How it’s calculated
An LED drops a nearly fixed forward voltage, so the series resistor sets the current. The resistor takes the remaining voltage, and Ohm’s law gives its value. Rounding up to the next standard value keeps the current at or below the target.
Example (Kuphaldt, Electric Circuits III §3.12): a red LED dropping 1.6 V at 20 mA from a 6 V supply. The resistor drops 6 − 1.6 = 4.4 V, so R = 4.4 ÷ 0.020 = 220 Ω and it dissipates 4.4 × 0.020 = 88 mW. From 24 V the resistor becomes 22.4 ÷ 0.020 = 1.12 kΩ (448 mW at exactly 20 mA); the next standard value up, 1.2 kΩ, gives 22.4 ÷ 1200 = 18.7 mA and 22.4² ÷ 1200 = 418 mW, which is what the calculator shows.
Leave some headroom: if the supply is barely above the LED string voltage, small changes in supply or forward voltage cause large current swings. For high-power LEDs use a constant-current driver.
Frequently asked questions
Can I use one resistor for several LEDs in parallel?
Not reliably. LEDs’ forward voltages differ slightly, so one LED hogs the current. Give each parallel LED (or each series string) its own resistor.
Why round the resistor up rather than to the nearest value?
A larger resistance gives slightly less current, which is safe; a smaller one could push the LED above its rated current.
What forward voltage should I use?
Use the datasheet’s typical value at your current. Without a datasheet, about 2 V for red/yellow/green and 3.2 V for blue/white is a reasonable starting point.
Embed this calculator
Add this free calculator to your own website. Copy the code below into your page’s HTML:
Sources
- Electric Circuits III (Kuphaldt), 3.12 Special-purpose Diodes (light-emitting diodes) — Workforce LibreTexts
- University Physics Vol. 2, 9.4 Ohm’s Law — OpenStax
Formulas are taken from the free public references above. Results are provided “as is” for informational and educational purposes only. See our disclaimer.
Spotted a mistake or missing option? Report a problem · GitHub issue· Suggest a calculator