Temperature coefficient of resistivity for semiconductors Consider intrinsic or lightly doped semiconductors around room temperature. Do semiconductors have a positive temperature coefficient of resistivity?

Introduction / Context:
The temperature dependence of resistivity differentiates conductors, semiconductors, and insulators. Metals usually show a positive temperature coefficient (resistivity increases with temperature), while semiconductors behave differently because carrier concentration changes dramatically with temperature.

Given Data / Assumptions:

• Intrinsic or lightly doped semiconductor near ambient conditions.
• Moderate temperature range where the bandgap remains fixed and no phase change occurs.
• Ohmic behavior (linear I–V) assumed.

Concept / Approach:
Resistivity ρ is inversely proportional to charge transport capacity: ρ ∝ 1 / (q (n μ_n + p μ_p)). In semiconductors, as temperature T increases, intrinsic carrier concentration n_i rises steeply (n_i ∝ T^(3/2) * exp(−E_g / (2 k T))). Even though carrier mobilities μ decrease with temperature due to increased phonon scattering, the exponential growth in n and p dominates, reducing ρ with temperature. Therefore, the temperature coefficient of resistivity is negative, not positive.

Step-by-Step Solution:
Note metal vs semiconductor mechanisms: metals mainly mobility-limited; semiconductors carrier-concentration-dominated.Increase T → n_i increases exponentially.Mobility decreases with T but more slowly than n_i increases.Net effect: conductivity increases, resistivity decreases → negative temperature coefficient.

Verification / Alternative check:
Measured I–T curves for silicon diodes/transistors show leakage and conductivity increasing with temperature, consistent with decreasing resistivity.

Why Other Options Are Wrong:
“True” would describe metals, not semiconductors in the intrinsic/lightly doped regime.

Common Pitfalls:
Extrapolating metal behavior to semiconductors; ignoring effects of heavy doping or freeze-out at very low temperatures.

Final Answer:
False