Heatsink thermal resistance calculator
The largest heatsink thermal resistance that keeps the junction at its limit, given the power, ambient, the package θJC and the interface material θCS.
- Total θJA allowed
- 7.5 °C/W
- Temperature budget
- 75 °C
- Case temperature at the limit
- 110 °C
- Heatsink temperature at the limit
- 105 °C
- Junction with the heatsink entered
- 110 °C
How this is calculated
θSA ≤ (T_J − T_A) / P_D − θJC − θCS θJA = θJC + θCS + θSA T_case = T_J − P_D · θJC T_sink = T_A + P_D · θSA
- Model
- Steady-state series resistances, the thermal analog of Ohm's law (SLVA118A Figure 6). All heat is assumed to leave through the heatsink.
- Interface
- SLVA118A gives 0.1 to 1 °C/W for common interface materials, up to 5 °C/W with none, and about 0 °C/W for an exposed pad soldered to copper. Use your material's datasheet figure for its thickness and pressure.
- Airflow
- Heatsink θSA changes with airflow (SLVA462). Compare against the sink's rating at your airflow, in still air if there is no fan.
- TI SLVA118A, Linear Regulator Design Guide for LDOsRevised June 2008, Equations 8 to 10
- TI SLVA462, Understanding Thermal Dissipation and Design of a HeatsinkMay 2011, Equations 3 and 4
Questions
Divide the temperature budget by the power, then subtract θJC and θCS. 10 W, 125 °C, 50 °C ambient, θJC 1.5 and θCS 0.5 °C/W leaves 5.5 °C/W for the heatsink.
Then no heatsink can hold the junction at the limit: the package and interface already use the whole budget. Reduce power or ambient, use a better interface, or choose a package with lower θJC.
SLVA118A puts common interface materials at 0.1 to 1 °C/W and a dry joint at up to 5 °C/W. At 10 W, the difference between 0.5 and 5 °C/W is 45 °C at the junction.
Use it as a ceiling, not a target. Many design rules derate it; SLVA118A Example 2 derates 125 °C to 115 °C, which cuts the allowed θJA from 39 to 34 °C/W at 1.9 W and 50 °C.