Size the series resistor for one LED from supply voltage, forward voltage and target current — exact ohms, the nearest E12 value, actual current and wattage. Runs entirely in your browser — your numbers never leave this device.
The recommended wattage is at least twice the power the resistor actually dissipates, the standard derating habit for a part that runs warm in an enclosed project.
An LED fixes its own forward voltage — about 2 V for a red part, about 3.2 V for a typical white or blue one — and will draw destructive current trying to hold it, so a series resistor must absorb the rest of the supply and set the current: R = (Vs − Vf) ÷ I. Enter the supply, the LED's forward voltage from its datasheet, and the target current in milliamps (20 mA is the standard rating for indicator LEDs), and the exact resistance comes back.
Exact values are rarely stocked values, so the calculator rounds up to the smallest E12 preferred value at or above the requirement — 290 Ω becomes a 330 Ω part. Rounding up, never down, guarantees the actual current lands at or below target: at 330 Ω the 9 V example runs at 17.58 mA, not over 20 mA. The result also shows the power the resistor dissipates and the rating to buy: the smallest standard rating, from 0.25 W to 5 W, that is at least twice the dissipation, the usual derating habit for a part that will run warm.
Because resistors are manufactured in preferred-value series, and E12 (twelve values per decade) is the common one for general parts. Rounding up is also the safe direction: a larger resistance can only reduce the LED current below target, while rounding down to 270 Ω in the 9 V example would push the current above the 20 mA the LED was specified for.
The value from your LED's datasheet at your target current — typical figures are about 1.8–2.2 V for red, 2–2.4 V for green and yellow, and 3–3.4 V for blue and white. Forward voltage varies between parts and rises slightly with current, which is another reason the E12 round-up margin is welcome rather than wasted.
For LEDs in series, yes: add their forward voltages together and enter the sum as the forward voltage — two white LEDs at 3.2 V each are a 6.4 V load, and the supply must exceed that total. LEDs in parallel should each get their own resistor, because no two share current exactly equally and one part will hog current and fail first.
A resistor run at its full rating runs very hot — hot enough to discolour a board and to drift in value — and its rating assumes free-air conditions that a closed project box does not provide. Doubling the rating (0.06 W dissipated, 0.25 W part) keeps the part cool, stable and long-lived for a cost of essentially nothing.
The LED cannot light: there is no positive voltage left for a resistor to drop, and this calculator says so rather than inventing a negative resistance. Use a higher supply, a lower-forward-voltage LED, or a boost driver circuit designed for the job.
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