A voltage divider is a quick way to estimate a lower voltage from two resistors. It is useful for understanding sensor outputs and setting a reference for a high-impedance input. This guide explains the calculation, the hidden effect of a connected load, and how to use the free calculator on PCB Electronics.

The basic calculation

Place R₁ between the input and the midpoint, and R₂ between the midpoint and the circuit’s 0 V reference. Measure the output at the midpoint relative to 0 V. For an unloaded divider, Vout = Vin × R₂ ÷ (R₁ + R₂). The current through both resistors is I = Vin ÷ (R₁ + R₂). Use the same resistance unit for both resistors.

Worked example: 5 V to about 3.3 V

Suppose Vin is 5 V, R₁ is 10 kΩ and R₂ is 20 kΩ. The unloaded result is 5 × 20 ÷ (10 + 20) = 3.333 V. Divider current is 5 V ÷ 30 kΩ = 0.167 mA. The top resistor dissipates about 0.278 mW and the bottom about 0.556 mW in this ideal example. Those small calculated values do not replace checking real resistor ratings, supply variation and tolerances.

Try those values in the voltage divider calculator. It also shows the divider’s output resistance, about 6.67 kΩ here. That figure helps you judge how much a connected circuit may pull the midpoint away from 3.333 V.

What changes when you connect a load?

A load from the midpoint to 0 V sits in parallel with R₂. Replace R₂ in the divider formula with the equivalent R₂ ∥ RL = (R₂ × RL) ÷ (R₂ + RL). For the same 10 kΩ and 20 kΩ divider, a 100 kΩ load makes the effective bottom resistance 16.667 kΩ. The output becomes 5 × 16.667 ÷ 26.667 = 3.125 V. That is 0.208 V below the unloaded estimate. The site calculator deliberately reports the unloaded result; it does not account for a load.

A useful first check is to compare the load resistance with the divider’s output resistance, R₁ ∥ R₂. If they are of similar size, loading matters. A very high input resistance reduces the error, but input leakage, component tolerances and changing loads can still matter.

Choose resistor values with a trade-off in mind

Keeping the same R₁:R₂ ratio gives the same ideal output voltage, but scaling both resistances up lowers wasted current and raises output resistance. Scaling both down makes the divider less sensitive to a given load, while drawing more current. Do not assume one pair is universally best. Check the connected input’s specified impedance and leakage, the supply’s actual range, resistor tolerances and the resulting power.

For example, 1 kΩ and 2 kΩ produce the same 3.333 V ideal output from 5 V as 10 kΩ and 20 kΩ, but draw 1.667 mA instead of 0.167 mA. Their output resistance is about 0.667 kΩ instead of 6.667 kΩ. With a 100 kΩ load, the lower-value pair changes much less. Whether that extra current is acceptable depends on the design.

A quick checking routine

1. Confirm the maximum and minimum input voltage, not just its nominal value.
2. Calculate the unloaded output and divider current.
3. Estimate the connected load and recalculate with R₂ ∥ RL.
4. Check the lowest and highest output caused by component tolerance and supply variation.
5. Compare the result with the connected device’s allowed input range and check resistor power ratings.

A divider is a passive ratio, not a regulated power supply. If the load current changes substantially, or the output must stay close to a target voltage, consider an appropriate regulated or buffered design instead.

Related learning

Start with Ohm’s law if current and resistance are new to you. The electronics calculators page also includes resistor networks, RC timing, LED resistors and frequency conversion.