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Electronic Properties of Materials: Conduction - Problems and Solutions

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The resistivity of aluminum at 25oC has been measured to be 2.72 x 10-8 ohm-m. The thermal coefficient of resistivity of aluminum at 0oC 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.

  1. Calculate the resistivity of aluminum at -40oC.
  2. What is the thermal coefficient of resistivity at -40oC?
  3. Estimate the mean free time between collisions for the conduction electrons in aluminum at 25oC, and hence estimate their drift mobility.
  4. If the mean speed of the conduction electrons is about 2.0 x 106 m/s, calculate the mean free path and compare this with the interatomic separation in Al (Al is FCC). What should be the thickness of an Al film that is deposited on an IC chip such that its resistivity is the same as that of bulk Al?
  5. What is the percentage change in the power loss due to Joule heating of the aluminum wire when the temperature drops from 25oC to -40oC?
       





Solution   Given:  {rho_25} = 2.72 x 10-8 Omegam and {alpha_0} = 4.29 x 10-3 K-1

We know that (for metals) rho={rho_0}(1+{alpha_0}{Delta}T), which is an equation of straight line (shown below) between rho and DeltaT, whose slope is given by {rho_0}{alpha_0} and the intercept by {rho_0}.

The slope is constant, so we can write (for all temperatures) {rho_0}{alpha_0} = {rho_T}{alpha_T}

     


  1. Using above equation for resistivity, we can write: {rho_25}={rho_0}(1+25{alpha_0}) and {rho_{-40}}={rho_0}(1-40{alpha_0})

    Or, {{rho_25}}/{{rho_{-40}}} = {1+25{alpha_0}}/{1-40{alpha_0}}
    Or, {rho_{-40}} = {1-40{alpha_0}}/{1+25{alpha_0}}{rho_25}
    doubleright   {rho_{-40}} = 2.03456 x 10-8 Omegam

    Thus, {rho_0} = {rho_{-40}}/{1-40{alpha_0}}
    doubleright   {rho_0} = 2.456 x 10-8 Omegam

  2. From {rho_0}{alpha_0} = {rho_T}{alpha_T}, we can determine {alpha_{-40}}
    doubleright   {alpha_{-40}} = 5.178 x 10-3 K-1

  3. To determine the mean free time between collisions for the conduction electrons (i.e., relaxation time tau), we use the expression for conductivity sigma = {n{e^2}{tau}}/{m}
    Or, {1/rho} = {n{e^2}{tau}}/{m}, where n is the concentration of electrons (i.e., the number of electrons per unit volume), m is the mass of electron (9.1 x 10-31 kg), and rho is the resistivity of aluminum.
    Here, we need to first calculate the concentration of electrons (n), i.e., the number of conduction electrons in one cubic centimeter. It can be very easily calculated by using the atomic mass (27 amu) and the density of aluminum (2.7 g/cm3). But the total number of conduction electrons is 3 times the total number of atoms because each atom contributes 3 conduction electrons (valency of aluminum is 3).

    27 gm = 6.023 x 1023 atoms
    2.7 gm = 6.023 x 1022 atoms

    The density of aluminum is 2.7 g/cm3, which means there are 6.023 x 1022 atoms in a cubic centimeter of aluminum.
    Thus, the total number of conduction electrons per cubic centimeter = 3 x total number of atoms per cubic centimeter
    doubleright  n = 3 x 6.023 x 1022, or n = 1.8069 x 1023 electrons per cubic centimeter or 1.8069 x 1029 electrons per cubic meter.

    Plugging the values in the above equation gives the mean free time (tau) = 7.2327 x 10-15 seconds.

    Drift mobility is given by the equation: {mu_d} = {e{tau}}/{m}
    Plugging the values gives, {mu_d} = 12.717 cm2/Vs.

  4. Mean free path = mean speed x relaxation time
    doubleright  {lambda_mfp} = 2 x 106 x 7.2327 x 10-15 m = 1.44 x 10-8 m or 144 {A}over{o}

    Let's determine the lattice parameter of aluminum (a) using the known information. Aluminum is face centered cubic, which means it has 4 atoms in its unit cell.
    Density rho = 4 x weight of one atom / a3 (density is 2.7 g/cc, and weight of one atom = 27 divided by 6.023 x 1023 = 4.483 x 10-23 gm)
    doublerighta3 = 6.64 x 10-23 cm3
    Or, a = 4.05 x 10-8 cm or 4.05 {A}over{o}

    Let's compare the mean free path ({lambda_mfp}) and the lattice parameter (a).
    The mean free path ({lambda_mfp}) = 144 {A}over{o}, whereas the lattice parameter (a) = 4.05 {A}over{o}, which means a conduction electron in aluminum at room temperature can move approximately 36 unit cells far (144/4.05 = 36) from its original position without scattering.

    Based on the above calculations, the thickness of an Al film deposited on an IC chip should be at least 144 {A}over{o} or approximately 15 nanometers. It should be noted that thin films have higher resistance compared to bulk materials because of more pronounced surface scatterings.


  5. Power (P) = VI = I2R
    Power loss from 25oC to -40oC = [ (I2R-40 - I2R25) / I2R25] x 100%
    or DeltaP = [ (R-40 - R25) / R25] x 100%
    Since R is propotional to rho, we can write DeltaP = [( rho-40 - rho25 ) / rho25] x 100%
    doubleright  DeltaP = [(2.03 - 2.72)/2.72] x 100%
    doubleright  DeltaP = -25.36 % (power loss reduces)