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Calculate Thermal Resistance
Thermal resistance Rth is the key parameter in heat sink design. Enter power dissipation and temperature limits: the calculator instantly determines the required Rth.
Why is Thermal Resistance Critical?
Every electronic component has a maximum operating temperature that must not be exceeded. The heat sink must dissipate the generated power loss so that this limit is maintained. Thermal resistance Rth (unit: K/W) describes how efficiently a heat sink achieves this: the smaller the Rth, the more heat can be dissipated at the same temperature difference.
Rth Calculator
Unit of Thermal Resistance: Kelvin per Watt (K/W)
Thermal resistance Rth is expressed in Kelvin per Watt (K/W), equivalent to °C/W. The value indicates how many Kelvin a component heats up per Watt of dissipated power loss. An Rth of 0.1 K/W means: at 100 W power dissipation, the temperature rises by 10 K relative to the cooling medium. The smaller the K/W value, the more capable the heat sink.
Formula and Derivation
ΔT = T_max − T_0 The temperature difference is the maximum component temperature minus the ambient or coolant inlet temperature.
R_th = ΔT / P_V The required thermal resistance is the temperature difference divided by the power dissipation. Unit: K/W.
Total Thermal Resistance: the Resistance Chain
The calculated Rth is the total thermal resistance from the component to the cooling medium. In practice it is made up of several partial resistances connected in series:
R_th,total = R_th,JC + R_th,CH + R_th,HA - R_th,JC – junction to case (junction-to-case, from the component datasheet)
- R_th,CH – case to heat sink (thermal paste, insulating pad, contact interface)
- R_th,HA – heat sink to ambient or cooling medium (the part you can design)
Each interface adds up. The heat sink must be designed so that the sum of all partial resistances does not exceed the required total Rth.
Practical Calculation Example
An IGBT module has a maximum junction temperature of 70 °C. The power dissipation is 2,000 W, the cooling water inlet temperature is 20 °C.
-
ΔT = 70 °C − 20 °C = 50 K -
R_th = 50 K / 2,000 W = 0.025 K/W
The required total thermal resistance is 0.025 K/W. This value is only achievable with liquid cooling.
Reference Values: Typical Thermal Resistance by Cooling Type
The following table maps typical Rth ranges to the appropriate cooling technology. It helps you quickly classify the calculated value and choose the right cooling strategy.
| Required Rth | Suitable Cooling Type | Typical Application |
|---|---|---|
| > 0.2 K/W | Standard air heat sink | Low power loss, sufficient installation space and airflow |
| 0.05 – 0.2 K/W | High-performance air cooling (Bondfin, Pinfin) | Medium power loss, limited installation space |
| < 0.05 K/W | Liquid cooling (Monopress, Structureflow) | High power density, IGBT modules, inverters |
Related calculators and topics
Frequently Asked Questions
Thermal resistance Rth is a system parameter that indicates how much a component heats up per Watt of power loss. It is measured in Kelvin per Watt (K/W). A low value means that a lot of heat is dissipated at a small temperature difference – the heat sink works efficiently.
Thermal resistance is expressed in Kelvin per Watt (K/W), occasionally also in °C/W – for temperature differences both units are equivalent. The value describes the temperature rise per Watt of dissipated power loss.
Thermal conductivity λ (W/mK) is a material property that indicates how well a material conducts heat. Thermal resistance Rth (K/W) is a system property that describes how much temperature difference arises for a given heat power in the overall system. Rth depends on material, geometry, contact interfaces and flow conditions.
No. The calculator computes the total required thermal resistance from the component to the cooling medium. In practice, this consists of several partial resistances: semiconductor → package → thermal paste → heat sink → cooling medium. Each of these interfaces must be taken into account in a detailed design.
A very small Rth (< 0.05 K/W) means that a large amount of heat must be dissipated over a small temperature difference, practically unachievable with air cooling. Liquid cooling is required in this case. COOLTEC offers various designs from Monopress to Structureflow.
The calculation is a first estimate based on simplified assumptions. It is suitable for preliminary sizing and assessment of the cooling strategy. Thermal simulations, measurements and specific geometry data are required for the final design.
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