Resistivity of Materials & Temperature Dependence

Why Resistivity Values Differ Across Materials

We have already derived that the resistivity of a material is $$\rho = \frac{m}{n e^2 \tau},$$ where $n$ is the density of free charge carriers (per m$^3$) and $\tau$ is the mean free time between collisions. Two materials can have very different resistivities for two reasons:

1. Different carrier densities ($n$) — Metals have roughly $n \sim 10^{28}$ electrons/m$^3$, semiconductors have $n \sim 10^{16}$ to $10^{19}$ at room temperature, and insulators have almost no free carriers.
2. Different relaxation times ($\tau$) — Even for two metals with similar $n$, scattering from impurities, lattice defects, and thermal vibrations can make $\tau$ very different.

That is why the resistivity of materials spans an astonishing 26 orders of magnitude, from superconductors ($\rho = 0$) to fused quartz ($\rho \approx 10^{16}\,\Omega\cdot\text{m}$).

[Memory hook] Remember the three classes:

• Conductors: $\rho \approx 10^{-8}$ to $10^{-6}\,\Omega\cdot\text{m}$ (Ag, Cu, Al, Fe)
• Semiconductors: $\rho \approx 10^{-5}$ to $10^{6}\,\Omega\cdot\text{m}$ (Ge, Si, C in graphite form)
• Insulators: $\rho \approx 10^{10}$ to $10^{16}\,\Omega\cdot\text{m}$ (glass, rubber, fused quartz)

Key Point: The large gap between conductors and insulators is dominated mainly by the huge difference in carrier density $n$.

Temperature Dependence — The Big Picture

Experiments show that for almost every material, resistivity changes with temperature. But how it changes depends on the material class:

Material class	Sign of $d\rho/dT$	Reason
Metals (conductors)	Positive ($\rho$ rises with $T$)	Lattice vibrations increase → smaller $\tau$; $n$ barely changes
Alloys (Nichrome, Manganin, Constantan)	Slightly positive, very small	Disordered lattice already produces strong scattering; extra thermal scattering adds little
Semiconductors	Negative ($\rho$ falls with $T$)	Thermal energy sets free many more carriers; rise in $n$ dominates over fall in $\tau$
Insulators	Negative (similar to semiconductors)	Carrier density is so low that any thermal generation has a huge fractional effect
Superconductors (below $T_c$)	$\rho$ = 0	Cooper pairs move without scattering

The equation $\rho = m/(n e^2 \tau)$ helps explain these trends:

• In a metal, $n$ is fixed and $\tau$ drops as $T$ rises, so $\rho$ increases.
• In a semiconductor, as $T$ rises $n$ increases exponentially while $\tau$ only falls slowly, so $\rho$ decreases.

Quantifying the Temperature Dependence in Metals

For metals in a limited range of temperatures, resistivity varies approximately linearly with temperature:

$$\boxed{\rho_T = \rho_0 \big[1 + \alpha (T - T_0)\big]}$$

where:

• $\rho_0$ is the resistivity at a reference temperature $T_0$ (usually $0^\circ$C or $20^\circ$C)
• $\rho_T$ is the resistivity at temperature $T$
• $\alpha$ is the temperature coefficient of resistivity, with SI unit K$^{-1}$ (or $^\circ$C$^{-1}$)

Since the dimensions of a conductor change only slightly with temperature, the same form also holds for resistance:

$$R_T = R_0 \big[1 + \alpha (T - T_0)\big]$$

Typical values of $\alpha$ at $20^\circ$C

Material	$\alpha$ (K$^{-1}$)
Silver (Ag)	$+4.1 \times 10^{-3}$
Copper (Cu)	$+4.0 \times 10^{-3}$
Aluminium (Al)	$+4.3 \times 10^{-3}$
Iron (Fe)	$+6.5 \times 10^{-3}$
Tungsten (W)	$+4.5 \times 10^{-3}$
Manganin (alloy)	$+0.002 \times 10^{-3}$
Nichrome (alloy)	$+0.4 \times 10^{-3}$
Carbon (graphite)	$-0.5 \times 10^{-3}$
Silicon (semiconductor)	$-70 \times 10^{-3}$

Two things jump out:

1. Pure metals all have similar $\alpha$ ($\sim 4 \times 10^{-3}$ K$^{-1}$) — about 0.4% change per kelvin.
2. Alloys have $\alpha$ hundreds of times smaller — that is exactly why Manganin and Constantan are used to make standard resistors and resistance boxes.

[JEE Tip] Over wide temperature ranges the relation $\rho_T = \rho_0[1+\alpha(T-T_0)]$ breaks down. For a clean metal at very low $T$, $\rho$ approaches a non-zero residual resistivity due to impurity scattering — it does not go to zero.

Why Do Semiconductors Have Negative $\alpha$?

In a semiconductor the number of free carriers depends strongly on temperature. The density of carriers obeys (approximately):

$$n(T) \propto T^{3/2}\, e^{-E_g / (2 k_B T)},$$

where $E_g$ is the energy band gap and $k_B$ is Boltzmann's constant.

As $T$ increases:

• The exponential factor grows very rapidly — many more electrons can jump across the band gap.
• Hence $n$ rises much faster than $\tau$ falls.
• From $\rho = m/(n e^2 \tau)$, the rise in $n$ dominates and $\rho$ decreases with $T$.

This explains why:

• A thermistor (semiconductor resistor) is used in circuits to compensate for temperature drift — its resistance falls as it heats up.
• Semiconductor thermometers exploit this strong $T$-dependence to measure small temperature changes.
• A semiconductor at very low temperatures can behave almost like an insulator — not enough thermal energy to lift electrons into the conduction band.

[Common trap] Students sometimes say "semiconductors have negative $\alpha$ because electrons collide less at high $T$." This is wrong — $\tau$ still decreases at higher $T$. The real reason is the huge rise in $n$.

Superconductivity — When Resistivity Becomes Exactly Zero

At very low temperatures, certain metals, alloys, and ceramic oxides lose all resistance — a phenomenon discovered by Kamerlingh Onnes in 1911 while cooling mercury below $4.2$ K. Below a material-specific critical temperature $T_c$, the resistivity drops sharply to exactly zero.

Key features:

• Below $T_c$, electrons form correlated pairs (Cooper pairs) that glide through the lattice without scattering.
• A current set up in a superconducting loop can persist for years with no measurable decay — a test of $\rho = 0$ to incredible precision.
• Superconductors expel magnetic fields from their interior (Meissner effect) — this is why they can levitate magnets.

Some important $T_c$ values

Material	$T_c$ (K)
Mercury (Hg)	4.2
Lead (Pb)	7.2
Niobium (Nb)	9.3
YBa$_2$Cu$_3$O$_7$ (YBCO)	92
HgBa$_2$Ca$_2$Cu$_3$O$_8$	134

Materials with $T_c > 77$ K (the boiling point of liquid nitrogen) are called high-temperature superconductors. Their discovery in 1986 was a huge leap — you no longer need expensive liquid helium to cool them.

Applications

• Superconducting magnets (MRI, particle accelerators, fusion reactors).
• SQUIDs — extremely sensitive magnetometers used in biomedical imaging.
• Maglev trains — frictionless magnetic levitation.
• Efficient power transmission (zero $I^2 R$ losses).

Key Point: A superconductor does not merely have low resistance — it has exactly zero resistance, and hence zero voltage drop for any finite current below $T_c$.

Solved Examples

Example 1: Resistance of Copper Wire at a High Temperature

The resistance of a copper coil at $20^\circ$C is $2.5\,\Omega$. Find its resistance at $100^\circ$C. ($\alpha_\text{Cu} = 4.0 \times 10^{-3}$ K$^{-1}$)

Solution: Using $$R_T = R_0[1 + \alpha(T - T_0)],$$ we get $$R = 2.5[1 + 4.0 \times 10^{-3}(100 - 20)] = 2.5(1 + 0.32) = 2.5 \times 1.32 = 3.30\,\Omega$$

Final Answer: $3.30\,\Omega$

Example 2: Finding the Temperature Coefficient

A platinum resistance thermometer reads $5.0\,\Omega$ at $0^\circ$C and $5.2\,\Omega$ at $50^\circ$C. Find $\alpha$ for platinum.

Solution: Using $$R_T = R_0[1 + \alpha(T - T_0)],$$ we get $$5.2 = 5.0[1 + 50\alpha]$$ $$1.04 = 1 + 50\alpha$$ $$50\alpha = 0.04$$ $$\alpha = 8.0 \times 10^{-4}\,\text{K}^{-1}$$

Final Answer: $8.0 \times 10^{-4}\,\text{K}^{-1}$

Example 3: Temperature from a Measured Resistance

A tungsten filament has $R_0 = 100\,\Omega$ at $20^\circ$C. When the bulb is lit, its resistance is $1100\,\Omega$. Estimate the operating temperature. ($\alpha_\text{W} = 4.5 \times 10^{-3}$ K$^{-1}$)

Solution: $$1100 = 100[1 + 4.5 \times 10^{-3}(T - 20)]$$ $$11 = 1 + 4.5 \times 10^{-3}(T - 20)$$ $$4.5 \times 10^{-3}(T - 20) = 10$$ $$T - 20 = \frac{10}{4.5 \times 10^{-3}} \approx 2222$$ $$T \approx 2242^\circ\text{C}$$

Final Answer: $\approx 2242^\circ\text{C}$

Example 4: Two Resistors in Series — Effective $\alpha$

A $40\,\Omega$ wire of copper ($\alpha_1 = 4.0 \times 10^{-3}$ K$^{-1}$) is joined in series with a $60\,\Omega$ wire of iron ($\alpha_2 = 5.0 \times 10^{-3}$ K$^{-1}$). Find the effective temperature coefficient of the combination.

Solution: For small temperature change, $$\alpha_\text{eff} = \frac{R_1\alpha_1 + R_2\alpha_2}{R_1 + R_2}$$ So, $$\alpha_\text{eff} = \frac{40(4.0 \times 10^{-3}) + 60(5.0 \times 10^{-3})}{100}$$ $$= \frac{0.16 + 0.30}{100} = 4.6 \times 10^{-3}\,\text{K}^{-1}$$

Final Answer: $4.6 \times 10^{-3}\,\text{K}^{-1}$

Example 5: Designing a Temperature-Stable Resistor

You want a series combination of a carbon resistor ($\alpha_C = -5 \times 10^{-4}$ K$^{-1}$) and a copper wire ($\alpha_\text{Cu} = +4.0 \times 10^{-3}$ K$^{-1}$) so that the total resistance is nearly temperature-independent. If the copper wire has resistance $R_\text{Cu} = 10\,\Omega$ at $20^\circ$C, how large should the carbon resistance be?

Solution: For zero net temperature coefficient, $$R_C\alpha_C + R_\text{Cu}\alpha_\text{Cu} = 0$$ $$R_C = -\frac{R_\text{Cu}\alpha_\text{Cu}}{\alpha_C} = -\frac{10 \times 4.0 \times 10^{-3}}{-5 \times 10^{-4}} = 80\,\Omega$$

Final Answer: $80\,\Omega$

Example 6: Silver Wire Near a Flame

A silver wire has resistivity $\rho_0 = 1.62 \times 10^{-8}\,\Omega\cdot\text{m}$ at $27.5^\circ$C. Find its resistivity at $100^\circ$C. ($\alpha_\text{Ag} = 4.1 \times 10^{-3}$ K$^{-1}$)

Solution: Using $$\rho_T = \rho_0[1 + \alpha(T - T_0)],$$ we get $$\rho = 1.62 \times 10^{-8}[1 + 4.1 \times 10^{-3}(100 - 27.5)]$$ $$= 1.62 \times 10^{-8}[1 + 0.29725]$$ $$= 1.62 \times 10^{-8} \times 1.29725 \approx 2.10 \times 10^{-8}\,\Omega\cdot\text{m}$$

Final Answer: $2.10 \times 10^{-8}\,\Omega\cdot\text{m}$

Example 7: Negative $\alpha$ — A Thermistor

A thermistor has resistance $R_0 = 2000\,\Omega$ at $20^\circ$C. Its temperature coefficient in this range is $\alpha = -0.045$ K$^{-1}$. Find its resistance at $40^\circ$C.

Solution: $$R = 2000[1 + (-0.045)(40 - 20)] = 2000(1 - 0.9) = 200\,\Omega$$

Final Answer: $200\,\Omega$

Note: This uses the linear approximation over a small range.

Example 8: A Heating Coil's Hot Resistance

A nichrome heating coil has resistance $50\,\Omega$ at $30^\circ$C. In operation, it reaches $830^\circ$C. If $\alpha_\text{Nichrome} = 4.0 \times 10^{-4}$ K$^{-1}$, find its hot resistance.

Solution: $$R = 50[1 + 4.0 \times 10^{-4}(830 - 30)] = 50[1 + 0.32] = 50 \times 1.32 = 66\,\Omega$$

Final Answer: $66\,\Omega$

Example 9: Solving for the Temperature Coefficient from Two Measurements

A wire has $R_1 = 2.0\,\Omega$ at $T_1 = 25^\circ$C and $R_2 = 2.5\,\Omega$ at $T_2 = 150^\circ$C. Find the temperature coefficient of the wire's material, taking $T_1$ as reference.

Solution: Using $$R_2 = R_1[1 + \alpha(T_2 - T_1)],$$ we get $$\alpha = \frac{R_2 - R_1}{R_1(T_2 - T_1)} = \frac{2.5 - 2.0}{2.0(150 - 25)} = \frac{0.5}{250} = 2.0 \times 10^{-3}\,\text{K}^{-1}$$

Final Answer: $2.0 \times 10^{-3}\,\text{K}^{-1}$

Example 10: Resistance Doubles at What Temperature?

A copper conductor has $R_0 = 5\,\Omega$ at $20^\circ$C. At what temperature will its resistance be twice its value at $20^\circ$C? ($\alpha_\text{Cu} = 4.0 \times 10^{-3}$ K$^{-1}$)

Solution: For doubling, $$2R_0 = R_0[1 + \alpha(T - 20)]$$ $$2 = 1 + \alpha(T - 20)$$ $$1 = \alpha(T - 20)$$ $$T - 20 = \frac{1}{4.0 \times 10^{-3}} = 250$$ $$T = 270^\circ\text{C}$$

Final Answer: $270^\circ\text{C}$

Example 11: Why $\alpha$ Depends on Reference Temperature

In a data table, the temperature coefficient of copper is listed as $\alpha_0 = 4.3 \times 10^{-3}$ K$^{-1}$ at $0^\circ$C and $\alpha_{20} = 4.0 \times 10^{-3}$ K$^{-1}$ at $20^\circ$C. Why are they different?

Solution: The coefficient $\alpha$ is defined relative to a reference temperature $T_0$: $$\alpha = \frac{1}{\rho_0}\frac{d\rho}{dT}\Bigg|_{T=T_0}$$ Since $\rho_0$ depends on the reference temperature, the quoted value of $\alpha$ changes slightly when the reference temperature changes.

Final Answer: $\alpha$ depends on the chosen reference temperature, so $\alpha_0$ and $\alpha_{20}$ need not be exactly equal.

Example 12: Manganin — Why It Is Special

Standard resistance coils are made of Manganin, whose temperature coefficient is about $2 \times 10^{-5}$ K$^{-1}$. Compare the fractional change in resistance over $\Delta T = 50^\circ$C for Manganin and pure copper.

Solution: For Manganin: $$\frac{\Delta R}{R} = \alpha \Delta T = 2 \times 10^{-5} \times 50 = 10^{-3} = 0.1\%$$

For copper: $$\frac{\Delta R}{R} = 4.0 \times 10^{-3} \times 50 = 0.20 = 20\%$$

Thus copper changes 200 times more for the same temperature rise.

Final Answer:

• Manganin: $0.1\%$
• Copper: $20\%$
• Copper changes 200 times more