Linear Regulator (LDO) Power & Temperature Calculator

Q: What does the "low dropout" in LDO mean?

It can keep regulating with only a small difference between input and output, often a few hundred millivolts. It does not reduce heat at a large drop.

Q: How do I lower θ_JA?

Use a package with an exposed thermal pad, add copper area and thermal vias, attach a heatsink, or improve airflow.

Q: Is 41.67% efficiency normal for a linear regulator?

Yes, efficiency is roughly V_out/V_in regardless of the part, so a 12 V to 5 V drop always wastes more than half the input power.

Understanding the LDO regulator heat calculator — concept and principle

A linear regulator, including a low-dropout (LDO) type, produces a steady lower voltage by dropping the difference across a pass transistor. That difference times the load current turns directly into heat. Use it to check that a regulator stays within its temperature limit before choosing a part, package or heatsink.

An LDO burns the voltage difference as heat: 1.4 W here, which lifts the chip to 95 °C.

How it is calculated

Power dissipated: P = (V_in − V_out) × I, in watts.
Junction temperature: T_j = T_a + P × θ_JA, where θ_JA (°C/W) is the junction-to-ambient thermal resistance from the datasheet and T_a is the surrounding air temperature.
Efficiency: η = V_out / V_in × 100%.

The small ground (quiescent) current of the regulator is ignored, which is reasonable when it is much smaller than the load current.

The principle

A linear regulator passes the full load current from input to output, so input power is V_in × I and output power is V_out × I; whatever is left over is heat. Heat flowing out of the silicon behaves like current through a resistor: the temperature rise equals power times thermal resistance. θ_JA lumps together the path from the chip through the package and the circuit board into the air, so it depends strongly on the package and the copper area around it.

Worked example

Inputs: V_in = 12 V, V_out = 5 V, I = 0.2 A, θ_JA = 50 °C/W, T_a = 25 °C.
P = (12 − 5) × 0.2 = 1.4 W.
Temperature rise = 1.4 × 50 = 70 °C.
T_j = 25 + 70 = 95 °C; thermal check: OK (< 125 °C).
Efficiency = 5 / 12 = 41.67%.

Each point is the calculator re-run with only “Input (V)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Input (V)	Power dissipated (W)
3	-0.4
6	0.2
9	0.8
12	1.4
15	2
18	2.6
24	3.8

How to read the result

A 95 °C junction is 30 °C below the roughly 125 °C maximum most regulators allow, so this design is acceptable at 25 °C air. Turn the formula around to find the power budget: P_max = (T_j,max − T_a) / θ_JA = (125 − 25) / 50 = 2 W, meaning only about 0.29 A at a 7 V drop. At 0.5 A it would dissipate 3.5 W and reach 200 °C. Fixes include a lower input voltage, a better heatsink or more copper (lower θ_JA), or a switching (buck) converter.

Common mistakes

Looking only at the dropout voltage and ignoring the heat at full input voltage.
Using θ_JA from the datasheet as if it applied to any board; it is measured on a specific test board and can be much higher on a small PCB.
Calculating at room temperature when the device will sit in a warm enclosure.

Keep in mind: This is a steady-state estimate. Real thermal behaviour depends on layout, airflow and neighbouring parts. Check the manufacturer's datasheet and measure the board temperature; safety-critical designs need a qualified engineer.

Frequently asked questions

What does the "low dropout" in LDO mean?