Enter any three values and click Calculate.
Note: The circuit diagram updates automatically as you enter values.
Auto-calculate as you type
Note: The circuit diagram updates automatically as you enter values.
| R1 | R2 | Actual Vout | Error (%) |
|---|
Given: Vin=12V, R1=6.8kΩ, R2=4.7kΩ
Given: Vin=5V, R1=350Ω, R2=350Ω
Given: Vin=5V, R1=2kΩ, R2=3kΩ
A voltage divider is the simplest and most-used building block in electronics: two resistors in series that split an input voltage into a smaller, predictable fraction. The output is taken from the junction between them and follows Vout = Vin × R2 / (R1 + R2), where R2 is the resistor across which the output is measured. Because the ratio — not the absolute values — sets the output, a 2 kΩ/3 kΩ pair and a 20 kΩ/30 kΩ pair both give the same 60% output.
The formula above assumes nothing is drawing current from the output. As soon as you connect a real load, it sits in parallel with R2, lowers the effective resistance, and pulls the output below the calculated value. The rule of thumb is to make the divider current at least 10× (ideally 100×) the load current, or to buffer the output with an op-amp follower. This is why dividers are perfect for high-impedance inputs (an ADC pin, a comparator, a transistor base bias) but a poor way to power anything.
Pick the ratio for the voltage you need, then scale both resistors together to trade current for noise: smaller resistors waste more power but are less sensitive to stray capacitance and input leakage; larger resistors save power but load more easily and pick up noise. For battery devices, keep the total resistance high (tens of kΩ to MΩ) so the divider itself sips only microamps.
Scaling a battery or supply voltage down so a microcontroller ADC can measure it, setting reference and threshold voltages, biasing transistor and op-amp stages, level-shifting logic signals, and building the resistive half of sensors such as thermistor and photoresistor circuits.
Can a divider supply current to a circuit? Only tiny amounts. For real loads use a proper regulator; a divider is for setting a reference voltage, not delivering power.
Why is my measured output lower than calculated? Almost always the loading effect — whatever you connected is drawing current. Increase the divider current or add a buffer.