LED Resistor Calculator — Series Resistor & Power Rating

One division, and then the part you can actually buy

The resistor takes whatever the LED does not, at the current the LED wants:

R = (Vs − n × Vf) / If

An LED is a diode, not a resistor. Its forward voltage barely moves as current changes, so nothing in the LED limits the current — past the knee, a tenth of a volt more across it can double the current through it. Connect one straight across a supply and it draws whatever the supply will give until something fails. The resistor is what turns a voltage source into something the diode can live on.

Round up, never to nearest

The division almost never lands on a value anybody makes. 350 Ω is not a stocked part; 330 and 360 are. Rounding up is the rule here, and it is not a style preference: rounding down raises the current above the figure the datasheet permits, and the datasheet figure is the entire reason there is a resistor in the circuit. Rounding up costs a fraction of a milliamp of brightness that no eye can detect.

Everything downstream is then recomputed from the value you will actually fit rather than the one the division produced — the real current, the real dissipation, the wattage to order. A calculator that reports 350 Ω and 20.0 mA is describing a circuit that will not exist.

Two times the power, not one

A resistor at its nameplate wattage is within specification and also running hot enough to discolour a board over a few years. The rating assumes free air at 25 °C, which the inside of an enclosure is not. The trade builds to twice the calculated figure, and that is what the recommendation here uses — advice, not a code rule.

Series, not parallel

Several LEDs in series share one resistor quite happily: the same current flows through all of them, and that current is what sets brightness. Add up the forward voltages, subtract once, divide once.

LEDs in parallel on a single resistor is the classic mistake. Forward voltages never match exactly, even between two parts from the same reel. The one with the lowest Vf takes the largest share of the current, runs hottest, and its forward voltage falls further as it heats — which draws yet more current. It fails, and the survivors then split its current between them. Give every parallel string its own resistor.

Where this stops being the right circuit

A series resistor burns the difference between the supply and the LED as heat. On a 12 V rail feeding one 2 V LED that is 83 % of the power wasted, which is fine for an indicator and absurd for lighting. Once you are past a few hundred milliwatts, or the supply voltage is well above the LED, use a constant-current driver — it regulates the current directly instead of dropping voltage across a heater, and it holds that current as the supply sags or the LED warms.

The resistor also does nothing about temperature. Forward voltage falls roughly 2 mV per °C, so a fixed resistor delivers more current to a hot LED than a cold one. For an indicator this does not matter. For anything driven hard, it is why constant-current drive exists.

Reading the value you land on

The answer comes out as a preferred value from the E12 or E24 series, which is what the colour code can express — ask for 350 Ω and no set of bands will code it. To check the part in your hand matches, decode the bands there; to work out the voltage across the resistor or the power in it from first principles, the Ohm's law solver takes any two of V, I, R and P.