About LED Resistor Calculator — Series Resistor Value & Wattage
An LED is a diode, not a lamp: above its forward voltage its current rises almost vertically, so something external must limit that current. The simplest limiter is a series resistor, and this calculator sizes it: enter the supply voltage, the forward voltage of one LED, how many LEDs are wired in series, and the current you want, and it computes R = (Vs − ΣVf)/If, then rounds up to the next standard E24 resistor value from the IEC 60063 preferred-number series — up, because a slightly larger resistor means slightly less current, which is the safe direction.
The tool also answers the question that burns fingers: how much power the resistor dissipates. P = (Vs − ΣVf)·If all turns into heat in the resistor, and the calculator recommends a power rating of at least twice that figure from the standard ⅛ W to 10 W ladder, so the part runs warm instead of at its limit. Typical forward voltages are offered as datasheet-verified hints (Kingbright WP7113 family: red ≈ 2.0 V, old GaP green ≈ 2.0 V, modern InGaN green, blue, and white ≈ 3.3 V typical) — always confirm against your own part's datasheet, because Vf varies by chemistry, bin, and current.
How It Works
- Enter the DC supply voltage. For battery projects use the real terminal voltage (a "9 V" alkaline starts near 9.5 V and dies around 6 V; a "12 V" car system sits near 13.8 V when charging) — the LED current follows the supply through the resistor equation.
- Enter the forward voltage of ONE LED, or pick a color to prefill a datasheet-typical value. The tool multiplies by the series count to get the string voltage ΣVf; the supply must exceed ΣVf or no resistor can set the current, and the calculator says so.
- Enter the LED current. Standard 5 mm indicators are rated 20 mA absolute-maximum; 5–10 mA is plenty bright for most indicators and extends life. The exact resistance R = (Vs − ΣVf)/If appears alongside the next E24 standard value at or above it.
- Read the wattage recommendation: dissipation P = (Vs − ΣVf)·If, doubled for headroom, then rounded up to the standard rating ladder. A resistor at 100% of its rating runs at its maximum surface temperature — the 2× margin keeps it merely warm.
- Check the "actual current" line: with the E24 value (slightly above the exact one) the current lands slightly below your target — e.g. 19.6 mA instead of 20 mA with 510 Ω instead of 500 Ω. If that matters, series-parallel combinations or the next series down (E48/E96) close the gap.
Worked Example
Wire one red LED (Vf ≈ 2.0 V) to a 12 V supply at 20 mA. The resistor must drop 12 − 2.0 = 10 V, so R = 10 / 0.020 = 500 Ω exactly. 500 Ω is not a standard value: the E24 series jumps 470 → 510, and rounding UP picks 510 Ω, giving an actual current of 10 / 510 = 19.6 mA — imperceptibly dimmer and safer than rounding down to 470 Ω (21.3 mA, over the 20 mA target). The resistor dissipates P = 10 V × 0.020 A = 0.2 W, so twice that is 0.4 W and the smallest standard rating at or above it is ½ W. A common ¼ W resistor would run at 80% of its rating — it works, but it runs hot; the ½ W part is the recommendation.
Formulas
- Series resistance
R = (Vs − n·Vf) / If- Standard value (round up)
R_E24 = min { e ∈ E24 : e ≥ R }- Resistor dissipation and rating
P = (Vs − n·Vf) · If; rating ≥ 2·P- Actual current with the standard value
I_actual = (Vs − n·Vf) / R_E24
Standards & References
- IEC 60063:2015 — Preferred number series for resistors and capacitors (the E24 series used for the standard-value snap is exact by definition)
- Typical forward voltages verified against Kingbright 5 mm WP7113-family datasheets: WP7113ID red 1.9 V typ / 2.3 V max, WP7113GD GaP green 2.0/2.4, WP7113ZGC InGaN green 3.3/4.1, WP7113QBC/D blue 3.3/4.0, WP7113QWC/D white 3.3/4.0 — presented as typicals; confirm your part's datasheet
- Rating recommendation uses a 2× dissipation margin — standard derating practice so the resistor runs well below its maximum surface temperature
Frequently Asked Questions
What resistor do I need for an LED on 12 V?
For a typical red LED (Vf ≈ 2.0 V) at 20 mA: R = (12 − 2)/0.02 = 500 Ω, rounded up to the standard 510 Ω E24 value, rated ½ W (dissipation 0.2 W, doubled for margin). At a gentler 10 mA the same LED needs 1 kΩ and a ¼ W part is ample. Blue and white LEDs (Vf ≈ 3.3 V) on 12 V at 20 mA need (12 − 3.3)/0.02 = 435 → 470 Ω. Enter your exact parts — the answer moves with Vf and current.
Why round the resistor UP to the next standard value instead of to the nearest?
Rounding up gives slightly less current than designed; rounding down gives more. Less current means an imperceptibly dimmer LED running further inside its ratings — strictly safe. More current eats into the LED's absolute maximum and the resistor's power margin. The brightness cost is trivial: 510 Ω instead of 500 Ω is a 2% current drop, invisible to the eye, which perceives brightness roughly logarithmically.
Can I put one resistor before several parallel LEDs?
It works on the bench and fails in production: parallel LEDs share current according to their individual forward voltages, and no two LEDs have exactly the same Vf. The lowest-Vf LED hogs current, runs hotter, drops further, and hogs more. Give each parallel branch its own resistor, or better, wire LEDs in series (one string, one resistor, identical current everywhere) — this calculator's series count handles exactly that.
What forward voltage should I type if I don't have a datasheet?
Use the tool's typical hints: about 2.0 V for red and the old yellow-green GaP parts, and about 3.3 V for modern InGaN green, blue, and white (values verified against Kingbright WP7113 datasheets). Beware the green trap: an old-style yellow-green indicator is a 2 V part, a modern pure-green is a 3.3 V part. If it matters, measure: drive the LED at the design current through any safe resistor and read the voltage across the LED.
Why does the calculator refuse when my LEDs total more than the supply?
Because a series resistor can only DROP voltage, not add it. Three white LEDs at 3.3 V need 9.9 V of string voltage; a 9 V supply can never push the design current through them, and the math would return a negative resistance. Fix it by using a higher supply, fewer LEDs per string (e.g. two strings of one and two), or a boost-type LED driver that steps the voltage up.
Is a series resistor the right way to drive power LEDs?
Up to roughly 1 W it is common and fine if the resistor rating and heat are respected. Beyond that it is wasteful and thermally fragile: the resistor burns (Vs − Vf)·If as heat, and Vf's temperature drift moves the operating point. High-power LEDs want a constant-current driver, which regulates If directly and stays efficient. The calculator's wattage line is your tell — when the recommended rating climbs past a few watts, switch to a driver.