Enter thermal and geometry values
Use steady-state values. Dimensions are entered in millimetres. The radiation value is an optional equivalent coefficient.
Example Data Table
| Parameter | Example value | Purpose |
|---|---|---|
| Heat load | 35 W | Heat entering the sink. |
| Air coefficient | 35 W/m²·K | Moderate forced airflow estimate. |
| Material conductivity | 205 W/m·K | Common aluminum alloy estimate. |
| Fin geometry | 12 × 30 mm × 1.5 mm | Straight rectangular fins. |
| Base footprint | 100 mm × 80 mm | Available fin mounting area. |
Formula Used
These equations estimate steady-state behavior for straight rectangular fins. They assume nearly uniform material properties and an equivalent heat-transfer coefficient.
How to Use This Calculator
- Enter the device heat load and nearby ambient temperature.
- Choose a target base temperature below the component limit.
- Set convection from expected natural or forced airflow.
- Add radiation only when you have a reasonable equivalent estimate.
- Enter the material conductivity and straight-fin dimensions.
- Enter heat-source dimensions for the base-thickness estimate.
- Add interface and spreading resistance from your mounting design.
- Calculate, then review resistance, fin efficiency, temperature, and spacing.
- Export the result and validate the final assembly with measured data.
Thermal Resistance Fundamentals
A heat sink moves energy from a hot device into surrounding air. Thermal resistance measures how strongly that path opposes heat flow. Lower resistance means a smaller temperature rise for the same load. It is commonly expressed in degrees Celsius per watt. A value of two degrees Celsius per watt creates a twenty degree rise at ten watts. The value depends on material, shape, airflow, mounting, and nearby obstructions. A realistic estimate prevents weak cooling choices before hardware is built. The calculator combines these effects into one practical resistance estimate. It also predicts the heat sink base temperature. It does not replace laboratory validation for safety-critical hardware.
Fin Geometry and Efficiency
Fins increase the area exposed to air. More area usually improves convection. However, a long thin fin may become inefficient. Its temperature falls from the base toward the tip. The outer section then transfers less heat than an equally hot surface. Fin efficiency accounts for this temperature change. It compares actual heat transfer with ideal transfer at base temperature. High conductivity materials keep more of each fin useful. Thicker fins also improve conduction. They reduce available spacing and sometimes restrict airflow. Fin count therefore needs balance. The calculator uses a corrected fin length and a perimeter based fin model. This approach includes tip behavior. It works best for straight rectangular fins with uniform airflow. Curved fins, pin fins, and vapor chambers require specialized correlations.
Airflow and Material Choices
Air movement often dominates heat sink performance. Natural convection usually provides modest heat transfer. Forced airflow can reduce resistance sharply. The heat transfer coefficient represents that effect. It changes with fan speed, fin spacing, orientation, and channel length. Radiation can provide an additional path. Dark finished surfaces may radiate better than bright untreated ones. The calculator lets you add an equivalent radiation coefficient. Aluminum is common because it is light, affordable, and conductive. Copper conducts better but adds mass and cost. The base thickness and source footprint also matter. A small heat source can create spreading resistance. Poor mounting pressure or a thermal pad adds contact resistance. These losses may outweigh a small improvement in fin area.
Reading Results for Design Decisions
Review total resistance before reviewing cosmetic details. Multiply total resistance by heat load to find temperature rise above ambient. Add ambient temperature to estimate base temperature. Compare that value with the device limit and desired reliability margin. Use the target-temperature result to estimate allowable heat load. Check fin efficiency as well. Very low efficiency suggests fins are too tall, thin, or poorly conducting. Check calculated fin spacing for a practical airflow channel. Negative spacing means the selected geometry cannot fit on the base. Increase base width, reduce fin count, or use thinner fins. Treat the base conduction and spreading terms as engineering estimates. Real assemblies include screws, interface materials, enclosure recirculation, and fan degradation. Prototype testing should confirm the design under worst-case ambient conditions. Good cooling decisions use conservative inputs and verification.
FAQs
What does total thermal resistance mean?
It is the total opposition to heat flow from the source path into surrounding air. Multiply it by heat load to estimate temperature rise above ambient.
What fin shape does this model use?
It models straight rectangular plate fins. Each fin runs along the entered heat sink length. Pin fins, folded fins, and curved profiles need different correlations.
Why does airflow change the answer so much?
Airflow changes the convection coefficient. Higher coefficients transfer more heat from the fin surface. Fan performance, channel blockage, and recirculation can change the real coefficient.
When should I include radiation?
Include it when surface finish, temperatures, and surrounding view allow meaningful radiative transfer. Use a defensible equivalent coefficient. It is often smaller than forced convection.
Can this predict junction temperature?
Not directly. Add the component junction-to-case and case-to-sink resistances to this heat-sink result. Then multiply the combined resistance by heat load.
How is allowable heat at target temperature calculated?
The calculator divides the available temperature difference by total resistance. The target must be warmer than ambient. This is a steady-state estimate.
What is a good fin efficiency?
Higher is generally better. Values above about 70 percent indicate useful fins in this simplified model. Very low values suggest excessive height, low conductivity, or thin fins.
What happens when fin spacing is too small?
Airflow can choke, dust can accumulate, and the assumed convection coefficient may be unrealistic. Increase spacing or confirm performance with airflow testing.
Does copper always make the best heat sink?
No. Copper conducts heat well but is heavier and often more expensive. Aluminum may give better system value when airflow and surface area dominate.
Why do heat-source dimensions matter?
A small source pushes heat through a smaller base area. That increases local conduction and spreading losses before heat reaches the fins.
How should I validate a finished design?
Measure temperatures at worst-case power, ambient conditions, airflow, and enclosure state. Good designs keep components cooler, safer, reliable, and efficient.