A voltage divider is two resistors in series across a supply, with the output taken from the point between them. It is the standard way to scale a voltage down for an analog input, to set a reference or bias point, or to shift a 5 V logic signal to 3.3 V. This calculator solves for any of the four quantities — output voltage, top resistor, bottom resistor or input voltage — and shows the divider current, power, output resistance and the effect of a load.
How to use the voltage divider calculator
- Pick what to solve for: Vout, R2 (bottom), R1 (top) or Vin.
- Enter the other three values with their units. R1 connects the input to the output node; R2 connects the output node to ground.
- Optionally enter a load resistance — the input resistance of whatever is connected to the output.
- When solving for a resistor, choose the E-series for standard-value suggestions.
- Read the result, the division ratio, the current through the divider and the power in each resistor. The schematic is labeled with your values.
Voltage divider formula
The same current flows through both resistors, so the input voltage splits in proportion to resistance:
Rearranged for the other unknowns:
| Solve for | Formula |
|---|---|
| R2 | R1 × Vout ÷ (Vin − Vout) |
| R1 | R2 × (Vin − Vout) ÷ Vout |
| Vin | Vout × (R1 + R2) ÷ R2 |
| Current | Vin ÷ (R1 + R2) |
A divider can only reduce a voltage, so Vout is always less than Vin.
Worked examples
12 V with R1 = 10 kΩ and R2 = 4.7 kΩ (the default)
Vout = 12 × 4.7 ÷ (10 + 4.7) = 3.837 V, a ratio of 0.3197.
The divider draws 12 V ÷ 14.7 kΩ = 816.3 µA, dissipating 6.664 mW in R1 and 3.132 mW in R2.
Shifting 5 V logic to 3.3 V. Solve for R2 with Vin = 5 V, R1 = 10 kΩ and Vout = 3.3 V: R2 = 10 × 3.3 ÷ 1.7 = 19.41 kΩ. The nearest E24 value is 20 kΩ, which gives 3.333 V — about 1% high and well within logic tolerances.
The same divider under load. Add a 100 kΩ load. R2 in parallel with the load becomes 16.26 kΩ, and the output falls to 3.096 V, about 6% below the unloaded value. The calculator flags any sag above 5% so you know to use smaller resistors or a buffer.
Loading and output resistance
Seen from the output, a divider behaves like a voltage source with an internal resistance equal to R1 and R2 in parallel. For 10 kΩ and 4.7 kΩ that is 3.197 kΩ. A load much larger than this — ten times or more — barely changes the output; a load near it changes it a lot. Analog-to-digital converters on microcontrollers often recommend a source resistance under 10 kΩ so their sampling capacitor charges fully, which sets an upper limit on divider values.
Common uses
- Battery monitoring. Scale a 12 V battery down to under 3.3 V for a microcontroller pin. With R1 = 39 kΩ and R2 = 10 kΩ, even a 14.4 V charging voltage reads only about 2.94 V.
- Sensor biasing. Thermistors and photoresistors form one half of a divider, so the output voltage tracks temperature or light.
- Reference and feedback networks. Adjustable regulators set their output with a divider in the feedback loop.
- Potentiometers. A pot is a divider with a sliding tap, so R1 and R2 change together while their sum stays constant.
To combine more than two resistors first, use the series and parallel resistor calculator; for current and power in a single resistor, the Ohm’s law calculator. Adding a capacitor across R2 turns the divider into a low-pass filter — see the RC time constant calculator.
For low-voltage signal circuits. Never use a resistor divider to measure or derive voltage from mains wiring; follow your local electrical code and use a licensed electrician for installations.
Frequently asked questions
What is the voltage divider formula?
Vout = Vin × R2 ÷ (R1 + R2), where R1 is the resistor from the input to the output node and R2 runs from the output to ground. With 12 V, R1 = 10 kΩ and R2 = 4.7 kΩ, Vout = 12 × 4.7 ÷ 14.7 ≈ 3.837 V.
How do I pick resistors to get a specific voltage?
Choose R1, then solve R2 = R1 × Vout ÷ (Vin − Vout). To get 3.3 V from 5 V with R1 = 10 kΩ, R2 = 10 × 3.3 ÷ 1.7 ≈ 19.41 kΩ. The nearest E24 value, 20 kΩ, gives about 3.333 V.
Why is my divider output lower than calculated?
Anything connected to the output draws current and acts as a resistor in parallel with R2, pulling the voltage down. Keep the load resistance at least ten times R2, use smaller divider resistors, or buffer the output with an op-amp.
Can I use a voltage divider to power a device?
No. A divider only holds its voltage when almost no current is drawn, and it wastes power continuously. To supply a load such as a microcontroller or motor, use a linear regulator or a switching converter.
How large should the divider resistors be?
It is a trade-off. Small values waste current (12 V across 1 kΩ total is 12 mA) but resist loading; large values save power but are easily disturbed by the load and by noise. For reading a battery voltage with a microcontroller, totals of 10 kΩ to 100 kΩ are common.