ch2sm

Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001) Chapter 2

2.1

Second Edition ( 2001 McGraw-Hill)

Chapter 2

2.1 Electrical conductionNa is a monovalent metal (BCC) with a density of 0.9712 g cm-3. Its atomic mass is 22.99 g mol–1.The drift mobility of electrons in Na is 53 cm2 V-1 s-1.

a. Consider the collection of conduction electrons in the solid. If each Na atom donates one electron tothe electron sea, estimate the mean separation between the electrons?

b. What is the approximate mean separation between an electron (e-) and a metal ion (Na+), assumingthat most of the time the electron prefers to be between two neighboring Na+ ions. What is theapproximate Coulombic interaction energy (in eV) between an electron and an Na+ ion?

c. How does this electron/metal-ion interaction energy compare with the average thermal energy perparticle, according to the kinetic molecular theory of matter? Do you expect the kinetic moleculartheory to be applicable to the conduction electrons in Na? If the mean electron/metal-ion interactionenergy is of the same order of magnitude as the mean KE of the electrons, what is the mean speed ofelectrons in Na? Why should the mean kinetic energy be comparable to the mean electron/metal-ioninteraction energy?

d. Calculate the electrical conductivity of Na and compare this with the experimental value of 2.1 × 107

Ω-1 m-1 and comment on the difference.

Solutiona If D is the density, Mat is the atomic mass and NA is Avogadro's number, then the atomicconcentration nat is

nDN

MatA

at

= = ××

−

− −( . )( . )

( . 971 2 6 022 10

22 99 10

23 1

3

kg m molkg mol )

-3

1

i.e. nat = 2.544 × 1028 m-3

which is also the electron concentration, given that each Na atom contributes 1 conduction electron.

If d is the mean separation between the electrons then d and nat are related by (see Chapter 1Solutions, Q1.5; this is only an estimate)

d ≈ =× −

1 12 544 101 3 28nat

/ ( . m )3 1/3 = 3.40 × 10-10 m or 0.34 nm

b Na is BCC with 2 atoms in the unit cell. So if a is the lattice constant (side of the cubic unit cell),the density is given by

D

M

N

a

at

A= =

(atoms in unit cell)(mass of 1 atom)

volume of unit cell

2

3

isolate for a: aM

DNat

A

=

= ×

× ×

− −

−2 2 22 99 10

0 9712 10 6 022 10

1 3 3

3 23 1

1 3/ /( .

( . )( . )kg mol )

kg m mol

1

-3

so that a = 4.284 × 10-10 m or 0.4284 nm


2.2

For the BCC structure, the radius of the metal ion R and the lattice parameter a are related by (4R)2

= 3a2, so that,

R = (1/4)√[3a2] = 1.855 × 10-10 m or 0.1855 nm

If the electron is somewhere roughly between two metal ions, then the mean electron to metal ionseparation delectron-ion is roughly R. If delectron-ion ≈ R, the electrostatic potential energy PE between aconduction electron and one metal ion is then

PEe e

do

= − + = − × + ×× ×−

− −

− − −( )( ) ( . ( .

( . )( . )41 602 10 1 602 10

4 8 854 10 1 855 10

19 19

12 1 10πε πelectron ion

C) C)F m m

(1)

∴ PE = -1.24 × 10-18 J or -7.76 eV

c This electron-ion PE is much larger than the average thermal energy expected from the kinetictheory for a collection of “free” particles, that is Eaverage = KEaverage = 3(kT/2) ≈ 0.039 eV at 300 K. In thecase of Na, the electron-ion interaction is very strong so we cannot assume that the electrons are movingaround freely as if in the case of free gas particles in a cylinder. If we assume that the mean KE is roughlythe same order of magnitude as the mean PE,

KE m u PEaverage e average= ≈ = − × −12

2 181 24 10. J (2)

where u is the mean speed (strictly, u = root mean square velocity) and me is the electron mass.

Thus, uPE

maverage

e

≈

= ×

×

−

−

2 2 1 24 109 109 10

1 2 18

31

1 2/ /( .

( . )J)

kg(3)

so that u = 1.65 × 106 m/s

There is a theorem in classical physics called the Virial theorem which states that if the interactionsbetween particles in a system obey the inverse square law (as in Coulombic interactions) then themagnitude of the mean KE is equal to the magnitude of the mean PE. The Virial Theorem states that:

KE PEaverage average= -12

Indeed, using this expression in Eqn. (2), we would find that u = 1.05 × 106 m/s. If theconduction electrons were moving around freely and obeying the kinetic theory, then we would expect(1/2)meu

2 = (3/2)kT and u = 1.1 × 105 m/s, a much lower mean speed. Further, kinetic theory predictsthat u increases as T1/2 whereas according to Eqns. (1) and (2), u is insensitive to the temperature. Theexperimental linear dependence between the resistivity ρ and the absolute temperature T for most metals(non-magnetic) can only be explained by taking u = constant as implied by Eqns. (1) and (2).

d If µ is the drift mobility of the conduction electrons and n is their concentration, then the electrical

conductivity of Na is σ = enµ. Assuming that each Na atom donates one conduction electron (n = nat), wehave

σ µ= = × × ×− − −en ( . 1 602 10 2 10 5 1019 28 3 4 2C)( .544 m )( 3 m V s )-1 -1

i.e. σ = 2.16 × 107 Ω -1 m-1

which is quite close to the experimental value.

Nota Bene: If one takes the Na+-Na+ separation 2R to be roughly the mean electron-electron separationthen this is 0.37 nm and close to d = 1/(n1/3) = 0.34 nm. In any event, all calculations are onlyapproximate to highlight the main point. The interaction PE is substantial compared with the mean thermalenergy and we cannot use (3/2)kT for the mean KE!


2.3

2.2 Electrical conduction

The resistivity of aluminum at 25 °C has been measured to be 2.72 × 10-8 Ω m. The thermal coefficient

of resistivity of aluminum at 0 °C is 4.29 × 10-3 K–1. Aluminum has a valency of 3, a density of 2.70 gcm-3, and an atomic mass of 27.

a. Calculate the resistivity of aluminum at –40 °C.

b. What is the thermal coefficient of resistivity at –40 °C?

c. Estimate the mean free time between collisions for the conduction electrons in aluminum at 25 °C,and hence estimate their drift mobility.

d. If the mean speed of the conduction electrons is ∼1.5 × 106 m s-1, calculate the mean free path andcompare this with the interatomic separation in Al (Al is FCC). What should be the thickness of anAl film that is deposited on an IC chip such that its resistivity is the same as that of bulk Al?

e. What is the percentage change in the power loss due to Joule heating of the aluminum wire when thetemperature drops from 25 °C to –40 °C?

Solution

a Apply the equation for temperature dependence of resistivity, ρ(T) = ρo[1 + αo(T-To)]. We have

the temperature coefficient of resistivity, αo, at To where To is the reference temperature. The two given

reference temperatures are 0 °C or 25 °C, depending on choice. Taking To = 0 °C + 273 = 273 K,

ρ(-40 °C + 273 = 233 K) = ρo[1 + αo(233 K - 273 K)]

ρ(25 °C + 273 = 298 K) = ρo[1 + αo(298 K - 273 K)]

Divide the above two equations to eliminate ρo,

ρ(-40 °C)/ρ(25 °C) = [1 + αo(-40 K)] / [1 + αo(25 K)]

Next, substitute the given values ρ(25 °C) = 2.72 × 10-8 Ω m and αo = 4.29 × 10-3 K-1 to obtain

ρ(-40 C) = (2.72 10 m)[1+ (4.29 10 K )(-40 K)][1+ (4.29 10 K )(25 K)]

-8-3 -1

-3 -1° × ××

Ω

i.e. ρ(-40 °C) = 2.03 × 10-8 Ω m

b In ρ(T) = ρo[1 + αo(T - To)] we have αo at To where To is the reference temperature, for example,

0° C or 25 °C depending on choice. We will choose To to be first at 0 °C = 273 K and then at -40 °C =233 K so that

ρ(-40 °C) = ρ(0 °C)[1 + αo(233 K - 273 K)]

and ρ(0 °C) = ρ(-40 °C)[1 + α-40(273 K - 233 K)]

Multiply and simplify the two equations above to obtain

[1 + αo(233 K - 273 K)][1 + α-40(273 K - 233 K)] = 1

or [1 - 40αo][1 + 40α-40] = 1

Rearranging,


2.4

α-40 = (1 / [1 - 40αo] - 1)(1 / 40)

∴ α-40 = αo / [1 - 40αo]

i.e. α-40 = (4.29 × 10-3 K-1) / [1 - (40 K)(4.29 × 10-3 K-1)] = 5.18 × 10-3 K-1

Alternatively, consider, ρ(25 °C) = ρ(-40 °C)[1 + α-40(298 K - 233 K)] so that

α-40

= [ρ(25 °C) - ρ(-40 °C)] / [ρ(-40 °C)(65 K)]

∴ α-40

= [2.72 × 10-8 Ω m - 2.03 × 10-8 Ω m] / [(2.03 × 10-8 Ω m)(65 K)]

∴ α-40

= 5.23 × 10-3 K-1

c We know that 1/ρ = σ = enµ where σ is the electrical conductivity, e is the electron charge, and µis the electron drift mobility. We also know that µ = eτ/me, where τ is the mean free time between electroncollisions and me is the electron mass. Therefore,

1/ρ = e2nτ/me

∴ τ = me/ρe2n (1)

Here n is the number of conduction electrons per unit volume. But, from the density d and atomicmass Mat, atomic concentration of Al is

nN d

MA

atAl

-- = =

6. mol kg/m

. kg/mol= . m

022 10 2700

0 0276 022 10

23 1 328 3

×( )( )( ) ×

so that n = 3nAl = 1.807 × 1029 m-3

assuming that each Al atom contributes 3 "free" conduction electrons to the metal. Substitute into Eqn.(1):

τρ

= = ×× × ×

m

e ne2

(9.109 10 kg)(2.72 10 m)(1.602 10 C) (1.807 10 m )

-31

-8 -19 2 29 -3Ω

∴ τ = 7.22 × 10-15 s

(Note: If you do not convert to meters and instead use centimeters you will not get the correctanswer because seconds is an SI unit.)

The relation between the drift mobility µd and the mean scattering time is given by Equation 2.5 (intextbook), so that

µ τd

e

e

m

C s

kg= =

×( ) ×( )×( )

− −

−

1 602 10 7 22 10

9 109 10

19 15

31

. .

.

∴ µd = 1.27 × 10-3 m2 V-1s-1 = 12.7 cm2 V-1s-1

d The mean free path is l = uτ where u is the mean speed. With u ≈ 1.5 × 106 m s-1 we find themean free path:

l = uτ = (1.5 × 106 m s-1)(7.22 × 10-15 s) ≈ 1.08 × 10-8 m ≈ 10.8 nm

A thin film of Al must have a much greater thickness than l to show bulk behavior. Otherwise,scattering from the surfaces will increase the resistivity by virtue of Matthiessen's rule.


2.5

e Power P = I2R and is proportional to resistivity ρ, assuming the rms current level stays relativelyconstant. Then we have

[P(-40 °C) - P(25 °C)] / P(25 °C) = P(-40 °C) / P(25 °C) - 1

∴ = ρ(-40 °C) / ρ(25 °C) - 1

∴ = (2.03 × 10-8 Ω m / 2.72 × 10-8 Ω m) - 1

∴ = -0.254, or -25.4%

(Negative sign means a reduction in the power loss).

2.3 TCR and Matthiessen’s ruleDetermine the temperature coefficient of resistivity of pure iron and of electrotechnical steel (Fe with 4%C), which are used in various electrical machinery, at two temperatures: 0 °C and 500 °C. Comment onthe similarities and differences in the resistivity versus temperature behavior shown in Figure 2Q3-1 for

the two materials.

Figure 2Q3-1 Resistivity versus temperaturefor pure iron and 4% C steel.

Solution

The temperature coefficient of resistivity αo (TCR) is defined as follows:

αρ

ρρo

o T

o

o

d

dTo

=

=1 Slope

where the slope is dρ/dT at T = To and ρo is the resistivity at T = To.

To find the slope, we draw a tangent to the curve at T = To (To = 0 °C and then To = 500 °C) and

obtain ∆ρ/∆T ≈ dρ/dT. One convenient way is to define ∆T = 400 °C and find ∆ρ on the tangent line

and then calculate ∆ρ/∆T.

Iron at 0 ˚C ,

Slopeo ≈ (0.23 × 10-6 Ω m - 0 Ω m) / (400 °C) = 5.75 × 10-10 Ω m °C -1

0

0.5

1

1.5

–400 0 400 800 1200

Pure Fe

Fe + 4%C

Temperature ( C)

Tangent

0.57

400 C0.23

0.11

0.96

0.53

0.85

0.68

1.05

400 C0.4

500 C

Resistivity ( m)


2.6

Since ρo ≈ 0.11 × 10-6 Ω m, αo = Slopeo/ρo ≈ 0.00523 °C -1

Fe + 4% C at 0 ˚C ,

Slopeo ≈ (0.57 × 10-6 Ω m - 0.4 × 10-6 Ω m) / (400 °C) = 4.25 × 10-10 Ω m °C -1


Iron at 500 ˚C ,

Slopeo ≈ (0.96 × 10-6 Ω m - 0.4 × 10-6 Ω m) / (400 °C) = 1.40 × 10-9 Ω m °C -1


Fe + 4% C at 500 ˚C ,

Slopeo ≈ (1.05 × 10-6 Ω m - 0.68 × 10-6 Ω m) / (400 °C) = 9.25 × 10-10 Ω m °C -1


*2.4 TCR of isomorphous alloysa. Show that for an isomorphous alloy A%-B% (B% solute in A% solvent), the temperature coefficient

of resistivity αAB is given by

α α ρρAB

A A

AB

≈

where ρAB is the resistivity of the alloy (AB) and ρA and αA are the resistivity and TCR of pure A .What are the assumptions behind this equation?

b. Estimate the composition of the Cu-Ni alloy that will have a TCR of 4 × 10–4 K–1, that is, a TCR thatis an order of magnitude less than that of Cu.

Solution

a By the Nordheim rule, the resistivity of the alloy is ρalloy = ρo + CX(1–X). We can find the TCRof the alloy from its definition

αρ

ρρ

ρalloyalloy

alloy

alloy

= = +1 11

d

dT

d

dTCX Xo[ ( ± )]

To obtain the desired equation, we must assume that C is temperature independent (i.e.the increase in the resistivity depends on the lattice distortion induced by the impurity) so that d[CX(1 -X)]/dT = 0, enabling us to substitute for dρo/dT using the definition of the TCR: αo =(dρo/dT)/ρo.Substituting into the above equation:

αρ

ρρ

α ρalloyalloy alloy

= =1 1d

dTo

o o

i.e. α ρ α ρalloy alloy = o o or α ρ α ρAB AB A A=

Remember that all values for the alloy and pure substance must all be taken at the sametemperature, or the equation is invalid.

b Assume room temperature T = 293 K. Using values for copper from Table 2.1 in Equation 2.17(both in the textbook), ρCu = 17.1 nΩ m and αCu = 4.0 × 10-3 K-1, and from Table 2.3 (in the textbook)


2.7

the Nordheim coefficient of Ni dissolved in Cu is C = 1570 nΩ m. We want to find the composition of

the alloy such that αCuNi = 4 × 10-4 K-1. Then,

ρ α ραalloyCu Cu

alloy

1

1

K n mK

= =−

−( . )( . )

( . )0 0040 17 1

0 0004Ω

= 171.0 nΩ m

Using Nordheim’s rule:

ρalloy = ρCu + CX(1 - X)

i.e. 171.0 nΩ m = 17.1 nΩ m + (1570 nΩ m)X(1 - X)

∴ X2 - X + 0.0879 = 0

solving the quadratic, we find X = 0.11.

Thus the composition is 89% Cu-11% Ni. However, this value is in atomic percent as theNordheim coefficient is in atomic percent. Note that as Cu and Ni are very close in the Periodic Table thiswould also be the weight percentage. Note: the quadratic will produce another value, namely X = 0.86.However, using this number to obtain a composition of 11% Cu-89% Ni is incorrect because the valueswe used in calculations corresponded to a solution of Ni dissolved in Cu, not vice-versa (i.e. Ni was takento be the impurity).

2.5 ConstantanConstantan has the composition 45% Ni-55% Cu. Cu-Ni alloys show complete solid solubility. As analloy, constantan is widely used in resistor applications (up to 500 °C), in strain gauges, and as one of

the thermocouple metal pairs. Given that the resistivity and TCR of copper at 20 °C are 17 nΩ m and0.004 K-1, respectively, and the Nordheim coefficient of Ni dissolved in Cu is 1570 nΩ m, calculate theresistivity ρ, TCR (α), and thermal conductivity κ of constantan and compare the values with the

experimental measurements: ρ(20 °C) = 5 × 10-7 Ω m, α (20 °C) = 2 × 10-5 K-1, κ = 21 W m–1 K–1.What are the reasons for the differences between the calculated and experimental values?

SolutionTake the given composition to be atomic percentage (approximately true for Cu and Ni in

constantan). Otherwise convert 45 wt. % to 47 at. % in calculations.

Given ρo = 17 nΩ m, C = 1570 nΩ m and X = 0.45, the alloy resistivity is

ρalloy = ρo + CX(1–X) = 17 nΩ m + (1570 nΩ m)(0.45)(1–0.45)

∴ ρalloy = 405.6 nΩ m

(X = 0.47 gives ρalloy = 408.1 nΩ m). The experimental value is 500 nΩ m. The listed value of C in thetables is therefore not very good in predicting the experimental value. In addition, C values are typicallydetermined for very dilute alloys.

αalloy = =−α ρ

ρCu Cu

alloy

1K n mn m

( . )( )( . )

0 004 17405 6

ΩΩ

= 1.68 × 10-4 K-1

This is about an order of magnitude greater than the experimental value. Using ρalloy = 500 nΩ m,

we obtain αalloy = 1.36 × 10-4 K-1, only about 20% lower than the theoretical value. Obviously the

expression ρalloyαalloy = αAρA does not work well in this case.


2.8

The conductivity is σalloy = 1/ ρalloy. The thermal conductivity of the alloy can be found from theWiedemann-Franz-Lorenz law,

κalloy = σalloyTCWFL = (405.6 × 10-9 Ω m)-1(20 + 273 K)(2.44 × 10-8 W Ω K-2)

i.e. κalloy = 17.6 W m-1 K-1

The experimental value is about 21 W K-1 m-1.

Addendum:

We know the resistivity to be ρalloy = ρo + CX(1–X). If we were to plot ρalloy -ρo versus X(1 - X),we would obtain a straight line with slope equal to C, the Nordheim coefficient. Table 2Q5-1 lists valuestaken from Figure 2Q5-1 for resistivity of Ni-Cu alloy in varying compositions, along with correspondingvalues of X(1 - X):

Table 2Q5-1

% of Ni in Ni-Cu

at. % of Ni inNi-Cu

Resistivity ofalloy (nΩ m)

Change inres is t iv i ty(ρalloy - ρo)

X(1-X)

0 0 1 7 0 02 2.162 5 0 3 3 0.02116 6.464 100 8 3 0.0605

1 0 10.74 191 174 0.09581 1 11.80 150 133 0.1042 0 21.30 266 249 0.1682 2 23.39 300 283 0.1793 0 31.69 375 358 0.2164 3 44.96 500 483 0.2474 5 46.97 500 483 0.249

Using these values, we obtain the following plot:

X (1 - X)

600

500

400

300

200

100

0

-1000 0.05 0.1 0.15 0.2 0.25 0.3

Cha

nge

in R

esis

tivity

(n

m)

Figure 2Q5-1 Plot of Change in resistivity versus X(1 - X)

The plot reveals a straight line with slope equal to 1882.3 nΩ m. This is close to the value of Cused before. Using this new value of C, the calculated values would come out closer to the experimentalvalues given.


2.9

2.6 Experimental Nordheim coefficient for Pd in AgSilver and palladium form a complete solid solution and, as alloys, they are widely used in variouselectrical switches. Pd improves the wear resistance of Ag and the alloy has good fabricability. Table2Q6-1 shows the resistivity of Ag-Pd alloys for various compositions. Using a suitable plot, obtain theNordheim coefficient for Pd in Ag. [Note: The atomic masses of Pd and Ag are very close, 106.42g/mol and 107.87 g/mol, respectively. You can therefore assume that the atomic % = weight %.]

Table 2Q6-1

wt.% Pd in Ag 0 1 3 10 30Resistivity, nΩ m 16.1 21.8 38.3 63.9 149.3

Solution

Using the data given in Table 2Q6-1, plot ∆ρ versus [X(1 - X)], where ∆ρ is the increase in

resistivity due to alloying, namely ∆ρ = ρ - ρo, and X is the atomic fraction of Pd ≈ weight fraction of Pd

(ρo = 16.1 nΩ m from Table 2Q6-1). Since the increase in resistivity is ∆ρ = C[X(1 - X)], plotting ∆ρversus [X(1 - X)] should yield a straight line graph with slope equal to the Nordheim coefficient C.

Residual Resistivity, n m

X(1 - X)

0 0.05 0.1 0.15 0.2 0.250

20

40

60

80

100

120

140

Figure 2Q6-1 Plot of change in resistivity versus [X(1 - X)]. The slope of the line is the Nordheimcoefficient.

The equation of the best fit line is

∆ρ = 623.76[X(1 - X)] - 0.5102

Therefore the Nordheim coefficient is C = 623 nΩ m or 62.3 µΩ cm.


2.10

2.7 Electrical and thermal conductivity of In

Electron drift mobility in indium has been measured to be 6 cm2 V-1 s-1. The room temperature (27 °C)

resistivity of In is 8.37 ×10 -8 Ω m, and its atomic mass and density are 114.82 amu or g mol–1 and 7.31g cm-3, respectively.

a. Based on the resistivity value, determine how many free electrons are donated by each In atom in thecrystal. How does this compare with the position of In in the Periodic Table (Group IIIB)?

b. If the mean speed of conduction electrons in In is 1.74 ×10 8 cm s-1, what is the mean free path?

c. Calculate the thermal conductivity of In. How does this compare with the experimental value of81.6 W m-1 K-1?

Solution

a From σ = enµd (σ is the conductivity of the metal, e is the electron charge, and µd is the electrondrift mobility) we can calculate the concentration of conduction electrons (n):

ne d

= = ×× ×

− −

− −σµ

( . )( .

8 37 101 602 10 10

8 1

19 4 2

Ω mC)(6 m V s )-1 -1

i.e. n = 1.243 × 1029 m-3

Atomic concentration nat is

ndN

MatA

at

= = × ××

−

− −( . )( . )

( . 7 31 10 6 022 10

114 82 10

3 23 1

3

kg m molkg mol )

-3

1

i.e. nat = 3.834 × 1028 m-3

Effective number of conduction electrons donated per In atom (neff) is:

neff = n / nat = (1.243 × 1029 m-3) / (3.834 × 1028 m-3) = 3.24

Conclusion: There are therefore about three electrons per atom donated to the conduction-electron sea in the metal. This is in good agreement with the position of the In element in the PeriodicTable (III) and its valency of 3.

b If τ is the mean scattering time of the conduction electrons, then from µd = eτ/me (me = electronmass) we have:

τ µ= = × ××

− −

−d em

e

( m V s )( .109 kg)C)

-1 -16 10 9 101 602 10

4 2 31

19

( .

= 3.412 × 10-15 s

Taking the mean speed u ≈ 1.74 × 106 m s-1, the mean free path (l) is given by

l = uτ = (1.74 × 106 m s-1)(3.412 × 10-15 s) = 5.94 × 10-9 m or 5.94 nm

c From the Wiedemann-Franz-Lorenz law, thermal conductivity is given as:

κ = σTCWFL = (8.37 × 10-8 Ω m)-1(20 ˚C + 273 K)(2.44 × 10-8 W Ω K-2)

i.e. κ = 85.4 W m-1 K-1

This value reasonably agrees with the experimental value.


2.11

2.8 Electrical and thermal conductivity of Ag

The electron drift mobility in silver has been measured to be 56 cm2 V-1 s-1 at 27 °C. The atomic massand density of Ag are given as 107.87 amu or g mol–1 and 10.50 g cm-3, respectively.

a. Assuming that each Ag atom contributes one conduction electron, calculate the resistivity of Ag at 27°C. Compare this value with the measured value of 1.6 × 10-8 Ω m at the same temperature andsuggest reasons for the difference.

b. Calculate the thermal conductivity of silver at 27 °C and at 0 °C.

Solutiona Atomic concentration nat is

ndN

MatA

at

= = × ××

−

− −( . )( . )

( . 10 50 10 6 022 10

107 87 10

3 23 1

3

kg m molkg mol )

-3

1

i.e. nat = 5.862 × 1028 m-3

If we assume there is one conduction electron per Ag atom, the concentration of conductionelectrons (n) is 5.862 × 1028 m-3, and the conductivity is therefore:

σ = enµd = (1.602 × 10-19 C)(5.862 × 1028 m-3)(56 × 10-4 m2 V-1s-1)

∴ σ = 5.259 × 107 Ω-1 m-1

and the resistivity ρ = 1/σ = 19.0 nΩ m

The experimental value of ρ is 16 nΩ m. We assumed that exactly 1 "free" electron per Ag atomcontributes to conduction - this is not necessarily true. We need to use energy bands to describeconduction more accurately and this is addressed in Chapter 4 (in the textbook).

b From the Wiedemann-Franz-Lorenz law at 27 °C,

κ = σTCWFL = (5.259 × 107 Ω-1 m-1)(27 + 273 K)(2.44 × 10-8 W Ω K-2)

i.e. κ = 385 W m-1 K-1

For pure metals such as Ag this is nearly independent of temperature (same at 0 °C).

2.9 Mixture rulesA 70% Cu - 30% Zn brass electrical component has been made of powdered metal and contains 15 vol.% porosity. Assume that the pores are dispersed randomly. Estimate the effective electrical resistivityof the brass component.

Solution:

Assume room temperature T = 293 K. From Table 2.1 and Equation 2.17 in the textbook, ρo for

Cu is 17 nΩ m, and from Table 2.3 (in the textbook) C = 300 nΩ m. The atomic fraction of the impurityZn is given as X = 0.3 (we can take the weight fraction given as the atomic fraction because Cu and Znhave very close atomic weights). The brass metal resistivity is then:

ρ = ρo + CX(1–X) = 17 nΩ m + (300 nΩ m)(0.3)(1 – 0.3)


2.12

∴ ρ = 80.0 nΩ m

The component has 15% air pores. Apply the empirical mixture rule in Equation 2.24 (in thetextbook). The fraction of volume with air pores is χ = 0.15. Then,

ρeff = +−

= +−

ρ χχ

( )( )

( . )( . )( . )

11

80 01 0 151 0 15

12

12n mΩ = 101 nΩ m

2.10 Thermal conductiona. An 80 at % Cu–20 at % Zn brass disk of 40 mm diameter and 5 mm thickness is used to conduct heat

from a heat source to a heat sink.

1. Calculate the thermal resistance of the brass disk.

2. If the disk is conducting heat at a rate of 100 W, calculate the temperature drop along the disk.

b. What should be the composition of brass if the temperature drop across the disk is to be halved?

Solutiona

(1) Assume T = 20 ˚C = 293 K. Apply Equation 2.20 (in the textbook) to find the resistivity of thebrass in the disk with ρCu = 17.1 nΩ m (Table 2.1 and Equation 2.17, in the textbook) and XZn = 0.20:

ρbrass = ρCu + CZn in CuXZn(1 – XZn)

i.e. ρbrass = 17.1 nΩ m + (300 nΩ m)(0.20)(1 – 0.20)

∴ ρbrass = 65.1 nΩ m

We know that the thermal conductivity is given by κ/σbrass = CFWLT where σbrass is the conductivityof the disk, CFWL is the Lorenz number and T is the temperature. This equation can also be written asκρbrass = CFWLT so that κ = CFWLT/ρ. Applying this equation,

κ(20 °C) = (2.44 × 10-8 W Ω K-2)(293 K) / (6.51 × 10-8 Ω m)

∴ κ(20 °C) = 109.8 W K-1 m-1

The thermal resistance is θ = L/(κA), where L is the thickness of the disk and A is the cross-sectional area of the disk.

θ = L/(κA) = (5 × 10-3 m)/[(109.8 W K-1 m-1)(π)(2 × 10-2 m)2] = 0.0362 K W-1

(2) From dQ/dt = Aκ∆T/∆x = ∆T/θ (∆x can be taken to be the same as L), and dQ/dt = P (powerconducted), we can substitute to obtain:

∆T = Pθ = (100 W)(3.62 × 10-2 K W-1) = 3.62 K or 3.62 °C

Note: Change in temperature is the same in either Kelvins or degrees Celsius - i.e. ∆T = T1 - T2 =(T1 + 273) - (T2 + 273).

b Since ∆T = Pθ, to get half ∆T, we need half θ or double κ or double σ or half ρ. We thus need1/2ρbrass or 1/2(65.1 nΩ m) which can be attained if the brass composition is Xnew so that

ρnew = ρCu + CZn in CuXnew(1 – Xnew)


2.13

i.e. 1/2(65.1 nΩ m) = 17 nΩ m + (300 nΩ m)Xnew(1 – Xnew)

Solving this quadratic equation we get Xnew = 0.0545, or 5.5% Zn. Thus we need 94.5% Cu-5.5% Zn brass.

2.11 Thermal resistanceConsider a thin insulating disc made of mica to electrically insulate a semiconductor device from a

conducting heat sink. Mica has κ = 0.75 W m-1 K

-1. The disk thickness is 0.1 mm, and the diameter is10 mm. What is the thermal resistance of the disk. What is the temperature drop across the disk if theheat current through it is 25W.

SolutionThe thermal resistance of the mica disk can be calculated directly from Equation 2.37 (in the

textbook)

θκ κπ π

= = =×( )

( ) ×( )−

− − −

L

A

L

d

4 4 1 10

0 75 1 102

3

1 1 2 2

m

W m K. = 1.698 K W-1

The temperature drop across the disk according to Equation 2.36 (in the textbook) is

∆T Q= ′θ = 42.4 °C

*2.12 Thermal resistanceConsider a coaxial cable operating under steady state conditions when the current flow through the

inner conductor generates Joule heat at a rate P = I2R. The heat generated per second by the coreconductor flows through the dielectric; Q’ = I2R. The inner conductor reaches a temperature Ti whereasthe outer conductor is at To. Show that the thermal resistance θ of the hollow cylindrical insulation is

θπκ

=−( ) =

T T

Q

b

aL

i o

©

ln

2Thermal resistance of hollow cylinder

where a is the inside (core conductor) radius, b is the outside radius (outer conductor), κ is the thermalconductivity of the insulation, and L is the cable length. Consider a coaxial cable that has a copper coreconductor and polyethylene (PE) dielectric with following properties: Core conductor resistivity ρ = 19

nΩ m, core radius, a = 4 mm, dielectric thickness, b -a = 3.5 mm, dielectric thermal conductivity κ =

0.3 W m-1 K-1. The outside temperature To is 25 °C. The cable is carrying a current of 500 A. What isthe temperature of the inner conductor?

SolutionConsider a thin cylindrical shell of thickness dr as shown in Figure 2Q12-1. The temperature

difference across dr is dT. The surface area of this shell is 2πrL. Thus, from Fourier’s law,

′ = −Q rLdT

dr( )2π κ

which we can integrate with respect to r from r = a where T = Ti to r = b where T = To,


2.14

′ = −∫ ∫Qdr

rL dT

a

b

T

T

i

o

2π κ

i.e. ′ = −

Q T TL

b

a

i o( )2πκ

ln

Thus the thermal resistance of the hollow cylindrical insulation is

θπκ

= −′

=

( )T T

Q

b

aL

i o

ln

2

Inner conductor

To ToTidrr

dT

a

b

I2R

L

Dielectric

Thin shell

Outer conductor

Q

Figure 2Q12-1 Thermal resistance of a hollow cylindrical shell. Consider aninfinitesimally thin cylindrical shell of radius r and thickness dr in the dielectric andconcentrically around the inner conductor. The surface area is 2πrL.

The actual length of the conductor does not affect the calculations as long as the length issufficiently long that there is no heat transfer along the length; heat flows radially from the inner to theouter conductor. We consider a portion of length L of a very long cable and we set L = 1 m so thatcalculations are per unit length. The joule heating per unit second (power) generated by the current Ithrough the core conductor is

′ = = ××

−

−Q IL

a2

22

9

3 250019 10 1

4 10ρπ π

( )( )( )

( ) = 94.5 W

The thermal resistance of the insulation is,

θπκ π

=

=

+ ××

−

−lnb

aL2

4 3 5 104 10

2 0 3 1

3

3ln( . )

( . )( ) = 0.33 °C/W

Thus, the temperature difference ∆T due to Q′ flowing through θ is,

∆T = Q′θ = (94.5 W)(0.33 °C/W) = 31.2 °C.

The inner temperature is therefore,

Ti = To + ∆T = 25 + 31.2 = 56.2 °C.


2.15

Note that for simplicity we assumed that the inner conductor resistivity ρ and thermal conductivity

κ are constant (do not change with temperature).

2.13 The Hall effectConsider a rectangular sample, a metal or an n-type semiconductor, with a length L, width W , andthickness D. A current I is passed along L, perpendicular to the cross-sectional area WD. The face W ×L is exposed to a magnetic field density B . A voltmeter is connected across the width, as shown inFigure 2Q13-1, to read the Hall voltage VH.

a. Show that the Hall voltage recorded by the voltmeter is

VIB

DenH = Hall voltage

b. Consider a 1-micron-thick strip of gold layer on an insulating substrate that is a candidate for a Hallprobe sensor. If the current through the film is maintained at constant 100 mA, what is the magneticfield that can be recorded per µV of Hall voltage?

LW

D

B

I

VH

Figure 2Q13-1 Hall effect in a rectangular material with length L, width W,

and thickness D. The voltmeter is across the width W.

Solutiona The Hall coefficient, RH, is related to the electron concentration, n, by RH = -1 / (en), and isdefined by RH = Ey / (JB), where Ey is the electric field in the y-direction, J is the current density and B isthe magnetic field. Equating these two equations:

− =1

en JByE

∴ Ey

JB

en= −

This electric field is in the opposite direction of the Hall field (EH) and therefore:

EH = -Ey =JB

en(1)

The current density perpendicular (going through) the plane W × D (width by depth) is:

JI

WD=

Isolating for W: WI

JD= (2)


2.16

The Hall voltage (VH) across W is:

V WH H= E

If we substitute expressions (1) and (2) into this equation, the following will be obtained:

VIB

DenH =

Note: this expression only depends on the thickness and not on the length of the sample.

In general, the Hall voltage will depend on the specimen shape. In the elementary treatment here,the current flow lines were assumed to be nearly parallel from one end to the other end of the sample. Inan irregularly shaped sample, one has to consider the current flow lines. However, if the specimenthickness is uniform, it is then possible to carry out meaningful Hall effect measurements using the vander Pauw technique as discussed in advanced textbooks.

b We are given the depth of the film D = 1 micron = 1 µm and the current through the film I = 100

mA = 0.1 A. The Hall voltage can be taken to be VH = 1 µV, since we are looking for the magnetic field B

per µV of Hall voltage. To be able to use the equation for Hall voltage in part (a), we must find theelectron concentration of gold. Appendix B in the textbook contains values for gold’s atomic mass (Mat=196.97 g/mol) and density (d = 19.3 g/cm3 = 19300 kg/m3). And since gold has a valency of 1 electron,the concentration of free electrons is equal to the concentration of Au atoms. Therefore:

ndN

MA

at

= =( ) ×( )

×( ) = ×−

− −−19300 6 022 10

196 97 105 901 10

3 23

3 128 3

.

. .

kg m mol

kg molm

-1

Now the magnetic field B can be found using the equation for Hall voltage:

VIB

DenH =

∴ BV Den

IH= =

×( ) ×( ) ×( ) ×( )( )

− − − −1 10 1 10 106 6 19 3 V m 1.602 C 5.901 10 m

0.1 A

28

∴ B = 0.0945 T

As a side note, the power (P) dissipated in the film could be found very easily. Using the valuefor resistivity of Au at T = 273 K, ρ = 22.8 nΩ m, the resistance of the film is:

RL

A

L

WD= = =

×( )( )( ) ×( ) =

−

−

ρ ρ 22 8 10 0 001

0 0001 1 100 228

9

6

. .

. .

ΩΩ

m m

m m

The power dissipated is then:

P = I2R = (0.1 A)2(0.228 Ω) = 0.00228 W

2.14 The strain gaugeA strain gauge is a transducer attached to a body to measure its fractional elongation ∆L/L under anapplied load (force) F. The gauge is a grid of many folded runs of a thin, resistive wire glued to aflexible backing, as depicted in Figure 2Q14-1. The gauge is attached to the body under test such thatthe resistive wire length is parallel to the strain.

a. Assume that the elongation does not change the resistivity and show that the change in the resistance∆R is related to the strain, ε = ∆L/L by


2.17

∆R ≈ R(1+2υ)ε Strain gauge equation

where υ is the Poisson ratio, which is defined by

v t

l

= − = −Transverse strainLongitudinal strain

εε

Position ratio

where εl is the strain along the applied load, that is, εl = ∆L/L = ε, and εt is the strain in the transverse

direction, that is, εt = ∆D/D, where D is the diameter (thickness) of the wire.

b. Explain why a nichrome wire would be a better choice than copper for the strain gauge (consider theTCR).

c. How do temperature changes affect the response of the gauge? Consider the effect of temperature onρ. Also consider the differential expansion of the specimen with respect to the gauge wire such thateven if there is no applied load, there is still strain, which is determined by the differential expansioncoefficient, λspecimen – λgauge, where λ is the thermal coefficient of linear expansion: L = Lo[1 + λ(T –To)], where To is the reference temperature.

d. The gauge factor for a transducer is defined as the fractional change in the measured property ∆R/Rper unit input signal (ε). What is the gauge factor for a metal-wire strain gauge, given that for most

metals, υ ≈ 1/3?

e. Consider a strain gauge that consists of a nichrome wire of resistivity 1 µΩ m, a total length of 1 m,

and a diameter of 25 µm. What is ∆R for a strain of 10–3? Assume that υ ≈ 1/3.

f. What will ∆R be if constantan wire is used (see Question 2.5)?

Gauge Length

Solder Tab

Grid of metal wires

Adhesive tape

Figure 2Q14-1 The strain gauge consists of a long, thin wire folded several timesalong its length to form a grid as shown and embedded in a self-adhesive tape. The

ends of the wire are attached to terminals (solder pads) for external connections. Thetape is stuck on the component for which the strain is to be measured.

Solutiona Consider the resistance R of the gauge wire,

RL

D=

ρ

π2

2 (1)

where L and D are the length and diameter of the wire, respectively. Suppose that the applied loadchanges L and D by δL and δD which change R by δR. The total derivative of a function R of twovariables L and D can be found by taking partial differentials (like those used for error calculations inphysics labs). Assuming that ρ is constant,


2.18

δ δ δ ρπ δ ρ

π δRR

LL

R

DD

DL

L

DD= ∂

∂

+ ∂

∂

=

−

4

2

42 3

so that the fractional change is

δ

ρπ

ρπ

δ

ρπ

ρπ

δR

R

D

L

D

L

L

D

L

D

D=

−

4

4

2 4

4

2

2

3

2

i.e.δ δ δR

R

L

L

D

D= − 2 (2)

We can now use the definitions of longitudinal and transverse strain,

δ εL

L l= andδ ε υεD

D t l= = −

in the expression for δR/R in Eqn. (2) to obtain,

δ ε υε υ εR

R l l= − − = +2 1 2( ) ( ) (3)

where ε = εl.

b The change in R was attributed to changes in L and D due to their extension by the applied load.There are two reasons why a nichrome wire will be a better choice. First is that nichrome has a higherresistivity ρ which means that its resistance R will be higher than that of a similar size copper wire and

hence nichrome wire will exhibit a greater change in R (δR is easier to measure). Secondly nichrome has

a very small TCR which means that ρ and hence R do not change significantly with temperature. If wewere to use a Cu wire, any small change in the temperature will most likely overwhelm any change due tostrain.

c Consider a change δR in R due to a change δT in the temperature. We can differentiate R with

respect to T by considering that ρ, L and D depend on T. It is not difficult to show (See Question 2.15)

that if α is the temperature coefficient of resistivity and λ is the linear expansion coefficient , then

1R

dR

dT

= −α λ (4)

Typically λ ≈ 2 × 10-5 K-1, and for pure metals α ≈ 1/273 K-1 or 3.6 × 10-3 K-1. A 1 °C fluctuation

in the temperature will result in δR/R = 3.6 × 10-3 which is about the same as δR/R from a strain of ε = 2

× 10-3 at a constant temperature. It is clear that temperature fluctuations would not allow sensible strainmeasurements if we were to use a pure metal wire. To reduce the temperature fluctuation effects, we needα ≈ λ which means we have to use a metal alloy such as a nichrome wire or a constantan wire.

Even if we make α - λ = 0, the temperature change still produces a change in the resistancebecause the metal wire and specimen expand by different amounts and this creates a strain and hence a


2.19

change in the resistance. Suppose that λgauge and λspecimen are the linear expansion coefficients of the gauge

wire and the specimen, then the differential expansion will be λspecimen - λspecimen and this can only be zero if

λgauge = λspecimen.

d The gauge factor is defined as

GFFractional change in gauge property

Input signal= = = +( )δ

ευ ε

εR R/ 1 2

i.e. GF = 1 + 2υ = 1 + 2(1/3) = 1.67

which applies to metals inasmuch as υ ≈ 1/3.

e For nichrome wire, given that ρ = 10-6 Ω m, L = 1 m, and D = 25 × 10-6 m,

we have, RL

D=

=×

−

−

ρ

π π2

10

25 102

2

6

6 2

(

Ω m)(1 m)

m= 2037 Ω

For a strain of ε = 10–3,

δR = R(1 + 2υ)ε = (2037 Ω)[1 + 2(1/3)](10-3) = 3.40 ΩThis can be easily measured with an ohm-meter, though instrumentation engineers would normally

use the gauge in an electrical bridge circuit. Note that δR/R = (3.4 Ω)/(2037 Ω) = 0.0017, or 0.17 %.

f For a constantan wire, given that ρ = 5 × 10-7 Ω m, L = 1 m, and D = 25 × 10-6 m,

we have, R = ××

−

−

(

5 10

25 102

7

6 2

Ω m)(1 m)

mπ= 1019 Ω

so that δR = R(1 + 2υ)ε = (1019 Ω)[1 + 2(1/3)](10-3) = 1.70 Ω

2.15 Thermal coefficients of expansion and resistivity

a. Consider a thin metal wire of length L and diameter D. Its resistance is R =ρL/A, where A = πD2/4.

By considering the temperature dependence of L, A, and ρ individually, show that

1

R

dR

dT o o= −α λ

where αo is the temperature coefficient of resistivity (TCR), and λo is the temperature coefficient oflinear expansion (thermal expansion coefficient or expansivity), that is,

λo oT T

LdL

dTo

=

−

=

1 or λo oT T

DdD

dTo

=

−

=

1

Note: Consider differentiating R = ρL/[(πD2)/4] with respect to T with each parameter, ρ, L, and D,having a temperature dependence.

Given that typically, for most pure metals, αo ≈ 1/273 K-1 and λo ≈ 2 × 10–5 K–1, confirm that the

temperature dependence of ρ controls R, rather than the temperature dependence of the geometry. Isit necessary to modify the given equation for a wire with a noncircular cross section?


2.20

b. Is it possible to design a resistor from a suitable alloy such that its temperature dependence is almostnil? Consider the TCR of an alloy of two metals A and B, for which αAB ≈ αAρA/ρAB.

Solutiona Consider the resistance R of the wire,

RL

D=

ρ

π2

2 (1)

Consider a change δR in R due to a change δT in the temperature. We can differentiate R with

respect to T by considering that ρ, L, and D depend on T,

dR

dT

d

dT

L

D

L

D

d

dT D

dL

dT

L

D

dD

dT=

=

+

−

ρπ π

ρ ρπ

ρπ

4 4 4

2

42 2 2 3

divide by R:1 1 1

21

R

dR

dT

d

dT L

dL

dT D

dD

dT

=

+

−

ρ

ρ(2)

At T = To we have,

1 1 12

1R

dR

dT

d

dT L

dL

dT D

dD

dTo o o

=

+

−

ρ

ρ(3)

where the derivatives are at T = To. Recall that the temperature coefficient of resistivity, TCR (αo), and

the linear expansion coefficient λo are defined as follows,

αρ

ρ0

0

1=

d

dTand λ0

0 0

1 1=

=

L

dL

dT D

dD

dT(4)

which reduces Eqn. (3) to

10 0R

dR

dT

= −α λ (5)

Typically λo ≈ 2 × 10-5 K-1, and for pure metals αo ≈ 1/273 K-1 or 3.6 × 10-3 K-1. Thus,

13 6 10 2 10 3 6 103 1 5 1 3 1

0R

dR

dT

= × − × ≈ × ≈− − − − − −. . K K K α

Since αo is much larger than λo, it dominates the change in R.


2.21

Af

Rf

Figure 2Q15-1 Wire with non-circular cross section.

There is no need to modify Eqn. (5) for a non-circular cross sectional area. You can derive thesame expression for a rectangular or an elliptic cross section, or, indeed, any arbitrary cross section. Onecan consider the wire to be made up of N thin fibers each of circular cross section Af. Imagine holding abunch of these in your hand and then sliding them into any cross section you like as in the Figure 2Q15-1.In all cases A = ΣAf. However, since the fibers are in parallel, the total resistance is given by R-1 = ΣRf

-1 =

NRf-1. Thus R = Rf / N. For each fiber, δRf = Rf(αo - λo)δT as we have derived above. Then,

δR = (δRf)/N = [Rf(αo - λo)δT]/N = R(αo - λo)δT

[There are a few assumptions such as the resistivity is homogeneous and the cross section does not changealong the wire!]

b For pure metals, αo > > λo. Alloying metal A with metal B reduces the TCR (temperature

coefficient of resistivity) αA of metal A to αAB = αA(ρA/ρAB). The αo - λo can be brought to zero by using a

suitable composition alloy for which αo = λo. Since strictly αo depends (however slightly) on T, the

condition αo - λo can only be exactly true at one temperature or approximately true in a small regionaround that temperature. In practice this temperature range may be sufficient to cover a typical applicationrange in which, for most practical purposes, αo - λo is negligible.

2.16 Einstein relation and ionic conductivityIn the case of ionic conduction, ions have to jump from one interstice to the neighboring one. This

process involves overcoming a potential energy barrier just like atomic diffusion, and drift and diffusionare related. The drift mobility µ of ions is proportional to the diffusion coefficient D because drift islimited by the atomic diffusion process. The Einstein relation relates the two by

D kT

eµ= Einstein relation

The diffusion coefficient of the Na+ ion in sodium silicate (Na2O-SiO2) glasses at 400 °C is 3.4 × 10-9

cm2 s-1. The density of such glasses is 2.4 g cm-3. Calculate the ionic conductivity and resistivity of(17.5 mol% Na2O)(82.5 mol% SiO2) sodium silicate glass at 400 °C and compare your result with the

experimental value of about 104 Ω cm for the resistivity.


2.22

SolutionWe can apply the conductivity expression σ µ= q ni i i , where qi is the charge of the ion, Na+, so

that it is +e, ni is the concentration of the Na+ ions in the glass and µi is their mobility.

We can think of the glass as built from [(Na2O)0.175(SiO2)0.825] units. The atomic masses of Na, Oand Si are 23, 16 and 28.1 g mole-1 respectively. The atomic mass of one unit is

Mat = × ( ) +[ ] + + × ( )[ ]0 175 2 23 16 0 825 28 1 2 16. . . = 60.43 g mol-1 of [(Na2O)0.175(SiO2)0.825]

The number of [(Na2O)0.175(SiO2)0.825] units per unit volume can be found from the density d by

nd N

MA

at

= =( ) ×( )

( )− −

−

2 4 6 02 10

60 43

3 23 1

1

. .

.

g cm mol

g mol = 2.39 × 1022 units cm-3

The concentration of Na+ ions is equal to the concentration of Na atoms and then

n n ni Na= = ( ) × =

= × ( )+ + +

×( ) = ×− −

atomic fraction of Na in the unit

cm

.. ( ) . ( )

. 0 175 2

0 175 2 1 0 825 1 22 39 1022 3 2.79 10 cm21 3

The mobility of sodium ions can be calculated from Einstein relation

µi

eD

kT= =

×( ) ×( )×( ) +( )( )

− − −

− −

1 602 10 3 4 10

1 38 10 400 273

19 9 2 1

23 1

. .

.

C cm s

J K K = 6.04 × 10-8 cm2 V-1 s-1

Thus the conductivity of the glass is

σ µ= = ×( ) ×( ) ×( )− − − −eni i 1 602 10 2 79 1019 21 3 2 1 1. .C cm 6.04 10 cm V s-8 = 2.7 × 10-5 Ω -1 cm-1

and its resistivity is

ρσ

= =×( )− − −

1 1

2 7 10 5 1 1. Ω cm = 3.8 × 104 Ω cm

2.17 Skin effecta. What is the skin depth for a copper wire carrying a current at 60 Hz? The resistivity of copper at 27

°C is 17 nΩ m. Its relative permeability is µr ≈ 1. Is there any sense in using a conductor for powertransmission with a diameter of more than 2 cm?

b. What is the skin depth for an iron wire carrying a current at 60 Hz? The resistivity of iron at 27 °C is

97 nΩ m. Assume that its relative permeability is µr ≈ 700. How does this compare with the copperwire? Discuss why copper is preferred over iron for power transmission even though the iron isnearly 100 times cheaper than copper.

Solution

a The conductivity is 1/ρ. The relative permeability (µr) for copper is 1, thus µCu = µo. The angular

frequency is ω = 2πf = 2π(60 Hz). Using these values in the equation for skin depth (δ):


2.23

δω

ρµ π

π= =

( ) ×( )×( )

− −

−

112

11

12

2 604 10

17 10

7 1

9o Cu -1s

H m

m

Ω

∴ δ = 0.00847 m or 8.47 mm

This is the depth of current flow. If the radius of wire is 10 mm or more, no current flows throughthe core region and it is wasted. There is no point in using wire much thicker than a radius of 10 mm(diameter of 20 mm).

b The conductivity is 1/ρ. The relative permeability (µr) for Iron is 700, thus µFe = 700µo. The

angular frequency is ω = 2πf = 2π(60 Hz). Using these values in the equation for skin depth (δ):

δω

ρµ π

πFe

Fe -1sH m

m

= =

( ) ( ) ×( )×( )

− −

−

112

11

12

2 60700 4 10

97 10

7 1

9o

Ω

∴ δ = 0.000765 m or 0.765 mm

Thus the skin depth is 0.765 mm, about 11 times less than that for copper.

To calculate the resistance we need the cross sectional area for conduction. The material crosssectional area is πr2 where r is the radius of the wire. But the current flow is within depth δ. We deduct

the area of the core, π (r - δ)2, from the overall area, πr2, to obtain the cross sectional area for conduction.

Comparison of Cu and Fe based on solid core wires

The resistance per unit length of the solid core Fe wire (RFe) is:

RA r r rFe

FeFe Fe

Fe Fe Fe

Fe

Fe Fe Fe

= =− −( )

=−

ρ ρπ π δ

ρπ δ πδ2 2 22

The resistance per unit length of solid core Cu wire is:

RA r r rCu

CuCu Cu

Cu Cu Cu

Cu

Cu Cu Cu

= =− −( )

=−

ρ ρπ π δ

ρπ δ πδ2 2 22

If we equate these two resistances, we can make a comparison between Fe and Cu:

R RFe Cu=

∴ ρπ δ πδ

ρπ δ πδ

Fe

Fe Fe Fe

Cu

Cu Cu Cu2 22 2r r−=

−

∴ 2 2 2πρ δ ρ π δ πδCu Fe Fe Fe Cu Cu Cur r= −( ) (Neglect the δFe2 term which is small)

∴ rr

FeFe Cu Cu Cu

Cu Fe

=−( )ρ π δ πδ

πρ δ2

2

2

We can assume a value for rCu for calculation purposes, rCu = 10 mm. The resistivity ρCu is given

as 17 nΩ m and the skin depth of Cu is known to be δCu = 8.47 mm. The resistivity of Fe is given as ρFe

= 97 nΩ m and its skin depth was just calculated to be δFe = 0.765 mm. We can substitute these valuesinto the above equation to determine the radius of Fe wire that would be equivalent to 10 mm diameter Cuwire.


2.24

rFe

m m m m

m m=

×( ) ( )( ) − ( )[ ]×( )( )

−

−

97 10 2 0 010 0 00847 0 00847

2 17 10 0 000765

9 2

9

. . .

.

ΩΩ

π ππ

∴ rFe = 0.364 m

Now compare the volume (V) of Fe per unit length to the volume of Cu per unit length:

V

V

r

rFe

Cu

Fe

Cu

mm

mm

= ( )( )

= ( )( )

=ππ

2

2

2

2

11

0 3640 010

.

. 1325

Even though Fe costs 100 times less than Cu, we need about 1300 times the volume of Cu if Fe isused. The cost disadvantage is 13 times in addition to weight disadvantage.

ADDENDUM JANUARY 2001

Submitted by George Belev

Comparison of Cu and Fe wires: any shape and any number

To determine if it is worthwhile to use iron rather than copper, we must compare the amount ofiron needed to perform the equivalent task of some amount of copper (i.e. have the same resistance).

Let us first assume that by choosing a proper shape for the conductors we can eliminate theinfluence of the skin effect on conduction.

The resistance per unit length of the Fe wire (RFe) is:

RAFe

FeFe= ρ

The resistance per unit length of Cu wire is

RACu

CuCu= ρ

If we equate these two resistances, we can make a direct comparison between Fe and Cu:

R RFe Cu=

∴ A AFeFe

CuCu= ρ

ρand the volumes of iron and copper per unit length will be in the same ratio

V VFeFe

CuCu= ρ

ρLet us compare the masses of Fe and Cu needed per unit length

M

M

V D

V D

D

DFe

Cu

Fe Fe

Cu Cu

Fe

Cu

Fe

Cu

= = = ⋅ ≈−

−ρρ

9717

7 868 96

53

3

n mn m

g cmg cm

ΩΩ

.

.

Since Fe costs 100 times less than Cu, if we use iron conductors, we will reduce the cost for wireby 100/5 = 20 times. So it seems that the use of Fe will have great economic advantage if we can find areasonable way to eliminate the influence of the skin effect on conduction.

There is no sense in making the conductor with a radius bigger than the skin depth, so let usconsider a single copper conductor with radius δCu, and using N iron conductors each with radius δCu asshown bellow:


2.25

2 δCu

2 δFe

N

As we have calculated above both conductors will have equal resistance per unit length if

A

A

NFe

Cu

Fe

Cu

Fe

Cu

= =πδπδ

ρρ

2

2

and we can calculate the number N of Fe wires which should run in parallel

N Fe

Cu

Cu

Fe

= ≈ρρ

δδ

2

2 700

Thus, copper as a single conductor has 700 times better performance than a single iron conductor.

It is not necessary to manufacture 700 Fe wires and run them in parallel. The iron conductor can beproduced more conveniently in the shapes shown bellow and it will be cheaper than the Cu conductor.

2 δFe = 1.53 mm

2 N δFe = 1071 mm

δFe = 0.763 mm

N δFe4π

= 682 mm

But it will be of impractical size, it will have poor mechanical properties and will be 5 times heaviercompared with the single Cu wire. A power grid based on Cu conductors will be much cheaper and muchsmaller than the one based on Fe conductors.

2.18 Temperature of a light bulba. Consider a 100 W, 120 V incandescent bulb (lamp). The tungsten filament has a length of 0.579 m

and a diameter of 63.5 µm. Its resistivity at room temperature is 56 nΩ m. Given that the resistivityof the filament can be represented as

ρ ρ=

0

0

T

T

n

Resistivity of W

where T is the temperature in K, ρo is the resistance of the filament at To K, and n = 1.2, estimate thetemperature of the bulb when it is operated at the rated voltage, that is, directly from the mains outlet.Note that the bulb dissipates 100 W at 120 V.

b. Suppose that the electrical power dissipated in the tungsten wire is totally radiated from the surface ofthe filament. The radiated power at the absolute temperature T can be described by Stefan's Law

Pradiated =∈σsA(T4 – To4) Radiated power

where σs is Stefan's constant (5.67 × 10–8 W m–2 K–4), ∈ is the emissivity of the surface (0.35 fortungsten), A is the surface area of the tungsten filament, and To is the room temperature (293 K).Obviously, for T > To, Pradiated = ∈σsAT4.


2.26

Assuming that all of the electrical power is radiated from the surface, estimate the temperature ofthe filament and compare it with your answer in part (a).

c. If the melting temperature of W is 3407 °C, what is the voltage that guarantees that the light bulb willblow?

Solutiona First, find the current through the bulb at 100 W and 120 V.

P = VI

∴ I = P/V = (100 W)/(120 V) = 0.8333 A

From Ohm’s law the resistance of the bulb can be found:

R = V/I = (120 V)/(0.8333 A) = 144.0 Ω

The values for length of the filament (L = 0.579 m) and diameter of the filament (D = 63.5 µm) atoperating temperature are given. Using these values we can find the resistivity of the filament when thebulb is on (ρ1).

RL

D= ρ

π1

2

4

∴ ρ

π π

1

2 6 2

74144 0

463 5 10

0 5797 876 10= =

( ) ×( )( )

= ×−

−R D

L

. .

. .

ΩΩ

m

mm

Now the bulb’s operating temperature (T1) can be found using our obtained values in the equationfor resistivity of W (assuming room temperature To = 293 K and given n = 1.2):

ρ ρ=

0

0

T

T

n

isolate: T1 2652 K=

= ( ) ×

×

=−

Tn

01

0

17

1

1 2

2937 876 10ρ

ρ

.

.

Km

56 10 m-9

ΩΩ

b First we need the surface area A of the Tungsten filament. Since it is cylindrical in shape:

A = L(πD) = (0.579 m)(π)(63.5 × 10-6 m) = 0.0001155 m2

Now the temperature of the filament T1 can be found by isolating it in Stefan’s equation andsubstituting in the given values for emissivity (∈ = 0.35), Stefan’s constant (σs = 5.67 × 10-8 W m-2 K-4)and room temperature (To = 293 K).

P A T Ts= −( )∈σ 14

04

∴ TP

AT

s1 0

4

1

4

= +

∈σ

∴ T1 8 2 1

4

1

4100

0 35 5 67 10293=

( ) ×( )( ) + ( )

− − −

. .

W

W m K 0.0001155 mK2


2.27

∴ T 1 = 2570 K

Note: We can even ignore To to get the same temperature since To << T1:

P ATs=∈σ 14

∴ TP

As1

1

4

8 2 1

1

4100

0 35 5 67 10=

=( ) ×( )( )

− − −∈σ

. .

W

W m K 0.0001155 m2

∴ T 1 = 2570 K

These values are fairly close to the answer obtained in part (a).

c Let V be the voltage and R be the resistance when the filament is at temperature Tm. We are giventhe temperature Tm = 3407 °C + 273 = 3680 K. Since we know the following:

RL

D= π ρ

42

ρ ρ=

0

0

T

Tm

n

We can make a substitution for ρ and use the values given for the light bulb filament to find theresistance of the filament at temperature Tm.

RL

D

T

Tm

n

=

= ( )

×( )×( )

−

−

π ρ π4

0 579

463 5 10

56 103680

20

0 6 2

91 2.

.

.m

mm

K293 K

Ω

∴ R = 213.3 ΩAssuming that all electrical power is radiated from the surface of the bulb, we can use Stefan’s law

and substitute in V2/R for power P:

V

RA T Ts

2

14

04= −( )∈σ

∴ V R A T Ts m2 4

04= −( )∈σ

∴V 2 8 2 4

4 4

213 3 0 35 5 67 10 0 0001155

3680 293

= ( )( ) ×( )( )

( ) − ( )[ ]− − −. . . .

Ω W m K m

K K

∴ V = 299 V

"I think there is a world market for maybe five computers"

Thomas Watson (Chairman of IBM, 1943)

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