Properties of Electrical Materials

4% of exam

Semiconductors, conductivity & resistivity, permittivity, permeability, thermal properties.

3 concepts

Exam areas

A. Semiconductor materials

Semiconductor Materials

Intrinsic vs. extrinsic silicon, n/p doping, the mass-action law, band gap and mobility, and the pn-junction built-in potential the FE expects you to compute.

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B. Electrical

Electrical Properties of Materials

Resistivity and conductivity, temperature coefficients, permittivity and dielectrics, permeability and magnetic cores, and the conductor-vs-insulator scale the FE leans on.

Every component on the exam is ultimately a piece of material with three electrical signatures: how it conducts current (resistivity

\rho

), how it stores charge (permittivity

\varepsilon

), and how it stores magnetic flux (permeability

\mu

). The FE Reference Handbook scatters these across the Materials Science and the Electrical and Computer Engineering chapters —

R = \rho L/A

, the linear temperature law

\rho = \rho_0[1+\alpha(T-T_0)]

, the capacitor relation

C=\varepsilon A/d

, and the coil relation

L = N^2\mu A/\ell

. Examinees lose points by treating

\varepsilon

and

\varepsilon_r

as interchangeable, by mishandling the temperature term, or by mixing meters and millimeters. This concept ties the material property directly to the device parameter so you can move between them in one step.

Resistivity, conductivity, and resistance

Resistivity

\rho

is the intrinsic property; resistance

R

is what a particular geometry gives. A longer wire has more resistance, a fatter one has less, and the proportionality constant is

\rho

. Conductivity

\sigma

is simply its reciprocal,

\sigma = 1/\rho

, and is handy when you think in terms of how well something conducts. Copper sits near

\rho = 1.72\times10^{-8}\,\Omega\!\cdot\!\text{m}

(equivalently

\sigma \approx 5.8\times10^{7}\,\text{S/m}

); aluminum is about

1.6\times

higher. For a thin film or diffused layer of uniform thickness

t

the geometry collapses to a sheet resistance

R_s = \rho/t

(units

\Omega/\square

), and the resistance of a patterned strip is just

R = R_s(L/W)

— the number of squares — which is how on-chip and PCB-trace resistances are estimated.

R = \frac{\rho L}{A}, \qquad \sigma = \frac{1}{\rho}, \qquad R_s = \frac{\rho}{t}

Temperature coefficient of resistance

For metallic conductors, resistivity and resistance rise nearly linearly with temperature because hotter lattice atoms scatter electrons more. The handbook gives both forms:

\rho = \rho_0[1+\alpha(T-T_0)]

and

R = R_0[1+\alpha(T-T_0)]

, where

\alpha

is the temperature coefficient referenced to

T_0

(usually

20\,^\circ\text{C}

). Copper's

\alpha \approx 0.00393\,/^\circ\text{C}

, so a copper winding running

55\,^\circ\text{C}

above ambient gains about

22\%

resistance. Semiconductors and carbon behave oppositely (

\alpha < 0

when intrinsic), which is exactly why thermistors are made from semiconductor material.

R = R_0\bigl[1 + \alpha(T - T_0)\bigr]

Permittivity and dielectrics

Permittivity

\varepsilon

measures how much electric flux (stored charge) a material supports for a given field, and it sets capacitance directly through

C = \varepsilon A/d

. It is almost always written as

\varepsilon = \varepsilon_r\varepsilon_0

, where

\varepsilon_0 = 8.85\times10^{-12}\,\text{F/m}

is the free-space value and the relative permittivity

\varepsilon_r

(also called the dielectric constant

\kappa

) is the dimensionless multiplier of the insulator. Vacuum/air is

\varepsilon_r \approx 1

, common plastic films

2

–

4

, and ceramics tens to thousands. A real dielectric also has a breakdown field above which it conducts; that, not capacitance, limits voltage rating.

C = \frac{\varepsilon A}{d} = \frac{\varepsilon_r \varepsilon_0 A}{d}, \qquad \varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}

Permeability and magnetic materials

Permeability

\mu

is the magnetic analog of permittivity: it sets how readily a material carries magnetic flux, and it governs inductance. The handbook's coil formula

L = N^2\mu A/\ell

shows that a high-permeability core multiplies inductance for the same winding. As with permittivity,

\mu = \mu_r\mu_0

with

\mu_0 = 4\pi\times10^{-7}\,\text{H/m}

and

\mu_r

the relative permeability — near

1

for air/copper/aluminum (non-magnetic) but

10^3

–

10^5

for ferromagnetic iron and ferrites. The magnetic-circuit shorthand is reluctance

\mathcal{R} = \ell/(\mu A)

, so

L = N^2/\mathcal{R}

L = \frac{N^2 \mu A}{\ell} = \frac{N^2}{\mathcal{R}}, \qquad \mathcal{R} = \frac{\ell}{\mu A}

Conductors, semiconductors, and insulators

The same three materials classes that appear in the resistivity figure are defined by band structure. Conductors have overlapping or partly filled bands, so

\rho

is tiny (

10^{-8}\,\Omega\!\cdot\!\text{m}

) and rises with temperature. Insulators have a wide band gap, so almost no carriers exist and

\rho

is enormous (

10^{10}

–

10^{16}\,\Omega\!\cdot\!\text{m}

). Semiconductors sit in between with a moderate gap and a strongly temperature- and doping-dependent

\rho

. On the exam, the sign of the temperature coefficient is often the giveaway: positive

\alpha

flags a metal, negative

\alpha

flags a semiconductor or carbon.

Exam strategy

Decide first whether you need the material property (

\rho

\varepsilon

\mu

) or the device parameter (

R

C

L

), then apply the single bridging formula. For temperature problems, always identify

T_0

before plugging in —

\alpha

is only valid relative to its reference, and a missing

(T-T_0)

subtraction is the most common slip. Never insert

\varepsilon_r

where the formula calls for

\varepsilon

: multiply by

\varepsilon_0

first. Keep every length in meters when working in SI so that

\varepsilon_0

and

\mu_0

come out right, and convert plate spacings and core lengths from mm before substituting.

Key equations

Resistance from resistivity

R = \frac{\rho L}{A}

$\rho$ = resistivity $(\Omega\!\cdot\!\text{m})$ , $L$ = length (m), $A$ = cross-section (m²). Gives $R$ in $\Omega$ .

Conductivity

\sigma = \frac{1}{\rho}

Reciprocal of resistivity, in S/m. Copper $\approx5.8\times10^7\,\text{S/m}$ .

Temperature dependence of resistivity

\rho = \rho_0\bigl[1 + \alpha(T - T_0)\bigr]

$\rho_0$ at reference $T_0$ ; $\alpha$ = temperature coefficient $(/^\circ\text{C})$ . Same form for $R$ .

Temperature dependence of resistance

R = R_0\bigl[1 + \alpha(T - T_0)\bigr]

Direct device form; copper $\alpha\approx0.00393\,/^\circ\text{C}$ referenced to $20\,^\circ\text{C}$ .

Parallel-plate capacitance

C = \frac{\varepsilon A}{d} = \frac{\varepsilon_r\varepsilon_0 A}{d}

$\varepsilon$ = permittivity (F/m), $A$ = plate area (m²), $d$ = separation (m).

Permittivity

\varepsilon = \varepsilon_r\varepsilon_0, \qquad \varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}

$\varepsilon_r$ (= dielectric constant $\kappa$ ) is dimensionless; air $\approx1$ .

Coil inductance

L = \frac{N^2\mu A}{\ell} = \frac{N^2}{\mathcal{R}}

$N$ = turns, $\mu$ = core permeability (H/m), $A$ = core area, $\ell$ = mean path length.

Permeability

\mu = \mu_r\mu_0, \qquad \mu_0 = 4\pi\times10^{-7}\ \text{H/m}

$\mu_r$ = relative permeability; $\approx1$ for non-magnetic, $10^3$ – $10^5$ for iron/ferrite.

Reluctance

\mathcal{R} = \frac{\ell}{\mu A}

Magnetic-circuit resistance $(\text{H}^{-1})$ ; flux $\phi = \mathcal{F}/\mathcal{R}$ with mmf $\mathcal{F}=NI$ .

Sheet resistance

R = R_s\frac{L}{W}, \qquad R_s = \frac{\rho}{t}

Thin-film/diffused/trace layer: $R_s$ in $\Omega/\square$ , $t$ = thickness, $L/W$ = number of squares.

Worked examples

Resistance of a copper feeder, hot and cold

Problem. A $200\,\text{m}$ run of copper wire has a diameter of $2.05\,\text{mm}$ . With $\rho_0 = 1.72\times10^{-8}\,\Omega\!\cdot\!\text{m}$ at $20\,^\circ\text{C}$ and $\alpha = 0.00393\,/^\circ\text{C}$ , find the resistance at $20\,^\circ\text{C}$ and again at $75\,^\circ\text{C}$ .

Solution. Area:

A = \tfrac{\pi}{4}d^2 = \tfrac{\pi}{4}(2.05\times10^{-3})^2 = 3.30\times10^{-6}\,\text{m}^2

. Cold resistance:

R_{20} = \rho_0 L/A = (1.72\times10^{-8})(200)/(3.30\times10^{-6}) = 1.04\,\Omega

. Hot:

R_{75} = R_{20}[1+\alpha(75-20)] = 1.04[1+0.00393(55)] = 1.04(1.216) = 1.27\,\Omega

. Sanity check: a

55\,^\circ\text{C}

rise on copper adds

\approx0.00393\times55 = 21.6\%

, matching the

1.04\to1.27\,\Omega

jump. Answer:

R_{20}=1.04\,\Omega

R_{75}=1.27\,\Omega

R_{75} = \frac{\rho_0 L}{A}\bigl[1+\alpha(T-T_0)\bigr] = 1.04\,(1.216) = 1.27\ \Omega

Capacitance of a film capacitor

Problem. A capacitor has plate area $A = 0.020\,\text{m}^2$ separated by a $0.10\,\text{mm}$ polyester film with $\varepsilon_r = 3.2$ . Find the capacitance.

Solution. Permittivity

\varepsilon = \varepsilon_r\varepsilon_0 = 3.2(8.85\times10^{-12}) = 2.83\times10^{-11}\,\text{F/m}

. Capacitance

C = \varepsilon A/d = (2.83\times10^{-11})(0.020)/(0.10\times10^{-3}) = 5.66\times10^{-9}\,\text{F}

. Sanity check: substituting directly,

C = 3.2(8.85\times10^{-12})(0.020)/(1\times10^{-4}) = 5.66\,\text{nF}

— a reasonable film-cap value. Answer:

C = 5.66\,\text{nF}

C = \frac{\varepsilon_r\varepsilon_0 A}{d} = \frac{3.2(8.85\times10^{-12})(0.020)}{1.0\times10^{-4}} = 5.66\ \text{nF}

Inductance of an iron-core coil

Problem. A coil of $N = 300$ turns is wound on a core of cross-section $A = 1.0\times10^{-4}\,\text{m}^2$ , mean path length $\ell = 0.15\,\text{m}$ , and relative permeability $\mu_r = 2000$ . Find the inductance.

Solution. Core permeability

\mu = \mu_r\mu_0 = 2000(4\pi\times10^{-7}) = 2.51\times10^{-3}\,\text{H/m}

. Then

L = N^2\mu A/\ell = (300)^2(2.51\times10^{-3})(1.0\times10^{-4})/0.15

. Numerator

= 9\times10^4 \times 2.51\times10^{-3}\times10^{-4} = 2.26\times10^{-2}

; divide by

0.15

to get

0.151\,\text{H}

. Sanity check via reluctance:

\mathcal{R}=\ell/(\mu A)=0.15/(2.51\times10^{-3}\cdot10^{-4})=5.97\times10^{5}\,\text{H}^{-1}

, and

L=N^2/\mathcal{R}=9\times10^4/5.97\times10^5=0.151\,\text{H}

— agrees. Answer:

L = 151\,\text{mH}

L = \frac{N^2\mu_r\mu_0 A}{\ell} = \frac{(300)^2(2000)(4\pi\times10^{-7})(10^{-4})}{0.15} = 151\ \text{mH}

Common pitfalls

•Substituting $\varepsilon_r$ where the formula needs $\varepsilon$ . Always form $\varepsilon = \varepsilon_r\varepsilon_0$ first; forgetting $\varepsilon_0$ makes $C$ too large by $\sim10^{11}$ .
•Same trap with permeability: use $\mu = \mu_r\mu_0$ in $L = N^2\mu A/\ell$ , not $\mu_r$ alone.
•Dropping the $(T-T_0)$ subtraction or using the wrong reference. $\alpha$ is defined relative to $T_0$ ; $R = R_0[1+\alpha(T-T_0)]$ , not $R_0(1+\alpha T)$ .
•Mixing mm and m. Plate spacing, core length, and wire diameter must be in meters before they meet $\varepsilon_0$ or $\mu_0$ ; a stray factor of $10^3$ is the most common numeric error.
•Using area = diameter² or $\pi d^2$ for a round conductor. The correct cross-section is $A = \pi d^2/4$ .
•Assuming every material has positive $\alpha$ . Semiconductors and carbon have negative coefficients; the sign of $\alpha$ alone distinguishes a metal from a semiconductor.
•Treating dielectric constant as the voltage limit. Breakdown field, not $\varepsilon_r$ , sets the voltage rating; a high- $\kappa$ material can have a low breakdown strength.

References

NCEES FE Reference Handbook — Electrical and Computer Engineering: Resistivity, Capacitors and Inductors
NCEES FE Reference Handbook — Materials Science/Structure of Matter: Electrical Properties
IEEE/ASTM standard conductor tables — copper/aluminum resistivity and temperature coefficients

C. Thermal

Thermal Properties of Materials

Thermal conductivity and Fourier's law, specific heat, thermal expansion, the thermal-resistance network for component cooling, and the self-heating that drives thermal runaway.

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