Semiconductor and superconductor physics %2528korjattu%2529 (1).pdf

8/10/2019 Semiconductor and superconductor physics %2528korjattu%2529 (1).pdf

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Semiconductor and SuperconductorPhysics

Erkki Lahderanta

Lectures follow the books:

Puolijohdeteknologian perusteet (Sinkkonen) and

A. L. Rose-Innes and E. H. Rhoderick: Introduction to Superconductivity


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Contents

1 Classical conductor 3

1.1 Electrical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Hall effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.1.3 Frequency dependence of conductivity . . . . . . . . . . . . . . 8

1.2 Optical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Quantum mechanics 15

2.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Schrodinger equation (1926) . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Classical limit (macroscopic particles) . . . . . . . . . . . . . . . . . . . 21

2.4 Solution of Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . 25

3 Structure of material 29

3.1 Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Atomic bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.3 Solid material and crystal structure . . . . . . . . . . . . . . . . . . . . 31

3.4 Lattice vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4 Free-electron model of metals 34

4.1 Free electron gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.2 Thermal emission current . . . . . . . . . . . . . . . . . . . . . . . . . 404.3 Field emission current . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

5 Energy bands 45

5.1 Formation of bands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5.1.1 Periodical potential . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.2 Models for free energy bands . . . . . . . . . . . . . . . . . . . . . . . . 52

5.2.1 Free electron model . . . . . . . . . . . . . . . . . . . . . . . . . 52

5.2.2 Kronig-Penney-model (1930) . . . . . . . . . . . . . . . . . . . . 54

5.3 Electrons movement in the energy band . . . . . . . . . . . . . . . . . . 57

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5.3.1 Semiclassical equation of motion . . . . . . . . . . . . . . . . . . 57

5.3.2 Electrical conductivity . . . . . . . . . . . . . . . . . . . . . . . 63

5.3.3 Charge carriers in semiconductor: Holes . . . . . . . . . . . . . 64

5.4 Energy bands for various materials . . . . . . . . . . . . . . . . . . . . 665.4.1 Metal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5.4.2 Semiconductors and insulators . . . . . . . . . . . . . . . . . . . 67

5.5 Interaction between electron and wave . . . . . . . . . . . . . . . . . . 68

5.5.1 Disturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.5.2 Optical properties of semiconductor . . . . . . . . . . . . . . . . 71

6 Semiconductor physics 74

6.1 Density of charge carriers . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6.1.1 Density of states in energy band . . . . . . . . . . . . . . . . . . 746.1.2 Electron and hole densities . . . . . . . . . . . . . . . . . . . . . 76

6.1.3 Doped semiconductor . . . . . . . . . . . . . . . . . . . . . . . . 78

6.1.4 Charge carrier density of doped semiconductor . . . . . . . . . . 80

6.2 Mobility, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6.3 Generation and recombination of charge carriers . . . . . . . . . . . . . 85

6.4 Diffusion current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

6.5 Current caused by temperature gradient . . . . . . . . . . . . . . . . . 90

6.5.1 Current density . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

6.5.2 Density of heat flow rate (lampovirrantiheys) . . . . . . . . . . . 91

7 Preparation of semiconducting materials 93

7.1 Crystal growing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

7.2 Selective doping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

7.3 Diffusion and ion-implantation . . . . . . . . . . . . . . . . . . . . . . . 95

7.4 Oxidation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

7.5 CVD-growing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

7.6 Metal growing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

7.7 MBE-method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

8 Spintronics 102

8.1 Magnetic semiconductors . . . . . . . . . . . . . . . . . . . . . . . . . . 102

8.2 CMR and GMR phenomenon . . . . . . . . . . . . . . . . . . . . . . . 103

8.3 Multilayer structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

8.4 Spintronics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

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Chapter 1

Classical conductor

Electrons in metal. Free electron gas and positive metal ions.Current carrier react on electromagnetic field ELECTRICAL and OPTICAL proper-ties.

Density of atoms 5 1022 atoms/cm3 dense medium.Medium Crystal lattice periodic potential modify properties of current carriersand gives energy bands.

Temperature Thermal vibrations.

1.1 Electrical properties1.1.1 Conductivity

Figure 1.1

E= 0motion + collision + motion + collision + . . .Short motion is linear and to arbitrary direction Brownian motion.

Thermodynamic stabilitystatistical methods, specially AVERAGE.Velocity distribution of a classic particle: Maxwell-Boltzmann distribution

f(vx, vy, vz) =Nexp

EkinkT

= Nexp

12

m(v2x+v2y+ v

2z)

kBT

(1.1)

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N = normalization constant;

vx, vy, vz = velocity inx, y and z direction;

m= mass of particle;

kB = Boltzmann constant = 1.380 622 1023 J K1 = 8.6173 105 eV K1.

IfT= 300 K kBT = 0.0258 eV (small).(1.1) is symmetricvelocity average = 0 particle drift and electric current = 0.Thermodynamics and statistical physicsAverage kinetic energy per direction

12

mv2 always positive

= Exkin = Eykin = Ezkin =1

2 kBT

3 dimensionEkin = 32 kBTmost commonspeed = thermal speed vT.

1

2 mv2T =

3

2 kBT vT =

3 kBT

m (1.2)

Numerical values m = 9.1 1031 kg

T= 293 K vT 107cm

s

= 105m

s

= 100km

s

(1mm in 10ns)

Electric field E= 0 F = qEEquation of motion

F =ma mdvdt

= qE (1.3)

E= constant v(t) =v(t0) qm

E(t t0) (1.4)t0= initial moment and v(t0) = initial velocity.

Speed depends linearly on Eand on (t t0).Ift0= previous collision, then (1.4) is working up to next collision. For every electron

we have own (1.4) and own t0. Average of (1.4) over collision times t gives

v

v (1.4)

= qm

E

t t0

= qm

E vD (1.5)

=

t t0

= average time between collision = relaxation time.

In (1.5) is taken into account v(t0) = random

v(t0)

= 0.

Average velocity vD, proportional to electrical field E

vD = drift velocity.

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ELECTRIC CURRENT

Figure 1.2

We investigate time period electrons move distance x= vDElectron density n across the area Ais moving amount of electric charge

Q= qnxA Electric current density J

J= I

A=

Q/

A =

qnxA

A =

qnvD

(1.5)=

q2 n

m E E (1.6)

Here is definition of conductivity

q2

m n (1.7)

This gives resistivity

1 (1.8)Ohmic law (1.6): Electric current density is directly proportional to electric field

strength

Proportionality constant depends on

n= density of current carriers and = relaxation time (= collision frequency)

Numerical values. Metal n amount of atoms 5 1022 electrons/cm3 106 1 cm1

(1.7) 1013 s.Free path, l = free distance = how long is average distance between the collisions.

Thermal speed =vT = 105 m s1 l= vT = 105 m s1 1013 s = 10 nm (293 K).

Much less than size of component.

MOBILITY,

Average velocity (1.5) vD can be written

|vD

|=E (1.9)

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(1.5) gives

=q

m (1.10)

which is MOBILITY of current carrier.Numerical values: 1013 s mobility 200 cm2 V1 s1.Average velocity, vD, is much less than thermal velocity, vT.

IfE= 10 V cm1 = 1kV m1 vD = 20 m s1, but vT 105 m s1.Now we try to take vD from equation of motion.

ma= F, i.e. m dvdt

= qEcould give vD, but collisions??Drude model: we take collision into account with additional friction term.Drude model. Collision (i.e. scattering) is limiting F = qEto times with t < Similar result if we add a term which is similar to friction, i.e. term proportional to

velocity,vDma= F Ffriction mdvD

dt = qE vD (1.11)

Friction-term taken from physics of fluid dynamics (viscosity). In steady state (tbig)

dvDdt

= 0 (1.5) = m

(1.12)

Solution of Eq. (1.11) ifEis switched ON when t= 0

vD(t) =qE

m 1 expt (1.13)

Figure 1.3

Now we see why is called as relaxation time:

After a change (e.g. field ON or OFF), the velocityvD becomes stable after one collision,

that is after t .Equation (1.11) is called Hydrodynamic model, or Drude model. Charges behave like

particles in liquidresult is obtained from average and viscosity.

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1.1.2 Hall effect

=(n, ), meaning depends on mobility, , and electron density, n.

How to measure n? Answer: B.

Lorentz-force

F = qv B Fv and FB (1.14)Total force F = qE+ qv BIn material appears small voltage perpendicular () to direction of current Fvand magnetic field

FB.

Figure 1.4

This voltage is called Hall-voltage.

Drude model m dvD

dt

(1.11)(1.14)= m

vD q(E+vD B) (1.15)

Steady state situation dvDdt

= 0 and current densityJ (1.6)

= qvDn

E= 1

J RHJ B (1.16)

where

RH= 1

nq = Hall-constant (1.17)

Eq. (1.16) has 2 parts

1. Usual ohmic voltage parallel to current

EL= 1

J=

vD

(1.18)

2. EH= field perpendicular to current:

EH= RHJ B = vD B (1.19)

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In steady state this EH compensates the Lorentz-ForceHall-voltage=

UH=EH H= 1qn

I BW

(1.20)

In measurements longitudinal voltage gives

UL= EL L= L I

W H (1.21)

In measurements perpendicular voltage, Hall-voltage, givesn, density of current carries.

These together gives mobility, .

Ratio ofEH and EL (1.18) and (1.19)

EHEL

=B (1.22)

Numerical values. IfB = 1 T = 1 V s m2 and = 600 cm2 V1 s1 EHEL

0.06.

1.1.3 Frequency dependence of conductivity

Electron collision to positive atoms different time windows give different results.Apply electric field

E=E0exp(it) (1.23)

E0= amplitude

= angular frequency (2 = )

this gives average velocity, drift velocity, from (1.4) and conductivity

vD=q

m

1

1 i

E (1.24)

() = DC1 i (1.25)

where DC is given in (1.7).

AC-behaviour by replacing 1i in DC-behaviour

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(1.25) in 2 parts

Re[()] =DC/(1 +22)

Im[()] = DC/(1 +2

2

)

(1.26)

Figure 1.5

Low frequency

< 1

= DCHigh frequency ( 1) Re 0 like 1

2.

Im-part grows linearily with , when 1

, maximum at = 1

.

Im> Re when > 1

wire behaves like impedance. Maxwell-equation

Jtot = J+ E

t (1.27)

J= usual current density with J=() E

E

t = dieletric displacement current (siirrosvirta)

(1.25) means that resistance and inductance are in series

R=

DC A (1.28)

L=

DC A = m

q2n

A (1.29)

A= cross-section; = length.

Second term in Maxwell equation (1.27) is connected to capacitance, so called geomet-

rical capacitancein parallel

C=A

(1.30)

Conductor, wire, is LC-circuit with losses

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Figure 1.6

Geometrical capacitance is charged when voltage is applied.

Electrons behave in inductive way. Resistance from collision (t ).Resonance frequency ofLC-circuit =

p = 1

LC=

q2n

m (1.31)

Name: PLASMA FREQUENCY

1.2 Optical properties

Real part of conductivity meanslosses and damping of the wave =() Differentfrequencies give different reflection and propagation

Maxwell E= Bt

(1.32)

1

B =J+ Et

(1.33)

= permittivity;

= permeability.

Ohm law

J=E (1.34)

If additional chargeElectric field

E= (1.35)

where = charge density.

4 equationsElectromagnetic fields.Wave equation. Start with neutral and homogenous conductor (= 0)

Calculation2E= E

t +

2E

t2 (1.38)

Solution is

E=E0exp[i(k r t)], which is a wave (1.39)

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k r t = phase angle (r, t), E0= amplitude, k= wave vector.Phase angleis constant ifrk, meaning in direction perpendicular to propagation ofthe wave.

In direction of wave propagation (rk) phase angle = const if

r=

k t r0 (t= time)

Interpretation: In direction of propagation the phase moves with velocity, so called

phase velocity (vaihenopeus) vp.

Figure 1.7

vp=r

t =

k (1.40)

During distance of wave length , phase angle change by 2 = 2 r k+ (t)

=0= 2 k= 2

k=2

(1.41)

i.e. lenght of wave vector is connected to

Question: When (1.39) is solution of Maxwell equations?

(1.32) B =B 0exp[i(k r t)] (1.42)with amplitudeB 0 =

1

k E0 (1.43)

In other words: Magnetic field component in plane wave is perpendicular to Eandk.

Moreover, because charge density = 0 (1.35) E= 0 and therefore also

k E= 0 (1.44)

In electromagnetic wave B and Earein comparison to propagationExist 2 directions of polarization.

Wave equation (1.38) requires (calculations)

k2 =i( i) (1.45)

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Figure 1.8

This is called dispersion relation and it gives k= k().

Dispersion = wave beam splits to wave components.

Tangential components(direction of surface) is continuous

incident angle = angle of reflection; =

snell lawk sin = k

sin

(1.46)

ifk

=k

() wave beam will split to components.

Figure 1.9

INSULATOR. = 0, and are real constants (non-imaginary)

(1.45) gives k=

1

v

(1.47)

where v 1

= velocity (1.48)

v usually is written as

v= c

n (1.49)

with c= velocity in vacuum; n= index of refraction (how thick optically).

n=

00=

c

v (1.50)

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(1.47) can be written also in the form

k= n

c (1.51)

Snell law (1.46) can be written in the form

n sin = n

sin

(1.52)

CONDUCTOR. = const, real number

k (1.45)

= 1

v

1 +

i

1/2(1.53)

Wave numberk: Imaginary part gives attenuation, damping (k2 =2 +i).

Low frequency

Imk=

1

21

(1.54)

whereis called skin depth (Finnish languagepenetration depth)High frequency is complex, similar to (1.25)

Figure 1.10

If 1 Eq. (1.53) (is the time between collision).If 1 Im ReConductor behave like insulator and wave numberbecomes

k=1

v

1

p

21/2(1.55)

wherep is plasma frequency (1.31).

If > p k is real (non-imaginary) and NO DAMPING of the wave.

Interpretation: Free electrons cannot follow the electromagnetic fieldsX-rayspenetrate the material.

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In metals p 1015 s1 (2000 nm?).

Reflectance. From air to material with index of refraction = n electric field components

E0 = incoming, E1= reflected, E2 = refracted. Tangential components continuous

E0+E1 = E2 (1.56)

Figure 1.11

Magnetic field components in similar way B0+B1 = B2

(1.43) B0 = 1

k E0

(1.50) k= n

c

B0 proportional to n E0 (1.56)=E0 E1 = nE2 (1.57)

calculations= E1E0 = 1n1+n and E2E0 = 2n+1

Coefficient of reflection =

R =intensity of reflecting waveintensity of incoming wave

=v0|E1|2v0|E0|2 =

1 n1 +n

2(1.58)

Coefficient of trasmission =

T =

intensity of trasmission

intensity of incoming wave

=v0n

n2|E2|2

v0|E0|2 =

4n

(1 +n)2(1.59)

(v0= velocity of incoming wave, in vacuum).

More high frequency X-rays ( 0.1 nm) Macroscopic quantities, such asdielectricity constant, not enough.

Reflection from atoms, not from surfaceBragg law

2d sin = integer (1.60)

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Chapter 2

Quantum mechanics

First experimental results, which were not possible to explain classical physics.

Figure 2.1

2.1 History

A. Black body radiation Classical model: Energy densisty of radiationE 2.Experimentally 2 dependence only with small , with high frequency intensity

0, when increasesClassical model: Radiation energyE is continuous. In temperatureT average

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energy for waveform=

E =

0 dE E exp(E/kt)

0 dEexp(E/kt) =kBT = const (2.1)

Density of waveform 2 Energy density 2.PLANCK(1901): Energy exchange between electromagnetic wave and material

happens as portions, quantas, not continuous and size of energy quanta .Meaning, energy of generated irradiation has form

E=n (2.2)

with n= integer and

= constant.In this case average will be discrete (

0 )

E= n=0 n exp(n/kt)n=0exp(n/kt)

=

exp

kBT

1

(2.3)

Energy quantization explained spectrum of radiation. = Planck constant = h2

.

Irradiation contain photons and each photon has energy = Amount ofphotons in a wave =

n = 1exp

kBT

1

(2.4)

This has a name Bose-Einstein distribution and it describe behaviour of Bosons

(spin is integer number. Spin halfFermion).

Figure 2.2: Experimental arrangement and explanation

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B. Photoelectric effect No current until > 0 = threshold frequency. Light

intensity influence only on size of current.

Threshold frequency not possible to understand with classical model.

Planck idea about quantization of energy in lightEinstein explained thresholdfrequency (1905):

When frequency is enough, only then energy is able to lift electron to height

ofW0

0= workfunction.

turned out to be the same constant as in black body radiation. Energy that

remain after 0, will be kinetic energy

W0 = 12

mv2 (2.5)

C. Specific heat of solid state materials (phonon) Classical physics: Atom vi-

brating around equilibrium positionEnergy of Harmonic oscillator

E=12

M v2x+1

2x2 (2.6)

M= mass, = string constant.

If classicalMaxwell-Boltzmann distribution gives energy averageE =kBT (2.7)

This is xdirection. 3-dimensional E= 3 kBTand final resultSpecific heat =cv =

ET

= 3 kB per particle (2.8)

In high temperatures OK, in low temperatures experiments show that cv 0Einstein explained (1906) low temperature behaviour: Energy for vibrating atom

is also quantized:

E=n (2.9)Quantum is called PHONON

D. Bohr model (1913) Classical mechanics: Particle energy is continuous, therefore

infinite amount of orbits.

Experiments: Emission and absorption spectrum not continuous, sharp lines.

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Figure 2.3

Bohr (1913): Angular momentum is quantized

L= n (2.11)

Only fixed orbits are accepted.Circular motion: v=r p= mv = mr L= pr = mr2

mr2 =n (2.12)

Additionally F =ma

14E0

q2r2

=m2r (2.13)

Possible orbits

En= e4m

2(4E02)2 1

n2 (2.14)

rn=4E02

e2m n2 (2.15)

Valid for atoms with one electron (hydrogen-like). n= quantum number.

E. Compton scattering (Phys. Rev. 22, 409 (1923))

Experiment: X-rays into electrons.

Interaction between photon and electron is like collision of two particles. Also

conservation of momentum.

Classical physics: light has energy (but no particles and therefore no momentum)

Quantum mechanics: light consists of photons and photons have momentum.

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For photon, with m0 = 0 and velocity =c, we haveE=pc andE=

p=

c =

h

with h= 2 (2.16)

Photoelectric effect: All photon energy to electron, photon disappears.

Compton scattering: Part of the photon energy to electron, amount of photons

remain constant.

F. Wave nature of electron (de Broglie, PhD-thesis 1924)

Photon has wave nature and also particle behaviour.

de Broglie: Also particle have both behaviours, including wave nature. Photon

p= h , de Broglie: this is valid also for particles

=h

p =

h

mv (2.17)

Also energy similar to photonE=

=E

(2.18)

Additional result: Simple interpretation of Bohr quantization rule: (2.17) and

(2.18)Length of the orbit must be

2r = n (2.19)

Figure 2.4

Question: de Broglie is theoretical idea. What about experiments?

Answer: 1927 Yes, Electron beam to crystal gave scattering according to Bragg

law

2d sin = n (2.20)

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2.2 Schrodinger equation (1926)

On atomic level the measuring accuracy is not enough for usualmeasuring.

Tunneling microscope: r is ok, but tis not ok.

Fast lasers (attolaser,


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2.3 Classical limit (macroscopic particles)

Quantum mechanics is for atom-size objects. For macroscopic particles the results must

be similar to classical mechanics. We describe macroscopic particles with wavefunction:

First we think about free microscopic particle. 1D Schrodinger gives wavefunction with

normalization:

(x, t) = 1

Lexp

i

kx E(k)t

(2.33)

where k = wavenumber, L = size of volume (1D) and has been replaced by E(k)

,

where

E(k) = energy = 2k2

2m (2.34)

Example:

Probability distribution = ||2 = = 1L

= const.

Linear combination of (2.33)

(x, t) =

dk a(k)exp

i

kx E(k)t

(2.35)

is also solution of Schrodinger. We call this wave packet.

What isa(k)?

It is simply multiplication function. Wave packet (2.35) is solution of Schrodinger and

it describes a particle with all a(k)functions.We can describe classical particle by choosing

a(k) =

1, when

k0 k2

< k V0, now V1< V0).

Intermediate result

=k tankL2 (2.81)

or

= k cotkL

2 (2.82)

where definitionsk=

2m2

E and =

2m2

(V0 E) (see (2.77)).V0 big kL2 = n2 (standing wave), meaning

k=n

L (n N) (2.83)

Final result: Energy levels

En= 22n2

2mL2 (n N) (2.84)

These are deep well energy levels (V0 very high)

IfV0< Energy levels little below (2.84).

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Chapter 3

Structure of material

3.1 Atom

Bohr model is rather simple. More detailed with Schrodinger equation. Hydrogen like

atom possible to describe. Schrodinger equation

2

2m2 e

2

40r

(r) = E(r) (3.1)

gives solution

(r) = nlml(r) =Rnl(r)Ymll (, ) (3.2)

In spherical coordinates (r) separates in two parts: one part depending on distance

and one part depending on direction. Variables for spherical coordinates r, and .

Figure 3.1

Electron energy in hydrogen atom in state n:

En= e2m

2(4E0)2 1

n2 = 13.6 eV

n2 (3.3)

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In solution(3.2)indicesn, l,ml are called quantum numbers, they characterize different

solutions.

For hydrogen the energy depends only on main quantum numbern. In more complicated

atoms the electron energy depends also on l and ml

l= orbital quantum number = 0, 1, 2, . . . , n 1 (3.4)

describes angular momentum of electron.

ml = magnetic quantum number = 0, 1, 2, . . . , l (3.5)

describeszcomponent of angular momentum of electron.Angular momentumL = p

ract as operator and spherical harmonic functionYml

l

(, )

fulfill:

L2Ymll = l(l+ 1)Ymll (3.6)

LzYmll = mlY

mll (3.7)

Here is more clearly seen that l is connected to magnitude ofL andml is connected to

zcomponent ofL.Question:Where iszaxis of an atom?

Answer: Need external interaction, meaning a magnetic field zaxis is definited indirection ofBName for ml = magnetic quantum number.More heavy atomsMany particle problem and correction terms. Energy dependsalso on l E= EnlCorrection terms: In Schrodinger equation is second derivate. Not exact solution, only

approximations. In magnetic field energy depends also on ml (Zeeman spliting).

Names and spectroscopic notations

Occupation of electron shell nlx

x= amount of electrons in subshell nl

n= main quantum number, n= 1, 2, 3...

Orbital quantum number notation with letter

l= 0 s-statel= 1 p-statel= 2 d-statel= 3

f-state

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Example 1.

1s2 describes electron shell of He

Example 2.

Every shell with own notation. Larger atommore shells 1s2

2s2

2p6

is spectroscopicnotation of electron shell for Ne.

3.2 Atomic bonds

A. Ionic bond. Coulomb interaction.

B. Covalent bond. Neighbouring atoms have common electrons.

C. Metallic bond. Electron gas, electron sea, about 1 electron from each atom.8-N-rule.

3.3 Solid material and crystal structure

Lattice points are similar: Each of them have similar environment. Lattice vector:

R= n1a1+n2a2+n3a3 (3.8)

ai = basis of lattice (i= 1, 2, 3) and ni are integers.Exist 14 different lattice structures, for example 3 different cubic structures

Miller indices. Macroscopically crystal looks like planes. Mathematical name for

different planes = Miller indices. Described in lectures.

Direction corresponding to plane: Similar numerical values for Miller indices, but

different kind of brackets; (nkl)= plane and [nkl] direction(nkl).

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3.4 Lattice vibrations

Figure 3.2

Average total energy per atom = 3kBT(3 dimensions). Vibration of atoms is connected

to each other, modelling with ideas from springs, collective behaviour. Nearest neighbour

interactionequation of motion for atom i (ma= F)

M d2ui

dt2 =[(ui+1 ui) (ui ui1)] =[ui+1+ui1 2ui] (3.9)

M= mass of atom i

ui= deviation from stable position

= elasticconstant = stringconstant

Figure 3.3

Differential equation (3.9) has solution as plane wave

un=A exp[i(kna t)] (3.10)

Inserting (3.10) into (3.9)It is required

=

M2sin

ka

2

(3.11)

This has a namedispersion relation, frequency as function ofk. With long wave lengths

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Figure 3.4

(3.10) is elastic wave, meaning sound wave. long k small. When k 0 we have(3.11) in form

= M

2sinka2 ka


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Chapter 4

Free-electron model of metals

In this chapter will be combined Fermi-Dirac distribution with Schrodinger equation.Results will support earlier results of classical conductor.

Figure 4.1

4.1 Free electron gas

Electrons are completely free inside the metal. No interactions. Infinite potential-energy

barriers in edge of the metal.Schrodinger equation has solution as plane wave

= 1

Vexp(i k r) (4.1)

whereV= volume of the metal. Putting (4.1)to Schrodinger equation will give equation

betweenE and k:E(h) =

2k2

2m (4.2)

HereE is kinetic energy.

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Figure 4.2

Figure 4.3

Size of the metalL L L, withL macroscopic. Periodic boundary condition: wave

function similar in opposite edges:

(x= 0, y , z ) = (x= L, y,z) kxL= 2 n (n N)

(In reality = 0 in the edges because infinite energy barrier in the edge. This is difficult

to calculate and it is enough to have periodic boundary condition).

Periodic boundary condition in 3D

k=2

L nxi +nyj+nzk

(4.3)

Difference between neighbouring k-values = 2Lonek-vector, meaning one STATE,

occupies in k-space a volume

Vk=

2

L

3=

(2)3

V (4.4)

Density of states is constant in kspace.Density of state g(E) (How many places for electrons?) Amount of states, dN,

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Figure 4.4

between energy valuesE E+ dEper unit volume:

dN=g(E) dE= 2V

VkVk

(4.4)=

1

2k2 dk

(4.2)=

1

22

2m

h2

3/2 EdE (4.5)

In calculations in more details.

Vk = volume in k-space betweenE andE+dE.Numerical values g(E) 1022 states/(eVcm3).(4.5)

g(

E) = 1

22 2m2 3/2

E

(density of states for free electron)

Figure 4.5

Fermi-Dirac distribution

Maxwell-Boltzmann distribution: Classical, particle density small, individual particle

possible to indentify.

Pauli exclusion principle

Electrons are identical (not possible to identify)

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In state with energy = E, the amount of electrons is on average

f(E) 1expEEF

kBT + 1

(4.6)

This is the possibility that state, with energy = E, is occupied by an electron.Fermi-energyEF

1. Parameter. Strong connection with particle density. Depends on temperature.

2. In zero temperature electron exist up toEF. Electrons take lowest energy.

3. New electron appearing in the systemIt has energy EF.Analysing the distribution

A. States with low energy are all occupied

E EF EF E>> kBT f(E) = 1 = 100%

B. States with high energy are occupied according to Maxwell-Boltzmann

E EF EEIkT = EkT f(E) = exp(E/kT)| exp(. . .) 1 exp(. . .) + 1exp(. . .).

Fermi-Dirac distribution changes strongly only nearEF:Practical estimation for width is

4kBT.

In metal n 1022 el/cm3 and density of states 1022 states/(eVcm3) EF feweVbutkT 0.02 Also in room temperature approximation

f(E) =1, whenE< EF0, whenE> EF (4.7)

Figure 4.6

Approximation(4.7)possible to calculate density of electrons, n, as a function ofenergyE

n=

0

g(E)f(E) dE (4.5)= 122

2m

2

3/2 0

Ef(E) dE (4.7)=

= 122

2m23/2 EF

0

EdE integrate= 132

2mh2

EF3/2 (4.8)37


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Information about where is Fermi-level, meaning we get Fermi-energy EF dependenceon electron density n

EF = 2

2m(32n)2/3 (4.9)

States are occupied up toEF.EF in other unit:

- Fermi wave vector, meaning radius os Fermi-sphere

kF = (32n)1/3 (4.10)

- Electron wave length on the Fermi-surface

F =2kF

= 2 3n1/3 (4.11)

- Fermi velocity, meaning electron velocity on Fermi-surface

vF = kF

m (4.12)

Numerical values:

vF 10classical thermal velocity and F 2distance between electrons.Example: Specific heat

C=U

T (4.13)

Classical U=3

2kBT and Cclassical =

3

2kB per electron (4.14)

Experimentally electron contribution smaller.

Explanation: Fermi-Dirac distribution Only those electrons that are nearEF canabsorb energy. Specific heat due to these electrons only. Width of the area where

changes 4kBT Classical result must be scaled by 4kBTEF

Cel 32

4kBTEF kB (4.15)

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kBT EF Cel Cclassical, meaning specific heat much less than classical prediction.

Figure 4.7

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4.2 Thermal emission current

Figure 4.8

Electrons in potential well. Zero level for potential energy: Electrons free from the well.

Photoelectric effect: Fermi level distance from zero level is =q =W= Workfunction.

Distribution of electron velocity

Electron has state (k) and momentum p= hk, meaning velocity =

v= p

m

k

m (4.16)

One state occupy in k-space a volume (Eq. (4.4)) Vk = (2)3

V Amount of states dN

per unit volume with wave vector in k-volume d3k

dN= 2 d3k

Vk/V = 2

d3k

(2)3/V/V =

1

43d3k (4.17)

Multiplication 2 from spin.

Fermi-Dirac distributionf(E) density dn of electrons with wave vector in k-volumed3k

dn= f(E) dN (4.17)

=

1

431

exp EEFkT+ 1 d3k (4.18)

v= m

k d3v= m

3d3k and kinetic energyE= 2k2

2m = 1

2mv2

dn= 1

43

m

3 d3vexp

12mv

2EFkBT

)

+ 1(4.19)

This is so called thermal emission = current of electrons from metal to vacuum = out

from the potential well.

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Electrons needE= 12

mv2 > EF+q Calculation of current density

J=

dn=

[Eq. (4.19)] dvxdvydvz

Integral from 12

mv2x= EF+q, meaning integral from = vx =

2(EF+q)m

.

Additionally approximationq kT Maxwell-Boltzmann Final result for currentdensity (thermal)

J=C T2 exp

qkBT

(4.24)

with

C= Richardson constant =qmk2B223

= 120A cm2 K2 (4.25)

More clear presentation for current density (4.24) with n= density of electrons

J= qveffn expq

kBT.

(4.26)

This equation has more clear interpretation:

Probability isexpqkBT

that electron has enough big energy, q. These electrons move

with effective velocity

veff=3

8 kBT

EF

2

vF (4.27)

where Fermi velocity is decreased by a factor

kBTEF2

. Probability expq

kBT

is very

small in room temperatureTo have current it is needed heating.Extreme example wolframq =4.5 eV Needs very high temperature On the otherhand it can stand high temperature (light bulb).

Numerical values for wolfram: T= 2500K J= 3 mA cm2.Quantum mechanical correction

Part of electrons are reflected even if enough energy. Current density even more small.

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Figure 4.9

4.3 Field emission current

TEMPERATUREElectric field instead.

(a) (b)

Figure 4.10

In chapter 4.2 electrons go out from metal with help of temperature. Now we see how

electrons go out form metal with help of electric field.

Electric field E >0 toxdirectionElectron has potential energy

V = qEx (4.29)

Potential barrier with heightEF+q.In quantum mechanics barrier replaced by but mathematics is similar Tunneling

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probability (2.79), meaning current, meaning field emission current

T(E) =exp(2L) = exp22m

2 1/2 (EF+q E)3/2

qE (4.30)whereE= kinetic energy.Remarks:

1 Tunneling probability increases whenE increases.

2 Tunneling only from those states that are occupied

2a Occupied states exist up toEF

2b AboveEFstates are occupied with probability exp EkBT= smallTunneling mainly form Fermi-level E= EF Current is

J T(E) exp2

2m

2

1/2(q)3/2

qE

(4.20)

Does not depend on temperature.

Question: When current flows?

Answer: Compare Eq. (4.20)to Eq. (4.26)where probability isexp qkBT Tunnelingcurrent if

2

2m

2

1/2 (q)3/2qE

>

qkBT

(4.21)

Need electric field

E >kBT

q 2

2mq

2

1/2(4.22)

Numerical values: Ifq =4.0 eV and T = 300KE >5 106 V cm1 (Comparablewith electric break down 3 MV m1 = 0.03 106 V cm1)

Practical example: Tunneling microscope. G. Binning and H. Rohrer (1982, Nobel1986).

(4.30) J exp(L)

Resolution

Vertical direction 0.1 AHorizontal direction 1 A

Tunneling, feed back circuit, scanning the sample surface. STM and AFM. Feedback

voltage is the measuring result.

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Figure 4.11

44


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Chapter 5

Energy bands

Up to now two extreme cases

A. Electrons inside the atom. Localized. Sharp energy levels and line spectrum.

Describes behaviour of insulators.

B. Free electrons in metal, chapter 4. Delocalized. EnergyE = 2k22m

, energy is

continuos meaning all energy values are permitted (if the metal piece is macroscopic.

See calculations).

Between these two extreme cases is SEMICONDUCTOR: The energy is continuos

only with some energy values= ENERGY BANDS. Between energy bands existFORBIDDEN BAND, energy gap, with no allowed states for electron.

Roughly:

Electric properties of semiconductor Free-electron behaviour.

Optical properties of semiconductor Behaviour inside energy gap.

5.1 Formation of bands

Figure 5.1

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Start with metal. One atom, for example Na. Then many Na atoms and crystalPotential is sum of individual atoms. When atoms come near, potential barrier between

the atoms is less than energy of 3s-electrons 3selectron is FREEMETAL.

Figure 5.2

In energy-axis appear band widening of sharp line. In this picture 5 atoms and 5

electrons with 3s energy. In reality approximately 1023 electrons (about 1 mol).

Spinevery electron has 2 possibilities, meaning 2 allowed states.Band has double amount of states and band is half filled.Fermi energyEF in the middle of the band, INSIDE THE BAND.

No 3s-electron in potential well Standing wave with0 = 2aand E0 =

22

2ma2 . TwoatomsTwo electrons and two potential wells.

Figure 5.3

If these two atoms are near each otherone energy well, standing wave2(2a) =n (5.1)

meaning 1 = 4a and 2 = 2a. Also two energy levelsEn E0n2 , meaningE1 =E0 andE2= 14E0.Finally, many Na atoms and many potential wells: Nwells near to each other.

46


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Figure 5.4

Standing wave N(2a) =n (n= 1, 2,...,N) (5.2)

Energy levels

En=

1

n2

E0 (n= 1, 2,...,N) (5.3)

Formation of bands in covalent material

Previous analysis was metal. Now we take covalent material. For example silicon

(1s22s22p6)3s23p2.

Outermost electrons are 3sand 3pstates. Totally 4 electrons4 wavefunctions.

A. First we think that Si-atoms are far from each other (but with correct crystal

structure, DIA)Atoms are separated and energy levels are sharp

Figure 5.5

1. B. Atoms closer, lattice constant smaller. Potential barrier between the atoms

reduces (compare with metal)electrons will spread to neighbouring atoms andappears covalent bond, valence bond. Sharp energy levels widen to be bands.

C. Atoms more closer 3s2 and 3p2 bands overlap giving one wide band similar tometal. Total amount of states = 8 = 2

spin

( 13s

+ 33p

) per atom.

D. Atoms more and more closer One wide band deviates to two parts. Theseparts have equal amount of states. Lower band is VALENCE BAND. Upper band

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Figure 5.6

is CONDUCTION BAND. In between NO ALLOWED STATES. Width = energy

gap =EgFORBIDDEN GAP. In reality the lattice constant gives situation D.

Figure 5.7

In valence band the total amount of states (with N= amount of atoms) = 12 (total)

= 12 [2 (1 + 3) N] = 4N 4Nelectrons and 4N statesvalence band is full

and conduction band is empty.

For siliconEg = 1.1 eV.Eg kBT at room temperatureno electrons in conduction band.Interesting electrons (for example possible current carriers) are near the edges, meaningnear to valence band maximum and conduction band minimum. This is correct

regardeless the bands are several eV wide.

5.1.1 Periodical potential

Lattice symmetry: Periodicity

V(r+R) =V(r) (5.4)

withR = n1a1 +n2a2 +n3a3 = lattice vector. a1, a2, a3 = unit vector, ni Z. Quantum

48


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Figure 5.8

number in the case of atom n,l,ml. Now in the case of periodical potential new quantum

numberk. Very useful when talking about energy bands. In physics sense k is wavenumber of electron.

Symmetry. Solution of Schrodinger equation, wavefunction (x), describes electron

and is connected to a specific energy valueE. We move the coordinate system by onestep (a): x =x+a and results should remain similar because V(x+a) =V(x)

d(x)dx

= d(x+a)

dx and

d2(x)dx2

= d2(x+a)

dx2 (5.6)

Also wave function (x) is solution of Schrodinger equation

(x) and (x) are

linearily dependent:

(x+a) = (x) =c(x) (5.7)

Normalization

||2 = 1 |c|2 = 1 c= exp(i) =ei (5.8)

Nonreal wave equation can be given in two parts

(x) =ei(x)

R(x) (5.9)

where Phase function, (x), Amplitude function, R(x), are real.

Periodical potential (symmetry)Physical properties are periodic|(x)|2 = ei(x)R(x)2 = |R(x)|2 R is periodic. Periodicity and (5.9) calculation=

(x+a) =ei[(x+a)(x)](x) (5.10)

Comparing this with (5.7) and (5.8) c has phase angle

= (x+a) (x) (5.11)

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This was result of moving one step. Then we move 2 steps: x=x + 2a result will be

2= (x+ 2a) (x) (5.12)

And general case with n steps

n= (x+na) (x) (5.13)

LINEARITY. Phase and phase function are linear vs. position

(x) =k x(k= const.) (5.14)

And finally, we write wave function (5.9) in general form

(x) =eikxR(x) (5.15)

This is Bloch theorem (1928). Wave function(5.15) has name Bloch function. Felix

Bloch, PhD 1928 about quantum theory of solids. Nobel 1952 about developing the

nuclear magnetic resonance method.

We put Bloch function (5.15) into Schrodinger equationwe get wave equation forR(x), for the periodic amplitude function

2

2md

2

R(x)dx2

ik2

mdRdx

+ 2k22m

+V(x)R(x) = ER(x) (5.16)Wave function depends on ksolutionR(x) depends on kenergyEdepends on k

(x) = nk=eikxRnk(x) (5.17)

E= En(k) (5.18)

k appears in a same way like quantum number of atom.

k = wave number. Very near to each other. Solution (orR) can be understoodanalyzed as a function ofk.

n is index for energy band. Energies with one nvalue form energy band curve as afunction ofk, meaning an energy band.

Next: in 3 dimensions.

3 dimensional analysis. Bloch function (5.15) will be in form

(r) = nk(r) =eikrUnk(r) (5.19)

k= wave vector.Unk = function with periodicity of the lattice: Unk(r+R) =Unk(r) with R= lattice

50


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vector.

Again move the coordinate system with a step of lattice vectorFinal result in 3Dfor Bloch function

nk(r+R) =eik

R

nk(r) (5.20)

and energy values becomes k-dependent

E= E(k) (5.21)

This was traslation symmetry, when moving the coordinate system. Similar way would

be obtained result when using lattice rotation symmetry or inversion symmetry:

En k En k (5.22)

Possiblek-values

First 1 dimension. Boundary conditions, size of the crystal = L = N aand wave function

must fulfill (x+N a) = (x) We get

k2

N a n(n= 0, 1, 2, . . . , N 1) (5.24)

k is wave number and (5.24) is similar to Bohr quantization for free electron.

Figure 5.9

Ifn in (5.24) gets values n= 0, 1, 2, . . . , N 1 All possible phase angles in (5.20).All possible phase anglesAll possible possibilitiesamount ofk-values =N=

amount of atoms. kmax=

2

a and distance between neighbouringkvalues = 2

L =

2

Na .UsuallyN= 0, 1, 2, . . . , N 1 is replaced by n= 0, 1, 2, . . . , N

2

a k

a .

3dimensionalFirst usual cubic structure (SC). All coordinate axis in the same way In k-spacepossiblekvalues form cubic lattice with length of crystal face = 2

L . Full length of the

crystal = 2a

. k-vector has totallyNdifferent values. This has a name First Brillouin

zone, 1st BZ. If original crystal is not simple cubic1st BZ will be different. Also inthese cases k-vector has N values. k-vector takes in k-space a volume

Vk=(2)3

V (5.25)

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Figure 5.10: 1st BZ of SC-crystal

Similar to (4.4) for free electron.

Figure 5.11: 1st BZ of FCC-crystal. Symmetry points and lines are noted in standard way

5.2 Models for free energy bands

5.2.1 Free electron model

Old information: wave function

(x) = 1

Leiqx (5.26)

Energy

E= 2k2

2m (5.27)

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For wave number we use symbol q. Divide in 2 parts:

q= k +K (5.28)

where k= wave number inside 1st BZ and

K= integer 2a

(5.29)

(x) =eikx 1

L eiKx = nk (5.30)

Meaning periodic amplitude function R(x) is described byeikx and index n, that labels

the bands, is in form ofk

Energy bands

E= En(q) = 2

2m(k+K)2 (5.31)

OriginalEn(k) values are moved by value ofK Energies are inside 1st BZ. Physicalresults remains unchanged.

Figure 5.12

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Figure 5.13

5.2.2 Kronig-Penney-model (1930)

We describe lattice by rectangle potential, name Kronig-penney-potential. Calculate

(x) in different parts. Result:

0=A exp(ik0x) +B exp(ik0x); 0 x a1=Cexp(ik1x) +D exp(ik1x); a x a +b (5.32)2=Eexp(ik2x) +Fexp(ik2x); a +b x 2a +b

k0 = k2=

2mE2

and k1 =

2m

(E V0)2

(5.33)

It is possible to calculate coefficients A, B , C, . . . and to obtain condition for existence.

We will not calculate here, but we make simplification

Barrier height V0 and width b 0 in such way that area remains: meaningdelta-functions, -function

V(x) =n

V0b(x na) (5.41)

Coefficient A, B , C , D, . . . will have solution only if

Figure 5.14

Psin aa

+ cos a= cos ka (5.42)

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where

P =ma

2V0b (5.43)

and

= 2mE2

Right-hand side of (5.42) only cos this is between [1, +1] also left-hand side

Figure 5.15: Left-hand side of (5.42)

must be between [1, +1].Exist forbidden values for and therefore for energy E, values where A, B , C , . . .notexist and not existENERGY GAPS.

Maximum of each band in place a = n

En= 22n22ma2

(5.44)

These maxima in same level as free electron energies. Large potential: Areafunction=V0b . Eq(5.42)has solutionsin = 0 and final result will be sharp energy levelslike isolated atoms.

Area of-function =V0blarge, but < Final result will be

En(k) = 22n2

2ma2 1 W

n[1 (1)n cos ka]

(5.47)

where width of the belt, W, should not be large.

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Figure 5.16

This shows that sharp energy levels start to develop to be wide, to be belt. This

approximation has name tight binding approximation.

Weak potential. Other extreme case, electrons almost free. V0b small. Approximations

and series expansion and mathematics

En(k) (5.43)= 22

2m =

2

2m(20+ 20) (5.50)

where0 =k+n 2

a

, meaning k when 1st BZ and n 2a

addition for other Brillouin

zone. In Eq. (5.50) is small correction because of the potential. Without potentialnoandEn=

2202m

= 2k2

2m. Values for in extremes of energy band (edge or center of

1st BZ)

= 02P0a2

(5.51)

Two values in edge of 1st BZ and two values in center of 1st BZ. These give corrections

to energy

E=022P

ma2(5.43)

= 2V0ba

(5.52)

In extremes of energy band, appears a gap 2V0 ba , meaning forbidden gap = Eg = 2V0 ba .Expanded description of band: wave number can change by multiple of 2

asimilar

result.

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Figure 5.17

Inversion symmetry:E(k) = E(k) because cos is symmetric.Derivative antisymmetric:

dE(k)dk

= dE(k)dk

(5.54)

Derivative=velocity Full energy band has antysimmetric velocity distribution (laterin details).

5.3 Electrons movement in the energy band

5.3.1 Semiclassical equation of motion

Bloch-function and quantum mechanics is difficultBetter to build semiclassical.Particle is decribed by wave packet. Electron stay inside one energy band. Wave packet:

- One energy band.

- From Bloch function.

(x, t) =Uk(x)

dka(k)exp

i

kx E(k

)t

(5.55)

General case for electron in energy band.

Simplifications

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Figure 5.18

A. Multiplication functiona(k) thin in k-space. Upper graph.

k2a

, meaning width of BZ (5.56)

B. In the area ofk, the amplitude functionUk , amplitude of

Uk Uk = const. (5.57)

wave packet

(x, t) =Uk(x)

dka(k)exp

i

kx E(k

)t

(5.58)

Important result

1. Wave packet is superposition of plane waves.

2. Same form as for free electron (Eq. (2.35))

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3. BUT. Wave packet must be multiplied with periodic amplitude function from

Bloch-function menaning envelope of plane wave superposition modulatesthe periodic partUk(x). Lower graph.

Width of the wave packet (compare with chapter 2) =

x =2

k (5.59)

and x lattice spacing. x a, but small in the sense that it describes point likeparticle.

Energy in wave packet is energy of energy band. Periodic potential changes dynamics

of particle E(k) no longer quadratic.Velocity will be similar to Eq. (2.39)

v=1

dE(k)dk

(5.60)

and in vector form in 3D

v= 1

kE(k) (5.61)

For equation of motion,F =ma, compare the equation of motion in electromagnetic

field

dk

dt =

q(E+v

B) (5.62)

Take derivative of (5.61)

a= dv

dt

(5.61)= k

1

kE(k)

dk

dt

(5.62)=

1

2kkE(k) (q)[E+v B] (5.63)

F =ma

a= 1m F gives idea about mass, effective mass m (reciprocal tensor)

1

m =

1

2

d2

dk2E(k) (5.64)

in 1D: 1m

= 12

d2

dk2E(k)

Mass, i.e. effective mass, depends on shape (curvature) of energy bandEffectivemass depends on material.

Eq. (5.61) and (5.63) give path of current carriers, if shape ofE(k) is knownSemiclassical equation of motion.

Not need to know periodical potential, nor Bloch function,E(k) is enough. UsuallyE(k) can be modelled using a few parameters.

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Example. Kronig-Penney tight binding approximation (compare (5.47))

E(k) =W(1 cos ka) (5.65)

v(k) (5.61)

= W a

sin ka (5.66)

1 For small k v(k) is linear like in the case of free electron (see Fig. 5.20).

2 There exist maximum in v(k) (see Fig. 5.20).

3 v(k) = 0 in the border of 1st BZ.

Reason: v(k) = 0 antisymmetric

v k=

a= v k=

a

v(k) = 0 in BZ border

Figure 5.19: a)Energy band b)Velocity v(k) c)Effective mass m(k) for tight binding ap-proximation

Effective mass (5.64)

m = 2d2E/dk2

= 2

W a2 cos ka (5.67)

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(See Fig 5.20)

with small k-values m = 2

Wa2

m can be different from m, depends on width Wof the energy band.

k increases m increases. m whenk

2a . m negative whenk >

2a (See Fig5.20) (mass


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Figure 5.21: Electron moving in electric field inside one energy band. Dashed and dottedline describes free electron

and kinetic energy

Ekin= 2q(t)2

2m (5.72)

Tight binding beltvelocity (5.66) v(k) = W

sin ka Integral gives location. k(t) istaken from (5.69)

x(t) =

t0

v[k(t)] dt (5.66)

= . . .= W

q|E|

1 cos

1

q|E|at

(5.73)

cos oscillates also in x-space

Frequency = =1

q

|E

|a=

2

T

(5.74)

Amplitude of oscillations =A = W

q|E| (5.75)

These are called Bloch oscillations:

E= 105 V cm1

W= 1 eV

Amplitude 0.1 m

Electric field causes (aiheuttaa) oscillating current. (Strange). This is because of

periodic potential, which makes electrons to behave atomic-like.Bloch oscillations never observed experimentally.

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Scattering cut electron movement before electron goes through all 1st BZ.

Scattering has relaxation time 1014 . . 1 013 s 1012 s.Extreme experiment: Artificial superlattice with a 5 nm Even in this case no clear

observation.

5.3.2 Electrical conductivity

ScatteringDrude model ((1.11), electrons flow like viscous liquid)Now we insert friction force to equation of motion of wave vector

(5.62) dkdt

= k q[E+v B] (5.76)

= relaxation time of scattering.If only electric field (B = 0)solution (with k= k0 when t = 0)

k(t) =k0exp

t

1

qE

1 exp

t

(5.77)

t < No scattering and behaviour similar to Eq. (5.69) that is k k0

t > k(t ) 1

qE (5.78)

All electrons in the energy band feel change ofk by an amount

k >1

q|E|

Total velocity and current = 0 (J= 0) in a full energy band.

Reason: Energy band inversion symmetric in k-space (E(k) =E(k)) Popula-tion probability (miehitystodennakoisyys) inversion symmetric. Additionally velocity

antisymmetric (v(k) = v(k))Total velocity, of all electron, is = 0.IfE

= 0

electrons move by k= 1

hqE. If energy band full

Population probability

does not change J= 0 Electrical conductivity exist only if there is energy band with incomplete occupation.

Only in this case inversion symmetry is lost and J= 0. J dissymmetry of velocitydistribution.

k small J q 2kv(k0) = 2q2E dE(k0)dk0

(5.79)

Based on energy band structure, it is possible to distinguish between conductor and

insulator.

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Figure 5.22: Electric field and scattering Population becomes dissymetric. Must havepartly filled energy band (metal)

5.3.3 Charge carriers in semiconductor: Holes

Maximum of valence band and minimum of conduction band are important: The charge

carriers are here.

In Fig 5.23 is direct energy gap: Maximum of valence band and the minimum of

conduction band in the same place in k-space. Other possibility: indirect energy gap.

Figure 5.23

Not big influence in electrical properties but optically important.

In the minimum of conduction band the amount of electrons 1020 cm2 non dense(harva) charge carrier gas Maxwell-Boltzmann distribution. Taylor series for energy

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band-curve

E(k) = E(k0) +12(k k0)

2E(k)2kk

k=k0

h(k k0) (5.80)

k0= location of energy minimum.

Electron motion with semiclassical equation of motion. For example effective mass m

with Eq. (5.64). Valence band negative curvature m negative. But it is enough tochange the sign of the chargemotion of hole. New particle: Positive chargePositive mass m (not need m negative).Hole. Properties opposite to electron, but moves like an electron.

E(k) graph for hole by changing E and k k.

Ehole= Ea= Eelectron= Ee (5.81)khole=ka = kelectron= ke (5.82)

For example velocity of hole =

Figure 5.24

va=1

Eaka

=1

Eeke =ve (5.83)

velocities are similar.Wave vector for hole (5.62)

dkadt

= dkedt

= |q|[E+V B]= |q|[E+V B] (5.84)65


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This shows that hole has positive charge. Moreover: (5.81) and (5.82) hole haspositive m.

Electric current in valence band

I= |q| i occupied

V(ki) (5.85)

i occupied statesI= |q|

holes

V(ki) (5.87)

Conclusion

1. For electric current it is enough to analyze empty state and to forget other states,

namely in valence band to forget states that are full.Holes in thin energy layerenough to know m.

2. Exist electrons and holes. Both form non-dense (harva) charge carrier gas. Charge

carriers behave like free particles, but m m

3. Classical free particle model (Returning back to simple ideas after long and

difficult discussion)

5.4 Energy bands for various materials

5.4.1 Metal

1Dsimple parabola (Fig 5.25) usually FCC. 1DMore complicated energy bandstructure (Fig 5.26). Scaling with

E0= 2k20

2m =

(2)2

2ma2 (5.88)

which is energy for electron having wave number k0= 2a.Numerical value: a 0.5 nm E 6.0 eV. In Fig 5.26 the horizontal lines showFermi-energy (EF) for different amount of electrons per unit cell (= valence-number)Analyze of Fig 5.26

Valence z= 1 lowest energy band half empty charge carriers are usual electrons.For example earth metals Li, Na, Ka,. . . and noble metals Cu, Ag, Au.

Valence z= 2 (Cd, Zn) EFin places where degeneracy and system is sensitive to

the influence of periodical potential. In kplace the lowest energy band is empty.

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Figure 5.25

In L-place next energy band is filled upCharge carriers are holes near k-placeof 1st BZ and charge carriers electrons near L-place of 1st BZ.

Earlier in lectures were described meaning ofk andL: Places in 3- dimensional

Brillouin zone, in 3-dimensional reciprocal space.

Valence z= 3 (Al)Lowest energy band is full. Second and third partly occupied.One example. Energy bands fo Cu. Electron structure (Ar)3d104s. Colour

comes from optical transition from 3d-belts toEF. Difference 2 . . 3 eV which isred-yellow wavelength.

5.4.2 Semiconductors and insulators

Ge, Si, GaAs. Most important semiconductors. Valence band similar: each has

maximum in the center of 1st BZ.

Two valence bands, different curvature (m) meaning heavy and light band.

GaAs direct energy gap. (Place of conduction band minimum)

Si has conduction band minimum near BZ border

Ge has conduction band minimum in BZ border

indirect

Rotation symmetry. Energy bands have similar rotational symmetry like the crystal

lattice.

Si has 6 minima (minimum in [100] direction) and Ge has 8 minima (minimumin [111] direction). Equal-energy surface: GaAs sphere but Si and Ge spheroid

Curvature different m has 2 components: Longitudinal and transversal.

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Figure 5.26

Insulator. Similar to semiconductor, butEg wide.SiO2 is most important in technical applications. Amorphic (non crystalline).

5.5 Interaction between electron and wave

5.5.1 Disturbation theory

Earlier electromagnetic field,Econstant,

optical and

kBT

Semiclassical

equations of motion.

Now we investigate eV (Semiconductor) kBT.Schrodinger equation ((2.27)) H0 =i

t

.

Here H0 is operator, H0= 22m 2

x2+V(x)

Giving for free electron

(r, t) = exp

i

ki r Eit

(5.91)

Interaction with wave means that now we have in Schrodinger-equation one additional

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Figure 5.27

INTERACTION-term(H0+ H

) =i

t (5.90)

This additional interaction term

H = A ei(qrt) +ei(qrt) (5.89)is like an disturbation for electron movement ((r, t)).

Eventually mathematics shows that disturbation H gives additional, new plane wave

components with

Ef= Ei (5.93)kf=ki q (5.94)

whereEf and kfare final energy and wave vector andEi and ki are initial, beforeinteraction. qis the wave vector of the wave (for example photon).

Disturbation makes electron to have transition from initial state i to final state f. Eq.

(5.93) and (5.94) are clearly the conservation laws for energy and momentum. Eq.

(5.94) gives momentum if multiplied by .

means disappearance (+) and appearance (-) of an energy quanta. These can becalled absorption/annihilation (+) and emission/creation (-).

For electromagnetic wave is photon. v = f Energy and momentum are con-nected

= vk (5.95)

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Figure 5.28

This can be understood as photon energy band curve v = 3 108 m s1 = 1 eVwhen = 1.24 m and q= 5.1 104 cm1Comparison to wave vector of an electron:

q2

=

2

a

=ka

=k a k (5.96)

Photon wave vector q 0 in comparison to electron wave vector k, becausea 5 104.

Electron wave vector remain unchanged in optical transition.

PHONON. v vsound 1000ms1 105 times less than light velocity.(5.95)Phonon energy very small (


72/108

Figure 5.29

5.5.2 Optical properties of semiconductor

Energy structure gives clear and easy explanation.

Direct energy gap (For example GaAs).

> Eg gives allways transition. One electron jump from valence band to conductionband and one photon disappears. Limit for optical absorption clearly from

Eg.

Inirect energy gap (Si, Ge, etc.)

For transition is needed also a phonon (k= 0) Transition probability smaller(because more complicated).

In practical situation light with Eg is absorbed to GaAs about 100 times strongerthan to Si. Many applications in components.

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Figure 5.30

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Figure 5.31: Energy-wave vector diagram for photon and phonon

Figure 5.32

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Chapter 6

Semiconductor physics

- Amount of charge carriers.

- How to influence to the amount of charge carriers.

- How well they move.

6.1 Density of charge carriers

6.1.1 Density of states in energy band

Charge carriers are near the max of valence band and min of conduction band. Equal

energy-surface is ellipsoid. Energy in conduction band

E(k) = Ec+3

i=1

2(ki koi)22mii

(6.1)

Ec= Energy of the minimum in conduction bandkoi= location of

Ec in k

axis

mii= effective mass in iaxis direction

Ifmii scalarEq. (6.1) looks like free electron energy.Density of states for free electron Eq. (4.5)Now density of states =

gc(E) = 122

2m

2

3/2

E Ec (6.2)

Here mis replaced by m andE

is replaced by (

E Ec).

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Figure 6.1: Near to the min of conduction band

Ifmii values are different m is average. One can get:

m= (m11m22m

33)

1/3 (6.3)

If more than one energy minimum (6.2)must be multiplied by the amount of minima.For valence band in similar way

gv(E) = 122

2m

2

3/2Ev E (6.4)

Ev = Energy of the maximum in valence band

mh = average effective mass for the hole

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Figure 6.2: Near to the max of valence band.

6.1.2 Electron and hole densities

Density of states increases asE Fermi energyEFdefines how big amount of charge

carriers fit into the energy band. UsuallyEF inside forbidden band and usually

Ec EF kBT (6.5)

EF

Ev

kBT (6.6)

This kind of semiconductor is called non-degenerated semiconductor (degeneroitumaton)

In this case charge carriers form non-dense (harva) gas Fermi-DiracMaxwell-

Figure 6.3: Density of states for valence band and conduction band

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Boltzmann distribution

f(E) = 1eEEFkBT

+ 1

calculations fe(E) = 1eEEFkBT

=e (EEF)

kBT (6.8)

Similar approximation in valence band fh(E) =e(EFE)

kBT .

Now we can calculate current carrier density with integral

n=

Ec

g(E)f(E) dE= (6.2)(6.8). . . =Nc e(EcEF)

kBT (6.12)

Nc= 2

mekBT

22

3/2= effective density of states in conduction band (6.13)

In integral is used 0

xex dx= 2

Nc Amount of statesvolume within thin energy layer ofkBT aboveEc.Numerical values:

m = m

T= 300 K

Nc= 10

19 cm3

For holes in similar way:

p= Ev

gv(E)fh(E) dE=Nve(EFEv)

kBT (6.14)

whereNv is similar to Nc but with mh (i.e. effective density of states in valence band).

(6.12) and (6.14) are basic equations for charge carrier densities for non-degenerated

semiconductors.

Multiply (6.12)(6.14)np- rule (EFdisappear, supistuu).

np= NcNvexp

Eg2kBT

= n2i (6.15)

For pure clean semiconductor (without doping): electrons are from valence bandn= p (6.16)

n= p = ni=

NcNvexp

Eg2kBT

(6.17)

For example Si: ni 1010 cm3. Depends exponentially on T. Very small.

(6.12)

(6.14) EF = EFi =1

2(Ec+ Ev) few eV

3

4 kBT0.02 eV ln

mem

h 1

2(Ec+ Ev) (6.18)

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MeaningEFapproximately in the middle of forbidden band.

6.1.3 Doped semiconductor

Changing of charge carrier concentration by doping. Basis for semiconductor technology.

Enough 106 doping, every 1 000 000th atom.

DONOR: Impurity that has one electron more than host atoms (other atoms) in crystal.

This electron becomes free, rise to conduction band.

ACCEPTOR: Impurity that has one electron less than host atoms. Takes one electron

(accepts electron) from enviroment and the result is a hole in valence band.

Donor increases density of electrons and acceptor increases density of holes.

For Si donor impurities are V-group atoms (P, As, Sb) and acceptor impurities are

III-group atoms (B, In, Ga, Al). For compound semiconductor (yhdistepuolijohde), e.g.GaAs, situation is more complex: Donor impurities are VI-group atoms and acceptor

impurities are II-group atoms. IV-group atoms can be donor or acceptor, depending on

if impurity replaces Ga or As. This depends on concentration.

Energy of electron of donor-atom Take one donor-atom. Donor will result in one elec-

tron plus positive im. Weak attraction force

Ed = Ec+ 2k2

2m e

2

4r (6.19)

Small = dielectric constant of the lattice.

2 first terms: Electron in conduction band.

Third term Coulomb potential, r > lattice constant.

Appears energy level belowEc+ 2k22m . System similar to Hydrogen atomEnergy

Figure 6.4: Energy level belowEc+ 2k22m

Ed Ec= 13.6 eV m

m

0

2(6.20)

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Radius

r = 0.53 Am

m

0

= average of distance (6.21)

Here 13.6 eV and 0.53A areE and r for ground state of hydrogen

Figure 6.5: Hydrogen atom

0 10

m

m 0.1

ionization energy = Ed Ec 0.01 eVand orbit radius 50 AkBT 0.025 eV >Ed Ec Electron rises from donor state to conduction band. and 0 are dielctric constant for the lattice and for vacuum.

If 0

only 5 Ed Ec 0.04 eV not suitable for electronics components.For acceptor: Similar hydrogen-like system, Eq. (6.20) and (6.21). Hole disen-

gages(vapautuu) to valence band.

Figure 6.6: Energy level aboveEv

Both energy levels are near to the energy bands. Important energy levels. Exist also

energy levels deep inside the forbidden band:

Reason: Large crystal imperfections. These energy levels are not good.

Deep levels and shallow levels. Impurity levels rather easy to measure (luminescence)

Exist good information.Purity is very important. CLEAN. Reason: 106 doping would be enough.

Density of crystal imperfections and unwanted impurities must bedoping. If notwrong energy levels (virhetiloja) inside energy gap.

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Figure 6.7: Impurity states and ionization energies for Ge,Si and GaAs. D and A meandonor/acceptor state.

Nowaday the best progress exist with Si: Unwanted impurities in the level of1013 at/cm3. This shows the level of quality.

In lectures analyze of Fig. 6.7.

6.1.4 Charge carrier density of doped semiconductor

Basic detail: in room temperature kT > Ed Ec and kT > Ea Ev

n= density of electrons = density of donor impurities = Nd

p= density of holes = density of acceptor impurities = Na

This is the influence of doping. Now we look this in more details.

In Fig 6.8 is shown energies as function of location. E= 0 Energy levels are descibedby horizontal lines.First: ntype semiconductors (with donors).Occupation probability by using Fermi-Dirac-distribution: Probability to have electron

in energy levelEd (i.e. donor state)

f(Ed) = 1eEdEFkT

+ 1(6.22)

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Figure 6.8: Energy band diagram

Ionizated donor atoms with density

N+d = [1 f(Ed)] Nd = Nd

eEdEFkT

+ 1(6.23)

where total amount of donor atoms = Nd. If Fermi-energy EFfar from Ed EFEd 0 N+d =Nd meaning all donor atoms ionizated.Acceptors in similar way. Ionizates acceptor atoms with density

Na =f(Ea)Na= Na

eEaEF

kT

+ 1(6.24)

IfEF far fromEa Ea EF 0 Na =Na meaning all acceptor are ionizated.Calculate Fermi-energyEF.Earlier n= p and Eq. (6.12) and (6.14). Now charge neutrality

n+Na =p+N+d (6.25)

Using this equation, it is possible to calculateEF and n, p, Na, Nd.Calculation rather complex. Approximative calculation in book and here. First n-type

semiconductor (Donors, and ifNa then anyhow Nd Na). In very high temperaturesnand pare high, meaning Na and Nd.In very high temperatures Na and Nd have no influence toEF.In high temperatureEF in center of forbidden gap.Temperature limit T0 where influence of doping disappears: Ni(T0) =

12

Nd. One get:

T0= Ed

2kBln 2NcNvNd

(6.28)

whereNc andNv are effective density of states in conduction band and valence band

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(Eq. (6.13)).

IfNd = 1017 cm3 T0 = 800 K for Si (far above room temperature).

Below this temperature limit one gets density of electrons and holes

n= Nd Na majority charge carriers

and p= n2i

Nd Na minority charge carriers

(6.29)

Result

1. Amount of dominant doping defines the density of dominant charge carriers (now

electrons).

2. Minority charge carrier density is proportional to 1doping density

and

ni(internal

density of charge carriers in undoped semiconductor).

3. Fermi energyEFmoves towards the majority band (now conduction band).

Now with n-type material one gets:

EF = Ec kBT ln NcNd Na (6.30)

New temperature limit T1: whenEF increases up toEc

T1= (Ec Ed)

kBln

NcNdNa

(6.31)IfNd = 10

17 cm3 T1 = 40 K for Si. Donors are no longer ionizated and electrondensityn Nd (frozen)ptype semiconductor. Similar results. Acceptor-doping is dominant. At very hightemperaturesp= n = ni.

Room temperature range: P =Na and n= n2iNa

.

Very low temperature: Charge carrier densities go to zero. Temperature dependence ofcharge carrier (hole) density and temperature dependence of Fermi-energyEF similarto Fig. 6.9.

Ifvery strong doping in semiconductor; Doping more than effective density of states

(Nc or Nv, Eq. (6.13)).

Must be doping 1019 . . 1 021 cm3.Usually is used sentence: Semiconductor is degenerated

Fermi-energy is inside a band Nd EFinside conduction bandNa EFinside valence band82


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(a) Fermi-energy in Si as a function

ofT

whenN

d =N

a = 1016

cm3

(b) Electrodensity in Si when Nd= 1016 cm3

Figure 6.9

In this case behaviour is like ntype orptype metal. This possibility is used in placeswith the contact.

6.2 Mobility,

Mobility of charge carriers is limited by scattering. In principle all disturbances inperiodic potential. Two most important: Phonon (lattice vibrations) and impurities in

doping.

Phonon scattering Similar to photon Fig. 5.30.

Figure 6.10: Phonon scattering

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Relaxation timephononeventually one gets

phonon=qphonon

m (m)5/2T3/2 (6.35)

phonon= relaxation time = time between collision to phonon.

Mobilityphonon increases when temperature decreases.

Mobilityphonon increases also when effective mass decreases.

Impurity scattering

Scattering from impurities is calculated like collision to obstacle (este) with an area, in

this case so called scattering cross-section.

Figure 6.11: Cross-section area of impurity scattering

Eventually one gets for mobility limited by impurities:

Imp= qImp

m T

3/2

mNImp

(6.39)

Imp= relaxation time = time between collisions to impurities.

NImp= Amount of impurities.

In this case mobility Imp increases when temperature increases.

MobilityImp increases when effective mass m decreases.

In practical cases exist both: Phonon and impurities. Indipendent of each othersScattering probabilities can added together

1

=

1

phonon+

1

Imp(6.40)

1

=

1

phonon+

1

Imp(6.41)

This has a name: Mathiessen rule. In low temperatures mobility is limited by impurity

scattering.

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Mobility increases with increasing temperature up to70 . . 100 Kwhen phonon scattering

becomes more important. In room temperature mobility is limited by phonon scattering

and decreases with temperature increases.

In Fig. 6.12 is mobility vs. concentration when T= 300K.

Figure 6.12

Impurity scattering starts to influence when concentration becomes 1016 cm3.Electron mobility in GaAs very good because effective mass m very small (In GaAsm

mc= 0.07).

6.3 Generation and recombination of charge carri-

ers

Amount of charge carriers can be influenced by temperature, doping AND BY LIGHT.

Electron is lifted to conduction band using photon energy Electron-hole-pair.This is GENERATION OF CHARGE CARRIERS

RECOMBINATION: Electron from conductivity band goes down to valence band into

an empty state.

Meaning electron goes to a hole.

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Figure 6.13

In this process, recombination, is emitted one photon; energy difference, E, whichbecomes free, is used for .

Recombination also through impurity levels which are inside energy gap. Specially deep

levels are important. Deep levels are called trapping states (loukkutilat), or sometimes

as recombination center.If energy gap is direct, then recombination is usually directly to valence band. Not

need deep levels. If the energy gap is indirect, then recombination takes place mainly

via deep level. For example in silicon:

G= generation rate (generaatiovilkkaus) =

= transition rate from valence band to conduction band =

= transition/time and volume

R= recombination rate == transition rate form conduction band to valence band

Generation because of external electromagnetic radiation, but also because of black

body radiation.

R= A np (6.42)Reason: Recombination proportional to occupied initial states in conduction band

(n) and to empty final states in valence band (p), meaning product np. A describes

transition probability. Material parameter.

In similar way

G= B NvNc (6.43)

B = transition probability, depending on excitation activity

Nv = density of states in full valence band

Nc= density of states in empty conduction band

Thermodynamic equilibrium: No external radiation, only black-body radiation

G0

and R0 where R0=A n0p0.

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Equilibrium G0=R0 and therefore

G0=R0=A n0p0 (6.15)= An2i (6.44)

In general case: Net recombination =

R= R R0 (6.44)= A(np n2i ) (6.45)

External optical radiationGeneration Goptcharge carrier density changes

n

t =Gopt R= Gopt A(np n2i ) =

p

t (6.46)

This gives movement (transition) of charge carriers between the energy bands. For

example ntype semiconductor;

n= n0+ n and p= p0+ p (6.47)

Recombination in n-p pairs, also generation in n-ppairs n= p. Additionally

n n0 and p n (6.48)

Recombination Rcan be approximated

R= A(np n2i ) =. . . An0p= 1

pp (6.49)

where time constant p = 1An0

= Recombination lifetime for minority charge carriers.

Radiationmore charge carriers.Steady state

tstabilizes. (6.46) and (6.49) Photoconductivity

=qpp+qnn= q(p+n)pGopt (6.52)

Switch OFF radiationRecombination of extra charge carriers

(6.46) p(t) = n(t) = p(0) expt

p

(6.53)

This shows that p is lifetime of minority charge carriers, can be measured using

resistivity (conductivity) measurement.

Fig. 6.14 Shows lifetime in silicon. With usual doping 1016 cm3 10 s.In GaAs direct gap direct recombination and 1 ns with doping level of

1017

cm3

.Unequilibrium between energy bands. Appears because of radiation. Time constant =

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Figure 6.14: Lifetime of minority, charge carriers vs. amount of majority charge carriers

lifetime of recombination. Many orders longer than scattering relaxation, meaning slow.

Unequilibrium is described by quasi Fermi level:

density of electrons and holes

n= Ncexp

Ec EFn

kT

(6.54)

p= Nvexp

EFp Ev

kT

(6.55)

whereEFn andEFp are quasi Fermi level for electrons and holes. If thermodynamicequilibrium EFn = EFp = EF.Usually for majority charge carriers quasi Fermi level not change;EFn = EF.But for minority charge carriers (p p0,p0 small) quasi Fermi level moves towardsthe bands (ifp valence band).

6.4 Diffusion current

In semiconductors often density gradient of current carrierssmall current, diffusioncurrent. Analysis of diffusion gives eventually total current density (left-right + right-

left)Jdiffusion=qDn

n

x (6.62)

where diffusion constant

Dn=1

3v2T=

kBT

m =

kBT

q n (6.63)

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Figure 6.15: Gradient of charge carrier density

here

vT = Thermal velocity =

3 kBT

m

= relaxation time = time between collisions

Dn=kBT

q n is called Einstein relation

One example: Electron density depends exponentially on location: n expxLn

,Ln=

profile parameterJdiffusion= qDn

n

x = qD

nLn n (6.64)

general formulation of current density = J=qvnn average velocity, now diffusionvelocity

vn =DnLn

(6.65)

If not ex

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