Power Dissipation Calculator — what rating do I actually need?

Component dissipation and the derated part rating you should actually specify.

Example: Dissipating 250 mW average at 25 °C means specifying at least a 0.5 W part.

Formula

P = I²R or V×I    P_avg = P_peak × duty    rating ≥ P_avg / derating × margin

Worked example

Dissipating 250 mW average at 25 °C means specifying at least a 0.5 W part.

  1. P = I² × R

    (50 mA)² × 100 Ω

    250 mW

  2. P_avg = P_peak × duty

    250 mW × 1

    250 mW

  3. rating ≥ P_avg / derating × margin

    250 mW / 1 × 2

    500 mW

Frequently asked questions

What power rating resistor do I need?

At least twice the dissipated power, and more if it runs hot. Compute P = I²R, apply your duty cycle, divide by the temperature derating factor, then double it. A resistor dissipating 0.2 W at room temperature wants a 0.5 W part.

Why derate for temperature?

Because the rating assumes the part can shed heat into cooler surroundings. Resistor ratings are quoted at 70 °C ambient and fall linearly to zero at 155 °C. Inside a warm enclosure a 1 W resistor may only be good for half a watt.

Can I use average power for a pulsed load?

Usually, but only if the pulses are much shorter than the part's thermal time constant — a few seconds for a small resistor. Slower pulses let it heat towards the peak between them, and then you must size for the peak instead.

Is power the only rating I need to check?

No. Resistors also have a maximum working voltage, and small SMD parts are often limited to 50–200 V regardless of power. A high-value resistor in a high-voltage divider can be well within its power rating and still arc over.

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