Real-time calculation of fin efficiency, thermal resistance network, and junction temperature. Supports forced and natural convection. Instantly optimize power device and electronics cooling design.
Thermal Design Parameters
Heat Sink Material
Cooling Method
Power Dissipation Q
W
Ambient Temperature T_a
℃
Fin Height H
mm
Number of Fins N
Fin Thickness t_f
mm
Base Width W
mm
Heat Transfer Coeff. h
W/m²K
θ_j-c (Device Spec.)
K/W
θ_c-s (Interface Resistance)
K/W
Results
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T_j Junction Temp. [°C]
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θ_total [K/W]
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Fin Efficiency η [%]
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θ_s-a [K/W]
Thermal Resistance Network
Resistance Breakdown
Resist
Number of Fins vs Junction Temperature
Fin
Fin efficiency (rectangular fin):
$$\eta_{fin}= \frac{\tanh(mH)}{mH}, \quad m = \sqrt{\frac{h \cdot P}{k \cdot A_c}}$$
$$\theta_{s\text{-}a}= \frac{1}{\eta_o \cdot h \cdot A_{total}}$$
Junction temperature:
$$T_j = T_a + Q \cdot (\theta_{j\text{-}c}+ \theta_{c\text{-}s}+ \theta_{s\text{-}a})$$
CFD Integration
This 1D thermal resistance model is used during pre-design stages before Icepak, FloTHERM, or similar CFD thermal analysis. For PCB-mounted devices, FEM analysis accounting for thermal spreading in the board is required. It is recommended to determine h from CFD analysis of the fin array airside, then verify using this tool.
What is Heat Sink Thermal Resistance?
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What exactly is "thermal resistance" for a heat sink? I hear engineers say a heat sink has a resistance of, say, 5°C/W, but heat isn't electricity...
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Great question! Basically, it's an analogy. Just like electrical resistance ($R = V/I$) opposes current flow, thermal resistance ($R_{th}= \Delta T / Q$) opposes heat flow. A value of 5°C/W means for every watt of heat you pump into it, the heat sink's temperature will rise 5°C above the ambient air. In this simulator, the main result, $R_{th,s-a}$, is calculated from your inputs. Try increasing the Power Dissipation Q slider and watch how the predicted Junction Temperature shoots up—that's the resistance in action!
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Wait, really? So the fins are like the main cooling part. But why does the calculator ask for "Fin Efficiency"? Aren't all fins 100% efficient?
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Not at all! In practice, the tip of a fin is cooler than its base because heat has to conduct along the fin's length. A fin's efficiency ($\eta_{fin}$) is the ratio of its actual heat transfer to the heat transfer if the entire fin were at the base temperature. A short, thick, highly conductive fin has high efficiency. A common case is an aluminum fin being much more efficient than a steel one. In the simulator, change the Heat Sink Material from Aluminum to Steel and see the efficiency drop, which increases the overall thermal resistance.
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That makes sense. So what's the deal with the "Heat Transfer Coefficient, h"? It seems like a magic number that changes everything.
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It's the key link between the solid and the fluid! Basically, *h* quantifies how well the air (or other coolant) whisks heat away from the fin surface. A tiny fan (forced convection) gives a high *h*, while still air (natural convection) gives a low *h*. For instance, in a CPU cooler, *h* might be 5x higher with the fan on. That's why this tool lets you pick Cooling Method—it sets a typical *h* range. Slide the *h* control and watch the required heat sink size shrink dramatically as you move from natural to forced convection.
Physical Model & Key Equations
The core of the model is calculating the efficiency of a single rectangular fin. The temperature drops along the fin's length, reducing its effectiveness. The parameter *m* captures the balance between convective cooling (h) and conductive ability (k).
$$\eta_{fin}= \frac{\tanh(mH)}{mH}, \quad m = \sqrt{\frac{h P}{k A_c}}$$
Where:
$\eta_{fin}$: Fin efficiency (0 to 1).
$H$: Fin height (m).
$h$: Heat transfer coefficient (W/m²·K).
$P$: Fin perimeter (m) = $2*(W + t_f)$ for a rectangular fin.
$k$: Thermal conductivity of the material (W/m·K).
$A_c$: Fin cross-sectional area (m²) = $W * t_f$.
Since a heat sink has many fins, we calculate an overall surface efficiency for the entire array. This accounts for the fact that the base plate between fins is 100% efficient, while the fins are less so.
Where:
$\eta_o$: Overall surface efficiency.
$N$: Number of fins.
$A_{fin}$: Surface area of a single fin (m²).
$A_{total}$: Total exposed surface area of the heat sink (fins + base) (m²).
The overall thermal resistance from sink to ambient is then $R_{th,s-a}= 1 / (h \eta_o A_{total})$.
Frequently Asked Questions
No. Fin efficiency η_fin takes a value in the range of 0 to 1. A value of 1 indicates that the entire fin is at the base temperature and achieving ideal heat dissipation. In reality, the temperature decreases toward the fin tip, so it is always less than 1. The calculated value of tanh(mH)/mH is also always ≤ 1.
Change the heat transfer coefficient h. For forced convection, a typical setting is h = 10–100 W/m²K, and for natural convection, h = 5–15 W/m²K. Higher wind speeds increase h, improving cooling performance.
First, consider increasing the fin height H or reducing the fin spacing to increase the number of fins. Next, changing the material to one with higher thermal conductivity k, such as from aluminum (approx. 200 W/mK) to copper (approx. 400 W/mK), is also effective. If still insufficient, try switching to forced convection or increasing the wind speed.
Yes, as long as the cross-sectional area Ac and perimeter P are correctly entered, it can be applied to circular pin fins and elliptical fins as well. However, since the efficiency calculation for the entire fin array simplifies the interference effects between fins, errors with actual measurements may occur in extremely dense arrangements.
Real-World Applications
Power Electronics Cooling: Inverters for electric vehicles or industrial motor drives use high-power IGBTs and MOSFETs that can dissipate hundreds of watts. Engineers use this exact 1D resistance model to select or design a heat sink that keeps the semiconductor junction below its maximum rated temperature (e.g., 150°C), often using forced air cooling from a fan or liquid cold plate.
CPU & GPU Coolers: The heat sink on your computer's processor is a classic application. Pre-design starts with estimating the CPU's TDP (Thermal Design Power) and using a model like this to determine the necessary fin density, height, and base thickness before running detailed 3D CFD simulations in tools like Icepak or FloTHERM to optimize airflow.
LED Lighting Systems: High-brightness LEDs are sensitive to temperature; their light output and lifespan drop as temperature rises. Heat sink design is critical. This calculation helps determine the minimal aluminum extrusion profile needed to cool an LED array by natural convection, which is often required for outdoor, sealed fixtures.
Telecom & Server Equipment: Router boards and server blades pack many heat-generating components in a confined space. This model is used for initial component placement and heat sink selection on voltage regulators and network processors, ensuring the system meets its thermal budget before committing to a costly prototype and full-system thermal testing.
Common Misconceptions and Points to Note
When you start using this simulator, watch out for a few common pitfalls. First, understand that the heat transfer coefficient h is not a constant. While the tool uses a fixed input value, the actual h varies depending on airflow velocity, temperature, and the fin geometry itself. For example, even in forced convection, if the flow path between fins is too narrow, airflow velocity can drop due to viscous resistance, leading to a phenomenon called "flow blockage" where h becomes lower than expected. You can experience this in the simulator by drastically increasing the "Number of Fins N"—you'll see that while surface area increases, the reduction in thermal resistance plateaus.
Next, remember that thermal resistance networks are not purely series-based in reality. This tool uses a simple series model (junction → case → heatsink → air), but in actual power devices, parallel paths exist where heat escapes directly from the case to the enclosure. Therefore, treat the simulator's "Total Thermal Resistance" as a worst-case value (heatsink path only). This is the correct approach for a conservative (higher temperature) estimate.
Finally, keep in mind that a material's thermal conductivity k changes with temperature. Especially for resin-encapsulated packages or low-cost heatsink materials, k can decrease by 10-20% at the intended operating temperature. The values you see when switching "Materials" in the simulator are representative values at room temperature, so it's safer to include a margin when designing for high-temperature environments.