LED series resistor

The formula

• R = (U_s − U_f) / I

U_s is the supply voltage, U_f the LED’s forward voltage, the drop across it when it is lit, and I the current the LED is to carry. The supply voltage splits between the two parts in series; the LED takes its forward voltage whatever the current, so the resistor has to take the rest at the chosen current. That is Ohm’s law applied to the resistor alone.

A worked example

A 9 V supply, a red LED with a forward voltage of 2 V, and 20 mA: R = (9 − 2) / 0.02 = 350 Ω. The nearest standard values are 330 Ω, which gives 21 mA, and 390 Ω, which gives 18 mA; both are fine for an LED rated at 20 mA, and the lower current is the safer side.

The resistor’s power

The resistor dissipates P = (U_s − U_f) · I = 7 · 0.02 = 0.14 W. A quarter-watt resistor carries it; with a 24 V supply the same LED would need 1,100 Ω dissipating 0.44 W, and a half-watt part. The electric power page has the formula.

Where the numbers come from

The forward voltage depends on the LED’s colour and chemistry, roughly 1.8 to 2.2 V for red, 3 to 3.4 V for blue and white, and the datasheet gives the exact figure and the rated current. The supply voltage is what the circuit actually provides under load, not what the battery says on its label. The resistor colour code calculator turns the result into bands.

Common mistakes

• Using the current in milliamps. The formula takes amperes: 0.02, not 20.
• Leaving the resistor out on a “low” supply. Without it the current is set by the supply’s and the LED’s internal resistance, and an LED on a 3 V cell can still burn out.
• Putting several LEDs in parallel on one resistor. Their forward voltages differ slightly and the lowest takes most of the current; one resistor per LED, or LEDs in series.

Frequently asked questions