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Voltage Divider Calculator

Solve a resistive divider for any unknown, handle a loaded output, and pick standard-value resistor pairs.

Calculator

Mode

Leave one field blank — the three you last edited drive it.

Check the inputs

Enter any three values to solve the fourth.

How Voltage Divider Calculator works

Two resistors in series across a supply divide the voltage between them in proportion to their resistances. The output taken at the junction is Vout = Vin × R2 / (R1 + R2), where R2 is the resistor between the tap and ground. The ratio alone sets the output — the absolute values do not.

What the absolute values do set is the current wasted through the chain and the divider’s stiffness. A pair of 100 Ω resistors across 12 V burns 60 mA continuously; a pair of 100 kΩ resistors burns 60 µA but is easily disturbed. The choice is a direct trade between quiescent current and how much load the output can drive.

That matters because any load connected to the tap appears in parallel with R2, lowering the effective resistance and pulling the output down. The rule of thumb is to make the divider current at least ten times the load current — then the loading error stays under about 10%, and at 100× it becomes negligible. The tool solves the loaded case directly so you do not have to guess.

Real resistors come from the E-series of preferred values: E12 at 10% tolerance, E24 at 5%, E96 at 1%. An exact computed ratio almost never lands on two available parts, so the tool also proposes standard pairs and shows the resulting error, including the worst case when both resistors sit at opposite tolerance extremes.

Reference

  • Vout = Vin × R2 / (R1 + R2)
  • Loaded: R2′ = (R2 × R_load) / (R2 + R_load), then Vout = Vin × R2′ / (R1 + R2′)
  • Divider current I = Vin / (R1 + R2)
  • Output impedance = R1 ∥ R2 = (R1 × R2) / (R1 + R2)

How to use this calculator

  1. Enter what you know

    Supply the input voltage plus any two of R1, R2, and the wanted output — the remaining unknown is solved.

  2. Add the load, if any

    Enter the load resistance to see the real loaded output rather than the ideal unloaded figure.

  3. Choose a standard pair

    Pick an E-series tolerance and take one of the suggested real-part pairs, with its error shown.

  4. Check current and dissipation

    Confirm the quiescent current is acceptable for your power budget and that neither resistor exceeds its rating.

Worked examples

Scaling 12 V for a 3.3 V ADC

Given
Vin 12 V, wanted Vout 3.0 V
Result
R1 = 30 kΩ, R2 = 10 kΩ, drawing 300 µA

The 3:1 ratio leaves headroom below the 3.3 V rail so a supply overshoot does not exceed the ADC input.

Loading error

Given
A 10 kΩ / 10 kΩ divider on 10 V, feeding a 10 kΩ load
Result
Output falls from 5.0 V to 3.33 V

The load halves the lower leg. This is the mistake behind most "my divider reads low" reports.

Tolerance stack-up

Given
A 2:1 divider built from two 5% resistors
Result
Output ranges roughly ±2.4% around nominal

Ratio errors partly cancel, so the output tolerance is better than 5% but not negligible. Use 1% parts for anything measured.

When to use it

  • Scaling a higher voltage down into the input range of an ADC or microcontroller pin.
  • Setting a reference or bias point in an analogue circuit.
  • Level-shifting a 5 V logic output down to a 3.3 V input.
  • Reading a battery voltage on a low-voltage microcontroller.
  • Building a potentiometer-style adjustment from two fixed resistors.

Things to watch out for

  • A voltage divider is not a power supply. It cannot deliver meaningful current, and its output sags with any load — for a regulated rail, use a regulator.
  • Any load in parallel with R2 changes the output. If the load is not at least ten times the divider’s resistance, buffer the tap with an op-amp follower.
  • For an ADC input, the divider’s output impedance must be low enough for the sample-and-hold capacitor to charge in the sampling window. Many microcontrollers specify 10 kΩ as the maximum.
  • High-value dividers are susceptible to noise pickup and to the input leakage current of whatever they feed. Above about 1 MΩ, both start to matter.

Frequently asked questions

What is the voltage divider formula?

Vout = Vin × R2 / (R1 + R2), where R1 sits between the input and the tap and R2 between the tap and ground. Only the ratio sets the output voltage; the absolute values set the current consumed and how much load the divider can tolerate.

Why is my output voltage lower than calculated?

Almost certainly loading. Whatever you connected to the tap sits in parallel with R2 and reduces the effective lower resistance. Either raise the load impedance, lower the divider resistances, or buffer the output with a unity-gain op-amp.

What resistor values should I use?

Pick the ratio for the voltage and the magnitude for the current. Values in the 1 kΩ to 100 kΩ range suit most low-power work: low enough to resist noise and loading, high enough to keep quiescent current small.

Can I power a circuit from a voltage divider?

No. The output impedance is R1 in parallel with R2, so the voltage collapses as soon as real current is drawn. Use a linear regulator or a switching converter for any actual supply rail.

How accurate is a divider built from 5% resistors?

Better than 5%, because ratio errors partially cancel — typically around ±2.4% worst case for a simple two-resistor divider. Use 1% E96 parts when the output feeds a measurement, and consider that both resistors drift with temperature too.

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