Thermal effects on conductivity

WHAT is happening (definitions)

WHY metals and semiconductors behave oppositely

The master equation $\sigma = nq\mu$ has two temperature-dependent parts: $n$ and $\mu$.

• Metals: $n$ is basically fixed — the "sea" of free electrons is already there and temperature barely changes it. Only mobility $\mu$ falls because hotter lattice = more vibrations = more scattering. Result: $\sigma \downarrow$, $\rho \uparrow$.
• Semiconductors: $\mu$ also falls a bit, BUT $n$ rises exponentially as electrons jump the band gap. The huge growth in $n$ dominates. Result: $\sigma \uparrow$, $\rho \downarrow$.

HOW to derive the metal law (from scratch)

Step 1 — Drift from Newton. Between collisions the field $E$ gives acceleration $a = qE/m$. Over average time $\tau$ the drift velocity gained is: $v_d = a\,\tau = \frac{qE\tau}{m}$ Why this step? $\tau$ is the "memory" the electron keeps before a collision randomizes it.

Step 2 — Define mobility. Mobility is drift per field: $\mu = \frac{v_d}{E} = \frac{q\tau}{m}$ Why this step? It isolates the material/temperature property from the applied field.

Step 3 — Plug into $\sigma$. $\boxed{\sigma = nq\mu = \frac{nq^2\tau}{m}}$ Why this step? Now the temperature story lives entirely in $\tau$ (and $n$).

Step 4 — How does $\tau$ depend on $T$? Collision rate $\propto$ vibration amplitude squared, and lattice thermal energy $\propto T$, so scattering rate $1/\tau \propto T$. Hence $\tau \propto 1/T$ and for a metal (constant $n$): $\rho = \frac{m}{nq^2\tau} \propto T$ This is why the metal resistivity is roughly linear in $T$ near room temperature: $\rho(T) = \rho_0\big[1 + \alpha(T - T_0)\big]$

HOW to derive the semiconductor law

Carrier creation needs energy $E_g$ (the band gap). The fraction of electrons with enough thermal energy follows Boltzmann statistics, giving: $n \propto e^{-E_g/(2k_BT)}$ Why the factor 2? Each excited electron leaves behind a hole; the pair-creation probability splits the gap energy across the two carriers, so the exponent carries $E_g/2$.

Therefore: $\boxed{\sigma \propto e^{-E_g/(2k_BT)}}$ As $T\uparrow$, the exponent's magnitude shrinks, $\sigma$ grows explosively. This exponential beats the mild $\mu$ decrease.

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Connections

• Free Electron Model of Metals
• Band Theory & Energy Gaps
• Intrinsic vs Extrinsic Semiconductors
• Mobility and Drift Velocity
• Superconductivity (τ → ∞ limit at low T)
• Doping and Carrier Concentration

Concept Map

Hinglish (regional understanding)