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Voltage divider calculator

Find the output voltage of a two-resistor divider, including current draw, output impedance and the drop caused by a load.

Leave blank for an unloaded divider.

Output voltage
2.5 V
Divider ratio0.5000
Current through chain250 µA
Total power1.25 mW
Output impedance5 kΩ

Two resistors in series across a supply split the voltage in proportion to their resistances. The point between them sits at a predictable fraction of the input, which makes the divider the cheapest way to scale a signal down.

V_out = V_in × R2 / (R1 + R2)

R1 is the resistor between the input and the tap; R2 is between the tap and ground. Only the ratio sets the output voltage — 10 kΩ / 10 kΩ and 100 Ω / 100 Ω both give half the input. What the absolute values change is current draw and how much the divider sags under load.

Choosing the resistor values

The whole chain draws I = V_in / (R1 + R2) continuously, whether anything is reading the output or not. On a 5 V rail, two 100 Ω resistors burn 25 mA and 125 mW forever. Two 100 kΩ resistors draw 25 µA. For battery-powered work, higher values are almost always right.

The counter-pressure is noise and loading. High-value dividers have high output impedance, which makes them slow to settle and easy to disturb. A practical band for most microcontroller work is 1 kΩ to 100 kΩ, with 10 kΩ as the default guess.

The loading problem

The formula above assumes nothing is drawing current from the tap. Connect a load and it sits in parallel with R2, pulling the effective bottom resistance down and the output voltage with it.

R2_eff = (R2 × R_load) / (R2 + R_load)

A 10 kΩ / 10 kΩ divider on 5 V should give 2.50 V. Hang a 10 kΩ load on it and the bottom leg becomes 5 kΩ, so the output drops to 1.67 V — a third off. The rule of thumb is to make the load at least ten times the divider impedance, which caps the error at roughly a few percent.

Where dividers work and where they do not

They are the right answer for:

  • Scaling a 12 V battery voltage down into a 3.3 V ADC range.
  • Setting a fixed bias point at a high-impedance input like an op-amp or a MOSFET gate.
  • Level-shifting a 5 V logic output down to 3.3 V on a slow signal.

They are the wrong answer for:

  • Powering anything. A divider is not a regulator. Its output moves with load and with the input rail. Use an LDO or a buck converter.
  • Fast digital signals. The output impedance and the stray capacitance of the target form a low-pass filter that rounds off edges. Above a few hundred kilohertz, use a proper level shifter.
  • Anything with varying current draw. The output voltage will move with it.

Common ratios

R1R2Ratio5 V in12 V in
10k10k0.5002.50 V6.00 V
10k20k0.6673.33 V8.00 V
20k10k0.3331.67 V4.00 V
10k4.7k0.3201.60 V3.84 V
100k10k0.0910.45 V1.09 V

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