Pick a resistor for an LED

LEDs burn out without a current limiter. R=(Vs−Vf)/I gives the value — then round to the E24 shelf and check the wattage.

Tools used in this guide

The job

A bare LED on a 5 V rail pulls all the current it can and dies. The series resistor burns off the voltage difference; Ohm's law gives the ohms, then you pick the closest real part.

Compute it

Enter supply voltage, the LED's forward voltage (≈1.8–2.2 V red, ≈3.0–3.4 V white/blue), and target current (20 mA for standard LEDs). The tool returns the ideal value, the next E24 value up, dissipation, and a suggested wattage rating ≥2×.

Series LEDs share one resistor: drop V = Vf × count.
Supply must exceed the total drop or there is nothing to limit — the tool flags that.

Worked example

9 V supply, one red LED (2.0 V, 20 mA): R = (9 − 2)/0.02 = 350 Ω. Nearest E24 up is 360 Ω; dissipation is 0.14 W, so a common 1/4 W part is comfortable. Actual current with the 360 Ω fitted: ~19.4 mA.

Limits

Assumes the LED behaves like a fixed voltage drop — near enough for design purposes. Real Vf varies with temperature and batch; the datasheet wins. For reading the color bands on the part you picked, use the resistor color tool.