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LED Series Resistor

Also known as LED resistor value

R=Vs−VfIR = \frac{V_{s} - V_{f}}{I}

Worked example: 5 V rail, 2.1 V LED, 20 mA → 145 Ω — press Try an example to run it live, then adjust anything.

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LED Series Resistor explained

VsRIVf

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

R=Vs−VfIR = \frac{V_{s} - V_{f}}{I}
Where
  • RR= Series resistance (Ω)
  • VsV_{s}= Supply voltage (V)
  • VfV_{f}= LED forward voltage (V)
  • II= LED current (A)

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