A voltage divider scales a voltage down using two resistors. Enter the input voltage and both resistors to find the output voltage and the current drawn.
Enter the input voltage.
Enter R1 (top) and R2 (bottom).
Read Vout and the current.
Output: Vout = Vin × R2 ÷ (R1 + R2)
A voltage divider is two resistors in series across a supply, with the output taken from the junction between them. It produces a fixed fraction of the input voltage. This calculator takes the input voltage and the two resistors, R1 on top and R2 at the bottom, and returns the output voltage across R2 and the current flowing through the divider.
Dividers are everywhere in electronics. They scale a 5 V signal down to 3.3 V for a microcontroller pin, set bias voltages for transistors, read sensors such as thermistors and light-dependent resistors, and let a microcontroller measure a battery voltage higher than its own supply. A potentiometer, the knob on a fan regulator or radio, is simply a divider with a sliding junction.
1. Note the input voltage Vin and the two resistors, R1 between the input and the output, R2 between the output and ground.
2. Add the resistors to get the total, R1 + R2.
3. Find the divider current: I = Vin ÷ (R1 + R2).
4. Find the output: Vout = I × R2 = Vin × R2 ÷ (R1 + R2).
5. Check the power in each resistor, I²R, against its rating.
6. If a load will be connected to the output, replace R2 with R2 in parallel with the load resistance and recalculate.
With nothing connected to the output, the same current flows through R1 and R2 because they are in series. That current is Vin divided by the total resistance, R1 + R2. The voltage across R2 is then that current times R2, giving Vout = Vin × R2 ÷ (R1 + R2). Only the ratio of the resistors matters for the output: 10 kΩ and 10 kΩ give half the input, and so do 1 kΩ and 1 kΩ. The absolute size affects only the current drawn and how well the divider copes with a load.
The formula assumes the output feeds something that draws almost no current. When you connect a load resistance R_L, it sits in parallel with R2, lowering the effective bottom resistance and pulling Vout down. Seen from the output, the divider behaves like a source with an internal resistance of R1 in parallel with R2. A rule of thumb is to keep the load at least ten times larger than that. This is also why a divider cannot power a motor or a phone: the output collapses under load. Use a voltage regulator for that job.
Small resistor values make the divider stiff against loading but waste current continuously, which matters in battery-powered devices. Very large values waste almost nothing but become sensitive to the input current of whatever reads the output, and to noise. For microcontroller inputs and sensors, values from about 1 kΩ to 100 kΩ are common. Tolerance matters too: with ±5% resistors, the output ratio can be several percent off, so precise references use ±1% parts or a trimmer.
Farhan wants to connect a 5 V sensor output to a 3.3 V ESP32 pin, and plans to use a 2.2 kΩ resistor on top and a 3.3 kΩ resistor to ground as a divider.
Vout = Vin × R2 ÷ (R1 + R2): = 5 × 3300 ÷ (2200 + 3300) = 3 V
Divider current: I = Vin ÷ (R1 + R2) = 0.9091 mA
Answer: Output voltage 3 V; Current through divider 0.9091 mA
Swapping R1 and R2 in the formula; the output is across the bottom resistor.
Using a divider to power a device that draws significant current.
Ignoring the loading effect of a meter or input connected to the output.
Choosing very low resistor values that drain a battery continuously.
Forgetting that resistor tolerance shifts the output voltage.
Level-shifting 5 V signals down to 3.3 V logic.
Measuring battery voltages above a microcontroller's input range.
Reading thermistors, LDRs and other resistive sensors.
Setting transistor bias points in amplifier circuits.
Class 12 potentiometer and potential divider experiments.
Can I power a device from a divider?
Not reliably; the output drops under load. Use a regulator instead.
How do I get exactly half the voltage?
Use two equal resistors.