Formula
The resistor drops the supply minus the sum of the LED drops, at the current you asked for. The suggested part is the next E24 value at or above that resistance, so the current does not go past the target. Power in the resistor is the current at that part, squared, times the resistance. The nearest E24 value is shown as well. It can sit slightly under the exact resistance, which lets the current rise.
R = (Vs − n Vf) / I
P = I² R
That current is the branch current for the wire size calculator and the voltage drop calculator. Supply watts are Vs times the current on the Ohm’s law and amp pages, then hours on battery runtime and kWh. Parallel strings each need their own resistor. This page is one string.
Worked examples
9 V, one LED at 2.0 V, 20 mA. Exact 350 Ω. Next E24 is 360 Ω. Current 19.44 mA. Resistor power 0.136 W.
Three of those LEDs. Exact 150 Ω, which is already E24. Power is 0.060 W.
5 V and two 3 V LEDs. No positive resistor. The readout is a dash.
Reference
E24 bases are 10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91.
Questions
What resistor feeds a 2 V LED from 9 V at 20 mA?
The exact value is (9 − 2) / 0.02 = 350 Ω. The next E24 value up is 360 Ω, so the current stays under the target. At 360 Ω the current is 19.44 mA and the resistor dissipates 0.136 W.
What about three LEDs in series?
(9 − 6) / 0.02 = 150 Ω, and 150 Ω is already an E24 value. Each LED still carries the same current.
What if the LEDs add up to more than the supply?
The readout is a dash. There is no positive resistor that will light that string from that supply.
Sources
- Series LED current I = (Vs − n Vf) / R.
- IEC 60063 E24 preferred numbers for the suggested resistor.