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Heat Sink Thermal Resistance: How to Calculate What You Need
Nov 11,2024

Heat Sink Thermal Resistance: How to Calculate What You Need

Short answer: thermal resistance R = ΔT / Q, where ΔT is the temperature rise you allow (in K or °C) and Q is the heat you must move (in watts). A 25 W module allowed to run 50 °C above ambient needs a sink of about 2 K/W or better. Rough sizing: natural convection moves 5–10 W per m² per degree, forced air 25–100 — so that 25 W sink needs roughly 0.06 m² of exposed surface with free air, less than half that with a fan. Send watts, temperatures, and airflow; the fin geometry follows.

Every heat sink question eventually becomes a math question, and the math is embarrassingly simple once you separate the two things people confuse: how much heat there is, and how hot things are allowed to get. Thermal resistance connects them. This guide runs the calculation with real numbers and tells you what to hand your factory so the quote you get is the sink you need.

What Is Thermal Resistance, Anyway?

Thermal resistance is the temperature rise you get per watt of heat, measured in kelvin per watt (K/W) — same scale as °C per watt. Think of it as the thermal version of electrical resistance: heat is current, temperature difference is voltage, and the sink is the resistor. A 2 K/W sink running 25 W rises 50 °C above its surroundings. Cut the resistance to 1 K/W and the same 25 W only lifts it 25 °C.

Heat flows through a chain of resistances: junction to case inside the component, case to sink through the thermal interface material, then sink to air. For buying and designing a heat sink, the number that matters is sink-to-ambient, Rth(s-a). It is the resistance you actually control with geometry and airflow.

SymbolMeaningTypical unit
QHeat dissipated by the componentW
T_maxMaximum allowed component/case temperature°C
T_ambAmbient air temperature°C
ΔTAllowable temperature rise (T_max − T_amb)K or °C
Rth(s-a)Thermal resistance, sink to ambientK/W
hConvection coefficient of the air sideW/m²·K
ATotal exposed surface area of the sink

How Do You Calculate the Thermal Resistance You Need?

The formula is R = ΔT / Q. Three steps, no more. First, decide the maximum temperature your component may reach — from the datasheet derating curve, not from hope. Second, subtract the worst-case ambient temperature the product will see in service. That difference is your total temperature budget. Third, divide by the heat load in watts. The result is the maximum total thermal resistance allowed from junction to ambient.

Hold on — one step is easy to skip and it hurts. The budget you just calculated covers the whole chain: junction-to-case, thermal interface material, and the sink. A real module might consume 1–2 K/W before heat even reaches the sink. If the datasheet says the junction may sit 60 °C above ambient and the internal chain eats 15 °C of that, the sink only gets to spend 45 °C. Subtract the known internal resistances first, then size the sink on what is left.

Worked Example: Sizing a Sink for a 25 W Module

Take a power module dissipating 25 W in an enclosure that may see 35 °C ambient. The datasheet allows the case to reach 85 °C. Budget: 85 − 35 = 50 °C. With a decent thermal pad adding roughly 0.5 K/W of interface resistance, the module case-to-air chain is already using 0.5 × 25 = 12.5 °C of the budget, leaving about 37 °C for the sink itself. The sink must therefore be around 37 / 25 ≈ 1.5 K/W or better.

Now rough-size the surface. Natural convection in still air moves about 5–10 W per square meter per kelvin. Take h = 8 W/m²·K as a middle value: A = Q / (h × ΔT) = 25 / (8 × 37) ≈ 0.084 m² of total finned surface. That is a real extruded heat sink profile — roughly 60 mm wide, 60 mm tall, 40 mm long with typical fins, a very ordinary part.

StepCalculationResult
1. Temperature budget85 °C − 35 °C50 K
2. Interface pad consumption0.5 K/W × 25 W12.5 K
3. Budget left for the sink50 − 12.537.5 K
4. Required Rth(s-a)37.5 / 25≈ 1.5 K/W
5. Surface area at h = 825 / (8 × 37.5)≈ 0.083 m²

Slap a fan on it and the story changes completely. Forced air at 2–3 m/s raises the convection coefficient into the 25–100 W/m²·K range. At h = 50, the same 25 W needs only 25 / (50 × 37.5) ≈ 0.013 m² of surface — the sink shrinks by more than 80%, which is exactly why every power supply has a fan in it.

What Air Flow Assumptions Should You Use?

The convection coefficient is where sizing lives or dies, and it is the number people guess. Use these typical values as a starting point, then let the geometry confirm them:

Cooling modeTypical h (W/m²·K)Notes
Natural convection, still air5–10Fins should stand vertical for best chimney effect
Gentle forced air, ~1 m/s10–25Small fans, some baffling
Moderate forced air, 2–3 m/s25–60Standard axial fan over a fin pack
Strong forced air, 5+ m/s60–100Ducted flow, high-pressure blowers

Three practical warnings. Natural convection hates horizontal fins — hot air cannot rise through them, and performance drops 20–40% versus vertical fins. Radiation matters more than textbooks admit for dark anodized sinks: it adds roughly the equivalent of 5–7 W/m²·K at typical electronics temperatures, so anodize is not just cosmetics. And the datasheet h values assume clean, unobstructed airflow — in a sealed box with no vents, natural convection numbers simply do not apply.

What Should You Send the Factory Instead of a CAD File?

A factory can machine any shape you draw. What we actually need to design the right fins is the thermal spec: heat load in watts, maximum allowed temperature, worst-case ambient, air flow mode and velocity if forced, the envelope you can afford, and how the sink mounts. If you already know the target Rth(s-a), send that too — it is the single most useful number.

InputWhy it matters
Heat load (W)Sets the size of everything
Max component/case temp (°C)Sets the temperature budget
Ambient temp (°C)Worst case, not typical
Airflow (natural / m/s)Picks h, changes area by 5–10×
Envelope (L×W×H)Sets fin length and count
Mounting and component footprintBase thickness, hole pattern, spreader layout

With those inputs we size the base thickness for heat spreading, choose fin pitch for the airflow you have, and pick the manufacturing process — thin dense fins want extrusion or skiving, prototypes and odd shapes want CNC-machined heat sinks, high volume folded packs want stamping. Which process wins at your volume is covered in our heat sink types guide, and the alloy choice is settled in our 6063 versus 6061 comparison. Either way, send the numbers to sc@bquq.com or WhatsApp +86 13713157787 and the quote comes back within 12 hours on working days.

Have a drawing? Get a factory quote within 12 hours.
Email sc@bquq.com or WhatsApp +86 137 1315 7787 with your PDF/DXF/STEP file. An engineer reviews it and replies with price, lead time and DFM notes on working days.

Which heat sink process fits? (Decision tree)

If your case...ChooseWhy
Uses a standard profile at 1,000+ pcsExtrudedLowest unit cost once the die exists
Is a prototype, low volume or odd shapeCNC machinedNo tooling, fast turnaround
Needs thin folded fins at high volumeStampedThin fins at low unit cost
Has high flux density at mid volumeSkivedDense fins, joint-free fin base
Dissipates more than about 300 WHeat pipe assemblySpreads heat beyond the base footprint
Is sealed with no airflowConduction to chassis or cold plateConvection is not available

Frequently Asked Questions

How do I calculate heat sink thermal resistance?

A: R = ΔT / Q. Divide the allowed temperature rise (max component temperature minus ambient) by the heat load in watts. Subtract internal resistances like junction-to-case and thermal pad first, then size the sink on the remaining budget.

What is a good thermal resistance for a heat sink?

A: There is no universal number. A typical extruded aluminum sink for a 10–50 W device lands between 0.5 and 3 K/W. Work backward from your temperature budget: if you must stay 40 °C above ambient at 20 W, you need 2 K/W or better.

How much surface area does a heat sink need?

A: With natural convection (h = 5–10 W/m²·K), about 0.02–0.04 m² per 10 W for a 40 K rise. Forced air at 2–3 m/s cuts the required area by 5–10 times.

Does anodizing improve heat sink performance?

A: Slightly. The coating adds radiation — worth roughly 5–7 W/m²·K at electronics temperatures — but a thick hard coat is insulating. Thin decorative anodize is the usual compromise, and it protects the surface.

What happens if I undersize my heat sink?

A: The component runs hotter than the datasheet allows, power derates, and life shortens — every 10 °C rise roughly halves the life of electrolytic capacitors and stresses semiconductors. Oversizing costs a few grams of aluminum; undersizing costs field failures.

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Data Sources and Verification

Tolerances, cycle times and price ranges in this guide come from BQUQ production records at our Dongguan plant, where CNC machining (±0.005 mm), stamping, custom springs and heat sinks run under one roof. BQUQ is an ISO 9001:2015 certified factory; the certificate and batch inspection reports are available on request with every quotation.

Related Resources

Authored by the BQUQ Engineering Team. BQUQ (Dongguan) runs CNC machining (±0.005 mm), metal stamping, custom springs, and heat sink production in one ISO9001 factory. Source-direct from Dongguan, China — quote in 12 hours: sc@bquq.com | WhatsApp +86 13713157787 | www.bquq.com



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