Voltage Divider Calculator

A ratio, until something is connected to it

Two resistors across a supply, output taken from the middle:

Vout = Vin × R2 / (R1 + R2)

Note what is not in that formula: the absolute values. 1 kΩ and 2 kΩ give the same output as 10 kΩ and 20 kΩ, and as 1 MΩ and 2 MΩ. Only the ratio sets the voltage. What the absolute values set is the current wasted down the chain, and how badly the output sags when you load it — and those two pull in opposite directions.

The loaded divider — the whole point of the page

The formula above describes a divider with nothing connected to its output. Connect something that draws current and that something sits in parallel with R2:

R2_eff = (R2 × RL) / (R2 + RL)
Vout = Vin × R2_eff / (R1 + R2_eff)

The effect is larger than people expect. A 12 V supply through 1 kΩ and 2 kΩ gives 8 V unloaded. Hang a 2 kΩ load on it — not an unreasonable input impedance — and R2 effectively becomes 1 kΩ, the ratio becomes one to one, and the output collapses to 6 V. A quarter of the voltage gone, and nothing in the circuit looks wrong.

This is the commonest disappointment with resistive dividers: the bench measurement agrees with the calculation exactly, because a multimeter has ten megohms of input impedance and loads nothing. Then the divider is connected to the actual circuit and the number moves. Measuring with a meter proves the divider works; it does not prove it works into your load.

The ten-times rule

Make the divider chain draw at least ten times the current the load does — equivalently, keep R2 at a tenth of the load impedance or less — and the ideal formula stays true to within a percent or so. Stiffer still is better, and costs only standing current.

The other way out is to stop the load drawing current at all: an op-amp voltage follower between the divider and the load presents megohms to the divider and drives the load from its own output. That is what a buffer is for, and it removes the problem rather than trading against it.

What a divider is and is not for

It is for signals: scaling a voltage down so an ADC or a comparator can read it, setting a reference, biasing a transistor base, making a feedback network for a regulator.

It is not a power supply. A divider has no regulation whatsoever — the output moves with the input, with the load, and with temperature. Using one to drop 12 V to 5 V for a circuit that draws real current wastes most of the power as heat and gives a rail that changes every time the load does. Use a regulator.

Power in the resistors

Easy to overlook, because dividers are usually low-power and then suddenly are not. With the chain current I:

P = I² × R

A divider across a 400 V DC bus with low-value resistors will cheerfully cook an eighth-watt part. The calculator flags anything past 125 mW. Raising both resistances fixes it without touching the output ratio — which is the one degree of freedom a divider gives you for free.

Getting a value you can buy

Solve for a resistor and the answer is exact and almost certainly not a stocked part. The calculator shows the nearest E24 value and what output that actually produces, which is the number worth checking — a 2 % shift in a resistor is a 2 % shift in a reference. For tighter divisions use E96 parts, or trim with a series combination. The colour code page lists which series each tolerance is stocked in.