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Voltage Divider Calculator

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Two resistors in series across a supply produce a voltage at their junction that is a fixed fraction of the input — the fraction being R2 over the total. It is the simplest way to scale a voltage down, and it is the right answer for feeding a measurement input and the wrong answer for powering anything. This calculates the output from a pair of resistors, or finds the resistor that produces the output you want, and reports the two things that decide whether the divider will actually work: how much current it wastes, and how heavy a load it can take before the answer stops being true.

What do you want to work out?
V
Ω

Between Vin and the output

Ω

Between the output and ground

How to use the voltage divider calculator

  1. 1Choose whether you are finding the output from two known resistors, or solving for R2 to hit a target.
  2. 2Enter the input voltage.
  3. 3Enter R1, the resistor between the input and the output node.
  4. 4Enter R2, the resistor between the output node and ground — or, in solve mode, the output voltage you want.
  5. 5Read the output voltage, then check the chain current and the minimum load figure before committing to those values.

Examples

An equal pair

Input
12 V, R1 = 10 kΩ, R2 = 10 kΩ
Result
6 V, drawing 0.6 mA

Equal resistors always halve the input, whatever their value. What their value changes is the current wasted and how much load the output can take.

Scaling a battery for an ADC

Input
12 V down to 3.3 V, R1 = 10 kΩ
Result
R2 = 3.793 kΩ; fit 3.9 kΩ for 3.37 V

A microcontroller input is high impedance, so it barely loads the divider — which is the case a divider is genuinely good at.

A divider that wastes too much

Input
12 V, R1 = 100 Ω, R2 = 100 Ω
Result
6 V, but drawing 60 mA and burning 720 mW continuously

The same 6 V as the first example, at a hundred times the current. On a battery this would be the largest load in the circuit.

About the voltage divider calculator

Where the formula comes from

With nothing connected to the output node, the two resistors form a single series loop, so the same current flows through both. That current is the input voltage divided by the sum of the resistances. The voltage at the junction, measured from ground, is whatever that current develops across R2 — the current times R2. Substituting one into the other gives Vin times R2 over R1 plus R2, and the current cancels out entirely.

The cancellation is the interesting part. It means the output depends only on the ratio of the resistors, not on their size. Ten ohms and ten ohms give the same voltage as ten megohms and ten megohms. Everything else about the circuit — how much power it wastes, how much load it tolerates, how susceptible it is to noise — depends on the size, and none of it shows up in the output voltage.

That is why this page reports the current and the loading limit alongside the voltage. The voltage answer alone is the easy half, and it is the half that makes two very different circuits look identical.

The four jobs a divider does well

Scaling a voltage for measurement is the main one. An analogue-to-digital converter typically accepts up to its own supply rail, so reading a 12 V battery means dividing it down first. The converter's input draws only leakage current, so it barely loads the divider, and the ratio holds.

Setting a reference is the second. Feedback networks in regulators and op-amp circuits use dividers to define a threshold, again driving a high-impedance input. The third is level shifting a logic signal downward — a 5 V output feeding a 3.3 V input — which works for slow signals but starts to fail at speed, because the divider and the input capacitance form a low-pass filter that rounds off the edges.

The fourth is biasing: putting the midpoint of a supply onto a node so an AC signal can swing both ways around it. Here the divider is usually paired with a capacitor to ground, which holds the bias point steady against the very load variations that would otherwise move it — an acknowledgement, built into the circuit, that a divider on its own cannot hold a voltage against a changing load.

Frequently asked questions

Why does my divider output sag when I connect something to it?
Because whatever you connected is now in parallel with R2, and a parallel combination is always lower than either part alone. A lower R2 means a smaller fraction of the input, so the output falls. The size of the effect depends entirely on the ratio: a load ten times R2 pulls the output down by roughly nine percent, while a load a hundred times R2 costs about one percent. That is the figure this page reports as the minimum load.
Can I use a divider to power a 3.3 V chip from 5 V?
No, and it is the most common misuse of this circuit. A chip's current draw changes constantly as it works, and since the divider's output moves with the load, the supply voltage would move with it — dropping under load and rising when idle, which is exactly what a supply must not do. Use a linear regulator or a buck converter, both of which hold their output steady regardless. A divider only suits inputs that draw negligible current.
How do I choose the resistor values rather than just the ratio?
The ratio sets the voltage; the absolute values set the trade-off between waste and stiffness. Small resistors draw more current, wasting power continuously, but they hold their voltage under a heavier load. Large resistors waste almost nothing but sag easily and pick up electrical noise. For a microcontroller ADC input, tens of kilohms is the usual compromise; battery gear goes higher, and anything driving a real load should not be a divider at all.
Does the divider still work with AC?
For a resistive divider at low frequencies, yes — the same ratio applies to the instantaneous voltage. Two things change as frequency rises. Stray capacitance across the resistors starts to matter, so the ratio becomes frequency dependent, which is why oscilloscope probes include a trimmer capacitor to compensate. And if what you are dividing has a DC offset you did not intend to keep, a resistive divider passes it through unchanged.
What does the tool do if I ask for an output above the input?
It refuses, with the reason. Two resistors can only ever share a voltage out between them, so the junction is always somewhere between the input and ground — there is no arrangement of passive resistances that produces more voltage than went in. Stepping up needs energy storage and switching, which is what a boost converter, a charge pump or a transformer does.

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