Thermal Resistance
Heat flow behaves like electric current: temperature difference is the 'voltage' and thermal resistance the 'resistor'. Learn series/parallel networks, U-values, composite walls, contact resistance, critical insulation radius and electronics junction-to-ambient budgets.
Real heat paths pass through several layers — air film, plaster, brick, insulation, air film — or chip, solder, spreader, paste, heat sink, air. Treating each as a thermal resistance and adding them like resistors makes complex problems simple.
This is how building U-values and electronics cooling budgets are calculated.
An analogy
Ohm's law for heat: ΔT = Q̇ × R, just like V = I × R. Resistances in series add; the biggest one controls the flow.
Series and parallel paths
Layers one after another are in series: R_total = R₁ + R₂ + …. Side-by-side paths (a stud and insulation in a wall) are in parallel: 1/R = 1/R₁ + 1/R₂. The largest series resistance dominates — improve that one first.
U-values in buildings
U = 1/(R_si + ΣL/k + R_se), with standard surface resistances (≈ 0.13 inside, 0.04 outside m²K/W). Modern codes demand wall U ≈ 0.15–0.3 W/m²K. Thermal bridges (steel, concrete edges) bypass insulation and must be detailed out.
Electronics thermal budget
Chip power flows junction → case → interface → heat sink → air. With a maximum junction temperature (e.g. 125 °C) and ambient, the allowed total resistance is (T_J − T_A)/P; subtract known resistances to get the heat-sink requirement θ_SA.
Critical radius and thermal networks
Adding insulation to a small cylinder increases surface area for convection; below r_c = k/h, more insulation increases heat loss (useful for electrical wires). Large thermal networks with capacitances (RC networks) model transient behaviour of batteries, motors and buildings.
Formulas
- Thermal Ohm's law:
Q̇ = ΔT / R_total - Conduction / convection / contact:
R = L/(kA), R = 1/(hA), R = R″_c/A - Cylinder:
R = ln(r₂/r₁) / (2πkL) - Critical insulation radius:
r_c = k/h (cylinder) - Electronics:
T_J = T_A + P (θ_JC + θ_CS + θ_SA)
Worked example: Composite wall U-value
Given: Inside film 0.10, brick 0.2/0.7, insulation 0.05/0.04, outside film 0.04 (m²K/W); ΔT = 25 K, area 10 m²
Find: U and heat loss; compare without insulation.
- ΣR″ = 0.10 + 0.286 + 1.25 + 0.04 = 1.676 → U = 0.597
- Q̇ = 0.597 × 10 × 25 = 149 W
- No insulation: ΣR″ = 0.426 → Q̇ = 587 W
Answer: 149 W vs 587 W — insulation cuts loss by 75 %
Worked example: Heat-sink requirement
Given: CPU 100 W, T_J,max = 90 °C, ambient 35 °C; θ_JC = 0.2 K/W, θ_CS (paste) = 0.1 K/W
Find: Maximum heat-sink resistance.
- Total allowed = (90 − 35)/100 = 0.55 K/W
- θ_SA ≤ 0.55 − 0.2 − 0.1 = 0.25 K/W
Answer: Heat sink ≤ 0.25 K/W (a large tower cooler or liquid cooler)
Common mistakes
- Wrong: Adding U-values of layers. — Right: Add resistances, then invert: U = 1/ΣR″.
- Wrong: Ignoring the interface (thermal paste) resistance. — Right: Include contact resistance — it can be a big share with small areas.
Applications
- Buildings: U-values, thermal bridging and energy-performance certificates.
- Electronics: Heat-sink selection, thermal pads and PCB copper design.
- Batteries & motors: Thermal networks predict hot-spot temperatures and derating.
Key takeaways
- ΔT = Q̇R; series add, parallel invert-add.
- U = 1/ΣR″ for walls.
- Electronics: T_J = T_A + PΣθ.
- Critical radius r_c = k/h for small cylinders.
Going further
- Spreading resistance: Heat from a small source into a large plate has extra resistance; vapour chambers minimise it.
- Thermal capacitance: RC thermal models give time constants for warm-up and transient overloads (motor and battery ratings).
- Measurement standards: JEDEC JESD51 defines θ_JA/θ_JC test boards so datasheet values are comparable.