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Linear Regulator Power Dissipation Calculator

Calculate the power a linear regulator or LDO must dissipate, its efficiency, junction temperature, and the heatsinking needed.

θ_JA depends on package and copper area — roughly 65 °C/W for a bare TO-220, 200+ °C/W for SOT-23, far lower with a heatsink. Dropout is from the datasheet: ~2 V for a classic 7805, 0.1–0.3 V for a modern LDO.
Power dissipated–
Efficiency–
Junction temperature–
Thermal verdict–
Max θ_JA to stay in spec–
Headroom vs dropout–
V_inREGV_outP_d
The regulator drops (V_in − V_out) at the load current — all of it as heat

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.

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