# LED resistor and voltage divider calculator

> Find the LED series resistor for Arduino or ESP32 and the output of a two-resistor voltage divider, with worked examples.

Canonical: https://roboticsengineering.in/led-resistor-voltage-divider-calculator/ | Published: 2026-11-03 | Updated: 2026-11-04

**Two formulas cover most beginner wiring: Ohm’s law for an LED resistor, and the divider rule for scaling a voltage down.** Both are below.

### LED series resistor

Supply voltage (V)LED forward voltage (V)LED current (mA)

Resistor: **–** ohmResistor power: **–** mW

### Voltage divider

Input voltage (V)R1 top (ohm)R2 bottom (ohm)

Output: **–** V

Choose the next standard resistor value up, and check the LED current is within its rating. A divider is for signals only, not for powering things.

## How it is calculated

LED resistor = (supply voltage − LED forward voltage) ÷ current. The divider output = input voltage × R2 ÷ (R1 + R2), where R2 is the resistor to ground.

## A worked example

A red LED drops about 2.0 V and you want 10 mA from a 5 V pin: (5 − 2.0) ÷ 0.010 = 300 ohm, so use 330 ohm. To read a 5 V signal on a 3.3 V ESP32 pin with R1 = 10 kohm and R2 = 20 kohm: 5 × 20000 ÷ 30000 = 3.33 V.

## Common mistakes

- Leaving out the resistor and burning the LED or the pin.
- Using a divider to carry current. Any load on the output changes the result.
- Assuming every LED drops 2 V. Blue and white LEDs drop about 3 V; check the datasheet.

## Related

[ESP32 pinout for robotics](https://roboticsengineering.in/esp32-pinout-for-robotics/), [I2C device not found](https://roboticsengineering.in/fix/i2c-device-not-found-arduino-esp32/), [Arduino IDE setup](https://roboticsengineering.in/arduino-ide-setup/).

Last reviewed: November 2026. The calculator runs in your browser and sends nothing anywhere. Results are ideal-case figures; add margin before you order parts.

## Another worked example: a white LED

A white LED drops roughly 3.0 V and you want about 15 mA from a 5 V supply. The resistor is (5 − 3.0) ÷ 0.015 = 133 ohm. A standard value of 150 ohm is the next one up. With 150 ohm the current is 2 ÷ 150 = 13.3 mA, and the resistor dissipates 0.0133² × 150 = 0.027 W, so a quarter-watt part is more than enough.

## A second divider example and the loading effect

To scale a 12 V signal to about 3.3 V, the ratio R2 ÷ (R1 + R2) must be 3.3 ÷ 12 = 0.275. Pick R2 = 10 kohm, so R1 + R2 = 36.4 kohm and R1 = 26.4 kohm. The nearest common value is 27 kohm, which gives 12 × 10 ÷ 37 = 3.24 V. The divider draws 12 ÷ 37,000 = 0.32 mA.

Now connect something to the output. If the input you feed has a resistance of only 10 kohm, it sits in parallel with R2, making it 5 kohm. The output falls to 12 × 5 ÷ 32 = 1.875 V. This is why dividers suit high-impedance inputs such as microcontroller ADC pins, and why you should keep R1 and R2 small compared with the load but large enough that the divider current is low.

## Assumptions and limits

- LED forward voltage depends on colour and current. Use the datasheet value at your chosen current.
- Brightness is not linear with current, and many indicator LEDs are bright at 2 to 5 mA.
- A divider only attenuates a signal. It does not isolate, protect against over-voltage or carry power. Check the input range of the pin you feed.

## Common mistakes

- Putting R1 and R2 the wrong way round. R2 is the resistor to ground.
- Rounding the LED resistor down. A lower value raises the current.