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Metals, Semiconductors, and Insulators Every Solid Has Its Own Characteristic Energy Band Structure

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Metals, Semiconductors, and Insulators Every Solid Has Its Own Characteristic Energy Band Structure Metals, Semiconductors, and Insulators Every solid has its own characteristic energy band structure. In order for a material to be conductive, both free electrons and empty states must be available. Metals have free electrons and partially filled valence bands, therefore they are highly conductive (a). Semimetals have their highest band filled. This filled band, however, overlaps with the next higher band, therefore they are conductive but with slightly higher resistivity than normal metals (b). Examples: arsenic, bismuth, and antimony. Insulators have filled valence bands and empty conduction bands, separated by a large band gap Eg(typically >4eV), they have high resistivity (c ). Semiconductors have similar band structure as insulators but with a much smaller band gap. Some electrons can jump to the empty conduction band by thermal or optical excitation (d). Eg=1.1 eV for Si, 0.67 eV for Ge and 1.43 eV for GaAs Conduction in Terms of Band Metals An energy band is a range of allowed electron energies. The energy band in a metal is only partially filled with electrons. Metals have overlapping valence and conduction bands Drude Model of Electrical Conduction in Metals Conduction of electrons in metals – A Classical Approach: In the absence of an applied electric field (ξ) the electrons move in random directions colliding with random impurities and/or lattice imperfections in the crystal arising from thermal motion of ions about their equilibrium positions. The frequency of electron-lattice imperfection collisions can be described by a mean free path λ -- the average distance an electron travels between collisions. When an electric field is applied the electron drift (on average) in the direction opposite to that of the field with drift velocity v The drift velocity is much less than the effective instantaneous speed (v) of the random motion −2 − 1 8− 1 1 2 3 In copperv ≈ 10 cm. swhilev≈10 cm. swherem v= k T 2 e 2 B The drift speed can be calculated in terms of the applied electric field ξ and of v and λ When an electric field is applied to an electron in the metal it experiences a force qξ qξ resulting in acceleration (a) a = me Then the electron collides with a lattice imperfection and changes its direction λ randomly. The mean time between collisions is τ = q ⋅ξ ⋅τ q ⋅ξ ⋅λ v The drift velocity isv = a ⋅τ = = me me ⋅ v If n is the number of conduction electrons per unit volume and J is the current density Combining with the definition of resistivity ivesJ g = nqν =σξ n⋅ q2 ⋅λ q ⋅λ q ⋅τ σ = μ = = me ⋅ v me ⋅ v me q=1.6x10-19C For an electron to become free to conduct, it must be promoted into an empty available energy state For metals, these empty states are adjacent to the filled states Generally, energy supplied by an electric field is enough to stimulate electrons into an empty state “Freedom” Empty States Energy Band States Filled with Electrons Electron Energy Distance Band Diagram: Metal T > 0 EC Fermi “filling” Energy band to function be “filled” EF Conduction band (Partially Filled) E = 0 At T = 0, all levels in conduction band below the Fermi energy EF are filled with electrons, while all levels above EF are empty. Electrons are free to move into “empty” states of conduction band with only a small electric field E, leading to high electrical conductivity! At T > 0, electrons have a probability to be thermally “excited” from below the Fermi energy to above it. Resistivity (ρ) in Metals Resistivity typically increases linearly with temperature: ρt = ρo + αT Phonons scatter electrons. Where ρo and α are constants for an specific material Impurities tend to increase resistivity: Impurities scatter electrons in metals Plastic Deformation tends to raise resistivity dislocations scatter electrons 1 σ= =nqμ ρ The electrical conductivity is controlled by controlling the number of charge carriers in the material (n) and the mobility or “ease of movement” of the charge carriers (μ) Temperature Dependence, Metals There are three contributions to ρ: ρt due to phonons (thermal) ρi due to impurities ρd due to deformation ρ = ρt + ρi+ ρd The number of electrons in the conduction band does not vary with temperature. All the observed temperature dependence of σ in metals arise from changes in μ ScatteringScattering byby ImpuritiesImpurities andand PhononsPhonons Thermal: Phonon scattering Proportional to temperature ρt= ρ o +a T Impurity or Composition scattering Independent of temperature Proportional to impurity concentration Solid Solution ρi=Ac( i 1− i c ) Two Phase ρt =αρVV α+ ρ β β Deformation ρd must= experiment be ally determined Insulator The valence band and conduction band are separated by a large (> 4eV) energy gap, which is a “forbidden” range of energies. Electrons must be promoted across the energy gap to conduct, but the energy gap is large. Energy gap º Eg “Conduction Band” Empty “Forbidden” Energy Gap Electron Energy “Valence Band” Filled with Electrons Distance Band Diagram: Insulator T > 0 Conduction band (Empty) EC Egap EF E Valence band V (Filled) At T = 0, lower valence band is filled with electrons and upper conduction band is empty, leading to zero conductivity. Fermi energy EF is at midpoint of large energy gap (2-10 eV) between conduction and valence bands. At T > 0, electrons are usually NOT thermally “excited” from valence to conduction band, leading to zero conductivity. Conduction in Ionic Materials (Insulators) Conduction by electrons (Electronic Conduction): In a ceramic, all the outer (valence) electrons are involved in ionic or covalent bonds and thus they are restricted to an ambit of one or two atoms. −Eg 2 BTk If Eg is the energy gap, the fraction of electrons in the conduction band is: e A good insulator will have a band gap >>5eV and kBT~0.025eV at room temperature As a result of thermal excitation, the fraction of electrons in the conduction band is ~e-200 or 10-80. There are other ways of changing the electrical conductivity in the ceramic which have a far greater effect than temperature. •Doping with an element whose valence is different from the atom it replaces. The doping levels in an insulator are generally greater than the ones used in semiconductors. Turning it around, material purity is important in making a good insulator. •If the valence of an ion can be variable (like iron), “hoping” of conduction can occur, also known as “polaron” conduction. Transition elements. •Transition elements: Empty or partially filled d or f orbitals can overlap providing a conduction network throughout the solid. Conduction by Ions: ionic conduction It often occurs by movement of entire ions, since the energy gap is too large for electrons to enter the conduction band. Z q.. D μ = The mobility of the ions (charge carriers) is given by: kB . T Where q is the electronic charge ; D is the diffusion coefficient ; kB is Boltzmann’s constant, T is the absolute temperature and Z is the valence of the ion. The mobility of the ions is many orders of magnitude lower than the mobility of the electrons, hence the conductivity is very small: σ n= Z... μ q Example: Suppose that the electrical conductivity of MgO is determined primarily by the diffusion of Mg2+ ions. Estimate the mobility of Mg2+ ions and calculate the electrical conductivity of MgO at 1800oC. Data: Diffusion Constant of Mg in MgO = 0.0249cm2/s ; lattice parameter of MgO a=0.396x10-7cm ; Activation Energy for the Diffusion of Mg2+ in MgO = -19 79,000cal/mol ; kB=1.987cal/K=k-mol; For MgO Z=2/ion; q=1.6x10 C; -23 kB=1.38x10 J/K-mol First, we need to calculate the diffusion coefficient D 2 ⎛ −QD ⎞ cm ⎛ − 79000cal/ mol ⎞ DD= o exp⎜ 0⎟ = . 0239 exp⎜ ⎟ ⎝ kT ⎠ s 1⎝./()cal 987 mol−1800 Kx + K 273⎠ D=1.119x10-10cm2/s Next, we need to find the mobility −19 −10 2 Z.. q( Dcarriers2 /1 6 )( . 10 ion× C )( 1× . 1 10 ) C−9 . cm μ= = −23 1= 12. × 10 k1B . T 38( . 10× )( 1800+ ) 273 J. s C ~ Amp . sec ; J ~ Amp . sec .Volt μ=1.12x10-9 cm2/V.s MgO has the NaCl structure (with 4 Mg2+ and 4O2- per cell) Thus, the Mg2+ ions per cubic cm is: Mg4 2+ ions/ cell n = 6= 4. × ions 1022 / 3 cm 0(. 396× −7cm 10) 3 6σ= 4nZq( 10 . μ = 2 )( ×22 1 )( 6 .× 10−19 )(× 1 .−9 12 ) 10 C. cm2 22σ=. 94 × −6 10 cm3..V s C ~ Amp.sec ; V ~ Amp.Ω σ = 2.294 x 10-5 (Ω.cm)-1 Example: The soda silicate glass of composition 20%Na2O-80%SiO2 and a density of approximately 2.4g.cm-3 has a conductivity of 8.25x10-6 (Ω-m)-1 at 150oC. If the conduction occurs by the diffusion of Na+ ions, what is their drift mobility? Data: Atomic masses of Na, O and Si are 23, 16 and 28.1 respectively Solution: We can calculate the drift mobility (μ) of the Na+ ions from the conductivity expression σ =ni × q× μi + Where ni is the concentration of Na ions in the structure. 20%Na2O-80%SiO2 can0 be 2written 2 as=.M 23At ( × ( 1 ) 16 (+ )) 0 + . 8 ( × (1 . 281 ) + ( 1 )) 2 16 (Na O) -(SiO ) . Its mass can be −1 2 0.2 2 0.8 M 60= ..g 48 mol calculated as: At The number of (Na2O)0.2-(SiO2)0.8 units per unit volume can be found from the density −3 23− 1 ρ2×NA 4(..)(.g cm 6× x 023 mol 10) n = = −1 M At 60..g 48 mol 22 −3 2n=.()()Na 39 × O 102 0 SiO 2.
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