Voltage divider calculator

Two resistors and a supply. And what happens the moment you connect something to the output, which is the part every other divider calculator leaves out. Nothing you type here leaves your browser — it is worked out on your own device, so there is nothing to send and nothing to log.

It is only a voltage divider while nothing is connected

Two resistors across a supply give you V_in × R2 ÷ (R1 + R2) at the join, and that is true right up until you attach something to it. Then the load sits in parallel with R2, the pair is smaller than R2 was, and the output drops.

Two 10 kΩ resistors on 9 V give 4.5 V. Attach a 10 kΩ load and you get 3 V — a third low. Every calculator on the internet gives you 4.5 and stops. This one asks what is connected, and shows both.

The fix costs power

Make the divider stiff — much smaller resistors than the load — and the loading error vanishes. The same 10 kΩ load on a 100 Ω divider costs 0.05% instead of 33%. But the divider now burns 0.4 W continuously, doing nothing, which on a battery is the whole design.

The usual rule of thumb is to make the divider current about ten times whatever the load draws. That is a compromise, and the tool above shows what it costs both ways so you can pick rather than guess.

What counts as a load

Anything you connect. A microcontroller's analogue input is a very light load — often megohms — and barely moves the answer, though its sample-and-hold capacitor briefly draws much more than its leakage suggests. An op-amp input is lighter still. A relay coil, a small motor or another circuit's input resistance can be a few hundred ohms and will flatten a high-value divider completely.

Tolerance stacks here

The output depends on the ratio of two resistors, so the errors do not cancel — but how badly they stack depends on the ratio itself, which is the part usually left out. The worst case is 2 × R1 ÷ (R1 + R2) × the tolerance: for two equal 5% resistors that is 5%, and for a divider where the top resistor dominates it approaches 10%. The tool above works it out for the values you actually typed rather than quoting either figure.

If the ratio matters — a reference, a sensor scaling network — use 1% parts, and remember they drift with temperature at different rates unless they came from the same array.

The ratio you want is usually not two preferred values

Resistors come in the E series, so R1 and R2 both have to be preferred values — twenty-four per decade in E24, ninety-six in E96. Want 3.3 V from 5 V and the ratio is 1 : 1.941. Put 10 kΩ at the top and the bottom one would have to be 19.41 kΩ — and E24 has 18 k and 20 k with nothing between them. 10 k with 18 k gives 3.214 V; 10 k with 20 k gives 3.333 V. Neither is 3.3.

E96 at 1% gets much closer, and so does putting two resistors in parallel to make a value that does not exist — see the series and parallel calculator. But the honest answer is usually that a divider is the wrong tool for a precise voltage, and a reference is the right one.

What it will not do

It will not make a power supply. A divider cannot hold a voltage under a changing load; that is a regulator's job.
It will not do alternating current, where the load's capacitance makes the division frequency-dependent.
It will not level-shift safely on its own for anything that matters. A divider dropping 5 V logic to 3.3 V works, and is slow, because the source resistance and the input capacitance form a filter.