JAIEsoE

lVand

andlx(wheredxdVAσIA

lV

AlIV

AlR

IRV

===

==

=≡==⇒

=⇒

=

==

σ

σρρ

ρ

ρ

yresistivitρandtyconductiviσ

)11

resistance*currentvoltage

Extra Notes

Energy Bands A solid can be thought of as being formed by bringing together isolated single atoms. Any single atom will possess a large number of discrete energy levels that can be occupied by the electrons of the atom as shown in Fig. 271(a). Normally the electrons exist in the ground state, occupying only the lowest-lying energy levels. It is, of course, possible to excite the electrons into higher energy levels. Usually only the highest-energy, or valence, electrons will participate in these excitations. Consider first the combination of two atoms. If there were no interaction between the two atoms, the value of each energy level would be the same as for each isolated atom, with the number of levels at a particular energy simply being doubled, as shown in Fig. 271(b).

Because of interactions, however, each previously single energy level is split into two levels, as shown in Fig. 271(c). In a similar fashion, if more atoms were brought together there would be a larger number of splitting of each energy level, one split for each additional atom. Figure 271(d)shows the splitting of energy levels when five atoms are brought together. Because there are of the order of 1023 atoms/ cm3 in solids, each previously single energy level of an isolated atom will be split into an enormous number of parts. Since the values of the energy levels remain approximately the same, the net effect of assembling a large number of interacting atoms is to form bands of practically continuous energy levels, separated by gaps where no electron states exist, as illustrated in Fig. 271(e). 3-Types of Materials …the bands in solids may be filled, partially filled or empty, as shown in Fig. 272. The highest energy band occupied by the valence electrons and the unoccupied band directly above it determine the conduction properties of a crystalline solid. If the band containing the valence electrons is filled, it will be referred to as the valence band, and the next-higher band will be referred to as the

conduction band; if the band containing the valence electrons is not filled, it will be called the conduction band.

A good conductor has a conduction band that is approximately half filled [Fig. 272(a)], or else the conduction band overlaps the next higher band [Fig. 272(b)]. In this situation it is very easy to raise a valence electron to a higher energy level, so these electrons can easily acquire energy from an electric field to participate in electrical conduction. An insulating material has a filled valence band, and the gap to the conduction band is large [Fig. 272(c)]. As a result, electrons cannot easily acquire energy from an electric field, so they cannot participate in electrical conduction. Some materials have a filled valence band, like an insulator, but a small gap to the conduction band [Fig. 272(d)]. At T =0 K the valence band is completely filled and the conduction band is empty, so the material behaves like an insulator. At room temperature, however, some of the electrons acquire sufficient thermal energy to be found in the conduction band, where they can participate in electrical conduction. In addition, these electrons leave behind unfilled “holes” into which other electrons in the valence band can move during electrical conduction. The excitation of electrons into these holes has the net effect of positive charge carriers supporting electrical conduction. The semiconductors just described are called intrinsic semiconductors. It is possible, however, by introducing proper impurities into a material, to control whether electrical conduction will be primarily by electron (negative) or hole (positive) charge carriers. Such “doped” semiconductors are called extrinsic semiconductors, and serve as the foundation for semiconductor devices. If the predominant charge carriers are electrons, the material is called an n-type (for “negative”) semiconductor; while if the holes are the predominant charge carriers, the material is called a p-type (for “positive”) semiconductor. GAUTREAU Ph.D, RONALD. Schaum's Outline of Modern Physics. Blacklick, OH, USA: McGraw-Hill Professional Book Group, 1999. p 311. http://site.ebrary.com/lib/auckland/Doc?id=10015352&ppg=317

Fermi level, Ef highest energy at 0K F(E) Fermi distribution, probability that an electron energy state is full. http://fiselect2.fceia.unr.edu.ar/fisica4/simbuffalo/education/semicon/fermi/functionAndStates/functionAndStates.html µ = Mobility , mean velocity in a field, E, of 1V vd= drift velocity

Evd=µ

µσσ

neEnevJ d

=⇒==

where n = number of charge carriers/volume, e= charge on electron =1.6x10-19C, J= current density, current /area, E= electric filed strength, volts/metre for semiconductor:

he pene µµσ += n= -electrons/volume, p= +holes /volume, µh=mobility of holes Mass Action Law: n.p=constant = ni

2 ni = no. intrinsic carriers /volume

)2

(exp0 kTE

nn gi −= n0=constant, Eg = energy of band gap, k=Boltzmann constant = 1.38x10-23 J/K

similarly, )2

(exp0 kTEg−= σσ σ = conductivity

Crystal structure for many semiconductors is diamond cubic, DC, 8 atoms/unit cell n-type semiconductor , major carrier = “negative“ electron, in exhaustion region n ≅ ND (no, donor atoms/volume & n>>p) p-type semiconductor, major carrier = “positive” hole, in exhaustion region p ≅ NA (no. acceptor atoms /volume & p>>n) pn-Junctions: see: http://fiselect2.fceia.unr.edu.ar/fisica4/simbuffalo/education/pn/biasedPN/index.html

1) open circuit: • get depletion zone, as holes & electrons cancel and left with minority carriers, width

of zone, W=Wp+Wn (NAWp=NDWn), • get barrier potential V0 (~0.7V for Si) • ID, Diffusion current = IS drift current

2) Forward Bias:

http://www-g.eng.cam.ac.uk/mmg/teaching/linearcircuits/diode.html • easier for current to flow, as bias, V, increases, depletion zone width, W, narrows and

barrier potential drops to (V0-V) • ID > Is

• Diode equation:

−= 1)exp(II s kT

eV

3) Reverse Bias: • harder for current to flow, only small drift current form thermally generated electron

hole pairs, until large at reverse bias, get breakdown (V=VZK breakdown Voltage, Zener)

• Barrier potential increase (V0+V) with reverse bias, V • depletion zone widens • IS>ID

Dielectrics p=q.d p=dipole moment, q=charge, d=distance charges move apart P=N.q.d P=Polarisation, N= no. of molecules,etc/volume

dA

CdA

C rεεε 000 & == ε0 = 8.85x10-12CV-1m-1

εr= dielectric constant or relative permittivity

C=capacitance, A=Area of plates, d = separation of plates

000

&CC

QQ

rr === εεεε

dtdVCI

IdtCC

QV

=⇒

== ∫1

tCVI

tVVωω

ωcos

sin

0

0

=∴=

Angle δ is a measure of dielectric power loss

Q=charge, C. related: δεεω tan/ 0

2rEvollossPower = (note: ω=2πf and ε0 = 8.85x10-12CV-1m-1)

f=frequency, E is electric field (V/m), Dielectric Strength: ξmax max voltage gradient a dielectric can withstand before breakdown

maxmax

=

dVξ

Important properties – basic: 1. Relative permittivity, εr Ability of polarisation, or energy storage εr<12 generally used as an insulator εr>12 generally used in capacitor or other charge storage 2. Tangent of lag (loss) angle, tan δ Energy loss 3. Dielectric strength, ξmax Ability to withstand high Voltage (insulation property)

cond

uctiv

ity

ρ ranging

~1024!

conducting materials

resistors/heating elements

semi-conductors

insulators

Intrinsic semiconductor materialsElemental semiconductors:

Si & Ge - most important semiconductor elementsBoron (B), tellurium (Te) and some types of tin (Sn) are also intrinsic semiconductors.

Comparison → Si and Ge______________________________________________Semiconductors Si GeEnergy gap, eV 1.06 (1.1) 0.67Application temp. °C ~200 ~100Electron mobility, μn 0.135 0.39Hole mobility, μp (m2/Vs) 0.048 0.19Intrinsic carrier density, n (/m3) 1.5×1016 2.4×1019

Resistivity, ρ (Ωm) 2300 0.46Density, g/cm3 2.33 5.32Melting point, °C 1410 937Lattice parameter, nm 0.543 0.566

Semiconductor Materials

LED?GaAsPIII-V,VInfrared photo-detector?HgCdTeII,II-VILaser crystals & LED?GaAlAsIII,III-VSolar cells2.4CdSII-VIX-ray photo detectors, solar cells1.5CdTeII-VISolar cells, substrates for other semi-1.3InPIII-VLED and lasers2.2GaPIII-VOptoelectronics, lasers, LED, solar cells1.4GaAsIII-VPhoto-detectors0.67 GeIV

Conduction control (transistors), IC, detectors, solar cells

1.06 eVSiIVApplicationsBandgapSemi-Group

? In ternary semiconductors, the bandgap depends on the ratio of Ga:Al.

Example:

Calculate the electrical conductivity of pure Si at 20°C and (b)

at 150°C.

The energy gap of Si Eg = 1.1 eV, intrinsic carrier concentration

at room temperature ni = 1.5x1016/m3, electron mobility μn =

0.135 m2/Vs, hole mobility μp = 0.048 m2/Vs.

(a) find σ at room temperature (293K):

(b) find σ at 150°C:

Example:

Calculate the electrical conductivity of pure Si at 20°C and (b) at

150°C.

The energy gap of Si Eg = 1.1 eV, intrinsic carrier concentration at

room temperature ni = 1.5x1016/m3, electron mobility μn = 0.135

m2/Vs, hole mobility μp = 0.048 m2/Vs.

(a)At room temperature (293K):

Directly use the relation σ = ni e (μn + μp)

(b) σ at 150°C:

First calculate σ0, than calculate σ423

Example:

Calculate the electrical conductivity of pure Si at 20°C and (b) at 150°C.

The energy gap of Si Eg = 1.1 eV, intrinsic carrier concentration

ni = 1.5x1016/m3, electron mobility μn = 0.135 m2/Vs, hole mobility

μp = 0.048 m2/Vs.

(a) At room temperature (293K):

σ = ni e (μn + μp) = 1.5×1016 × 1.6×10-19 × 0.183 = 4.39×10-4/Ωm

(b) σ at 150°C:

σ = σ0 exp(-Eg/2KT)

σ0 = σ/exp(-Eg/2KT) = 4.39x10-4/exp(-1.1 x 1.6x10-19/2 × 1.38x10-23

× 293) = 1.24x106/Ωm

σ423 = σ0 exp(-Eg/2KT) = 1.24x106 × exp(-1.1 x 1.6x10-19/2 ×

1.38x10-23 × 423) = 3.52 × 10-1 /Ωm

Conductivity of III-V compounds:

(1) wt.% → at.% (or mol.%)(2) 49.99%A + 50.01%B

49.99%A + 49.99%B = 99.98%(A+B) + 0.02% B≡ 99.98%Si + 0.02% doping of B

(3) Use unit cell to calculate number of atoms of Si(4) Use the ratio to calculate the doping atoms(5) Calculate the conductivity

Doping in compound semiconductors:

(1)Compound semi. → melting III and V elements with (slightly) different amounts:

50.1 at.%Ga + 49.9 %As: p-type – more Grp III Ga49.9 at.%Ga + 50.1 %As: n-type - more Grp V As

(NOTE: at%!!! if wt% need to convert to at%)How to calculate the resulting conductivity?

49.9 at.%Ga + 50.1 %As: = 49.9%Ga + 49.9%As + 0.2%AsEquivalent to 49.9%Si + 0.2%As

Example: In a sample of X-type Si, the doping is such that there are 1022 e/m3 and 2.3×1010 h/m3.

(a)What type of this extrinsic semi-?

(b)What are the minority carriers?

(c) Calculate the conductivity of this sample, and σ that is contributed by the majority & minority carriers, respectively.

μn and μp are known.

What type?

Majority carriers?

Minority carriers?

• What type? = n• Majority carriers? = electrons• Minority carriers? = holes

Total conductivity σ = n e μn + p e μp

σn = n e μn = 1022 x 0.135 x 1.6x10-19 = 2.16x102 /Ωm

σp = p e μp = 2.3x1010 x 0.048 x 1,6x10-19 = 1.8x10-10 /Ωm

σ = σn + σp