Voltage Divider Calculator

Compute Vout, divider current, and per-resistor power for any R1 / R2 pair — with an optional load to model the real world.

Vin5.00 VR110.0 kΩVout2.02 VR26.80 kΩ

Input voltage

R1 (top)

R2 (bottom)

Load (optional)

Vout

2.02 V

Current (R1)

298 µA

Current (R2)

298 µA

P · R1

886 µW

P · R2

602 µW

Recommended rating: ≥ 1/8 W (2× derating)

Practical Examples

  1. 5 V to 3.3 V level shift — Vin 5 V · R1 1.8 kΩ · R2 3.3 kΩ → Vout ≈ 3.24 V. Suitable for an ADC input where the source is push-pull and the signal is slower than ~10 kHz.
  2. Battery monitor — Vin 12 V · R1 100 kΩ · R2 33 kΩ → Vout ≈ 2.98 V on a 3.3 V ADC. Total current draw is just 90 µA — negligible for battery life.
  3. Loaded divider warning — Vin 9 V · R1 10 kΩ · R2 10 kΩ → Vout = 4.5 V unloaded. Add a 5 kΩ load and Vout drops to 3.0 V — a 33 % error. Lower R values or buffer with an op-amp.

How a voltage divider works

Two resistors in series share the same current. As that current flows through them, each resistor drops a share of the voltage proportional to its size. Vout is the voltage at the point between them, and it works out to Vin × R2 / (R1 + R2). That's the whole formula.

The formula assumes nothing is connected at Vout. As soon as you hook something up — say, an Arduino input pin — that thing pulls a little current and drags Vout down. The calculator's optional Rload field shows you the real Vout once a load is attached, along with the power each resistor is burning.

When the load barely sips current (an ADC pin, an op-amp input, a MOSFET gate), the drop is tiny and you can ignore it. When the load draws meaningful current, you have two choices: use smaller resistors so the divider is “stiffer”, or feed Vout into a unity-gain op-amp that copies the voltage but supplies all the load current itself.

Design rules of thumb

  • Make the load at least 10× bigger than your resistors — If R1 and R2 are 10 kΩ each, the load should be 100 kΩ or more. Otherwise Vout will sag noticeably.
  • For sensing, use values between 10 kΩ and 100 kΩ — Smaller wastes battery; larger gets noisy and picks up interference. This range is the sweet spot for most projects.
  • Check the power before you build it — A 12 V divider with two 100 Ω resistors burns over half a watt nonstop. Tiny resistors will scorch in a sealed enclosure.
  • Use 1 % resistors when precision matters — Cheap 5 % resistors are fine for pull-ups and LED dimming. For ADC inputs and reference voltages, spend a few cents on 1 % parts.
  • Buffer the output if anything fast is connected — Audio, logic signals, or any moving load will distort through a bare divider. Add a unity-gain op-amp after Vout — it copies the voltage and supplies the current.
  • Never power a circuit from a divider — It is not a regulator. The moment your circuit draws current, Vout collapses. Use an LDO, a buck converter, or a Zener instead.

FAQ

What is a voltage divider?

A voltage divider is two resistors in series across a voltage source, with the output taken from the junction between them. The output voltage is a fixed fraction of the input: Vout = Vin × R2 / (R1 + R2). It's one of the most-used patterns in analog electronics — for biasing transistors, scaling sensor outputs, generating reference voltages, and shifting logic levels.

Why does loading change Vout?

The simple formula assumes nothing draws current at Vout. As soon as you connect a real load (an ADC input, an op-amp, an LED string), that load forms a parallel combination with R2, lowering the effective bottom-leg resistance and pulling Vout down. The error is small when Rload ≫ R2, but becomes severe when they're comparable. Always check Vout under load — that's why this calculator includes an Rload field.

How do I choose R1 and R2 values?

Pick the ratio first: R2 / (R1 + R2) sets the divider ratio. Then pick the magnitude based on tradeoffs. Higher resistances waste less power but have higher output impedance, making them sensitive to loading and noise. Lower resistances are stiffer but burn more power. For most signal-path uses, the total R1 + R2 lands between 10 kΩ and 100 kΩ.

Can I use a voltage divider to power a circuit?

Almost never. Voltage dividers are not regulators — Vout drops as soon as the load draws meaningful current. They work for high-impedance inputs (ADC, op-amp, gate of a MOSFET) but fail for any load that draws current comparable to the divider current. For powering a circuit, use a linear regulator, a Zener reference, or a buck converter.

What is the output impedance of a voltage divider?

Looking back into Vout from the load, you see R1 in parallel with R2. This is the Thevenin equivalent output impedance. It must be much smaller than the load impedance for the divider to behave as the simple formula predicts. As a rule of thumb, keep Rload ≥ 10 × (R1 ∥ R2) for less than ~10% error.

How do I level-shift 5 V down to 3.3 V?

A divider with R1 = 1.7 kΩ and R2 = 3.3 kΩ gives exactly 3.3 V from 5 V. Practically you'd use R1 = 1.8 kΩ and R2 = 3.3 kΩ (E12 values), giving Vout ≈ 3.24 V. This works well for slow signals into a high-impedance input, but for fast logic signals use a dedicated level shifter — divider RC time constant will round off the edges.