Linear Regulator Power Dissipation Calculator
Calculate the power a linear regulator or LDO must dissipate, its efficiency, junction temperature, and the heatsinking needed.
Why linear regulators get hot
A linear regulator drops the difference between input and output as heat — it passes the same current straight through, so every volt of headroom costs power. Dropping 12 V to 5 V at 500 mA burns 3.5 W in the regulator while delivering only 2.5 W to the load.
P_d = (V_in − V_out) × I_out + V_in × I_q
Efficiency ≈ V_out / V_in
T_j = T_ambient + P_d × θ_JA
θ_JA is the thermal resistance from the silicon junction to the surrounding air. Multiply it by the watts being dissipated and you get the temperature rise above ambient. If the resulting junction temperature exceeds the part's maximum (typically 125 °C), it will thermally shut down or fail — you need a bigger package, more copper, a heatsink, or less headroom.
Related
For a small reference load a Zener shunt regulator may do; for efficiency at higher currents use abuck converter, which switches rather than burning the difference.
Thermal resistance depends on your board
The junction temperature calculated here is only as good as the θJA figure you supply, and that number is a property of the installationrather than the package. Datasheet values assume a specified copper area on a specified test board; a part soldered to a small pad on a thin board can run far hotter than the same part with a generous pour beneath it. Treat the result as a comparison between options rather than a prediction, and confirm with a thermocouple or an infrared camera on real hardware before relying on the margin.