Introduction to Engineering Materials ENGR2000 Chapter 18: Electrical Properties Dr.Coates
18.2 Ohm s Law V = IR where R is the resistance of the material, V is the voltage and I is the current. l R A l R = ρ A where ρ is the resistivity of the material, l is the length of the specimen and A is the area of cross - section. Measurement of Electrical Resistivity
Recall from Physics 2211 What are units for I, V, R? Does voltage flow through a material? Why/why not? Units of ρ? Does shape of cross-section affect ρ?
18.3 Electrical Conductivity The ease with which a material is capable of conducting an electrical current. 1 σ = ρ whereσ is the electrical conductivity of the material. Units of σ?
Ohm s law in a different form J = σε where J is the current density and Ε is the electric field intensity. J = Ε = I A V l Prove equivalence to V = IR for Class work!
Energy levels in atoms (Review) In a single isolated atom only certain discrete electron energy levels are allowed
18.5 Energy Band Structures in Solids As atoms are brought close together, the electrons are perturbed by the electrons and nuclei of adjacent atoms. Each atomic state is split into a series of closely spaced electron states in the solid electron energy band
Energy bands in solids The energy band structure as a function of the interatomic separation distance wider energy band wider energy band
Energy bands in solids For large separation distances the electrons associated with any atom are independent of those of the other atoms.
Energy bands in solids There is an energy gap E g known as a band gap when the energy levels do not overlap.
Energy bands in solids Energy levels overlap to form an extended energy band.
Four Types of Energy Bands Valence band the highest energy band that is at least partially occupied (eg. Fig 18.4c) Core bands all the bands below the valence band Conduction band the energy band above the valence band Band gap or energy gap forbidden energy range between the valence and conduction bands
Fermi energy The energy corresponding to the highest filled state at 0 K is called the Fermi energy, E f.
Classification of solid materials based on electrical conductivity Conductors Semiconductors Insulators What are conductivity ranges for each? See sec. 18.3!
18.6 Conduction in terms of band and atomic bonding models Free electrons For an electron to become free it must be excited or promoted into one of the empty and available electron states above E f Only electrons with energies >E f may be accelerated in the presence of an electric field and participate in the conduction process. Holes Electronic entity found in semiconductors and insulators, have energies < E f
Metals Band structures in Figs. 18.5a, 18.5b Very little energy is required to promote the Very little energy is required to promote the electrons into the empty states because?
For a metal, occupancy of electron states before and after an electron excitation Fermi energy Free electron
Insulators and Semiconductors Electrons must be promoted across the energy band gap into the empty states to become free Excitation energy is most often in the form of a non-electrical source such as heat or light
For an insulator or semiconductor, occupancy of electron states before and after an electron excitation from the valence band into the conduction band
Insulators and semiconductors Energy band gap > 2 ev Ionic or covalent bonding Valence electrons are tightly bound to individual atoms Energy band gap < 2 ev Covalent bonding Valence electrons are not as strongly bound to the atoms Electrons are easily made free by thermal excitation
Energy bands and charge carriers Electrical conduction requires the presence of empty energy levels that are not too different in energy levels currently occupied by the electrons.
Energy bands and charge carriers An electron jumping from a filled level into a nearby empty level An empty level or a hole is located near the bottom of the band.
Energy bands and charge carriers Transition can be viewed as either 13 electrons each moving up one energy level or the empty level moving down 13 levels.
18.7 Electron Mobility The ease with which the free electrons move through the solid in response to an electric field. Electric field => force on electron Why doesn t electron continually accelerate? Frictional forces-scattering of electrons due to imperfections in crystal lattice, impurity atoms,vacancies, interstitial atoms, dislocations, thermal vibrations
Drift velocity - Average electron velocity in the direction of the force imposed by the electric field. Drift velocity : v d = µ Ε and e e where µ is the electron mobility (frequency of scattering events). e Electrical conductivity : σ = n e µ e where n is the number of free electrons per unit volume = 1.6 10 19 C is the electrical charge on an electron.
18.8 Electrical conductivity of metals Electron mobility (or the electrical conductivity) depends on the nature of the charge carriers (the smaller size of electrons permits them to move easily through the solid) temperature Defects in the crystal structure
Model of an electron moving through a crystal structure v d = at where v d is the drift velocity a is the acceleration & t is the mean time between collisions. constant acceleration
Influence of temperature As temperature is increased atoms gain thermal and kinetic energy mean time between collisions decreases decrease in electron mobility Decrease in electrical conductivity ρ = ρ + at ρ t 0, a 0 = cons. for particular metal
Influence of impurities In the presence of impurities mean time between collisions is decreased decrease in electron mobility Decrease in electrical conductivity ρ = c i i Ac i ( 1 c ) i = impurity concetration ( at% /100) A = composition independent cons.
For a two-phase alloy ρ V ρ i s s = = = ρ V α α + ρ V β β volume fraction individual res. For a metal, the total electrical resistivity equals the sum of thermal, impurity and deformation contributions See Figure 18.8 ρ = ρ + ρ + ρ total i Mattthiessen ' s rule t d
Semiconductors Intrinsic semiconductors Electrical behavior is based on the electronic structure inherent to the pure material. Elemental Si, Ge Extrinsic semiconductors Electrical behavior is dictated by impurity (external) atoms.
Intrinsic Characterized by band structure 18.4b At 0K, completely filled valence band Band gap < 2eV Groups III-V compounds, ex. Gallium Arsenide (GaA) Groups IIB-VIA ex. Cadmium Suplhide (Cds) For these compounds, how might wider separation in electronegativity influence the type of bond and band gap energy? Which of ZnS and CdSe will have a larger band gap energy, E g? why?
Concept of a hole For every electron excited into the conduction band, there is left behind a missing electron in one of the covalent bonds. This missing electron is treated as a positively charged particle called a hole. A hole has the same magnitude of charge as that of an electron.
18.10 Intrinsic Semiconduction
Electron bonding model of electrical conduction in intrinsic silicon - before excitation
Electron bonding model of electrical conduction in intrinsic silicon - after excitation
Intrinsic conductivity Electrical conductivity : σ = n e µ + e p e µ h where p is the number of holes per cubic meter and µ is the hole mobility. σ = n i h For intrinsic semiconductors : n = p = where n e n i i is the intrinsic carrier concentration. Hence, the electrical conductivity : ( µ + µ ) e h
Example 18.1 For intrinsic gallium arsenide, the room temperature electrical conductivity is 10-6 (Ωm) -1 ; the electron and hole mobilities are 0.85 and 0.04 m 2 /Vs respectively. Compute the intrinsic carrier concentration n i at room temperature.
Problem statement : σ = 10 µ µ n i e h =? 6 ( Ωm) = 0.85m = 0.04 m 2 2 1 / Vs / Vs - intrinsic semiconductor Theory : σ = n = p = σ = n e µ + n i e e n i ( µ + µ ) e p e µ For intrinsic semiconductors : h h Hence, the electrical conductivity : Solution : n i = e σ ( µ + µ ) e = 7.0 10 12 h m 3
18.11 Extrinsic Semiconductors - n-type extrinsic semiconduction The addition of a Group V atom, such as P into a Si crystal
n-type extrinsic semiconduction The addition of a Group V atom, such as P into a Si crystal donor level
Extrinsic n-type semiconduction using electron bonding model - before excitation
Extrinsic n-type semiconduction using electron bonding model - after excitation
Energy band for a donor impurity level
Extrinsic n-type conductivity Electrical conductivity : σ = n e µ + p e µ e h For extrinsic n - type semiconductors, # of electrons in the conduction band >> n >> p Hence, the electrical conductivity : # of holes in the valence band : σ n e µ e
18.11 Extrinsic Semiconductors - p-type extrinsic semiconduction The addition of a Group III atom, such as B into a Si crystal
p-type extrinsic semiconduction The addition of a Group III atom, such as B into a Si crystal acceptor level
Extrinsic p-type semiconduction using electron bonding model
Energy band for an acceptor impurity level
Extrinsic p-type conductivity Electrical conductivity : σ = n e µ + p e µ e h For extrinsic p - type semiconductors, # of electrons in the conduction band << n << p Hence, the electrical conductivity : # of holes in the valence band : σ p e µ h
Doping Extrinsic semiconductors are produced from materials that are initially extremely pure. Controlled concentrations of specific donors or acceptors are added.
Temperature Dependence of Carrier Concentration How would band gap affect carrier concentration? Ge vs Si? Why concentrations increase with temperature for intrinsic?
Temperature Dependence of Carrier Concentration Why concentrations constant with temperature in extrinsic region? As dopant level is increased would you expect the temperature at which a semiconductor becomes in trinsic to increase, to remain essentially the same, or to decrease? Why?
Temperature Dependence of Carrier Mobility
Example 18.2 Calculate the electrical conductivity of intrinsic silicon at 423 K.
Problem statement : σ =? T = 423 K - intrinsic silicon Theory : σ = n = p = σ = n e µ + n i e e n i ( µ + µ ) e p e µ For intrinsic semiconductors : h h Hence, the electrical conductivity :
Solution : n i 19 3 ( 423 K ) = 4 10 m
µ µ e h ( 423K ) ( 423 K ) Electrical conductivity of σ = n = i e 0.52 = 0.06 m ( µ e + µ h ) ( Ωm) 1 2 = 0.022 m / Vs 2 / Vs intrinsic Si at 423 K :
Example 18.3 To high-purity silicon is added 10 23 m -3 arsenic atoms. Is this material n-type or p-type semiconductor? Calculate the room temperature electrical conductivity of this material. Compute the conductivity at 100 C.
Arsenic is a Group V element - n-type semiconductor
Problem statement σ =? : n = 10 23 m T = 298 K 3 ( room temperature) - n - type extrinsic semiconductor Theory : σ = n e µ + e p e µ h For n - type extrinsic semiconductors : n >> p Hence, the electrical conductivity : σ n e µ e
Solution : µ e ( 23 3 10 m ) = 0.07 m σ = n e µ = 1120 e 2 / Vs ( Ωm) 1
Problem statement σ =? : n = 10 23 m 3 T = 373 K - n - type extrinsic semiconductor Theory : σ = n e µ + e p e µ h For n - type extrinsic semiconductors : n >> p Hence, the electrical conductivity : σ n e µ e
Solution : µ e ( 373 K ) Electrical conductivity : σ = n e µ e = 0.04m = 640 2 / Vs ( Ωm) 1