LED Series Resistor
Also known as LED resistor value
Worked example: 5 V rail, 2.1 V LED, 20 mA → 145 Ω. Press Try an example to run it live, then adjust anything.
Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Change any unit and the answer follows.
Where LED Series Resistor fits
See also
Percent Voltage Drop: Both use supply voltage.RC Capacitor Charging: Both use supply voltage.555 Astable Frequency: Both use series resistance.Current Divider: Both use series resistance.Q Factor of a Series Resonant Circuit: Both use series resistance.Series RL or RC Impedance: Both use series resistance.Design the drop →
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LED Series Resistor explained
An LED is not a resistor: its current climbs almost vertically once forward voltage is reached, so a tiny voltage excess becomes a huge, fatal current. The resistor is what makes the circuit predictable: it eats the difference between supply and forward voltage, and Ohm's law sets the current. Running a red LED (Vf ≈ 2.1 V) at 20 mA from 5 V needs (5 − 2.1)/0.02 = 145 Ω, so you fit a 150 Ω standard value.
Two things beginners miss. First, check the resistor's power: here it dissipates (5 − 2.1) × 0.02 ≈ 58 mW, comfortable for a quarter-watt part, but a 12 V supply on the same LED wastes 198 mW and argues for a proper constant-current driver. Second, forward voltage depends on colour and on temperature (roughly 1.8–2.2 V for red, 3.0–3.4 V for blue and white) and it drops about 2 mV per °C as the die heats, which is precisely why high-power LEDs must never be run from a resistor alone.
LED Series Resistor formula
- = Series resistance (Ω)
- = Supply voltage (V)
- = LED forward voltage (V)
- = LED current (A)
Cite this page
formula.expert. (2026, October 16). LED Series Resistor Formula & Calculator. https://formula.expert/solvers/led-series-resistor