LED Resistor Calculator: Exact Value, Nearest E12/E24 Standard Resistor and Power Rating
Every LED needs something to limit its current, and for small indicator LEDs that's a single resistor. Enter your supply voltage, the LED's forward voltage, the current you want and how many LEDs are in series. The calculator gives the exact resistance, picks the nearest standard E12 or E24 value that keeps the current at or below your target, and suggests a power rating.
The formula
The resistor takes whatever voltage the LEDs don't:
R = (Vs − n × Vf) ÷ I
Example: a 5 V supply, one red LED (Vf ≈ 2 V) and a 10 mA target gives R = (5 − 2) ÷ 0.010 = 300 Ω. 300 Ω happens to be an E24 value, so you can use it directly. With E12 parts you'd use 330 Ω, for about 9.1 mA.
The resistor dissipates P = I² × R: about 30 mW in this example, so any common 1/4 W resistor is fine.
Typical forward voltages
| Color | Typical Vf |
|---|---|
| Red, yellow, orange | about 1.8–2.2 V |
| Green (standard) | about 2.0–2.2 V |
| Blue, white, true green | about 2.8–3.4 V |
These are common ranges. Check your LED's datasheet, because parts vary a lot between manufacturers.
Practical notes
- Why round up? Picking the next standard value above the calculated one guarantees the current never exceeds your target. A slightly lower current is rarely visible.
- Driving LEDs from Arduino pins: on the ATmega chips in the Arduino Uno and Mega, 40 mA per I/O pin is the absolute maximum. Aim for 20 mA or less per pin, and check the total current if many LEDs are on at once.
- One resistor per LED in parallel. If several LEDs share one resistor in parallel, small differences in Vf make them share current unevenly.
- High-power LEDs: when the resistor has to burn a lot of power, it's inefficient and hot. Use a constant-current LED driver instead.
Worked examples
Example 1: one red LED on a 5 V supply
Supply Vs = 5 V, red LED with Vf = 2 V, target current 20 mA. Convert first: 20 mA = 0.020 A. The resistor drops 5 − 2 = 3 V. Plugging into R = (Vs − n × Vf) ÷ I gives R = (5 − 1 × 2) ÷ 0.020 = 150 Ω.
150 Ω is a standard value in both E12 and E24, so the nominal current is exactly 3 ÷ 150 = 20 mA (0% error). Resistor power is P = I² × R = 0.020² × 150 = 0.060 W. With a 2× margin you need at least 0.12 W, so a 1/8 W resistor works and a common 1/4 W part is comfortable.
If an Arduino pin supplies the 5 V, check that board's per-pin and total current limits, and remember a loaded pin won't sit at exactly 5 V. Recalculate with the real numbers.
Example 2: three white LEDs in series on 12 V
Assume Vf = 3.2 V per LED and a target of 15 mA. The LEDs drop 3 × 3.2 = 9.6 V, leaving 12 − 9.6 = 2.4 V across the resistor. Ideal value: 2.4 ÷ 0.015 = 160 Ω.
With the E12 series selected, the calculator rounds up to the next standard value, 180 Ω. Actual current is 2.4 ÷ 180 = 13.3 mA, about 11% below target, which is the safe direction. Power is 2.4² ÷ 180 = 0.032 W, so 1/8 W is plenty. With E24 selected you'd get 160 Ω and exactly 15 mA.
Common mistakes
- Forgetting to convert units: the formula needs amps. Typing 20 instead of 0.020 gives a resistor 1000× too small.
- Mixing up series and parallel: for LEDs in series, subtract the total forward voltage; LEDs in parallel each need their own resistor.
- Checking resistance but not power: confirm the wattage rating too, and leave margin if the resistor sits in a warm enclosure.
- Treating the nominal result as exact: supply tolerance, the real Vf of your LED and resistor tolerance all shift the current a little.
Building something bigger? My 4×4×4 LED cube with an Arduino Mega2560 drives 64 LEDs by multiplexing, which is how you light many LEDs without one pin per LED.
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