Electrical Conduction in Solids What is Electric Current? What is current density? dq q i dt t J dq q Adt At What is Drift? … motion of charges under the influence of an Electric Field What is drift velocity ? … average velocity of charges under an applied E-field NOTE: many of these quantities are VECTORS !! Electrical Conduction in Solids • Which way do negative charges move with respect to the applied E-field? • Which way do positive charges move with respect to the applied E-field? • E A x responsible What type of charges are for current flow in conductors (metals)? v x dx • What type of charges are responsible for current flow in J x semiconductors? • What is mobility μ ? Voltage Gradient and E-Field Ex u x + Vibrating Cu ions (a) V (b) Quick Derivation of Current Density vdx 1 v x1 v x2 v x3 ... vxN N E x n N/V x Δx vdx Δt v A dx Δq e n A Δx Δq e n A vdx Δt Jx (t) e n vdx (t) J x Important Current Density Relationships! • Jx (t) e n vdx (t) e vdx μE μ me • Jx (t) e n μ E Jx (t) σ E σ e n μ • Mobility … expresses how “easy” electrons can move (drift) under the influence of an electric field; it is directly related to relaxation (or mean scattering) time Conductivity … the ability of a solid to conduct electricity (cousin of mobility!) ‘Mean Free Time’ or σ e n μn e p μp J = σE, Ohm’s Law • OHM’S LAW (V=IR!) Temperature Dependence of Resistivity • Electrons are scattered by the vibrating metal ions • Therefore the time between collisions (scattering time) can be affected by the vibration frequency … which implies that the mobility and therefore the conductivity and resistivity are also affected by the vibrating atoms (ions in metals) • What should be the relationship between frequency of vibration and time between collisions ? – • • The higher the freq. the shorter the time because the metal ions are vibrating faster (i.e. higher KE) and therefore colliding more often with moving electrons Shorter scattering times imply lower mobility … lower conductivity and therefore higher resistivity! (the math/derivation is in the book! Let’s look at the expressions…) Temperature Dependence of Mobility and Conductivity eC μd me T me T 1 1 ρ 2 σ e n μd e nC • IMPORTANT to remember the dependence of conductivity … mobility … etc. on T! … i.e. directly/inversely proportional Alloys … and Matthiessen’s Rule • Previous discussion was based on the ASSUMPTION that the material was a pure metal & perfect crystal! … i.e. no impurities no point defects etc. • When impurities are present then one needs to consider two scattering mechanisms: (1) host atoms (2) impurity atoms • The result – Strained 1 region by1 impurity1exerts a F = d(PE) /dx is: scattering force μd μL μI where subscript L is for Lattice and I is for Impurity I • Therefore the effective mobility is lower (than that of a pure metal) • 1 1 ρT ρI The “net” resistivity is: ρ e n μL e n μI T which is known as Matthiessen’s Rule Resistivity versus temperature for pure iron and 4% C steel. Nichrome ?? Temperature Dependence of Resistivity • All non-intrinsic effects (impurities, crystal defects etc.) on the reisstivity can be “summed” up in a “residual” reisistivity term and the resistivity can be re-written as: ρ ρT ρR AT B • NOTE that the thermal vibrations term is temperature dependent!! but the extrinsic component is not. • TEMPERATURE COEFFICIENT OF RESISTIVITY α is the fractional change in resistivity per unit temperature at the reference temperature TO 1 αO ρo • δρ δT To which leads to ρ ρo 1 αo (T - To ) Valid over a narrow temperature range… not a bad approximation BUT be cautious ! Temperature Dependence of Resistivity 100 2000 rT Inconel-825 NiCr Heating Wire 1000 10 Iron Tungsten r T Resistivity(nWm) Resistivity (nWm) Monel-400 Tin 100 Platinum 1 0.1 r T5 0.01 Copper Nickel 0.001 Silver r rR 0.0001 10 100 r (nW m) 3.5 rT 3 2.5 2 1.5 r T5 1 0.5 r = rR 0 0 20 40 60 80 100 T (K) 1000 Temperature (K) 10000 0.00001 1 10 100 Temperature (K) 1000 10000 Solid Solutions • What is a solid solution ? ρ ρT ρR • where ρR includes effect of impurities defects etc. 600 The second term in the above equation is NOT temperature Cdependent u -N i A llo y s • • R e s is tiv ity ( n W m ) 500 Therefore 4 0 0 when forming a solid solution of two metals … i.e. one metal is the “host” (and the other the “impurity”, 3 0 0 addition of the “impurity” metal will cause an then the increase in the resistivity and make the total resistivity 200 less and less temperature dependent ! 100 NOTE: at large “impurity” amounts the material will 0 become an ALLOY! And we need to consider “alloy 0 20 40 60 80 100 effects” 100% C u a t.% N i 100% N i Nordheim’s Rule of Solid Solutions • A simple way to determine the effect on resistivity component – the one due to “alloying impurities” ρI CX(1- X) where X is the fraction of the solute and C is a constant - the Nordheim coefficien t • … and combining Matthiessen’s and Nordheim’s rules then we get ρ ρMATRIX CX(1- X) Solid Solutions vs. Mixtures • Solid solutions are homogenous and mixing takes place at the atomic level … Nordheim’s rule applies. • What happens when the mixture is not homogenous? i.e. 2 phases L – First case: series mixture ρeff χa ρa χb ρb – Second case: parallel mixture A Jx a b Jy σ eff χa σa χb σb L A Mixtures – Continuous Phase w Dispersed 2nd Phase • If we have a mixture where the “host” material is continuous (c) and the added material is dispersed (d) therein … then the following two empirical expressions apply: ρeff 1 1 χd 2 ρc 1 χd 1 χd ρeff ρc 1 2χd for ρd 10 ρc Continuous phase Dispersed phase y A Jx for ρd 0.1 ρc σd σC σ σC Or the general expression : χ σ 2σ C σ d 2σ C L Temperature Rule ??? TA (a) Liquid, L TE a T1 100%A X1 a+L TB b +L b Two phase region ab X (% B) X2 One phase region: b only 100%B Resistivity (b) Mixture Rule Nordheim's Rule rA 0 X1 Fig 2.15 Composition X (% B) rB X 2 100%B (a) The phase diagram for a binary, eutectic forming alloy. (b) The resistivity vs composition for the binary alloy. Thermal Conduction HOT COLD HEAT Electron Gas + Vibrating Cu ions Thermal conduction in a metal involves transferring energy from the hot region to the cold region by conduction electrons. More energetic electrons (shown with longer velocity vectors) from the hotter regions arrive at cooler regions and collide there with lattice vibrations and transfer their energy. Lengths of arrowed lines on atoms represent the magnitudes of atomic vibrations. Thermal Conductivity • • • • • Good Electrical Conductors (like metals) are also good Thermal Conductors … Why ?? … electrons are responsible for the electrical and thermal conductivity … … they pick up the thermal energy from vibrating atoms and transfer to atoms elsewhere … It is easy to think of Thermal Conductivity the same way we think of Electrical Conductivity … dq δV dQ δT J σE ... or I -Aσ ... similarly Q' -Ak dt δx dt δx L 1L 1L Rρ ... θ A σA kA Thermal Conductivity Q = T/ T Hot Cold T Q Q Q A L (b) (a) Conduction of heat through a component in (a) can be modeled as a thermal resistance shown in (b) where Q = T/. • • • σ is the electrical conductivity and k the thermal conductivity Since electrons are responsible for both … the two are related i.e. k CWFL σT … where CWFL is the Weidemann-Franz-Lorenz coefficient Thermal Conductivity • k CWFL σT σ is related to temperature (in a certain range!) as 1/T … the above relationship suggests that k is temperature INDEPENDENT 450 • Since there are no free electrons in non-metals … how does the heat transfer take place ? Thermal conductivity, k (WK-1 m-1) • VIBRATIONS … vibrations will increase at the hot end of a material and the vibrational energy will be transferred along the material depending on the type of bonding Ag 400 Ag-3Cu Cu Ag-20Cu k s = TCWFL 300 Au Al 200 W Be Mg Mo Brass (Cu-30Zn) Ni Bronze (95Cu-5Sn) Steel (1080) Pd-40Ag Hg 100 0 0 10 20 30 40 6 50 60 -1 -1 Electrical conductivity, s, 10 W m 70 Electrical Conductivity Insulators Semiconductors Conductors Many ceramics Superconductors Alumina Diamond Inorganic Glasses Mica Polypropylene PVDF Soda silica glass Borosilicate Pure SnO2 PET SiO2 10-18 Metals Degenerately Doped Si Alloys Intrinsic Si Amorphous Intrinsic GaAs As2Se3 10-15 10-12 10-9 10-6 10-3 100 Conductivity(Wm)-1 Te Graphite NiCr Ag 103 106 109 1012 Charges under the influence ! • What happens when a charge finds itself under the influence of an electric field ? E Charges under the influence ! • What happens when a charge finds itself under the influence of a magnetic field ? B Right vs. Left-Handed Oriented xyz System …. And the Cross product! In order .... i.e. x y z xˆ yˆ zˆ yˆ zˆ xˆ In reverse order zˆ xˆ yˆ i.e. z y x yˆ xˆ zˆ zˆ yˆ xˆ xˆ zˆ yˆ The Hall Effect Example with p-type semiconductor; i.e. holes are the majority charge carriers; apply voltage in x direction i.e. current Ix. apply a B-field in the z-direction. Total Force on the charge carriers due to E and B fields is F q(E v B) The y-component of the force is Fy q(Ey v x Bz ) As the holes flow in the x-direction they experience a force in the y-direction due to the B-field. Holes will accumulate in the -y end of the bar setting up an electric field, i.e. a voltage VAB in the y direction. The net force in the y-direction becomes zero when the two components of the force i.e. due to the electric and due to the magnetic field are equal. The Hall Effect The “setting up” of the E-field in the ydirection is known as the Hall Effect. The voltage VAB is known as the Hall Voltage. This experiment is used to measure the mobility of the charge carriers as explained below: VAB Ey w JX qpνX νX JX qp Fy q(Ey v x Bz ) Fy 0 Ey v xBz 1 RH qp Jx Bz RH JxBz qp ( Ix wt)Bz JxBz IxBz p qEy q(VAB /w) qtVAB The Hall Effect Note: the current and magnetic fields are known quantities since they are externally applied; the hall voltage can be measured. The resistivity r of the sample can also be calculated by measuring the resistance R of the bar. ρL R wt σ σ qpμp μp qp 1 q RH ρ 1 qRH ρ Hall Effect Right hand rule • Index finger -> Current / charge velocity • Thumb -> Lorentz force • Rest of the fingers -> Magnetic filed Key Points • Direction of electric field is also the direction of conventional current. • Hall electric field is set by the movement charge due to Lorentz force in the presence of a magnetic field. B Bz Jy = 0 + + x Ey eE y Jx Ex v hx eE y VAB>0 >0 + + A Jx z vex evhxB z + y evexB z + A Bz V VAB>0 <0 Application of Hall Effect V12<0, Charge? Charge (-)ve V12 ? Charge (-)ve, V12 <0 B? Example #1 The resistivity of an alloy of Cu is 10-7 Ω-m; the dimensions of a rectangular bar of this material are w=4 mm, t=10 mm, and l=50 mm; the thermal conductivity of this alloy is: k = CWFL = 2.44x10 -8 éëWWK -2 ùû sT (300)(2.44x10 -8 )(K )(WWK -2 ) -1 -1 k= = 73.2Wm K -7 (10 )(W - m) Example #2 • Points A, B, and C – – – – What phases are present What is the composition of each phase What is the fraction of each phase present Max solubility of C in Fe to maintain the Ferrite phase Example #3 The resistivity of aluminum at 25 °C has been measured to be 2.72x10-8 Ω-m. The thermal coefficient of resistivity of aluminum at 0 °C is 4.29 x 10-3 K-1. Aluminum has a valency of 3, a density of 2.70 g cm-3, and an atomic mass of 27. (a) what is the resistivity @ -40 °C r (@25°C) = ro (@0°C)[1+ a o (@0°C)(25 - 0)] r (@- 40°C) = ro (@0°C)[1+ a o (@0°C)(-40 - 0)] r (@25°C) [1+ a o (@0°C)(25 - 0)] = r (@- 40°C) [1+ a o (@0°C)(-40 - 0)] Example #4 The resistivity of aluminum at 25 °C has been measured to be 2.72x10-8 Ω-m. The thermal coefficient of resistivity of aluminum at 0 °C is 4.29 x 10-3 K-1. Aluminum has a valency of 3, a density of 2.70 g cm-3, and an atomic mass of 27. (b) what is the thermal coefficient of resistivity @ -40 °C Example #4 The resistivity of aluminum at 25 °C has been measured to be 2.72x10-8 Ω-m. The thermal coefficient of resistivity of aluminum at 0 °C is 4.29 x 10-3 K-1. Aluminum has a valency of 3, a density of 2.70 g cm-3, and an atomic mass of 27. (c) Estimate the mean free time between collisions for the conduction electrons in aluminum at 25 °C, and hence estimate their drift mobility. s nq n n 1 r q me q nq 2 (nq ) r me me 1 ....we need n ... electron density kg 6.022 x10 23 (atoms ) 1[mol ] 2.7[ g ] 103 g 6.022 x10 28 atoms / m 3 3 3 mol 0.027[kg ] [cm ] m 10 2 cm 3electrons . atom 3 x6.022 x10 28 electrons / m 3 18.066 x10 28 electrons / m 3 1 nq 2 r 18.066 x10 29 (1 / 602 x10 19 ) 2 (2.72 x10 8 ) m (9.109 x10 31 ) e 1