What sets heat sink performance?

Surface area, heat transfer coefficient, fin efficiency and airflow. More fins add area but narrow the gaps, resisting flow and reducing fan airflow. Compare performance by thermal resistance (temperature rise ÷ heat) at the operating point where fan curve and pressure drop balance.

How is fin count chosen?

Find airflow at the intersection of fan curve and heat-sink pressure drop, and choose the count minimising thermal resistance there. In the example, resistance fell from 0.68 K/W at 10 fins to 0.21 at 25 and 0.16 at 35, but rose to 0.18 at 40, and at 50 airflow fell to 2.3 L/s, worsening it to 0.39 K/W. The optimum is around 35.

What is easily overlooked in heat sink calculations?

The air heating up. At low airflow, air temperature rises toward the outlet, shrinking the difference to the surface. In the example at 50 fins, surface heat transfer alone suggests an excellent 0.034 K/W, but including air temperature rise gives 0.39 K/W. Compare using resistance referenced to inlet air temperature.

Heat Sink Design

Category: Electronics Cooling | Revised 2026-10-01

Heat Sink Design

Theory: trading surface area against airflow

Overview

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I thought more fins always meant better cooling, but in the prototype temperature actually rose. Why?

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More fins add area but narrow the gaps, making airflow harder. A fan delivers less air against more resistance, so beyond some count the airflow loss outweighs the area gain. And with little airflow the air itself heats toward the outlet, shrinking the difference to the fins. The basics: set airflow by balancing fan curve and pressure drop, and compare thermal resistance including air temperature rise.

Thermal resistance to inlet air

$$ R_{th} = \frac{1}{\dot m c_p\left(1 - e^{-NTU}\right)},\qquad NTU = \frac{h\,A_{eff}}{\dot m c_p},\qquad A_{eff} = 2nHL\,\eta_f + A_{base} $$

Using heat-exchanger effectiveness (NTU), thermal resistance including air temperature rise is obtained. $\dot m c_p$ is air capacity rate and $A_{eff}$ the effective area including fin efficiency.

Fin efficiency and operating point

$$ \eta_f = \frac{\tanh(mH)}{mH},\ \ m = \sqrt{\frac{2h}{k t}},\qquad \Delta p_{hs}(Q) = \Delta p_{fan}(Q) $$

$H$ is fin height, $t$ thickness and $k$ material conductivity. Airflow $Q$ is where heat-sink pressure drop meets the fan's static-pressure curve.

Coffee Break Trivia

From vacuum tubes to CPUs

Cooling electronics has been a challenge since the vacuum-tube era. Transistors cut heat, but as integrated circuits sped up, heat rose again, and by the 1990s PC CPUs needed fan-equipped heat sinks. Research deriving optimal fin spacing theoretically (such as Bar-Cohen and Rohsenow) became a design guide. Today heat pipes and vapour chambers spread heat into fins, and CFD optimising fan and heat sink together is standard.

Worked example: varying fin count

Aluminium (200 W/mK), 100×100 mm base, 30 mm tall 1.2 mm fins, 40 °C air. Fan curve linear from 80 Pa static pressure to 15 L/s maximum flow. Temperature rise at 60 W is shown too (base spreading resistance omitted):

Fins	Gap	Airflow	Pressure drop	h	Resistance	Rise at 60 W
10	9.8 mm	12.4 L/s	14 Pa	24 W/m²K	0.677 K/W	40.6 K
20	4.0 mm	11.1 L/s	21 Pa	35 W/m²K	0.285 K/W	17.1 K
30	2.2 mm	8.8 L/s	33 Pa	52 W/m²K	0.173 K/W	10.4 K
35	1.7 mm	7.1 L/s	42 Pa	64 W/m²K	0.160 K/W	9.6 K
40	1.3 mm	5.4 L/s	51 Pa	79 W/m²K	0.176 K/W	10.5 K
50	0.8 mm	2.3 L/s	68 Pa	126 W/m²K	0.385 K/W	23.1 K

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The heat transfer coefficient keeps rising with fin count, yet resistance bottoms out around 35 fins.

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Narrower gaps raise h, but airflow falls and the air heats. At 50 fins airflow drops to 2.3 L/s and 60 W heats the air by about 23 K. Surface heat transfer alone makes 50 fins look superb at 0.034 K/W, but including air temperature rise gives 0.39 K/W. Miss that and you over-pack the fins. In practice base spreading resistance and dust clogging also matter, so leaving the design slightly coarser than optimal is sensible.

Design workflow

Set the required resistance from heat load, allowable rise and ambient air temperature.
Obtain the fan curve (static pressure versus airflow).
Vary fin count, height and thickness, comparing resistance at the operating point including air temperature rise.
Add base spreading and interface-material resistances.
Check with CFD for flow maldistribution in the enclosure (air bypassing the sink).

Coffee Break Trivia

“The airflow never arrives”

A heat sink predicted to cool well on its own ran components over 10 K hotter once installed. Whole-device CFD revealed gaps between sink and enclosure through which nearly half the fan's air escaped by the low-resistance path instead of between the fins. Adding a duct-like baffle to guide air through the fins brought temperatures near prediction. Heat-sink performance is set by the whole air path, not the sink alone.

Common mistakes

Mistakes and fixes

Mistake	Effect	Fix
Assuming fixed airflow	Over-packed fins	Use fan curve operating point
Ignoring air temperature rise	Resistance underestimated	Reference inlet air
Fin efficiency of 1	Tall fins overrated	Compute efficiency
Ignoring bypass flow	Hotter in the product	Ducting and system CFD
Ignoring base spreading	Hot spot above source	Add spreading resistance

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I'd like to try related calculations.

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Try the heat sink designer and finned heat sink calculator. Related pages include junction temperature, thermal resistance, thermal interface materials, power module cooling and pin-fin arrays.

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About the Authors

Heat Sink Design

Theory: trading surface area against airflow

Overview

Thermal resistance to inlet air

Fin efficiency and operating point

From vacuum tubes to CPUs

Worked example: varying fin count

Design workflow

“The airflow never arrives”

Common mistakes

Mistakes and fixes

Related Topics

Related fields