LED Series Resistor Calculator
Answer
R = (VS − n × VF) / IF. The resistor drops the leftover voltage at the LED current you chose. Power in that resistor is P = (VS − n × VF) × IF. Pick the nearest E12, E24 or E96 value and recompute the current; a smaller resistor means a brighter, hotter LED.
Calculator
How it works
An LED is a diode. Once it is forward-biased, current rises steeply with voltage, so a resistor (or a current source) has to set IF. The resistor sees whatever voltage is left after n × VF.
The calculator solves that leftover voltage for R, then snaps to the nearest preferred value you select and shows the current that value actually produces. A Fail on headroom means the supply cannot forward-bias the string. A Warn or Fail on power means the leftover voltage times IF is too much for a small chip resistor.
Parameters used on this page
| Parameter | Value | Source |
|---|---|---|
| Series resistor | R = (VS − n·VF) / IF | Ohm's law around one LED string |
| Resistor power | P = (VS − n·VF) × IF | Joule's law; the leftover voltage times the string current |
| Current with a preferred value | IF,actual = (VS − n·VF) / Rnearest | Same loop, with the E-series part you can buy |
| Typical red / green / blue VF | ≈ 1.8–2.2 V / 2.0–3.2 V / 2.8–3.4 V at rated current | Order-of-magnitude ranges from common indicator LED datasheets; use the VF from your part |
| Preferred values | IEC 60063 E12 / E24 / E96 | IEC 60063 |
| Power-rating rule of thumb | Use a resistor rated at least 2× the calculated dissipation | Allows for VF spread, supply tolerance and a hot board |
Worked example
A 5.0 V supply, one red LED with VF = 2.0 V, IF = 20 mA. Headroom = 3.0 V, so R = 3.0 / 0.020 = 150 Ω, which is already an E24 value. The resistor dissipates 60 mW; a 125 mW 0603 or a 250 mW through-hole part is enough. Two of those LEDs in series would leave only 1.0 V of headroom and need 50 Ω.
Pitfalls
- Putting LEDs in parallel on one resistor. The LED with the lower VF hogs the current. Give each string its own resistor.
- Trusting a website's generic VF instead of the datasheet at your current. VF moves with current and temperature.
- Forgetting resistor power on a 12 V sign. A 20 mA string dropping 10 V dissipates 200 mW; a 0603 will cook.
- Driving an LED from a 3.3 V pin when VF is already 3.2 V. The resistor becomes a few ohms and the current is set by the pin, not by you.
FAQ
How do I calculate an LED resistor?
Subtract the LED forward voltage (times the number of LEDs in series) from the supply, then divide by the current in amperes. That resistance goes in series with the string.
What resistor do I need for a 5 V red LED at 20 mA?
With VF = 2.0 V the exact value is 150 Ω. That is a standard E24 part and dissipates 60 mW.
Can I run several LEDs from one resistor?
Yes if they are in series and the supply is high enough. Do not parallel LEDs on a single resistor unless they are a matched array designed for it.
Why is my LED dimmer than the calculation?
The nearest preferred value may be larger than the exact resistance, VF may be higher than you assumed, or the supply may be sagging.
Related
Sources
- Ohm's law applied to a single LED string.
- IEC 60063 preferred values for the nearest-resistor step.