Voltage Divider Calculator — with load effects

Compute divider output voltage, or pick the best E24/E96 resistor pair for a target ratio — including the error your load introduces.

Example: A 10 kΩ / 10 kΩ divider on 12 V outputs 6 V, drawing 600 µA.

Formula

V_out = V_in × R2 / (R1 + R2)

Worked example

A 10 kΩ / 10 kΩ divider on 12 V outputs 6 V, drawing 600 µA.

  1. V_out = V_in × R2 / (R1 + R2)

    V_out = 12 V × 10 kΩ / 20 kΩ

    V_out = 6 V

  2. I = V_in / (R1 + R2)

    I = 12 V / 20 kΩ

    I = 600 µA

    Flows continuously, whether or not anything is connected to the output.

  3. R_source = R1 ∥ R2

    R_source = 10 kΩ ∥ 10 kΩ

    R_source = 5 kΩ

    How stiff the output is. A load must be well above this to read correctly.

Frequently asked questions

How do I make 5 V from 12 V with a voltage divider?

Pick R1 and R2 so R2 / (R1 + R2) = 5/12. A 10 kΩ top and 7.15 kΩ bottom gets close. But a divider is only suitable for sensing a voltage, not powering anything — its output collapses as soon as it has to supply real current. Use a regulator or buck converter to power a 5 V load.

Why is my divider output lower than calculated?

Something is loading it. The classic formula assumes nothing is connected to the output. Whatever you attach appears in parallel with R2 and pulls the voltage down. Enter your load resistance above to see the real figure.

What resistor values should I choose?

The ratio sets the voltage; the absolute values set the trade-off. Small resistors waste current continuously but give a stiff, low-impedance output. Large resistors save power but make the output easy to disturb and more susceptible to noise. For ADC sensing, 10 kΩ to 100 kΩ total is a common compromise.

Can I use a voltage divider to power a microcontroller?

No. A divider's output impedance is R1 in parallel with R2, so any meaningful current draw collapses the voltage — and the draw changes as the chip switches, so the supply would move constantly. Dividers are for sensing and biasing. Use a regulator to supply power.

Why does my ADC read the wrong value from a high-impedance divider?

The ADC's sample-and-hold capacitor has to charge through your divider's output impedance. If that impedance is too high, the capacitor does not reach the true voltage before sampling ends, and the reading comes in low. Either lower the resistor values, lengthen the sampling time, or add an op-amp buffer.

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