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Application: Power and status indicators in electronic devices
Application: Under-cabinet and accent lighting
Application: Turn signals and brake lights
An LED is a current-driven device, not a voltage-driven one: past its turn-on voltage its current rises almost vertically, so connecting it straight across a supply will let the current run away and destroy it. A series resistor is the simplest way to set a safe, steady current. Its value comes straight from Ohm's law: R = (Vsupply − Vf) / ILED, where Vf is the LED's forward voltage and ILED is the desired current.
The resistor's job is to "absorb" whatever supply voltage the LED doesn't. For a red LED (Vf ≈ 2 V) at 20 mA on a 5 V supply, the resistor drops 5 − 2 = 3 V, so R = 3 V / 0.02 A = 150 Ω. Always round up to the next standard value (e.g. 150 Ω → 150 or 160 Ω) so the current stays at or below target.
The resistor also dissipates power: P = (Vsupply − Vf) × ILED. In the example that is 3 V × 0.02 A = 60 mW, comfortably within a ¼ W part. At higher currents or bigger voltage drops the dissipation grows quickly, so always check it and choose a resistor rated at least twice the calculated power.
Forward voltage depends on LED colour: red/amber ≈ 1.8–2.2 V, green ≈ 2.0–3.0 V, blue/white ≈ 3.0–3.4 V. For several LEDs, wiring them in series and using one resistor is more efficient than one resistor each, as long as the supply exceeds the sum of the forward voltages. For many LEDs or for maximum efficiency, a constant-current driver beats a resistor.
Why not connect an LED directly to a battery? With no current limit the LED's current spikes and it burns out (or the battery does). Always limit the current.
Can one resistor drive LEDs in parallel? It's discouraged — small Vf differences make one LED hog the current. Give each parallel LED its own resistor, or wire them in series with a single resistor.
An LED is a diode, and its current rises exponentially with voltage. Past its forward knee, a few tens of millivolts more produces a very large increase in current — so connecting an LED directly across a voltage source means the current is set by nothing but the source impedance and the diode’s own tolerances. Either it barely lights or it destroys itself; there is no stable middle ground.
The series resistor converts a voltage source into an approximate current source:
R = (Vsupply − Vf) / If
A 5 V supply driving a red LED with Vf ≈ 2.0 V at 20 mA needs (5 − 2)/0.02 = 150 Ω, dissipating 3 V × 20 mA = 60 mW.
| Colour | Typical Vf |
|---|---|
| Red | 1.8–2.2 V |
| Yellow / amber | 2.0–2.4 V |
| Green | 2.0–3.0 V |
| Blue / white | 2.8–3.6 V |
Higher-energy photons require a larger band gap and therefore a higher forward voltage, which is why blue and white LEDs will not light at all from a supply that runs a red one comfortably.
Forward voltage varies between parts even from the same reel, and the LED with the lowest Vf takes a disproportionate share of the current. It then runs hotter, which lowers its forward voltage further and draws still more current. Each LED needs its own resistor; a shared one guarantees uneven brightness and shortens the life of whichever device happens to be the greediest.
The resistor wastes (Vsupply − Vf) × If as heat, which is acceptable for an indicator but not for illumination. High-power LEDs use a constant-current driver instead, which both wastes far less and holds brightness steady as the supply voltage and the LED’s temperature change.
R = (Vsupply − Vf)/If. For a 5 V supply, a 2 V red LED and 20 mA, that gives 150 Ω.
Because its current rises exponentially with voltage past the forward knee, so a small voltage change produces a large current change. The resistor turns a voltage source into an approximate current source.
Roughly 1.8–2.2 V for red, 2.0–2.4 V for yellow, 2.0–3.0 V for green and 2.8–3.6 V for blue or white. Higher-energy colours need a larger band gap and therefore a higher forward voltage.
No. Forward voltage varies between parts, so the lowest-Vf LED takes most of the current, heats up, and takes even more. Each LED needs its own resistor.
(Vsupply − Vf) × If. In the 5 V red LED example that is 3 V × 20 mA = 60 mW, comfortably within a standard quarter-watt part.
For high-power or lighting LEDs, where resistor losses become significant and brightness must stay constant despite supply and temperature variation. For small indicators a resistor remains the simplest answer.