogeneous elastic strain.

Therefore, the strain components can, like the stress components, be written asa (3× 3) matrix (symmetric) with again only six of the nine components beingindependent because of the requirement that the off-diagonal components obeythe condition ei j = e ji in order to exclude rigid rotations.

Special cases of strain include uniaxial strain in the x-direction with onlyone non-zero strain component being exx = e; uniform dilation or compression,resulting from hydrostatic stress, with ei j = e (i = x,y,z); and pure shear withei j = e ji = e (i = x, j = y) or other orthogonal pair.

As with a general stress, a general strain can be separated into a dilation-strain component (volume change) and a pure strain (deviatoric-strain) component(shape change). For a general strain component

ei j =e0

3δi j +(ei j−

e0

3δi j)

where the Kronecker delta symbol has the properties:

δi j = 1, i = jδi j = 0, i 6= j

Since there are only six independent stress and strain components, a convenientshort-hand notation is to relabel the component indices as follows:

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xx→ 1yy→ 2zz→ 3yz→ 4zx→ 5xy→ 6

Hooke’s law states that the stress and strain are directly proportional to eachother. Thus, in terms of elastic-stiffness coefficients, Ci j, a stress coefficient canbe written as a function of a strain as

σi =6

∑j=1

Ci je j

The relationship between stress and strain coefficients can also be written in ma-trix form as

σ1σ2σ3σ4σ5σ6

=

C11 C12 C13 C14 C15 C16C21 C22 C23 C24 C25 C26C31 C32 C33 C34 C35 C36C41 C42 C43 C44 C45 C46C51 C52 C53 C54 C55 C56C61 C62 C63 C64 C65 C66

e1e2e3e4e5e6

Alternatively, a strain coefficient can be expressed in terms of the elastic-compliancecoefficients, Si j, as a function of the stress:

ei =6

∑j=1

Si jσ j

where both elastic compliance and stiffness are quantities describing a material asan elastic continuum.

1.1.1 Physical MeaningPhysical meaning of ei j

e11 =∂ux

∂x

e22 =∂uy

∂y

e33 =∂uz

∂z

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Figure 1.4: Stress Strain Matrix

Simple tensile strains along the x,y,z axes are positive for tension and negative forcompression. Deformation in the xy plane of a rectangular thin film correspondsto:

e12 6= 0;e21 6= 0

e11 = e22 = 0

e3i = ei3 = 0; i = x,y,z

With angles greatly exaggerated, this corresponds to:

e12 ≈∂ux

∂y

e21 ≈∂uy

∂xwhich equals the angles respectively.

Figure 1.5: Pure Shear

Possible combinations of e12 and e21 are:

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Pure Shear e12 = e21

Pure Rotation e12 =−e21

Simple Shear e12 6= 0;e21 = 0

Figure 1.6: A pure rotation.

Figure 1.7: Simple Shear

e =

0 γ/2 0γ/2 0 00 0 0

+

0 γ/2 0−γ/2 0 0

0 0 0

ei j = εi j +ωi j

εi j = 1/2(ei j + e ji)

ωi j = 1/2(ei j− e ji)

ei j = e ji - symmetric with respect to i and j: measures shape change!

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Figure 1.8: Comparing pure strain and rotation.

ωi j =−ω ji - antisymmetric with respect to i and j: measures rotation.ωi j = 0(i = j)Therefore ωi j has three independent elements implying that three independent

rotations are possible, one about each axis.

1.1.2 Simplification of the ’Matrix’Of the 36 elastic stiffness (or compliance) coefficients in the most general caseof triclinic crystals, 21 are independent and non zero as a result of the generalcondition

Ci j =C ji

The presence of higher symmetry reduces the number of coefficients further.For cubic crystals, just three components are independent, C11,C12 and C44 with

C11 =C22 =C33,C12 =C13 =C23,C44 =C55 =C66,

and all other coefficients being zero.C11 C12 C12 0 0 0C12 C11 C12 0 0 0C12 C12 C11 0 0 00 0 0 C44 0 00 0 0 0 C44 00 0 0 0 0 C44

These elastic stiffness constants are related to the corresponding elastic compli-ance coefficients via the relationships

S11 =C11 +C12

(C11−C12)(C11 +2C12)> 0,

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S12 =−C12

(C11−C12)(C11 +2C12)≤ 0,

S44 =1

C44

as obtained from inversion of the stiffness matrix.In the case of elastically isotropic solids (amorphous solids), the number of

independent elastic coefficients decreases to two since, in addition to the equalitiesapplicable to cubic materials, the coefficients are further linearly related to eachother by the equation

C11 =C12 +2C44

Lame Constants: For such isotropic materials, the two independent elasticstiffness coefficients are conventionally referred to as the Lame constants:

λ≡C12

and

µ =C44

with

C11 ≡ (λ+2µ)

Uniaxial Stress: For uniaxial stress,Young’s modulus is defined as

E =σxx

exx

and so

E =1

S11=

(C11−C12)(C11 +2C12)

C11 +C12= µ

(3λ+2µ)(λ+µ)

for the case of isotropic media. Under such stress loading, the sample also deformsin directions perpendicular to the stress direction. This behaviour is quantified byPoisson’s ratio, defined as

ν =|transverse strain|

normal strain=|eyy|exx

with

ν =−S12

S11=

C12

(C11 +C12)=

λ

2(λ+µ)

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where 0≤ ν≤ 0.5.Hydrostatic Stress: In the case of hydrostatic stress (pressure), the bulk mo-

dulus B (equal to the inverse of the compressibility, κ = 1B relates the pressure to

the dilation (fractional change in volume) e0,

B =Pe0

with

B =1

3(S11 +2S12)=

(C11 +2C12)

3=

(3λ+2µ)3

The bulk modulus is connected to microscopic quantities relating to the interato-mic potential.

Pure Shear: Finally, for the case of pure shear, the shear (or rigidity) modulusis defined as:

G =shear stressshear strain

=σxy

exy

giving for isotropic solids

G =1

2(S11−S12)=

(C11−C12)

2≡C44 = µ

Some Interrelationships: The various elastic moduli are inter-related; forexample

G =E

2(1+ν

and

B =E

3(1−2ν)

1.1.3 Deformation Modes of Cubic Crystals

Each independent elastic constant is associated with a fundamental mode of de-formation. For an isotropic material there are two modes: dilation and shear. Thethree independent constants in a cubic crystal require there to be three independentmodes of deformation.

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(a) Dilation by hydrostatic stress

Under a hydrostatic pressure P:

σ1 = σ2 = σ3 =−P

σ4 = σ5 = σ6 =−0

and the dilation4= e1 + e2 + e3 is

4= e1 + e2 + e3

without rotation. Using these relationships we can show that the bulk modulus Bis given by:

B =−p4

=13(C11 +2C12)

Figure 1.9: Two shear modes for a cubic crystal.

Shear on a cube face parallel to a cube axis.

For example on the (010) plane in the [001] direction:

σ4 =C44e4

C44 is a shear modulus, which by convention we name µ0.

Shear at 45 to a cube axis

For example, on the (110) plane in the [110] direction. In this case the shearmodulus is:

µ1 =12(C11−C12)

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This allows us define an anisotropy factor,

µ0

µ1=

2C44

(C11−C12)

which of course will be one for an isotropic material.In terms of the Lame constants:

λ =C12;µ =C44;λ+2µ =C11

which explains why cubic crystals are not in general isotropic.

W Al Fe Cu Naµ0µ1

1.0 1.2 2.4 3.2 7.5Structure bcc fcc bcc fcc bcc

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Chapter 2

Elastic Waves in Solids

15

2.1. ELASTIC PROPERTIES SSP1

Introduction to Lattice Dynamics, Martin T. Dove, Cambridge topics in mine-ral physics and chemistry.

2.1 Elastic PropertiesForces acting on atoms disturb them elastically from their equilibrium positions.A Taylor series expansion of the energy near the minimum (equilibrium position)yields:

Figure 2.1: Equilibrium interatomic distance according to Madelung: energy mi-nimum U0 at R0.

U(R) =U0 +∂U∂R

∣∣∣∣R0

(R−R0)+12

∂2U∂R2

∣∣∣∣R0

(R−R0)2 + ....

For small displacements, neglect terms of order 3 and higher. At equilibrium,

∂U∂R

∣∣∣∣R0

so:

U(R) =U0 +12

∂2U∂R2

∣∣∣∣R0

(R−R0)2 =U0 +

ku2

2

where:

k =∂2U∂R2

∣∣∣∣R0

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SSP1 2.1. ELASTIC PROPERTIES

and

u = R−R0

following Hooke’s law in simplest form:

F =−∂2U∂R2 =−ku

The elastic properties are described by considering a crystal as a homogeneouscontinuum medium rather than a periodic array of atoms. In general, the problemis formulated in the following terms:

• Applied forces are described in terms of stress σ = F/A

• Displacements of atoms are described in terms of strain ε = δl/l = ∂u∂x

• Elastic constants C relate stress σ and strain ε so that σ =Cε

For an arbitrary 3D crystal the stress and the strain are tensors. For a local hydro-static pressure due to force F(Fx,Fy,Fz):

• Stress components, σi j,(i, j = 1,2,3) where x≡ 1,y≡ 2,z≡ 3.

• For compressive stress, i = j, (σ11,σ22,σ33) and for shear stress, i 6= j,(σ12,σ21,σ13,σ31,σ23,σ32).

• The shear forces must come in pairs: σi j = σ ji, i.e. no rotation/angularacceleration so the stress tensor is diagonal with six components.

The displacement produced by the vibration is represented by the vector u =uxx,uyy,uzz. The strain tensor components are defined as:

εi j =∂ui

∂x j

Compressive strain:

εxx =∂ux

∂x,εyy =

∂uy

∂y,εzz =

∂uz

∂z

Shear strain:

εxy =∂ux

∂y,εyx =

∂uy

∂x,εxz =

∂ux

∂z,εzx =

∂uz

∂x,εyz =

∂uy

∂z,εzy =

∂uz

∂y,

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2.1. ELASTIC PROPERTIES SSP1

Since σi j and σ ji are always applied together, we can define the shear strainssymmetrically:

εi j = ε ji =12

(∂ui

∂x j+

∂u j

∂xi

)so the strain tensor is diagonal and has six components.

σxxσyyσzzσyzσzxσxy

=

C11 C12 C13 C14 C15 C16C21 C22 C23 C24 C25 C26C31 C32 C33 C34 C35 C36C41 C42 C43 C44 C45 C46C51 C52 C53 C54 C55 C56C61 C62 C63 C64 C65 C66

εxxεyyεzzεyzεzxεxy

Figure 2.2: General matrix form of Hooke’s Law highlighting compression, shearand mixed coefficients.

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SSP1 2.2. WAVES IN CRYSTALS

2.2 Waves in CrystalsWe will now examine the vibrational behaviour of atoms in solids where λ ∼ a.The vibrations are thermally activated with a characteristic activation energy kBT .

Firstly, vibrational excitations are collective modes: all atoms in the materialtake part in the vibrational mode. The influence of the translational periodicity ofthe structure of crystals has a dramatic effect on the vibrational behaviour whenthe wavelength of the vibrations becomes comparable to the size of the unit cell.When the vibration wavelength is much larger than the structural variation of thematerial, the solid may be considered as an elastic continuum (continuum approx-imation, see above).

How can we visualise a vibration wave traveling through a crystal, where thespace that vibrates is not continuous (like a string on a musical instrument) but iscomposed of discrete atoms? The answer is to think of the wave as representingdisplacements, u(x, t), of the atoms from their equilibrium position.

Considering lattice vibrations three major approximations are made:

• atomic displacements are small, u << a, where a is the lattice constant

• forces acting on atoms are assumed to be harmonic/Hookian, F =−Cu

• adiabatic approximation is valid - electrons follow atoms so that the natureof the bond is not affected by vibrations

Figure 2.3: Longitudinal vibration (force) on an element of an elastic continuum.

The discreteness of the lattice must be taken into account where λ ∼ a. Forlong waves λ >> a, one may disregard the discrete/atomic nature of the solidand treat it as a continuous medium, i.e. a continuum. In this case vibrationspropagate as ordinary acoustic waves, i.e. elastic waves, at the speed of sound inthe material.

md2udt2 = ∑F → (ρAdx)

d2udt2 = F(x+dx)−F(x)

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2.2. WAVES IN CRYSTALS SSP1

⇒ ρ∂2u∂t2 =

1A

∂F∂x

=∂σxx

∂x

Assuming that the wave propagates along the [100] direction, and writing Hooke’slaw in the form σxx =C11εxx where σ is the stress and ε is the strain εxx =

∂ux∂x , we

can rearrange the above into wave equation form:

∂2u∂t2 =

C11

ρ

∂2ux

∂x2 (2.1)

which has the general solution:

u(x, t) = u0ei(kx−ωt) (2.2)

where k is the wavevector (momentum vector) given by k= 2π

λand ω is the angular

frequency ω = 2π f . The coefficient of the term on the right of the wave equationis 1/v2 and so:

vL =

√C11

ρ=

ω

k

This is the longitudinal sound velocity in the continuum.We can also have transverse vibrations in the medium. In this case, we use the

shear coefficients and the shear stress and strain. Simple analysis gives: ρ∂2u∂t2 =

∂σxy∂x , σxy =C44εxy and εxy =

∂u∂x .

Figure 2.4: Transverse vibration (force) on an element of an elastic continuum.

The transverse wave equation is:

∂2u∂t2 =

C44

ρ

∂2ux

∂x2 (2.3)

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SSP1 2.2. WAVES IN CRYSTALS

where the transverse sound velocity is now determined by the shear coefficientand the density:

vT =

√C44

ρ.

There are two independent transverse modes: displacements along y and z. Fork in the [100] direction in cubic crystals, by symmetry, the velocities of the twomodes will be the same, i.e. degenerate. Normally C11 >C44⇒ vL > vTConsidering waves propagating in different directions in a cubic crystal. In gene-ral, the sound velocity will depend on a combination of elastic constants.

v =

√Ce f f

ρ

where Ce f f is an effective lattice constant.

Mode k ‖[100] k ‖[110] k ‖[111]L C11

12(C11 +C12 +2C44)

13(C11 +2C12 +4C44)

T1 C44 C4413(C11−C12 +C44)

T2 C4412(C11−C12)

13(C11−C12 +C44)

2.2.1 VelocityThe wave (or phase) velocity , defined as

v =ω

k=

√C11

ρ

gives a simple linear relationship between ω and k. Sound waves of differingfrequency/wavelength propagate with the same velocity: the medium is said to benon-dispersive.

In general, another velocity associated with a traveling wave can be defined,and this is important for waves traveling in dispersive media where the linearitybetween ω and k breaks down. The group velocity, defined as

vg =∂ω

∂k,

or, in general,

vg = ∇kω(k)

21 V Casey

2.2. WAVES IN CRYSTALS SSP1

is a measure of the velocity of a wave packet, composed of a group of plane wa-ves, and having a narrow spread of frequencies about some mean value, ω. Foracoustic waves with long wavelengths (k ≈ 0), i.e. in the elastic continuum limit,the phase and group velocities are equal. In a liquid, only longitudinal vibrations(modes) are supported (shear modulus is zero). The situation is more complicatedin solids, where more than one elastic modulus is non-zero. As a consequence,both longitudinal and transverse acoustic modes exist even in isotropic solids, ha-ving in general different sound velocities. The situation is even more complicatedfor anisotropic crystals.

2.2.2 Static and Dynamic ModelsThe static lattice model which is only concerned with the average positions ofatoms and neglects their motions can explain a large number of material featuressuch as:

• chemical properties;

• material hardness;

• shapes of crystals;

• optical properties;

• Bragg scattering of X-ray, electron and neutron beams;

• electronic structure as well as electrical properties.

There are, however, a number of properties that cannot be explained by a staticmodel. These include:

• thermal properties such as heat capacity;

• effects of temperature on the lattice, e.g. thermal expansion;

• the existence of phase transitions, including melting;

• transport properties, e.g. thermal conductivity, sound propagation;

• the existence of fluctuations, e.g. the temperature factor;

• certain electrical properties, e.g. superconductivity;

• dielectric phenomenon at low frequencies;

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SSP1 2.3. LINEAR LATTICE MODELS

• interaction of radiation with matter, e.g. light and thermal neutrons.

Are the atomic motions that are revealed by these factors random, or can we finda good description for the dynamics of the crystal lattice? The answer is that themotions are not random but are constrained and determined by the forces thatatoms exert on each other.

2.3 Linear Lattice Models

2.3.1 Linear 1D Chain Model• all atoms identical (mass m)

• lattice spacing ‘a’

Figure 2.5: A linear monatomic lattice: mass and spring model.

For small vibrations, the force on any one atom is proportional to its displace-ment relative to all the other atoms. Choose atom s:

Fs = ∑p

Cp(us+p−us)

Here, p takes on both positive and negative values, C is the force constant whichfor a given direction now also depends on p, i.e. is large for p = 1, smaller forp = 2, etc..

Fs = ma = md2us

dt2 = ∑p

Cp(us+p−us)

We should expect harmonic vibrations represented by a traveling wave of ampli-tude u, where xs=sa:

us = uei(kxs−ωt)

23 V Casey

2.3. LINEAR LATTICE MODELS SSP1

and

us+p = uei[k(s+p)a−ωt].

Substituting for us and us+p gives

−mω2 = ∑

pCp(eikpa−1).

Now Cp =C−p, therefore:

−mω2 = ∑

p>0Cp(eikpa + e−ikpa−2)

−ω2 =

1m ∑

p>0Cp[2cos(kpa)−2]

ω2 = 2m ∑

p>0Cp[1− cos(kpa)]

= 4m ∑

p>0Cp cos2

(kpa

2

)For simplicity set p = 1, i.e. nearest neighbour interaction only:

ω2 = 4C1m sin2 ka

2

ω =√

4C1m sin2 ka

2

[Note that C1 is the effective spring constant for nearest neighbour interaction andis not necessarily the principal stress coefficient C11 which we have used in ourelastic theory.]

ωmax = ω

(±π

a

)= 2

√C1

m

which is periodic in k with period 2π/a.It is clear from the dispersion relationship that lattice waves in a one dimen-

sional lattice constitute a dispersive system for which the velocity varies withfrequency and wavelength. Only for very small k, i.e. for very long wavelengths,does the velocity become a constant.

The periodic form of ω(k) shown above suggests that some simplification innotation should be possible. Consider, for instance, a wave with wavenumber kcompared to a wave with wave number k′ = k+n(2π

/a).

V Casey 24


Figure 2.6: Dispersion curve for a linear monatomic lattice.

u′ = u0ei(k′x−ωt) = u0ei[(k+n2π/a)x−ωt]

= uei(n2π/a)x = uei(n2π/a)sa = uein2πs = u

The displacement for k′ is therefore the same as for k. k′ consists of a wave ofsmaller wavelength than that corresponding to k, passing through all the atoms,but containing more oscillations than needed for the description.

We can describe the displacement of the atoms in these vibrations most easilyby looking at the limiting cases k = 0 and k = π

a . The situation at k = 0 corre-sponds to an infinite wavelength; this means that all of the atoms of the lattice aredisplaced in the same direction from their rest position by the same displacementmagnitude. For long wavelength vibrations neighbouring atoms are displaced bythe same amount in the same direction. Since the long-wavelength longitudinalvibrations correspond to sound waves in the crystal, all of these vibrations witha similarly shaped dispersion curve, whether transverse or longitudinal vibrationsare involved, are called acoustical branches of the vibration spectrum.

[When k ∼= 0, dω/

dk = ω/

k = velocity of sound].

The atomic displacements at k = π/a can be seen by substituting this value ofk and x = sa into the general form of the solution for the displacement of a wave:

25 V Casey


us = uei(kx−ωt)+ue−i(kx+ωt)

= 2u(−1)se−iωt

showing that neighbouring atoms are displaced by the same distance in oppositedirections, giving rise to a minimum physically meaningful wavelength λmin =2a. This is equivalent to the Bragg reflection condition nλ = 2d sinθ in its onedimensional equivalent form λ = a with n = 1 and d = a. Waves with k = π/aare unable to propagate through the crystal; this is consistent with the fact that thegroup velocity at k = π/a is equal to zero. At k = π/a, ω has a maximum andtherefore dω

/dk = 0.

Since ω=0 when k = 0 and the latter corresponds to an infinite wavelength;i.e. represents one extreme of the spectrum and is therefore a good choice as theorigin. There is no additional information available by extending the plot outsideof the region −π

/a≤ k ≤ π/a; we may place a physical interpretation on this

maximum value of |k| = π/a - this represents the smallest wavelength that canpropagate in the lattice, i.e. λ = 2a.

Figure 2.7: Longitudinal and transverse modes supported by a linear monatomiclattice.

In addition to longitudinal vibrations, the linear lattice supports transverse dis-placements leading to two independent sets (in mutually perpendicular planes) ofvibrations that can propagate along the lattice. The forces acting in a transversedisplacement are weaker than those in a longitudinal one. They give rise to a newbranch of dispersive modes lying below the longitudinal branch.

V Casey 26


2.3.2 Diatomic Linear Chain ModelIf there are different kinds of atom in the structure, e.g. CsCl, then we must allowfor different displacements of the different masses in, for instance, a diatomiclinear chain. A similar analysis to the above leads to the following equations fornearest neighbour interaction:

M1us =C1(vs+1/2 + vs−1/2−2us)

M2vs+1/2 =C1(us+1 +us−2vs+1/2)

Figure 2.8: A linear diatomic lattice.

If we assume that

us = uei(ksa−ω t)

vs+1/2 = vei[k(s+1/2)a−ω t)

then we obtain the following dispersion relation:

M1M2ω4−2C1(M1 +M2)ω

2 +4C21 sin2 ka

2= 0

which when solved gives:

ω2 =C1

(M1 +M2

M1M2

)±

[C2

1

(M1 +M2

M1M2

)2

−4C2

1M1M2

sin2 ka2

]1/2

or

ω2± = A±

(A2−Bsin2 ka

2

)1/2

27 V Casey


Figure 2.9: The two groups of vibrational modes supported by a linear diatomiclattice: acoustical modes and optical modes.

There are now two solutions for ω2, providing two distinct groups of vibratio-nal modes. The first group, associated with ω2

−, contains the acoustic modes dealtwith above. The second group arises with ω2

+, and contains the optical modes.These correspond to the movement of the different atom sorts in opposite directi-ons. The name arises because in ionic crystals, they cause an electric polarizationand can therefore be excited by light, which as a result is strongly absorbed. Ho-wever, these optical modes occur in all crystals with two or more different atoms.

V Casey 28

Chapter 3

Specific Heat Capacity

29

3.1. SPECIFIC HEAT CAPACITIES SSP1

3.1 Specific Heat Capacities

3.1.1 Historical• Classical: Dulong and Petit (1819, Cv=3Nk, Correct at high temperature!)

• Einstein: Based on Planck’s quantum hypothesis (1901; Quantised energy,Showed exponential dependence of Cv).

• Debye: Showed complete dependence (1912)

3.1.2 Specific Heat: Classical ModelThe classical model for specific heats considered the atoms as being simple har-monic oscillators vibrating about a mean position in the lattice. Each atom couldbe simulated by three simple harmonic oscillators (SHOs) vibrating in mutuallyperpendicular directions.

For a classical SHO:

• Average kinetic energy = 1/2 kT

• Average potential energy = 1/2 kT

• Total average energy per oscillator = kT

• Total average energy per atom = 3kT

• For N atoms the total average energy = U = 3NkT

• The specific heat capacity is Cv =(dU

dT

)v = 3Nk = 3R

• Independent of T

Classical treatment - Dulong and Petit - the specific heat capacity of a givennumber of atoms of a given solid is independent of T and is the same for all solids.

3.1.3 Einstein ModelThe above treatment of the vibrational behaviour of materials has been entirelyclassical. For a harmonic solid, the vibrational excitations are the collective inde-pendent normal modes, having frequencies ω determined by the dispersion rela-tionship ω(k) with the allowed values of k set by the boundary conditions. In theclassical limit, the energy of a given mode with frequency ω, determined by thewave amplitude, can take any value.

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SSP1 3.1. SPECIFIC HEAT CAPACITIES

Figure 3.1: Experimentally measured specific heat and model values.

For the Boltzmann energy distribution: as T is raised P(Ehigh) increases andso the energy of the atomic vibrations becomes greater as we go from low to highT .

Einstein produced a theory of heat capacity based upon Planck’s quantumhypothesis (Planck’s treatment of electromagnetic radiation). He assumed that:

• each atom is in an identical quantum harmonic oscillator potential well;

• each atom of the solid vibrates about its equilibrium position with an angu-lar frequency ω;

• each atom has the same frequency and vibrates independently of other atoms.

The quantum mechanical result (can be treated analytically), treating each nor-mal mode as an independent harmonic oscillator with frequency ω. The energy isquantised and can only take values characterised by the quantum number n(k, p)for a particular branch p. A vibrational state of the whole crystal is thus specifiedby giving the excitation numbers n(k, p) for each of the 3N normal modes. Insteadof describing the vibrational state of a crystal in terms of this number, it is moreconvenient and conventional to say, equivalently, that there are n(k, p) phonons(i.e. particle like entities representing the quantised elastic waves).Einstein:-

• replaced the classical SHO with a Quantum Harmonic Oscillator, QHO;

• where energy does not increase continuously but in discrete steps;

• and the corresponding quantum particle is the phonon!

31 V Casey


Figure 3.2: The QHO potential and eigenstates.

For a 1D QHO:

En = (n+ 1/2)~ω (3.1)

n=0,1,2. . . ..The quantum mechanical expression for the energy implies that:

• the vibrational energy of a solid is non zero even when there are no phononspresent

• the residual energy of a given mode, 1/2 ~ω, is the zero point energy

The one dimensional energy partition function takes the form:

Z1D =∞

∑n≥0

e−β ~ω(n+1/2)

where β = 1/kBT . This may be rearranged to a more revealing form:

Z1D = e−β ~ω

2

∞

∑n≥0

e−nβ ~ω

where the summation is an infinite geometric series and so:

Z1D = e−β ~ω

21

1− e−β ~ω

Therefore:

Z1D =e−

β ~ω

2

1− e−β ~ω(3.2)

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SSP1 3.1. SPECIFIC HEAT CAPACITIES

Now we would like to use the energy partition function to get the energy ex-pectation value. We have:

〈E〉=− 1Z

∂Z∂β

=−∂ln Z∂β

(3.3)

Show this as an exercise!so with lnZ taking the form:

lnZ =−β ~ω

2− ln

[1− e−β ~ω

]

〈E〉=−∂ln Z∂β

=−

[−~ω

2− ~ωe−β ~ω

1− e−β ~ω

]= ~ω

[12+

1eβ ~ω−1

]

〈E〉= ~ω

[12+nB(β ~ω)

](3.4)

where,

nB(x) =1

ex−1is the Bose occupation factor.

The mode ω is an excitation that is excited on average up to the nthB level!

Alternatively, there is a boson orbital which is occupied by nB bosons (obey Bose-Einstein statistics).

C =∂〈E〉∂T

=∂

∂T(~ωnB(β ~ω)) = ~ω

∂

∂T

[1

eβ ~ω−1

]

C1D = kB(β ~ω)2 eβ ~ω

(eβ ~ω−1)2 (3.5)

Show this.

C3D = 3kB(β ~ω)2 eβ ~ω

(eβ ~ω−1)2 (3.6)

(a) High Tβ~ω << 1 or ~ω << kT and e

~ω

kT ≈ 1+ ~ω/kT

C = 3kB(β ~ω)2 1+β ~ω

(1+β ~ω−1)2 ' 3kB

33 V Casey


independent of ω. This is the classical limit because the energy steps, ~ω, are nowsmall compared with the mean energy of the oscillator.

(b) Intermediate TThe full Einstein formula must be used.

C3D = 3kB(β ~ω)2 eβ ~ω

(eβ ~ω−1)2

where θE = ~ω/k, the so called Einstein temperature (this is an ad-hoc parameter!).

Figure 3.3: Einstein model specific heat capacity general form.

(c) Low Tβ~ω >> 1 and e

~ω

kT >> 1

C3D ' 3kB(β ~ω)2 1(eβ ~ω)

' 3R(

θET

)2e−

θET

where θE = ~ω/kB. The system gets locked into the zero point energy state andthe system specific heat C approaches zero exponentially as T → 0.

V Casey 34

SSP1 3.2. DEBYE MODEL

Einstein model

• successfully accounts for high temperature specific heat, C, i.e. classicallimit 3kB;

• successfully predicts that C falls with decreasing T ;

• however, exponential decrease is not observed; if low frequencies are pre-sent, then ~ω will be small, much smaller than kT even at low temperatures;C will remain at 3kBT to much lower frequencies and the fall off is not asdramatic as predicted by the Einstein model;

• assumption of ‘an average’ single frequency ω is too simplistic;

• need a spread of frequencies - a spread of states!

3.2 Debye Model

3.2.1 Key Features• C does not go to zero as T goes to zero as rapidly as predicted by the Einstein

model.

• Reason - the overly restrictive requirement that all normal modes shouldhave the same frequency.

• Long λ modes have lower frequency than short λ modes.

• Long λ modes are much harder to freeze out than short λ modes.

• Energy spacing for long λ modes are much closer together than short λ

modes.

• Therefore C does not decrease as rapidly with T as suggested by the Emodel.

• The long λ modes are able to contribute to C even at low T

3.2.2 Debye ModelThe frequencies of the normal modes are estimated by treating the solid as a con-tinuous isotropic medium. This approach is reasonable because the only modesexcited at low T are the long λ modes, i.e. λ > r0. Assume a linear dispersionrelationship: v = f λ = ω/k where k = 2 π/λ or ω(k) = v|k|.

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3.2. DEBYE MODEL SSP1

〈E〉= 3 ∑k~ω(k)

[12+nB(β ~ω(k))

](3.7)

Unlike photons, phonons have three polarizations: one longitudinal and twotorsional (transverse). Hence the factor of 3! We also must sum over a range of k.Let L be the macroscopic dimension of the system.

eikr = eik(r+Lx)

⇒ k = n2π

Lx

where n is integer. What are the possible values of k? For large L, 2π/L will bevery small and the sum over all k can be replaced with an integral:

∑k→ L

2π

∫∞

−∞

dk

In 3D wrap crystal into hypertorous in 4D: let Lx = Ly = Lz = L. Each k point willoccupy a volume of k space given by:

(2π)3

L3

Therefore:

∑k3D

→ L3

(2π)3

∫3D

dk

The step or interval in k space is so small that we may integrate over k space.

〈E〉= 3L3

2π3

∫dk ~ω(k)

[12+nB(β ~ω(k))

](3.8)

Each excitation mode has a boson frequency ω(k) which is occupied on averagenB(β ~ω(k)) times.

Consider k space as a spherical shell (recall medium is isotropic). Can considerthe surface of this shell at arbitrary k and work out the volume of k space in theinterval dk from the surface area:∫

dkx dky dkz→ 4π

∫∞

0k2dk

〈E〉= 34πL3

2π3

∫∞

0ω

2dω1v3 ( ~ω)

[12+nB(β ~ω)

]V Casey 36


Figure 3.4: 3D to 1D integral

〈E〉=∫

∞

0dω g(ω)( ~ω)

[12+nB(β ~ω)

]

g(ω) = 34πL3

2π3

ω2

v3

= L3[

12 π ω2

(2π)3v3

]= N

[12 π ω2

(2π)3v3n

]= N

(9 ω2

ω3d

)

ω3d = 6π

2nv3 = 6π2 NV

v3

ωd =

(6π

2 NV

)1/3

v (3.9)

λd '2πvωd

r0 '(

VN

)1/3

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3.2. DEBYE MODEL SSP1

〈E〉=∫

∞

0dω g(ω)( ~ω)

[12+nB(β ~ω)

]g(ω) dω is the total number of oscillation modes between ω and ω+dω

The density of allowed states or just density of states.Count the number of states in an interval and multiply by the expected energy permode and integrate over all frequencies.

〈E〉= 9N ~ω3

d

∫∞

0dω ω

3[

12+nB(β ~ω)

]〈E〉= 9N ~

ω3d

∫∞

0dω ω

3 1eβ ~ω−1

Let x = β ~ω

〈E〉= 9 N ~ω3

d(β ~)4

∫∞

0dx ω

3 1ex−1∫

∞

0dx ω

3 1ex−1

Can be written as a special case of the Riemann Zeta Function

ζ(4) =π4

15

〈E〉= 9 N k4B T 4

( ~ωd)3π4

15(3.10)

C =3615

N k4B T 3 π4

( ~ωd)3 (3.11)

C =N kB (kB T )3

( ~ωd)312 π4

5(3.12)

C =N kB (kB T )3

( ~ωd)312 π4

5

Shows T 3 dependence!Prefactor of the T 3 term can be calculated in terms of known quantities such asthe velocity of sound in the crystal - i.e. is physical!Often we replace ~ωd by kBθd where:

θd =~ωd

kB

V Casey 38


C = N kB

(Tθd

)3 12 π4

5(3.13)

Material θd(K)Diamond 1850Berrilium 1000Silicon 625Copper 315Silver 215Lead 88

There is no upper limit on C as T increases. Should level off to 3 kB N at hightemperature.Physically this means that there is no upper limit for k and the number of os-cillation modes. Since ω = v k we need to find an upper limit for ω - a cut-offfrequency - ωc. Debye imposed an upper limit by insisting that the total numberof degrees of freedom, i.e. modes, should be limited to 3N, the number of atomsin the system. This neatly sets ωc.

3 N =∫

ωc

0dω g(ω)

〈E〉=∫

ωc

0dω g(ω)( ~ω)nB(β ~ω)

Low T implies large β. Therefore ωc has no impact - activity is all down near thezero point energy.

nB(β ~ω) =1

eβ ~ω−1' 1

1+β ~ω−1=

kB T~ ω

〈E〉= kB T∫

ωc

0dω g(ω)

For high temperature all modes will be accessible and so the integral evaluatesto 3 N and so:

〈E〉= 3 N kB T

and, C = 3 N kB.

3 N =∫

ωc

0dω g(ω) = 9 N

∫ωc

0dω

ω2

ω3d= 3 N

ω3c

ω3d

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3.3. PHONONS SSP1

Therefore ωc = ωdThe cut-off frequency is exactly the Debye frequency.

There is a physical basis for ωd and so also for θd!

3.3 Phonons

3.3.1 Actual Phonon Density of StatesNeutron scattering data may be used to elucidate the phonon density of statesand dispersion relation for a crystal. Multiple branches adds complexity to theactual distribution since these will superimpose on one another. Since the groupvelocity must disappear at some frequencies and it appears in the denominator ofthe density of states function, there will be singularities in the distribution - knownas van Hove singularities.

Figure 3.5: Actual density of states (upper curve) for aluminium compared toDebye model. The dashed curves show the individual transverse and longitudinalbranches.

3.3.2 Phonon InteractionsIn pure defect free crystals (if such existed) phonons will pass through each otherand reflect from the crystal surfaces without internal scattering or energy loss. Inreal crystals phonons interact, scatter and decay due to imperfections, impurities

V Casey 40

SSP1 3.3. PHONONS

and anharmonicity. It is useful to think of them as being grouped together to formparticles, i.e. a group of waves with differing wavelengths and phases but havinga reasonably well defined group velocity. At a given temperature we may expect a‘kinetic equilibrium’ to establish itself and so simple Kinetic Theory might yieldsome useful insights to the thermal properties of solids.

Figure 3.6: As the crystal expands the equilibrium spacing shifts to the right re-ducing the symmetry of the potential profile, ie. less well approximated by anharmonic potential.

Considering non-metals first, heat would have to be carried by phonons alonesince there are no free electrons available. We know from every day experiencethat heat does not travel through insulators at the speed of sound! Therefore theremust be some ‘retarding’ mechanism operative to reduce the effective phononvelocity - phonon interactions. Experimentally heat is found to diffuse throughinsulating solids with a mean free path on the order of 10 nm, i.e. much less thanthe macroscopic dimensions of the solid under test. This is similar to values forordinary gases and so provides encouragement to try out the kinetic theory withthese solids. The thermal conductivity of an ideal gas is given by:

K =13

C〈v〉Λ

where 〈v〉 is the RMS velocity, C is the specific heat capacity and Λ is the meanfree path.

At low temperature there are relatively few phonons excited and so phononinteraction is limited leading to large Λ’s (mean free paths). The specific heat

41 V Casey

3.3. PHONONS SSP1

Figure 3.7: Kinetic theory of gases applied to phonons.

capacity C will be zero at absolute zero C = 0, T = 0 and will increase as T 3 atlow temperature. The thermal conductivity will do likewise at low temperature.

At the other extreme of temperature, i.e. high temperature, there will be lotsof phonons and so lots of interactions: Λ will decrease; C will have saturated ata constant value and so the thermal conductivity decreases roughly in proportionto the number density of phonons 〈n〉 which will in turn be proportional to thetemperature.

K ∝ Λ ∝1〈n〉

∝1T

At intermediate temperatures the thermal conductivity will be set by the meanfree path and the size of the sample.

3.3.3 Anharmonicity and Thermal Expansion

At higher temperature more phonons are excited and the anharmonicity of thepotential increases. The effects of anharmonicity are more pronounced at highertemperatures. Phonons transfer energy to the lattice. It gives rise to thermal ex-pansion - equilibrium position shifts to the right. Not surprisingly, the thermalexpansion coefficient has a similar temperature profile to specific heat capacity.

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SSP1 3.3. PHONONS

3.3.4 Normal and Umklapp Phonon InteractionsWe can use the Brillouin zone representation to examine simple phonon combi-nation. For two phonons normal combination occurs where their vector sum isless than half a reciprocal lattice vector, i.e. < GGG///222, and so the limiting caseis defined by |qqq111| = |qqq222| ≤ |GGG///222|. The resultant phonon will be within the firstBrillouin zone, see Figure3.8, and this interaction is referred to as normal phononscattering.

Figure 3.8: Normal and Umklapp phonon scattering.

If however, the phonons combine to give a resultant phonon qqq333 > GGG/2 whichis outside the first Brillouin zone, qqq333 becomes oppositely directed to qqq111 and qqq222,see Fig.3.8. It is effectively reflected at the zone boundary in a process knownas Umklapping. Therefore the crystal structure is both a source and a sink forcrystal momentum in units of the reciprocal lattice vector GGG. The crystal acts as amomentum buffer enabling both energy and momentum balance to be achieved.

3.3.5 Neutron Scattering by a CrystalConsider a neutron that is incident upon a crystal. The neutron interacts stronglywith only the atomic nuclei in the crystal. It may pass right through the crystal ab-sorbing and emitting phonons. The phonon occupation number in the crystal maychange due to this interaction. Conservation of energy requires that the change inthe energy of the neutron is equal to the energy of the phonons it absorbed less theenergy of the phonons it emitted. Therefore the change in energy of the neutron

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3.3. PHONONS SSP1

contains information about the phonon frequencies. However, a second conserva-tion law is required to disentangle this information from the scattering data. Thesecond law is known as the conservation of crystal momentum (symmetries of theHamiltonian imply conservation laws). If we define the crystal momentum of aphonon to be ~ times its wavevector, then the change in neutron momentum is justthe negative of the chnage in total phonon crystal momentum, to within an addi-tive reciprocal lattice vector. This crystal momentum of a phonon is not, however,accompanied by any real momentum of the ionic system.

Because there are two conservation laws, it turns out to be possible to extractthe explicit forms of the ω(kkk) from the neutron scattering data in a simple way.

Zero-Phonon Scattering

The crystal initial and final states are identical. Energy conservation implies thatthe energy of the neutron is unchanged - the scattering is elastic. Crystal mo-mentum conservation implies that the neutron momentum can only change by~GGG where GGG is a reciprocal lattice vector. If we write the incident and scatteredmomenta as:

ppp = ~qqq, ppp′′′ = ~qqq′′′,

then these restrictions become:

q′ = q, qqq′′′ = qqq+++GGG

These are the Laue conditions that the incident and scattered X-ray wave vectorsmust satisfy in order for elastically scattered X-rays to produce a Bragg peak.Since a neutron with momentum ppp = ~qqq can be viewed as a plane wave withwave vector qqq, this is not surprising. Elastically scattered neutrons, which donot create or destroy phonons, are found only in directions that satisfy the Braggcondition, and give precisely the same structural information about the crystal aselastic X-ray scattering.

One-Phonon Scattering

One phonon scattering provides the most accessible spectroscopic information. Inthe case of the more important absorption mechanism, conservation of energy andmomentum implies that

E ′ = E +~ω(kkk)ppp′ = ppp+~kkk+~GGG

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SSP1 3.3. PHONONS

where kkk is the wave vector for a particular absorbed phonon in a specific branch.In the case of emission

E ′ = E−~ω(kkk)ppp′ = ppp−~kkk+~GGG

where the phonon has been emitted into a branch with wave vector kkk.

Figure 3.9: Neutron incident on a lattice and event diagram for one-phonon inte-raction.

In either case the crystal momentum law may be used to represent kkk in termsof the neutron momentum transfer, ppp′′′− ppp. Furthermore, the additive reciprocallattice vector that appears in this relation can be ignored, when the resulting ex-pression for kkk is substituted into the energy conservation law, since each ω(kkk) is aperiodic function in the reciprocal lattice:

ω(kkk±GGG) = ω(kkk).

As a result the two conservation laws yield one equation or:

p′2

2Mn=

p2

2Mn+~ω

(ppp′′′− ppp~

)p′2

2Mn=

p2

2Mn−~ω

(ppp− ppp′′′

~

)In a given experiment, the incident neutron momentum and energy are usually

specified. Thus for a given phonon dispersion relation ω(kkk) the only unknownsare the three components of the final neutron momentum ppp′′′. Generally, a singleequation relating the three components of a vector ppp′′′ will, if it has any solutions

45 V Casey

3.3. PHONONS SSP1

at all, specify a surface in three dimensional ω(ppp′′′) space. If we only examineneutrons emerging in definite direction we will specify the direction of ppp′′′, andcan therefore expect to find solutions at only a single point on the surface.

If we select a general direction we will see neutrons scattered by one-phononprocesses only at a few discrete values of p′, and correspondingly only at a fewdiscrete energies E ′ = p′2/2Mn. Knowing the energy and the direction in whichthe scattered neutron emerges, we can construct ppp′′′− ppp and E ′−E, and can the-refore conclude that the crystal has a normal mode with frequency (E ′−E)/~and wave vector ±(ppp′′′− ppp)/~. A point in the crystal’s phonon spectrum has beenestablished. By varying the controllable parameters such as incident energy, orien-tation of the crystal and direction of detection, a large number of such points maybe collected in order to map out the entire phonon spectrum (assuming it is pos-sible to distinguish the phonons scattered in one-phonon processes from all otherprocesses).

3.3.6 Inelastic Neutron SpectroscopyThe neutron matter interaction is weak. Typical parameters are:

• m f p v 1 cm

• λ v 10−10 m

• E v 0.08 eV

• kT v 0.025eV

Neutrons do have an intrinsic magnetic moment, i.e. spin, and so will interactwith atoms and ions in a crystal with magnetic moments.

Neutrons are neutral sub-atomic particles with no electrical charge. Because ofthis, these unassuming particles are non-destructive and can penetrate into mattermuch deeper than charged particles such as electrons. In addition, because of theirspin, neutrons can be used to probe magnetism on an atomic scale. There are twomain methods of producing neutrons for materials research. One is by splittinguranium atoms in a nuclear fission reactor. The other, called spallation, involvesfiring high-energy protons into a metal target, such as mercury or tungsten, toinduce a nuclear reaction that produces neutron beams. ILL is the most intensereactor neutron source in the world. ISIS is the most productive spallation neutronsource in the world. Neutron sources play a crucial role in research across thescientific spectrum, from nuclear and elementary particle physics, chemistry andmaterials science to engineering and life sciences.Neutrons have unique advantages as a probe of atomic-level properties:

V Casey 46

SSP1 3.3. PHONONS

Figure 3.10: European Spallation Source (ESS) is due to complete in Lund, Swe-dan by 2020, https://europeanspallationsource.se/

Figure 3.11: Three axis neutron spectrometer.

47 V Casey

3.3. PHONONS SSP1

• The process of neutron scattering is non-destructive, so that delicate or va-luable samples can be studied.

• Neutrons are penetrating, so that they can look deep inside engineering sam-ples to study, for example, welds.

• Neutrons with energies in the range of atomic motions have wavelengthsof the order of the distances between atoms making them very good atstudying both where atoms are and how they are moving.

• Neutrons are good at seeing light atoms, such as hydrogen, in the presenceof heavier ones.

• Neutrons are good at distinguishing neighbouring elements in the periodictable.

• Different isotopes of the same element scatter neutrons differently. For ex-ample, extra information can be gained by swapping hydrogen atoms withtheir deuterium isotopes in part of a sample.

• Neutrons have a magnetic moment, meaning that they can be used to studythe magnetic properties of materials.

V Casey 48

Chapter 4

Drude Free Electron Theory

49

4.1. CLASSICAL THEORY AND CONDUCTION IN SOLIDS SSP1

4.1 Classical Theory and Conduction in SolidsTexts

Simon The Oxford Solid State Basics

Myers Introductory Solid State Physics

Kittel Introduction to Solid State Physics

Ashcroft & Mermin Solid State Physics

4.1.1 Free Electron Theory (Drude’s Free Electron Theory)

Figure 4.1: Inorganic crystalline solids.

Before 1900 it was known that most conductive materials obeyed Ohm’s LawI =V/R.In 1897 J. J. Thompson discovers the electron as the smallest charge carryingconstituent of matter with a charge equal to “-q”

q = 1.6×10−19C

In 1900 Paul Drude formulated a theory for conduction in metals using theelectron concept. The theory assumed:

1. Metals have a large density of “free electrons”.

V Casey 50

SSP1 4.1. CLASSICAL THEORY AND CONDUCTION IN SOLIDS

2. The electrons move according to Newton’s laws until they scatter from ions,defects,etc.

3. After a scattering event the momentum of the electron is completely random(i.e. has no relation to its momentum before scattering - “memory loss”)

Figure 4.2: Electron scattering without and with an electric field.

Thermal velocity:

vth =

√3kTm0

(4.1)

Average thermal energy is 32kT and kinetic energy is 1

2mv2th!

Free electrons obey Maxwell-Boltzmann statistics and Newton’s Laws.Parameters:

• n free electron density (number per unit volume)

• ~E electric field

• -q electron charge

• m0 electron rest mass

Let τ be the scattering time and 1/τ be the scattering rate. This means that theprobability of scattering in a small time interval dt is: dt/τ. The probability of notscattering in time dt is then:

(1− dt

τ

). Let ~p(t) be the average electron momentum

at time t, then we have:

~p(t +dt) =(

1− dtτ

)(~p(t)−q~E(t)dt

)+

(dtτ

)(0) (4.2)

Scattering - average momentum after scattering is zero - dtτ(0) term!

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No scattering - motion follows Newton’s laws - first term on the right above!(1− dt

τ

)(~p(t)−q~E(t)dt

)Therefore:

⇒ d~p(t)dt

=−q~E(t)− ~p(t)τ

(4.3)

showing the interplay between the external force and the internal frictional dam-ping type term. (Eq of motion, free electron: F = m0a =−qE .)Three scenarios to deal with:

1. No electric field.

2. Constant uniform electric field.

3. Time dependent sinusoidal electric field.

4.1.2 Case I: No electric fieldd~p(t)

dt=−~p(t)

τ

~p(t) = ~p(0) e−tτ

Steady state solution: ~p(t) = 0. Random scattering averages momentum to zero.If you impart momentum to the electrons, on average, they will relax back to zeromomentum exponentially with a time constant τ.

4.1.3 Case II: Constant uniform electric fieldThe exponential term becomes negligible with time leaving only the field term~p(t) =−qτ~E . The electron drift velocity is defined as:

~vd =~pm0

=− qτ

m0~E =−µn~E

whereµn =

qτ

m0(4.4)

is the electron mobility (drift velocity per unit electric field, units: cm2/V s).The electron current density ~J (units: Amps/cm2) is:

~J = n×−q×~vd = nqµn~E = σ~E

V Casey 52


where n is the free electron density (units: cm−3)

σ0 = nqµn =nq2τ

m0(4.5)

(dc conductivity, units: Siemens/cm)

4.1.4 Case III: Time dependent sinusoidal electric fieldd~p(t)

dt=−q~E(t)− ~p(t)

τ

There is no steady state solution in this case. Assume the electric field, averagemomentum and currents are all sinusoidal with phasors given as follows:

~E(t) = Re[~E(ω)e−iωt

]~p(t) = Re

[~p(ω)e−iωt]

~J(t) = Re[~J(ω)e−iωt

]

d~p(t)dt

=−q~E(t)− ~p(t)τ

⇒−iω~p(ω) =−q~E(ω)− ~p(ω)τ

⇒ ~p(ω) =− qτ

1− iωτ

~E(ω)

~vd(ω) =~p(ω)m0

=−qτ

m0

1− iωτ

~E(ω)

The electron current density is:

~J(ω) = n(−q)~vd(ω) = σ(ω)~E(ω)

σ(ω) =

nq2τ

m0

1− iωτ=

σ0

1− iωτ(4.6)

- a very important result!Result is applied to propagation of em radiation in metals.

• Gives right result at dc ω = 0.

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• Ignores magnetic field effects (B only adds a term which is v/c smaller thanthe electric field term).

• Must assume that the electric field does not vary significantly locally, i.e.the wavelength of the em wave is much greater than the mean free path ofthe electron λ >> l. The electric field has impact only since last collision.

• Holds for visible light since wavelengths are 103 to 104A.

Drude theory accounts for Ohm’s Law

Considering unit volume of the metal, which contains n free electrons, each ofcharge −q, then the total charge will be −nq. The total charge crossing unit areaper unit time will be −nqvd . This is the definition of current density and so:

~J =−nq~vd = nqµn~E

~J = σ~E

whereσ = nqµn

is the sample conductivity which is reciprocally related to the resistivity:

σ =1ρ

By simple geometric considerations it is easy to show that the current densityleads to:

I =AρL

V

R =ρLA

i.e. Ohm’s law, which is one of the successes of the Drude theory. Since latticescattering is expected to increase with increasing temperature the theory also ex-plains why many metals have positive temperature coefficients of resistance, i.e.resistance increases with temperature. The current density equation is known asthe microscopic version of Ohm’s Law. Note also that

σ =nq2τ

m0

using the expression for drift velocity in the current density equation.

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Matthiessen’s Rule

According to the Drude theory of conduction we would expect the resistivity ofa metal to become smaller and smaller as the temperature is reduced reaching avalue of zero at absolute zero, due to reduced lattice vibration. However, real cry-stals always contain defects such as vacancies, dislocations and impurities. Thesewill scatter electrons also and contribute to a background resistivity which wouldremain even if absolute zero could be reached. We can attribute a characteristictime τi –the relaxation time – to the electron inter-collision time for each type ofimpurity and defect. Considering impurities only, the number of collisions for theith impurity is 1/τi and the total number of collisions for impurities is ∑1/τi. IfτL is the relaxation time for lattice scattering, then the total number of scatters persecond is

1τT

= ∑1τi+

1τL

(4.7)

where τT is the overall relaxation time. However, from (??) and using (??),

1τ=

nq2ρ

m0(4.8)

and so (4.7) becomesρT = ∑ρi +ρL (4.9)

after canceling the common factor, nq2/m. Calling ∑ρi, ρ0 gives

ρT = ρ0 +ρL (4.10)

where ρ0 is the residual resistivity which is independent of temperature and roughlyproportional to the amount of impurities present. ρL is the lattice scattering resisti-vity which approaches zero at zero Kelvin and rises approximately linearly (basisof platinum resistance thermometer) above a characteristic temperature for eachmetal (the Debye Temperature). Clearly, we can include additional terms to takedislocations, defects, etc., into account as appropriate. Equation (4.10) is knownas Matthiessen’s rule. It is important because it leads to the idea that, in an ide-ally pure metal, the resistivity is due solely to the thermal vibrations of the lattice.The purity of a metal may be expressed in terms of the ratio of its resistivity atroom temperature (293 or 300K) and its resistivity at liquid helium temperature(4.2K), i.e. the residual resistivity. However, it must be used with care since onlyimpurities in solid solution scatter electrons well: precipitates make only a smallcontribution.

55 V Casey


Mean Free Path

The average distance traveled by an electron between collisions is:

λ = τv (4.11)

where v is the average velocity of electrons made up of thermal and drift velocities.Normally |vth/vd|> 106 and so

λ' τvth (4.12)

Considering copper at room temperature (300K), σ = 6× 107Sm−1, n = 8.5×1022cm−3, τ = 2.3× 10−14s, ⇒ λ = 2.3nm. The atomic spacing in copper isabout 0.2nm. Therefore, using classical theory, the electron appears to travel tentimes the average distance between atoms before colliding with one. However,the experimentally determined value for the mean free path in copper at roomtemperature is about 53nm — much larger than the classically predicted value

Figure 4.3: Charge particles in magnetic fields: cyclotron/betatron/mass spectro-meter.

4.1.5 Hall Effect• A transverse voltage develops across a conductor that is carrying current

whilst in a magnetic field.

• The Lorentz force on the carriers is given by the cross product of the currentdensity and the magnetic field.

d~p(t)dt

+~p(t)

τ=−q

(~E +

~p×~Bm0

)

V Casey 56


Figure 4.4: Schematic of Hall effect experimental arrangement.

In steady state, the current is time independent, i.e. constant, and hence:

~p(t)τ

=−q

(~E +

~p×~Bm0

)

px

τ=−qEx−

qpyBz

mo

py

τ=−qEy +

qpxBz

mo

pz

τ=−qEz

Multiply both across by −nqτ/mo noting that ωc = qB/m (the cyclotron fre-quency) and simplifying:

σoEx = jx +ωcτ jy

σoEy =−ωcτ jx + jy(ExEy

)=

1σ0

(1 ωcτ

−ωcτ 1

)(jxjy

)For the Hall effect jy = 0 implying that:

Ey =−ωcτ

σ0jx

RH =Ey

jxBz=−ωcτ

σ0Bz=−1nq

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Figure 4.5: Hall probe.

Figure 4.6: The Rev counter/Pulse Counter.

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SSP1 4.2. DRUDE MODEL AND METAL REFLECTIVITY

Table 4.1: Hall Coefficients for some metals.

Metal RHnqNa -1.15Cu -0.68Ag -0.8Pd -0.73Pt -0.21Cd +0.5W +1.2Be +5.0

Hall voltage, VH = w×Ey may be measured directly by placing contact elec-trodes on the sides of the specimen:

VH =RH I B

t=−1nq

IBt

(4.13)

where I, B and sample thickness t are easily determined experimentally. ThereforeRH and/or n can be calculated.

Show how t rather than w enters the above relationships!Importance:

• Carrier type by sign of Hall Coefficient −1/nq, +1/pq.

• Carrier density.

Not in agreement with experiment in relation to:

• Positive Hall coefficients found in some metals.

• Magnetoresistance

The magnetoresistance is zero according to Drude since:

jx(B) = σ0Ex = jx

.However, metals do display magnetoresistive effects.

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4.2. DRUDE MODEL AND METAL REFLECTIVITY SSP1

Figure 4.7: Incident, transmitted and reflected electromagnetic waves - air to me-tal.

4.2 Drude Model and Metal ReflectivityWhen electromagnetic waves are incident on an air-metal interface there is a re-flected wave:The reflection coefficient is:

Γ =~ER

~EI=

√ε0−

√ε(ω)

√ε0 +

√ε(ω)

What is ε(ω) for metals?From Maxwell’s equation:Ampere’s law:

∇× ~H (~r, t) = ~J(~r, t)+ ε0∂~E(~r, t)

∂t

Phasor form:

∇× ~H (~r, t) = ~J(~r)− iωε0~E(~r)

= σ(ω)~E(~r)− iωε0~E(~r)

= −iωεe f f (ω)~E(~r)

where

εe f f (ω) = ε0

(1+ i

σ(ω)

ωε0

)(4.14)

is the effective (complex) dielectric constant for metals.The metal reflection coefficient becomes:

Γ =~ER

~EI=

√ε0−

√εe f f (ω)

√ε0 +

√εe f f (ω)

V Casey 60


Using the Drude expression for frequency dependent conductivity, Eqn.4.6,the frequency dependence of the reflection coefficient of metals can be explainedadequately all the way from radio frequencies (RF) to optical frequencies.

4.2.1 Drude model and plasma frequency of metalsFor small frequencies (ωτ)<< 1):

σ(ω)≈ σ0 =nq2τ

m0

⇒ εe f f (ω)≈ ε0

(1+ i

σ0

ωε0

)For large frequencies (ωτ)>> 1) (collision-less plasma regime):

σ(ω)≈ σ0

−iωτ= i

nq2

m0ω

⇒ εe f f (ω)≈ ε0

(1−

ω2p

ω2

)where the plasma frequency is:

ωp =

√nq2

ε0m0(4.15)

Electrons behave like a collision-less plasma.

• For ωp > ω >> 1τ, ε real, negative; decay - no propagation!

• When ε is positive (ω < ωp), oscillatory; radiation propagates; transparent.

4.2.2 Plasma oscillations in metalsThe electric field generated by the charge separation:

E =nquε0

The force on the electrons is:

F =−qE =−nq2uε0

61 V Casey


Figure 4.8: The electric field generated by charge movement from one region(positive) to another region of extent u.

As a result of this force the electron displacement u will obey Newton’s secondlaw:

F = m0d2u(t)

dt2 =−qE =−nq2u(t)ε0

⇒ d2u(t)dt2 =−ω

2pu(t)

A second order system with solution:

u(t) = Acos(ωpt)+Bsin(ωpt)

Plasma oscillations are charge density oscillations.

4.2.3 Plasma oscillations in metals with scatteringFrom Eqn.(4.3)we know in the presence of scattering:

d~p(t)dt

=−q~E(t)− ~p(t)τ

⇒ m0d2u(t)

dt2 =−q~E(t)− m0

τ

du(t)dt

~E(t) =nqu(t)

ε0

Combining these yields the differential equation:

d2u(t)dt2 =−ω

2pu(t)− 1

τ

du(t)dt

or

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d2u(t)dt2 +

1τ

du(t)dt

+ω2pu(t) = 0

A second order system with damping!

Case I: Underdamped

ωp >12τ

u(t) = e−γt [Acos(Ωt)+Bsin(Ωt)]

γ =12τ

Ω =√

ω2p− γ2

Damped plasma oscillations.

Case II: Overdamped

ωp <12τ

u(t) = Ae−γ1t +Be−γ2t

γ1 =12τ

+

√1

4τ2 −ω2p

γ2 =12τ−√

14τ2 −ω2

p

No oscillations.

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V Casey 64

Chapter 5

Band Theory of Solids

65

5.1. BAND THEORY OF SOLIDS SSP1

5.1 Band Theory of Solids

Application of Schrodinger’s Equation to a one dimensional crystal.

5.1.1 Introduction

In the free electron model of a metal the effects of the positive ions, i.e. the lattice,are neglected. We saw earlier that all energies were allowed in the free electroncase. The free electron theory (U = 0 or constant) in the form of the Drude theorywas reasonably successful in explaining many metallic properties. However, itwas not possible to account for the differences between metals, insulators andsemiconductors using this theory. In addition, the specific heat capacity of solidswas found to be much smaller than that predicted by the free electron theory.

Figure 5.1: A realistic periodic potential for a 1D crystal.(’Teeth shaped curvescorrespond to potential along a line of ions; continuous curve corresponds to po-tential along a line between planes of atoms; broken dashed curves are for isolatedions).

In order to account for the differences between electronic materials, the in-fluence of the crystal lattice on the electrons must be taken into account whenapplying Schrodinger’s equation to the crystal. The results, even for a very simplemodel of the lattice potential, are surprising. We find that a whole new theory ofsolids emerges, the Band Theory of Solids. In this theory, entities such as band-gaps, positive and negative electron masses, positive and negative electrons, i.e.holes and electrons, emerge as natural consequences of the fact that the crystallattice gives rise to a periodic potential energy profile within the crystal.

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SSP1 5.1. BAND THEORY OF SOLIDS

5.1.2 Bloch’s TheoremFor a 1D (monatomic) crystal the potential repeats itself with spatial period givenby the lattice constant a

U(x+a) =U(x)

Bloch’s theorem tells us that for such a potential the solutions to the Schrodingerequation can be taken to satisfy the condition:

Ψ(x+a) = eiKaΨ(x)

for some constant K.Let D be the “displacement operator”:D f (x) = f (x+a) For a periodic poten-

tial such as U(x), D commutes with the Hamiltonian operator:

[D,H] = 0

This means we are free to choose eigenfunctions of H that are simultaneous ei-genfunctions of D: DΨ = λΨ or,

Ψ(x+a) = λΨ(x)

Now since this relationship holds for all of x, λ cannot be zero. It is complex andso can be expressed in exponential form:

λ = eiKa

for some constant K.It follows from the above and using the Born-von Karmen boundary condition

that:

Ψ(x+Na) = eiKNaΨ(x)

eiKNaΨ(x) = Ψ(x)

so eiKNa = 1 or NKa = n2π, and hence

K =n2π

Na,(n = 0,±1,±2, .....)

and K is real. The important outcome of Bloch’s theorem is that we need onlysolve the Schrodinger equation within a single cell (0 ≤ x ≤ a). We may theneasily generate all other (infinite number?) of solutions.Note that since K is real, this means that:

|Ψ(x+a)|2 = |Ψ(x)|2

and so while Ψ(x) is not periodic its expectation value is.

67 V Casey


5.1.3 The Nature of the Crystal Potential Field

X-ray diffraction has shown that crystals consist of well defined periodic arraysof atoms. Let us consider a one dimensional crystal with atoms which consistof positively charged cores surrounded by loosely bound outer electrons (the ou-ter electrons are loosely bound due to the screening effects of the inner electronshells). Since the atoms are in a periodic arrangement, it is only reasonable toassume that the potential field arising from the atom cores will also be periodic.The outer electrons, i.e. the conduction electrons, move in this periodic poten-tial. A realistic potential profile, based upon Coulomb’s potential, i.e. U(x) α 1/x,might look like that shown in the Fig. 5.1. However, it is extremely difficult tosolve Schrodinger’s equation for a realistic periodic potential. Kronig and Penneysuggested a simplified model consisting of a 1D array of square well potentials ofwidth a, separated by potential barriers of height U0 and width b. It is assumed thatfor any electron, everything else in the crystal can be represented by this effectivepotential. This one electron is then considered to be representative of all otherelectrons in the system. The value of this model is that Schrodinger’s equationmay be solved individually for both regions. By establishing continuity equationsat the boundaries and using Bloch’s theorem, the form of overall solutions arearrived at.

Figure 5.2: Square Well Potential representation of the lattice potential.

5.1.4 Solving Schrodinger’s equation for the K-P 1D model

~2

2md2Ψ(x)

dx2 +(E−U(x))Ψ(x) = 0 (5.1)

d2Ψ(x)dx2 − 2m

~2 (U(x)−E)Ψ(x) = 0 (5.2)

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Figure 5.3: Kronig Penney Model of a 1-D crystal.

Region I

0≤ x≤ a, U(x) = 0

d2Ψ(x)dx2 +

2mE~2 Ψ(x) = 0 (5.3)

Solution:Ψ(x) = Aeiαx +Be−iαx (5.4)

where α2 = 2mE~2 . . . . an energy term

Region II

a≤ x≤ l, U(x) =U0

d2Ψ(x)dx2 − 2m

~2 (U0−E)Ψ(x) = 0 (5.5)

SolutionΨ(x) =Ceγx +De−γx (5.6)

γ2 =

2m~2 (U0−E) (5.7)

Now since the barrier is finite, there must be some probability of penetrationby the electrons. Therefore, both the wave function Ψ and its first derivative Ψ′

must be continuous at points such as x = 0, a, l, etc.. This allows us establishrelationships between the constants A,B,C and D.

Continuity at,x = 0

69 V Casey


Ψ1(0) = Ψ2(0)A+B =C+D (5.8)

Ψ′1(0) = Ψ

′2(0)

iα[A−B] = γ[C−D](5.9)

and at x = aΨ1(a) = Ψ2(a)

Aeiαa +B−iαa =Ceγa +De−γa (5.10)

Ψ′1(a) = Ψ

′2(a)

iα[Aeiαa−B−iαa] = γ[Ceγa−De−γa](5.11)

and at x = l, using Bloch’s theorem

Ψ1(l) = Ψ1(0)eikl = Ψ2(l)(A+B)eikl =Ceγl +De−γl (5.12)

Ψ′1(l) = Ψ

′1(0)e

ikl = Ψ′2(l)

iα(A−B)eikl = γ[Ceγl−De−γl](5.13)

Equations 5.10-5.13 have a solution only if the determinant of the coefficients ofA,B,C and D vanish or if,

α2− γ2

2αγSinh(γb) Sinh(αa)+Cosh(γb) Cosh(αa) =Cos(kl) (5.14)

for E < U0. For E > Uo, γ becomes purely imaginary. Equation 5.14 does notchange significantly if we replace γ by iγ:

γ2−α2

2αγSin(γb) Sin(αa)+Cos(γb) Cos(αa) =Cos(kl) (5.15)

(Cosh x = Cos ix; Sinh x = 1/i Sin ix )We could plot the left hand side of the last two equations versus E/U0, (see

Bar-Lev p69). However, to obtain a more convenient form Kroonig and Penneyconsidered the case where the potential barrier becomes a delta function, thatis, the case where U0 is infinitely large, over an infinitesimal distance b, but theproduct U0b remains finite and the same, i.e. as U0→ ∞,b→ 0, U0b = constant.Now γ2 ∝ U0 and also goes to infinity as U0.

Therefore

Lim(γ2−α2)U0→∞

∼= γ2

What happens the product γb as U0 goes to infinity?

V Casey 70


• b becomes infinitesmal as U0 becomes infinite;

• however, since γ is only proportional to√

U0 it does not go to infinity as fastas b goes to zero so,

• the product γb goes to zero as U0 goes to infinity.

γb→ 0 as U0→ ∞

Therefore:Sin(γb)→ γbCos(γb)→ 1as U0→ ∞

Also,

as b → 0a → l

since l = (a + b).Equation 5.15 then reduces to

γ2b2α

sin(αl)+ cos(αl) = cos(kl) (5.16)

orγ2ab

2Sin(αl)(αl)

+ cos(αl) = cos(kl) (5.17)

If we define

P = Lim(

γ2ab2

)|Uo|→∞

b→0

which is a measure of the barrier strength or stopping power, then

PSin(αl)

αl+Cos(αl) =Cos(kl) (5.18)

or

PSinc(αl)+Cos(αl) =Cos(kl) (5.19)

This is the dispersion relationship for electrons in a periodic 1D crystal.

71 V Casey


–1

1

2

3

4

–15 –10 –5 5 10 15

theta

Figure 5.4: Dispersion curve for K-P model with a barrier stopping power ofP=3.11.

–2

0

2

4

6

8

10

–15 –10 –5 5 10 15

Figure 5.5: Dispersion curve for K-P models with increasing barrier stoppingpower, 0,1,4,10.

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5.1.5 Dispersion Relation for the Kronig-Penney Model

When Schrodinger’s equation is applied to the K-P model as outlined above, solu-tions are only possible which satisfy the following dispersion relationship betweenenergy (alpha term) and momentum (k term):

PSin α l

α l+ Cos α l = Cos kl

where l = a + b, the lattice spacing;

α2 =

2mE~2

P =mabUo

~2

Clearly the right hand side of this relationship has limits of +1 and −1. Thisimposes bounds upon the left hand side of the equation. In other words, thisrelationship allows one determine whether a given α value is allowed or forbidden.The best way to proceed is to plot the left hand side of the equation as a function ofα. The parameter P is a measure of the stopping power of the barrier; it involvesthe product Uob. P has the value 3.11 for the simple crystal K-P potential shownearlier, figure5.1.4.

The plot of the LHS of the dispersion relation for a range of values of P isshown in figure5.1.4. The bounds imposed by the RHS of the relation are alsoshown. There are regions of αl space and hence energy which are allowed andother regions which are forbidden. A clearer picture emerges when energy E isplotted versus wave vector k as shown in figure5.6.

This latter plot is obtained directly from the plot of the dispersion relation.Consider, for instance, the case where P = 3.11, figure5.1.4. As αl increasesfrom zero, the LHS of the function is out of bounds. It comes into bounds at α1l,and hence energy E1, where Cos(kl) = +1. This will occur for kl = 0, 2π, etc.,(k = n 2π/l). Let us take k = 0 at this point, for convenience. The function remainswithin bounds until α2l (energy E2), at which point Cos(kl) =−1. Here kl = π ork = π/l and the first allowed energy band is defined. As αl increases further, thefunction goes out of bounds again. The next allowed zone starts at α2l (energyE2), where Cos(kl) is still equal to −1. Thus, in terms of the E versus k diagram,there is a discontinuity in energy at k = π/l; an energy gap occurs. The top of thesecond allowed band occurs where Cos(kl) is again equal to +1 (here k = 2π/l).The process may be repeated to give as many allowed and forbidden bands asrequired. The function is symmetric in k space so it is really only necessary toplot for positive k values and then map the curves into negative k space.

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Since the dispersion relation will have the same value if Cos(kl) is replaced byCos(k+n2π/l)l, the energy curves are unchanged if shifted in k space by n2π/l.We can therefore represent all the information in the region−π/l << k <<+π/l.This is known as the reduced zone scheme, figure5.7

Figure 5.6: Extended zone scheme

Figure 5.7: Reduced zone scheme

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SSP1 5.2. SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS

5.2 Some consequences of the band theory of solids

5.2.1 Band FillingAufbau Principle:

Electrons fill the energy bands starting with the lowest available level andworking upwards through the band!

Conductor (Metal)Conductivity occurs if:-the upper conduction band is partially filled,-a full band overlaps an empty band.

Figure 5.8: Conductor

Insulator:Each band is either full or empty and no overlap occurs. Also the bandgap is large,i.e. Eg > 3eV .

Semiconductor:Full valence band.Small energy gap, i.e. Eg ∼ 1eV.

5.2.2 Band Shapes of Real SemiconductorsThe interatomic distance and potential will depend upon direction within the cry-stal. This implies that the shape of the E versus k diagrams will depend upon thedirection of k, and so one may expect much more complicated band shapes than

Figure 5.9: Insulator

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5.2. SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS SSP1

Figure 5.10: Semiconductor

those obtained earlier for the simple 1D case. There will in general be a numberof minima (depending upon the k direction) in the conduction band. The valenceband maximum is always at k = 0. Arising from these considerations, two generalcategories of semiconductor may be defined. These are:

-direct bandgap semiconductors, and-indirect bandgap semiconductors.

Figure 5.11: Direct bandgap semiconductor

Direct Bandgap Semiconductors: GaAs; InSb; InP; CdSIndirect Bandgap Semiconductors: Si; Ge; GaP; AlSb.For direct band gap semiconductors such as GaAs, the lowest conduction band

minimum occurs at k = 0. For indirect bandgap semiconductors such as Si andGe, the lowest conduction band minimum occurs near the zone edge. An electron

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Figure 5.12: Indirect bandgap semiconductor

transition between the bands in direct bandgap semiconductors is essentially asingle event involving only an energy change. However, in the case of an indirectbandgap semiconductor, an electron transition between, for instance, the top of thevalence band and the minimum of the conduction band involves both an energychange step, ∆E, and a momentum change, ∆k, which of course is less likely tooccur.

5.2.3 Effective Mass and HolesWhat happens when an electron is accelerated in a lattice with a periodic poten-tial?

p =hλ= ~k

F = ma = qφ

a =qφ

m

a =dvg

dt

vg =dω

dk=

1~

dEdk

a =dvg

dt=

1~

d2Edk2

dkdt

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Figure 5.13: Group velocity and effective mass for the valence band of a semi-conductor

dE = qφdx = qφvgdt =qφ

~dEdk

dt

dkdt

=qE~

a =qφ

~2d2Edk2

m∗ =~2

d2E/dk2(5.20)

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Figure 5.14: Parabolic approximation for the valence and conduction band edges

When an electron is accelerated in a crystal, its response is no longer deter-mined by the electron rest mass. Instead the electron behaves as though it has aneffective mass given by 5.20 above. The effective mass of the electron is determi-ned by the curvature of the E versus k diagram. This is illustrated in the diagrambelow. Near the bottom of the band, the effective mass is positive and here elec-trons behave as normal except that the mass is no longer constant or equal to therest mass. Near the top of a given band we see that the mass of the electron may benegative. In simple terms, this means that the carriers here move in the oppositedirection to the applied force. The simplest interpretation of this is to assume thatwe are now dealing with positive electrons or holes as they are almost universallyreferred to.

5.2.4 Filling of the energy bands with electrons.

Parabolic Approximation

For semiconductor work we are interested mainly in the top of the valence bandand the bottom of the conduction band where the E versus k relationship is para-bolic. m∗ is therefore constant. Energies in these regions may be approximatedby:

E ∼= EC +~2k2

2m∗n(5.21)

E ∼= EV −~2k2

2m∗p(5.22)

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Density of States in Semiconductors

Once the band-structure has been determined, the next task is to determine thenumber of modes or states in the band or more importantly, the density of statesper unit energy within a given band. Once we have this we can then use appropri-ate statistics to determine how the bands are filled. Since we are interested in theelectronic properties of solids, we will be interested in working out how the bandsare filled with electrons. Our interest will be confined to semiconductors witheither an almost empty conduction band (n-type) or an almost full valence band(p-type). Therefore we need only consider the bottom of the conduction band orthe top of the valence band and so we may use the parabolic approximation tosimplify the analysis.

Consider a box shaped crystal of sides Lx,Ly,Lz with interatomic spacingslx, ly, lz respectively. There are Nx = Lx/lx, Ny = Ly/ly, and Nz = Lz/lz correspon-ding values of kx,ky,kz. A single level, therefore, requires a section of Brillouinzone in the kx direction of width

∆kx =2π/lxNx

=2π

Lx(5.23)

and the volume inkspace taken by a single level is ∆kx∆ky∆kz, since two stateswith opposing spin can have the same energy, a single state requires only half thisvolume. Therefore a single state volume is:

12∆kx∆ky∆kz =

(2π)3

2LxLyLz

The number of allowed states N in k space in a spherical shell of radius k andthickness dk is:

N =4πk2dk

(2π)3/2LxLyLz

=

(kπ

)2

LxLyLzdk

The density of states per unit volume is:

dN =

(kπ

)2

dk

For the bottom of the conduction band, E ∼= EC + ~2k2

2m∗n, implying:

k2 =2m∗n~2 (E−EC) (5.24)

and also,

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Figure 5.15: Density of states for parabolic valence and conduction bands

dk =m∗n~2k

dE

Therefore, the density of states in the conduction band is,

dN = 4π

h3 (2m∗n)3/2(E−EC)

1/2dE= NC(E)dE

where the density of states function is given by

NC(E) =4π

h3 (2m∗n)3/2(E−EC)

1/2 (5.25)

Similarly for the valence band, the density of states function has the form,

dN = 4π

h3 (2m∗p)3/2(E−EV )

1/2dE= NV (E)dE

NV (E) =4π

h3 (2m∗p)3/2(E−EV )

1/2 (5.26)

5.2.5 Occupation of allowed states.For collections of particles, e.g. atoms, molecules, electrons, a statistical treatmentwhich describes the average rather than detailed properties of a typical componentof the complete assembly of particles is most useful since the behaviour of thegroup of particles can then be deduced directly. The type of statistics used dependson -

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Figure 5.16: Maxwell-Boltzmann distribution for a classical ensemble of particles

• the type of particle present (neutral, charged, mass, etc..)• the possible interactions between them.

Classical particles obey Maxwell-Boltzmann statistics where there is no restrictionon the energy of the particles, e.g. ideal gas,

dN = N FMB(E) dEwhere N is the number of neutral molecules per cubic meter. FMB is the distri-bution function and gives the fraction of the total number of molecules per unitvolume in the energy range dE. In the high energy region,

FMB(E)≈ e−EkBT

Particles which obey the exclusion principle, e.g. electrons, interact quantummechanically in such a way that the occupancy of a particular state is restrictedby the Pauli exclusion principle. For such particles, Fermi-Dirac statistics apply.The Fermi-Dirac distribution function has the form,

FFD =1

1+ e(

E−EFkBT

) (5.27)

This gives for any ensemble obeying the exclusion principle, the probability thata particular stateEis occupied.

For high energy states, i.e. E >>EF , FFD∼=FMB. At high energies, the numberof electrons distributed over many available states is small and there are manymore energy levels than electrons to occupy them. Under these conditions, thereis little chance of two or more electrons occupying the same state and whether

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Figure 5.17: Fermi-Dirac distribution function for electrons at temperatures 2, 50and 100 K

the exclusion principle is included in the statistics or not becomes irrelevant to theform of the distribution.

At very low temperatures there is a small but finite probability that electronswill occupy available states for which E >EF but the probability rapidly decreaseswith increasing energy. As the temperature is increased, this tail of the probabilityfunction becomes more pronounced and the probability of occupancy of higherenergy states is correspondingly increased.

Note that the probability that an electron occupies the Fermi Energy Level,EF , (referred to simply as the Fermi level) is always 1/2 independent of the actualtemperature. Also, the probability function is symmetric about the Fermi level.

5.2.6 Properties of Semiconductors

RevisionIntrinsic SemiconductorsExtrinsic SemiconductorsDoping

5.2.7 Carrier Density

A Conduction Band

N(E)dE = NC(E)FFD(E)dE =4π

h3 (2m∗n)3/2(E−EC)

1/2

[1

1+ eE−EF

kBT

]dE

The density of electrons in the whole conduction band is,

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n =∫ ETop

EC

NC(E)FFD(E)dE (5.28)

Since EF is located at least a few kBT below EC, we can make the followingapproximations,We can replace FFD(E) by FMB(E) since (E−EF/kBT )>> 1Recall FMB(E) goes to zero as E goes to infinity. Therefore, we can replace thetop limit of the integral by ∞ without changing the result.Therefore,

n =4π

h3 (2m∗n)3/2

∫∞

EC

(E−EC)1/2e−

(E−EF

kBT

)dE

Now

e−(

E−EFkBT

)= e−

(E−ECkBT

).e−

(EC−EF

kBT

)

n =4π

h3 (2m∗n)3/2e−

(EC−EF

kBT

)∫∞

EC

(E−EC)1/2e−

(E−ECkBT

)dE

(E−EC)1/2 = kT

1/2(

E−EC

kBT

)1/2

Let x = (E-EC)/kBT ; also dE=kBTdxSubstituting,

n =4π

h3 (2m∗nkBT )12 e−

(EC−EF

kBT

)∫∞

0x

12 e−xdx

∫∞

0x

1/2dx =√

π

2

n =2h3 (2πm∗nkBT )

3/2e−(

EC−EFkBT

)(5.29)

n = NCe−(

EC−EFkBT

)(5.30)

where NC is a constant and is known as the Effective Density of States in theconduction band.

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B Valence Band

Similarly for holes in the valence band,

p =2h3 (2πm∗pkBT )

3/2e−(

EF−EVkBT

)(5.31)

p = NV e−(

EF−EVkBT

)(5.32)

where NV is a constant and is known as the Effective Density of States in thevalence band.

At 300K Si GaAsNC 2.8x1019cm−3 4.7x1017cm−3

NV 1.04 x 1019 cm−3 7.0x1018 cm−3

5.2.8 Fermi Level in SemiconductorsIntrinsic semiconductors

In intrinsic semiconductors, n = p = ni where ni is the intrinsic carrier density.

NCe−(

EC−EFikBT

)= NV e

−(

EFi−EV

kBT

)

NC

NV= e

EC−EV−2EFikBT =

(m∗pm∗n

)3/2

EFi =EC +EV

2+

3kBT4

Ln(

m∗pm∗n

)(5.33)

In both Si and Ge,m∗n ≈ m∗p. Therefore, EFi is practically mid-way between ECand EV , i.e. at EG/2.

5.2.9 Law of Mass Action

np = NCNV e−(

EC−EFikBT

)e−(

EFi−EV

kBT

)

np = NCNV e−(

EGkBT

)= n2

i

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np = AT 3e−(

EGkBT

)(5.34)

where A is a constant. The electron-hole product in semiconductors in thermalequilibrium is a function of T only since mn∗ and mp∗ and EG are relatively in-dependent of T . This is a general result holding for both intrinsic and extrinsicsemiconductors and is known as the law of mass action; the electron-hole productis a constant at a given temperature.

ni = 1.45 x 1010cm−3(Si)ni = 1.79 x 106cm−3(GaAs)

Fermi Level in Extrinsic Semiconductors

Doping

Figure 5.18: Symbols used with extrinsic semiconductors

n-type Addition of Gr V donor Impurity level just below EC!p-type Addition of Gr IV acceptor Impurity level just above EV !

NDDensity of donor atomsN+

D Density of ionised donorsNADensity of acceptor atomsN+

A Density of ionised acceptorsnFree electron densitypFree hole densityniIntrinsic electron/hole carrier density.Extrinsic Semiconductor

n 6= p

Charge Neutrality must prevailp+N+

D = n+N−A (5.35)

np=

NC

NVe−(

EC−2EF+EVkBT

)

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Figure 5.19: Summary diagram

EF =EC +EV

2+

kBT2

Ln(

NV

NC

)+

kBT2

Ln(

np

)

EF = EFi +kBT

2Ln(

np

)(5.36)

The Fermi level moves up towards EC when donor impurities are added. Thereverse happens when acceptors are added, i.e. EF moves down towards EV .

Figure 5.20: Free carrier densities in a semiconductor energy band, fromhttp://ece-www.colorado.edu/ bart/book/carriers.htm.

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5.3. CURRENT TRANSPORT AND CURRENT DENSITY EQUATIONSSSP1

Figure 5.21: Free carrier densities in semiconductor valence and conductionbands, from http://ece-www.colorado.edu/ bart/book/carriers.htm.

5.2.10 Temperature Dependence of Carrier Density

Taken from SZE - Physics of Semiconductor Devices, pp26)At high temperatures we have the intrinsic range since intrinsically generated

carriers will vastly outnumber any extrinsic carriers. At very low temperaturesmost impurities are frozen out and the slope is determined by the activation energyof the donor level. The electron density remains essentially constant over a widerange of intermediate temperatures (100-500K).

The diagram below shows the Fermi level for silicon as a function of tempe-rature and impurity concentration and also shows the dependence (small) of thebandgap on temperature.

5.3 Current Transport and Current Density Equa-tions

Current transport in semiconductors is a little more complicated than in simplemetals. Firstly, we must deal with two types of carriers. Secondly, in addition tocarrier transport due to electric field induced drift we must deal with an additio-nal current transport mechanism - diffusion - due to carrier density concentrationgradients which we will see can be induced by very many different ’stimuli’. Re-call that both the classical theory (Drude Model for drift velocity/drift current)and the quantum theory of conduction in metals can be used to arrive at Ohm’slaw. In the familiar form this defines the relationship between applied voltage andcurrent: V = RI where R is the resistance. R = ρL

A where ρ is the material resis-tivity. In dealing with microscopic models of solids the electric field φ current

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SSP15.3. CURRENT TRANSPORT AND CURRENT DENSITY EQUATIONS

Figure 5.22: Fermi energy of n-type and p-type silicon, EF,n and EF,p, as afunction of doping density at 100, 200, 300, 400 and 500 K. Shown are the con-duction and valence band edges, Ec and Ev. The midgap energy is set to zero,from http://ece-www.colorado.edu/ bart/book/

density J version is more useful, i.e. J = σφ where σ is the sample conductivity.For the classical model σ = qµn where µn is the electron mobility, assuming onlyelectrons are present and dealing with drift current only.

If a concentration gradient in the electron distribution exists at any point thenthere will be a diffusion flux of carriers given by Fick’s law, which for a onedimensional electron gradient gives a flux, Φ,

Φn ∝−dndx

which simply states that the flux of carriers is directed in the direction opposite tothat of increasing gradient and is proportional to the gradient. In a given materialand for a specific carrier type, i.e. electron/hole, the constant of proportionality isthe diffusion constant Dn for electrons and Dp for holes. Therefore, the electronflux is

Φn =−Dndndx

and since each electron carries −q charge the charge flux or electron current den-sity will be

Jn = qDndndx

The total electron current density, combining both drift and diffusion components,is therefore

Jn = q(nµnφ+Dndndx

). (5.37)

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5.3. CURRENT TRANSPORT AND CURRENT DENSITY EQUATIONSSSP1

By similar argument the total hole current density,

Jp = q(pµpφ−Dpd pdx

) (5.38)

(show this) and the total current density for a semiconductor, in the most gene-ral case, i.e. allowing for both the presence of free electrons and free holes andconcentration gradients in both carrier types with an electric field applied is

J = Jn + Jp. (5.39)

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Chapter 6

Unipolar Devices

91

6.1. HIGH FIELD EFFECTS SSP1

6.1 High Field EffectsFor low fields vd , the drift velocity, is proportional to ε, the electric field and ismuch less in general than the thermal velocity, vt (∼107cm/s for Si at 300K). Asε increases and vd approaches vt , a linear relationships no longer exists. In siliconthe drift velocity saturates at high fields.

Figure 6.1: High Field Effects in Silicon

Figure 6.2: High Field Effects in GaAs and InP

However, negative differential resistance (NDR) occurs above a threshold fieldεT for both GaAs and InP (n-type).

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SSP1 6.2. EXPLANATION OF NDR BASED ON BAND STRUCTURE

εT (kV/cm) vd(peak) x107 (cm/s)GaAs 3.2 2.2InP 10.5 2.5

6.2 Explanation of NDR based on band structureConsider GaAs. At room temperature, practically all the conduction electrons willoccupy states in the lower valence band and will therefore have high mobility andlow effective mass. By applying an electric field, electrons will initially transferthe energy gained from the field to the lattice.

Figure 6.3: Band structure details and transport properties for GaAs

Eventually however, for large values of ε, the electrons gain energy from thefield at a rate greater than they can loose it to the lattice through collisions. Theaverage energy of the conduction band electrons increases therefore. This energyincrease continues with increased field until such time as the electrons have gainedon average, 0.31eV . At this point they start transferring to adjacent energy states inthe satellite valleys. In these valleys, the electron mass increases, and the mobilitydecreases. All the conduction electrons may be transferred to the high mass lowmobility valley in this way, i.e. by applying an electric field above this criticalvalue required to allow transfer. Devices that exploit this effect are often calledTEDs, i.e. transfer electron devices.

6.3 Negative Differential ResistanceAt sufficiently high fields, the drift velocity increases linearly again in both InPand GaAs.

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6.3. NEGATIVE DIFFERENTIAL RESISTANCE SSP1

Figure 6.4: Transfer electron phenomenon in GaAs

Figure 6.5: Negative differential resistance (NDR)

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SSP1 6.4. TRANSFER ELECTRON DEVICES TEDS

Outside of the NDR region, the effective drift velocity takes on the asymptotevalues:

vd ∼ µ1ε for 0 < ε < εA

vd ∼ µ2ε for ε > εB

Since µ1εA > µ2εB, there exists a region in which the drift velocity decreaseswith increasing electric field.

Also since J = qnµε, this implies that the current decreases with increasingvoltage, i.e. a Negative Differential Resistance (NDR) region exists for the GaAsdevice.

6.3.1 NDR and InstabilityA semiconductor exhibiting NDR is inherently unstable because a random fluctu-ation of carrier density at any point in the semiconductor produces a momentaryspace charge that will grow exponentially with time.

_n = n− n = (n− n)t=0e−t

τR

where the pre-exponential factor is the initial carrier density fluctuation, i.e. attime zero – the nucleation event. Now,

τR =εS

qnµ

is the dielectric relaxation time: - the time constant for the decay of normal fluctu-ations (εS is the semiconductor dielectric constant). With NDR, µ is negative, i.e.the differential mobility dv/dε is negative. The average mobility for NDR willtherefore be negative and so the argument of the exponential term is positive. Thespace charge grows with time constant |τR|.

6.4 Transfer Electron Devices TEDs

6.4.1 Modes of OperationThere are many modes of operation for TED devices (i.e. Gunn diodes). The twomost common modes are:

• Accumulation Mode

• Domain Mode

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6.4. TRANSFER ELECTRON DEVICES TEDS SSP1

Figure 6.6: Accumulation mode.

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6.4.2 Accumulation

The largest space charge fluctuations usually occur near to the cathode (carriersare injected here). Therefore this region will be the space charge nucleation site.Biasing the device to εA makes the electric field distribution within the crystalunstable. As a charge fluctuation occurs at the nucleation site, the electric fieldε to the left of the space charge will drop while the electric field to the right willincrease. Thus carriers on the LHS will be traveling up to the space charge withhigh velocity (since this is a low field region) while the carriers on the RHS will bemoving away from the space charge with lower velocity (since this side is the highfield side). The space charge therefore accumulates more charge and in actual factit grows exponentially.

Figure 6.7: Electric field stabilisation in Accumulation Mode

A stable space charge pulse will be reached (t2) when the field in the low fieldregion has reduced (and consequently the low field velocity has reduced) and thefield in the high field region has increased (with a corresponding increase in highfield velocity) to such an extent that the rate of delivery of carriers from the LHSis equal to the rate of removal from the RHS.

The overall space charge pulse moves to the right with a velocity vA, the accu-mulation layer velocity. The pulse is discharged at the anode. The electric field inthe crystal is again instantaneously equal to εA; a space charge is again initiatedat the cathode and the process repeats.

Pulse velocity vp ∼ 107 cm/s

Crystal Length L ∼ 10µm

Transit Time τ = L/vp = 10−10s

Pulse Frequency = 10 GHz i.e. Microwave

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6.4.3 Domain ModeThis is an alternative stabilisation mode to accumulation layer formation. In ntype GaAs or InP, a dipole layer may easily be formed from a slight redistributionof the free carriers. Thus an accumulation layer of electrons at a particular sitewill come about due to the creation of an adjacent depletion region; a dipole layeris formed. The accumulation-depletion combination is known as a domain.

Figure 6.8: Domain Mode

The electric field will increase within the domain region and will be loweredon both sides of the domain. More electrons will accumulate on the LHS of thedomain due to the high velocity of the carriers in this region. The depletion willalso increase on the RHS of the domain due to the loss of free carriers again dueto their high velocity.

Since the formation of the domain reduces the field outside the domain, onlyone domain forms at a time. The domain travels to the anode and is absorbed,i.e. it gives rise to a current pulse. The field increases again; a new domain is

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Figure 6.9: Domain formation, growth and stabilisation.

nucleated at the cathode and the process repeats. This again is a Transit TimeMode with the oscillation frequency given by f = L

vDwhere vD is the domain

velocity. One can vary the frequency using an RF resonant cavity as a tuner!

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Chapter 7

pn Junction Diode

101

7.1. INTRODUCTION SSP1

7.1 IntroductionJunctions between p-doped and n-doped semiconductors are important becauseof the wide range of device applications, e.g. diodes, LEDs, photodiodes, Laserdiodes but also because the theory of pn junctions serves as the foundation of thephysics of general semiconductor devices. A pn junction occurs in a semicon-ductor when the impurity concentration changes from acceptor impurities (NA) todonor impurities (ND).

7.2 FabricationAlloyed junction: the simplest and original means of fabricating pn junctions wassimply to melt a small pellet of metal on the semiconductor surface which washeated so that the metal alloyed with the semiconductor producing an oppositelydoped region to that of the original.

Figure 7.1: Alloy pn junction

Epitaxial growth: the growth of a crystal layer on top of a single crystal semi-conductor in a gas reactor. This approach allows the incorporation and changingof impurities as the layer is grown.

Diffusion: a silicon dioxide layer acts as a diffusion barrier to the dopant spe-cies. The silicon wafer is placed in a furnace. Generally a gas dopant is carriedin an inert gas into the furnace. Windows pre-etched in the silicon dioxide definewhere diodes are to be formed. The depth of the junction is a function of thetime and temperature program used. The diffusion front advances into the waferconverting the semiconductor type in the process.

Figure 7.2: Diffused pn junction

Ion implantation: introduce the impurities by direct bombardment. This is the

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SSP1 7.3. IDEALIZED JUNCTION FORMATION

most versatile process which allows precise control over the dose and the location.It also has the advantage of being a low temperature process.

7.3 Idealized Junction FormationWe will consider the bringing together of two pieces of silicon to form a pnjunction, one heavily doped p-type and the other normally doped n-type. This isnot a realistic scenario but is useful in explaining junction behaviour. Both piecesof semiconductor are uniformly doped. The situation before contact is representedschematically in the diagram.

Figure 7.3: Free carrier density.

The contact surface between the two semiconductors is known as the metalur-gical junction and it is convenient to consider this as the origin for a one dimensio-nal axis through the junction. On going from the p-side to the n-side at the instantof contact, there is a very large hole concentration gradient. Similarly, there is alarge electron concentration gradient crossing from the n-side to the p-side of thejunction. Diffusion currents are therefore established which drive holes into the n-side of the junction and electrons into the p-side. When electrons and holes meet,they recombine. However, the ionised impurity atoms retain their charge. Con-sequently, a space charge builds up around the junction and a region is formed,known as the depletion region, which is devoid of majority carriers. The spacecharge creates an electric field which drives minority carrier drift currents whichare in opposition to the majority carrier currents. The depletion region extendson either side of the junction until the electric field has built up to such a levelthat it reduces the diffusion currents for electrons and holes until they are equal

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7.3. IDEALIZED JUNCTION FORMATION SSP1

and opposite to the minority electron and hole currents. This is the equilibriumcondition and no net current flows.

Since an electron and hole is lost in each recombination, the total space chargeon the p side of the junction must equal the total space charge on the n-side of thejunction. Sine the p side is more heavily doped than the n-side, the depletionregion extends further into the lower doped n-side. The space charge gives rise toan in-built potential, VB , otherwise known as the barrier potential. It is importantto note the polarity of this potential and also that it exists without any externalbias.

Figure 7.4: Depletion region space charge, electric field and potential at equili-brium.

7.3.1 Energy band diagramsAt equilibrium:

• No applied voltages

• No light shining on the device

• No thermal gradients

• No magnetic fields

Consider two electron states, A and B.At equilibrium the probability of an electron drifting from A to B is proportionalto F(EA)[1−F(EB)], i.e. =C F(EA)[1−F(EB)]. For the reverse process theprobability is: =C F(EB)[1−F(EA)]. These probabilities must be equal (sinceat equilibrium no net currents flow). The only solution occurs for F(EA) = F(EB).

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SSP1 7.3. IDEALIZED JUNCTION FORMATION

This implies that the Fermi levels for A and B must be equal. In terms of energyband diagrams, this means that the Fermi levels for the two systems must be alig-ned and flat. This is the key to drawing the energy band diagrams for junctiondevices in general.

7.3.2 Pn Junction Equilibrium Band Diagram

If EF is constant, the conduction and valence bands have to bend in the vicinity ofthe junction.

Figure 7.5: Isolated p and n type semiconductors

The equilibrium energy band diagram is constructed by first drawing a com-mon Fermi energy level. The bulk p and n regions and then constructed aroundthis Fermi level as in the isolated energy band diagrams. Finally, the conductionand valence band edge energy levels are bent in the vicinity of the junction in orderto link across the junction region. The band bending occurs within the depletionregion.

Figure 7.6: Energy band diagram for pn junction at equilibrium.

The energy shifts of the Fermi level on the two sides of the junction withrespect to the intrinsic Fermi level are qV p and qVn. The sum of these two energiesgives the equivalent barrier energy.

105 V Casey

7.3. IDEALIZED JUNCTION FORMATION SSP1

7.3.3 Currents flowing at equilibrium.Consider the electron currents flowing across the junction first. There will be adiffusion current of electrons from the n-side to the p-side because of the concen-tration gradient. Electrons are the majority carriers on the n-side, i.e. there aremany more electrons than holes on this side. However, there is an energy barrierwhich electrons must overcome in going from the n conduction band across tothe p conduction band (electron energy increase upwards in the diagram). Thisbarrier arises because of the space charge. Recall that the space charge is nega-tive on the p side and therefore will tend to repulse electrons crossing from the pside. Therefore, there exists a relatively plentiful supply of electrons but there is alarge energy barrier to be overcome. This barrier reduces the number of electronscrossing to a small number.

Hole energy increases downwards in the diagram. There is therefore a similarbarrier to holes crossing from the p-side to the n-side of the junction as there is forelectrons crossing in the opposite direction. Thus the barrier acts against majoritycarrier flow across the junction.

The depletion region is depleted of majority carriers. However, thermal gene-ration can create electron-hole pairs in this region, i.e. so called pair-production.The energy gradient is such as to sweep the pair apart as soon as it is formed: theelectron falls down the electron energy gradient and is therefore swept into theconduction band on the n side. This constitutes an electron drift current since itis driven by the in-built field. There may be an additional component to this driftcurrent due to minority carriers from the bulk of the p region being scattered intothe depletion region. If a p-side minority carrier is scattered to the edge of thedepletion region, it gets swept across the junction. Since both components of thiselectron drift current are minority carrier related, this current will be very small atordinary temperatures. At equilibrium, no net current flows. Therefore, the equi-librium condition corresponds to a situation where the majority diffusion currentof electrons is reduced to the minority drift current for electrons, each being in theopposite direction to the other.

A similar scenario holds for the holes. Minority carrier holes scattered upto the edge of the depletion region on the n-side of the junction will be sweptacross to the p-side (negative space charge here) as will holes generated by pairproduction within the depletion region. The hole diffusion current from the p-side to the n-side is reduced by the barrier. At equilibrium the drift and diffusioncurrents for holes cancel each other just as for the electrons.

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SSP1 7.4. BARRIER POTENTIAL VB.

7.4 Barrier potential VB.

nn = NCe−(

ECn−EFkBT

)

np = NCe−(ECp−EF

kBT

)

nn

np= e

(−ECn+EF+ECp−EFkBT

)

lnnn

np=

ECp−ECn

kBT=

qVB

kBT

VB =kBT

qln(

nn

np

)(7.1)

Also since,

nn

np=

nn pp

np pp=

nn pp

n2i

VB =kBT

qln(

nn pp

n2i

)(7.2)

Now at room temperature, nn ≈ ND and pp ≈ NA so,

VB =kBT

qln(

NDNA

n2i

)(7.3)

The number of silicon atoms per cubic centimeter is about 5×1022. Note that VBdepends upon the concentration of majority carriers on both sides of the junctionand n2

i for each material.

Semiconductor n2i (cm−3)

Silicon 2×1020

Germanium 6×1026

Gallium Arsenide 4×1012

107 V Casey

7.5. DEVICE EQUATIONS AT EQUILIBRIUM SSP1

7.5 Device Equations at equilibriumWe require quantitative solutions for the electric field, E , the potential V , thedepletion region widths dn, and dp, and the charge density ρ, across a pn junction.Since there is a space charge region around the junction we should try to solvePoisson’s equation,

∇2V =− ρ

εεo(7.4)

where ρ is the charge density. In one dimension, we get,

d2Vdx2 =−ρ(x)

εεo=− q

εεo[p−n+ND(x)−NA(x)] (7.5)

where ND(x) and NA(x) are the impurity profiles, and in the most general sense E ,V , p, n, etc. are functions of x.

It is not easy to solve Poisson’s equation in exact form for most devices, howe-ver, because n and p are in turn functions of V and E , the unknowns. To solve theequation in closed form, i.e. V, E , etc. as an explicit function of x, it is necessaryto make several simplifications.

The set of simplifications we will use are collectively known as the ‘depletionapproximation’. We can justify its use since the solutions obtained can explain theexperimental data for real diodes.

7.5.1 The Depletion Approximation.Consider a uniformly doped pn step (abrupt) junction.

Figure 7.7: Junction/Depletion region.

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SSP1 7.5. DEVICE EQUATIONS AT EQUILIBRIUM

1. For −dp ≤ x ≤ 0, NA >> np or pp. Hence ρ = -qN−A in this region.

2. For 0 ≤ x ≤ dn, ND >> nn or pn. Hence ρ = qN+D in this region.

3. ρ is zero in the bulk regions of the semiconductor, i.e. x > dn and x <−dp.The bulk is charge neutral.

4. Choose V to be zero at the junction.

5. The total voltage VT is the sum of the internal potential VB and the externalbias voltage VA and appears entirely across the junction, i.e. between x = dnand −dp.

Solving Poisson’s Equation using the Depletion approximation.

p-side−dp ≤ x ≤ 0

d2V (x)dx2 =

qNA

εεo(7.6)

Integrating gives,

dV (x)dx

=qNA

εεox+C1

Now E = -dV(x)/dx = 0 at x =−dp. Therefore,

C1 =qNA

εεodp (7.7)

and

ϕ =−dVdx

=−qNA

εεo(x+dp) (7.8)

which is linear with negative slope.n-side

0 ≤ x ≤ dn

d2V (x)dx2 =−qND

εεo(7.9)

Integrating gives,

dV (x)dx

=−qND

εεox+C2

109 V Casey

7.5. DEVICE EQUATIONS AT EQUILIBRIUM SSP1

Now E = -dV(x)/dx = 0 at x = dn. Therefore,

C2 =qND

εεodn (7.10)

and

E =−dVdx

=qND

εεo(x−dn) (7.11)

which is linear with positive slope.Now E must be continuous at x = 0.

−qND

εε0dn =

−qNA

εε0dp

or

NDdn = NAdp (7.12)

and multiplying across by qA, where A is the junction area,

qANDdn = qANAdp

which confirms that the total positive space charge will be equal to the total nega-tive space charge.

If it is a so called one-sided pn junction, e.g. p>> n, then NA >>ND implyingthat dn >> dp. The depletion region extends mainly into the lower doped side. Asecond integration yields,

V =qNA

εεo

(x2

2+dpx

)+D1,

for the p-side and,

V =−qND

εεo

(x2

2−dnx

)+D2

for the n-side of the junction.At x = 0, V = 0⇒ D1 = D2 = 0

VT =V (dn)−V (−dp)

VT =q

2εε0

(d2

nND +d2pNA)

(7.13)

The depletion region width will be proportional to V 1/2T .

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SSP1 7.5. DEVICE EQUATIONS AT EQUILIBRIUM

7.5.2 Depletion Region WidthWe already have qNAdp = qNDdn. Using the expression for VT above,

dp =

(2εεoVT

qND

NA(NA +ND)

)1/2

(7.14)

dn =

(2εεoVT

qNA

ND(NA +ND)

)1/2

(7.15)

d = dn +dp

d =

√2εεoVT

q

[1

ND+

1NA

](7.16)

7.5.3 Maximum Electric FieldFrom the equations for electric field developed above, we see that E is a maximumwhen x = 0. Also at x = 0, the two fields, i.e. at either side of the junction, areequal,

Emax =qNDdn

εεo=

qNAdp

εεo.

Substituting for dn (or dp),

Emax =√

2qVTεεo

NANDNA+ND

=

√2qVTεεo

[1

ND+ 1

NA

]−1(7.17)

If NA >> ND then,

dn ≈√

2εεoVTqND

dp ≈ 0

Emax =

√2qNDVT

εεo(7.18)

If ND >> NA, then,

dn ≈ 0

dp ≈√

2εεoVTqNA

111 V Casey

7.6. JUNCTION CAPACITANCE SSP1

Emax =

√2qNAVT

εεo(7.19)

Note Emax is very high in junctions where both p and n regions are heavily dopedand much smaller in junctions where one or both regions are lightly doped.

For reverse bias conditions (see later) the space charge regions extend them-selves outwards into the crystal. For large reverse bias (i.e. VA >>VB) the extentof the space charge region is proportional to

√VA, also ϕmax ∝

√VA.

The results obtained above for depletion region width and maximum electricfield are accurate for reverse bias conditions where the depletion approximation istrue. However, for high forward bias, there may be voltage drops in the bulk dueto the large currents flowing and the reduced barrier height. Clearly, the depletioncondition which requires all the voltage drop to occur across the junction does nothold and the equations obtained above should not be expected to provide a faithfulrepresentation of the behaviour.

7.6 Junction Capacitance

Consider what happens to the depletion region when the reverse bias on a junctionis increased.

Figure 7.8: Junction capacitor.

The depletion region contains a space charge but is devoid of mobile carriers. It istherefore somewhat like a dielectric. Once the reverse bias increases, the depletionregion extends since d is proportional to V 1/2

A . This means that free or mobilecharge is removed from the edges of the depletion region ‘exposing’ increasedfixed charge. Sincedhas increased the effective dielectric thickness has increased.Therefore, we would expect the capacitance to be associated with the junction todecrease with increasing reverse bias. We may obtain an exact expression for thejunction capacitance CJ using our earlier results for d.

First we define a depletion layer capacitance (junction capacitance) as

V Casey 112

SSP1 7.6. JUNCTION CAPACITANCE

CJ =−dQSC

dVwhere QSC is the total space charge on either side of the junction.For an abrupt junction, QSC=qANAdn=qANDdp, which by substitution for dn/p

QSC = A

√2εεoqVT

NAND

NA +ND

CJ =−dQSC

dV= A

√εεoq2VT

NAND

NA +ND. (7.20)

Therefore, if there is heavy doping, CJ will be large whereas if there is light dopingon either side, CJ will be small.For a linearly graded junction, the junction capacitance can be shown to have avalue of,

CJ == A 3

√εεoq2VT

NAND

NA +ND. (7.21)

7.6.1 Diode Information from CJ measurements.It is possible to derive information on doping levels and profiles as well as a valuefor the barrier potential from capacitance-voltage (CV) plots for pn junctions.(i) Barrier Voltage (Contact Potential), VB.

C2J = A2 εεoq

2VT

NAND

NA +ND

VT =VB +VA = A2[

εεoq2

NAND

NA +ND

]1

C2J

VA = A2[

εεoq2

NAND

NA +ND

]1

C2J−VB (7.22)

Therefore, a plot of 1/C2J versus VA, where VA is the reverse bias voltage will give

a straight line with intercept equal to −VB, e.g.(ii) Doping Levels

Consider a one sided pn junction, e.g. p+n.NA >> ND. The slope of the aboveplot will be,

Slope≈ A2εεoq2 ND

For an n+p junction,

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7.6. JUNCTION CAPACITANCE SSP1

Figure 7.9: Reciprocal capacitance plot versus voltage.

Slope≈ A2εεoq2

NA

(iii) Doping ProfilesIf we plot log CJ versus log |VA|, the slope gives an indication of the doping

profile at the junction. If it is an abrupt junction, the slope will be −1/2, whereasa linearly graded junction will give a slope of −1/3. Real pn junctions will giveslopes which are in between these two simple cases.

Figure 7.10: Log-Log plot to establish doping profile in semiconductor

Varactor

The voltage dependence of junction capacitance is exploited in electronic tuningcircuits using varactors or voltage variable capacitors. The Philips BB105B is ageneral purpose varactor used in TV tuning.

V Casey 114

SSP1 7.7. PN JUNCTIONS UNDER EXTERNAL BIAS.

Figure 7.11: Philips BB105B varactor diode characteristic.

7.7 Pn Junctions under external bias.

7.7.1 Reverse BiasWe have shown that d and Emax increase with reverse bias. This is true for alltypes of junction. What about drift and diffusion currents? We have for reversebias VT = VB +VA, VA has the same polarity as VB. VB represents the internalbarrier to diffusion current flow at equilibrium. Reverse bias increases this barrierheight and so the diffusion current components of the current density equationsreduce and with sufficient reverse bias these approach zero values.

Figure 7.12: Energy band diagrams for equilibrium, reverse and forward bais.

The drift components of current do not change. We are left with a net currentflow which depends on the minority carrier densities which will be small. This iscalled the reverse leakage or saturation current.

7.7.2 Forward BiasThe external bias, VA is now of opposite polarity to VB. Therefore, VT =VB−VA,i.e. the barrier height is reduced. VB will always be greater than VA since as VA

115 V Casey

7.7. PN JUNCTIONS UNDER EXTERNAL BIAS. SSP1

approaches VB the forward current increases dramatically causing ohmic dropsin the bulk and thermal runaway. Under forward bias, the lower barrier allowslarge number of holes to flow from the p side to the n-side of the junction. Onthe n-side holes are minority carriers. In addition, electrons flow from thento thep-side where they are minority carriers. This process is known as minority carrierinjection. The drift currents will again be unchanged.

For a p+n junction, hole injection predominates since pp >> nn, figure7.7.2.

Figure 7.13: Free carrier densities at the depletion region for equilibrium andforward bias

The bar denotes equilibrium values.Recall that the total electron current density is given by the sum of the drift

and diffusion components,

Jn = q(µnnE +Dndndx

)

as is the total hole current density,

Jp = q(µp pE −Dpd pdx

).

For a p+n junction, Jn involves relatively small quantities, i.e. n and dn/dx.However, Jp involves p and dp/dx which are many orders of magnitude greaterthan n and dn/dx. Therefore any small imbalance in Jdri f t and Jdi f f for holescan result in large currents flowing. For instance, diffusion current densities aretypically of the order of 104− 105Acm−2. We may therefore conclude that thesystem is only very slightly disturbed from equilibrium under normal forward bias.Consequently, it is possible to assume that the relationships between E , p and nare the same under forward bias as at equilibrium. In particular,

Jn = Jp = 0.

Considering hole current only.

Jp = qµp pE −qDpd pdx

= 0

V Casey 116


E =Dp

µp

1p

d pdx

=kTq

1p

d pdx

VT =VB−VA =−n∫

p

Edx =+kTq

p∫n

1p

d p

VT =kTq

lnpp

pn

Now pp = NA and pn(0) is the minority carrier density at the depletion boundary.

pn(0) = ppe−qVT

kT = NAe−qVB

kT .eqVAkT

For VA = 0, i.e. equilibrium, we get,

pn = NAe−qVB

kT

which is the equilibrium minority carrier density. Thus the injected minority car-rier density at the depletion edge is,

pn(0) = pneqVAkT . (7.23)

The excess minority carrier density, i.e. excess over equilibrium value, is denotedby a hat symbol,

pn(0) = pn(0)− pn = pn(eqVAkT −1)

pn(0) = pn(eqVAkT −1) (7.24)

For positive VA carrier injection increases exponentially.

Current

pn decreases with distance from the junction due to recombination. As each holerecombines, an electron ‘disappears’ too, so new electrons must flow from theohmic contact on the n-side. The current therefore gradually changes from beingcarried by holes to being carried by electrons.

The current is constant at any cross section. Therefore, in order to obtain I, itis enough to find IP(0) since at x = 0, all the current is carried by holes.

I = Ip(0) = Ip(x)+ In(x)

117 V Casey


Figure 7.14: Current flowing in a pn junction

Since there is a concentration gradient of holes in the x direction, the holecurrent Ip will be due to diffusion (no voltage drop and hence no electric field inthe bulk semiconductor region which would give rise to a drift current).

7.7.3 The diode or rectifier equation.Before obtaining the diode equation, it is necessary to define a diffusion lengthwhich is related to carrier lifetime. Specifically, the diffusion length for a particu-lar carrier type is the distance over which the carrier density will have decreasedby e−1.

L2 = Dτ (7.25)

where D is the diffusion constant and τ is the carrier lifetime. τn and τp for siliconare of the order of 1µs.

The analysis presented here will assume a wide diode. This means that L ismuch shorter than the region into which the carriers are injected. In terms of thediagrams used earlier the diffusion length for this type of diode is as shown below.

Figure 7.15: Carrier injection profile under forward bias

pn(x) = pn + pn(x)

pn(L) = pn(0)e−1 + pn

V Casey 118


Under steady state conditions the current of carriers entering the region dx differsfrom that leaving the region by the amount of recombination of minority holes inthe element.

−1q

dJp

dx=

pn(x)τp

dIp(x)dx

=−qApn(x)

τp(7.26)

where A is the junction area.However, Ip(x) is a diffusion current and hence,

Ip(x) =−qADpd pn(x)

dx,

which gives us by differentiation,

dIp(x)dx

=−qADpd2 pn(x)

dx2 .

Therefore, we may conclude that

d2 pn(x)dx2 =

pn(x)Dpτp

=pn(x)

L2p

(7.27)

The general solution has the form,

pn(x) =C1e−x

Lp +C2ex

Lp .

Imposing boundary conditions:For x >> LP,

pn(x)→ 0⇒C2 = 0

Therefore,

pn(x) =C1e−x

Lp

pn(x) = pn(0)e− x

Lp

By substituting in Ip(x),we get,

Ip(x) = qADppn(0)

Lpe−

xLp

119 V Casey


At x = 0,

Ip(0) = qADppn(0)

Lp.

Recall, pn(0) = pn(eqVAkT −1), therefore,

Ip(0) = qApn

√Dp

τp(e

qVAkT −1) (7.28)

In(0) is found in exactly the same way but with reference to the p-side of thejunction.

In(0) = qAnp

√Dnτn

(eqVAkT −1) (7.29)

I = Ip + In = qA

(pn

√Dpτp

+ np

√Dnτn

)[e

qVAkT −1

](7.30)

At room temperature,

pn =n2

iND

np =n2

iNA

I = qAn2i

(1

ND

√Dpτp

+1

NA

√Dnτn

)[e

qVAkT −1

](7.31)

I = I0

[e

qVAkT −1

](7.32)

where,

I0 = qAn2i

(1

ND

√Dpτp

+1

NA

√Dnτn

)(7.33)

The reverse saturation current, I0, is determined primarily by minority carriersgenerated in the bulk region. I0 is small if ni is small or if NA and ND are large.

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