Switch between natural and forced convection, then tune fin geometry to compute fin efficiency η, thermal resistance Rth, and junction temperature rise in real time. Find the optimal fin spacing interactively.
Convection Mode
Heat Sink Geometry
Base width W (mm)
mm
Fin length L (mm)
mm
Fin height H (mm)
mm
Fin thickness t (mm)
mm
Fin spacing s (mm)
mm
Conductivity k (W/mK)
W/mK
Air velocity U (m/s)
m/s
Heat load Q (W)
W
Results
While paused, move the sliders to update the result instantly.
Dissipation Visualizer — air sweeping through the fin channels carries heat away
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Dissipated Qdiss [W]
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Base temp Tbase [°C]
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Thermal Rth [°C/W]
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Fin efficiency η
Fin temperature (hot base → cooler tip) Inlet air (cool) Exhaust air (heated)
Results
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Number of fins N
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Fin efficiency η
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Thermal resistance (°C/W)
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Temp. rise ΔT (°C)
Thermal Resistance vs Fin Spacing & Fin Efficiency vs Fin Height
Rth
Fin Efficiency η
Theory & Key Formulas
Fin efficiency: $\eta = \dfrac{\tanh(mH)}{mH}$, $m=\sqrt{\dfrac{2h}{kt}}$
Total thermal resistance: $R_{th}= \dfrac{1}{\eta N h \cdot 2HL + h \cdot A_{base}}$
What is Finned Heat Sink Design?
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What exactly is a "fin efficiency," and why is it so important for this heat sink calculator?
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Basically, it's a measure of how good a fin is at its job. A perfect fin would be the same temperature as the hot base it's attached to, giving 100% efficiency. In practice, the fin tip is cooler, so efficiency is less than 1. In this simulator, if you set the fin height (H) very high or the conductivity (k) very low, you'll see the efficiency drop because the heat can't travel to the tip effectively.
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Wait, really? So just adding more fins or making them taller doesn't always help? What's the trade-off?
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Exactly! That's the key design challenge. Adding more fins (by reducing fin spacing, s) increases surface area but can choke the airflow, lowering the convective heat transfer coefficient, h. Try it: set the air velocity to a low value for natural convection and slide the fin spacing. You'll see the total thermal resistance, R_th, reach a minimum at an optimal spacing.
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So the goal is to minimize that R_th value. How do the parameters for a forced air cooling system, like in a gaming PC, differ from a passively cooled one?
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Great question. Forced convection (high U) gives you a much larger h. In the simulator, crank up the air velocity and watch R_th plummet. This allows you to use tighter fin spacing and taller fins effectively. A common case is a server CPU heatsink versus a silent TV set-top box heatsink. The former uses high velocity and dense fins; the latter uses wider spacing for buoyancy-driven air flow.
Physical Model & Key Equations
The core model assumes a rectangular fin with an insulated tip. The fin efficiency (η) calculates how effectively the fin's surface area transfers heat compared to a fin at a uniform base temperature.
Where: η = Fin efficiency (0 to 1). m = Fin performance parameter (1/m). h = Convective heat transfer coefficient (W/m²K), dependent on air flow. k = Thermal conductivity of the fin material (W/mK). t = Fin thickness (m). H = Fin height (m).
The total thermal resistance from the heat sink base to the ambient air combines the resistance of the finned area and the exposed base area.
$$R_{th}= \dfrac{1}{\eta N h \cdot (2HL) + h \cdot A_{base}}$$
Where: Rth = Total thermal resistance (K/W). A lower value means better cooling. N = Number of fins, determined by base width (W), fin thickness (t), and fin spacing (s). L = Fin length (m). Abase = Area of the base not covered by fins (m²).
The junction temperature rise is then simply: ΔT = Q × Rth.
Frequently Asked Questions
If η is low, the fins may be too tall or too thin. Reducing the fin height H or increasing the thickness t can lower the mH value and improve η. Additionally, switching to a material with higher thermal conductivity k (such as copper) is also effective.
In natural convection mode, the optimal fin spacing becomes wider, and a design with reduced fin height is suitable. In forced convection mode, narrowing the fin spacing and increasing the height can maximize heat dissipation performance. Switching modes allows comparison of optimal solutions under both conditions.
To check the junction temperature rise, in addition to the thermal resistance Rth, please input the power consumption Q (W) of the semiconductor device and the ambient temperature Ta (°C). The junction temperature Tj = Ta + Rth × Q is calculated instantly.
The optimal spacing is automatically calculated based on fluid dynamics from the convective heat transfer coefficient h and the fin geometry. If the result is extremely narrow, forced convection may not be assumed, or the fin height may be excessive. Switch to natural convection or reduce the fin height and recalculate.
Real-World Applications
Consumer Electronics (Phones, Laptops): Here, space is the ultimate constraint. Designers use this exact analysis to choose between a copper spreader (high k) or an aluminum fin array, optimizing fin density to work with tiny, low-speed fans without causing audible turbulence.
Power Electronics & EV Chargers: These components handle high heat loads (Q) in rugged environments. Engineers use these calculations to design extruded aluminum heatsinks, often in natural convection, ensuring reliability by keeping junction temperatures well below the semiconductor's maximum rating.
LED Lighting Fixtures: High-power LEDs generate significant heat that must be dissipated to prevent light output degradation and color shift. Passive finned heatsinks are common, and their design directly uses fin efficiency theory to maximize lifespan while meeting aesthetic and size requirements.
Automotive & Aerospace Electronics: In these fields, forced convection is available but comes with strict limits on weight, volume, and power for fans. CAE tools using this model help find the lightest, most compact heatsink that can survive the vibration and temperature extremes of the application.
Common Misconceptions and Points to Note
First, the idea that "selecting a material with high thermal conductivity (k) solves everything" is dangerous. While copper (k≈400 W/mK) is indeed superior to aluminum (k≈200 W/mK), its cost and weight can be more than double. What's crucial is the overall system performance, characterized by the "thermal resistance, Rth". For instance, a design using aluminum with a fin efficiency η=0.8 can outperform a more expensive copper design with η=0.5. Material selection should be a comprehensive judgment considering cost, weight, and manufacturability.
Next, the misconception that "with forced convection, you should just pack the fins as densely as possible". While a narrow fin pitch (e.g., s=1mm) can be effective if the fan's air velocity U is sufficiently high, real-world issues like fan performance variation and "clogging" from dust accumulation between fins are major problems. For example, in industrial equipment meant for long-term operation, it's standard practice to design the fin pitch 1.5 to 2 times wider than the calculated optimum to ensure reliability.
Finally, remember that the "junction temperature" calculated by tools is ultimately an ideal value. In reality, "contact thermal resistance" always occurs at the interface between the heat sink and the heat source. For instance, surface roughness or insufficient mounting pressure can easily cause temperatures to be 10°C to 30°C higher than calculated values. After simulation, a phase of temperature verification using thermography or actual measurement is essential.