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EnergyCalcHQ
Vout = Vin · R2 / (R1 + R2)

Voltage Divider Calculator

Output voltage from two resistors — and what happens to it the moment you connect a load, which is the part the textbook formula leaves out.

What you know

Mode

What the output feeds (optional)

Output voltage
8.000V
12.00 V × 2,000 / (1,000 + 2,000). Nothing connected to the output.
Divider current4.000mA
Power in R116.00mW
Power in R232.00mW

This is the unloaded figure. Connect anything that draws current — an ADC input, a transistor base, a meter — and the output falls. Enter that load's resistance above to see by how much; it is the commonest surprise with resistive dividers.

For page numbers, keep Headers and footers ticked under More settings in the print dialog.

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.

Tolerance is the error you actually get

The ratio above is exact arithmetic on nominal values, and you will not buy nominal resistors. A pair of 5 per cent parts can land at opposite ends of their tolerance bands, and the ratio error that produces is close to the sum of the two — around 10 per cent worst case, not 5. Move to 1 per cent parts and the same reasoning gives roughly 2 per cent.

That matters most when the divider is feeding something that measures. A 10-bit ADC resolves about 0.1 per cent per step, so a 5 per cent divider throws away almost the entire useful precision of the conversion — the reading is stable and repeatable and wrong by fifty counts. Nothing in the firmware will reveal it, because the error is constant.

Two ways out. Use 1 per cent or 0.1 per cent parts where the absolute value matters, or calibrate the divider in software by measuring one known input and storing the correction. The second is free and is what most production designs actually do, and it is the reason a divider feeding a microcontroller can be built from ordinary parts while one feeding a panel meter cannot.

Where the divider drifts

Resistors change value with temperature, and the specification that describes it is the temperature coefficient, quoted in parts per million per degree. A common thick-film part is 100 to 200 ppm per degree Celsius. Over a 40-degree rise inside an enclosure that is up to 0.8 per cent of drift — comparable to the tolerance you paid for.

A divider is partly protected from this, and it is worth knowing why: the output depends on the ratio of the two resistors, so if both drift by the same fraction the ratio does not move at all. That only holds while both parts are the same type, from the same family, at the same temperature. Mixing a thick-film part with a metal-film one, or placing one resistor next to a hot regulator and the other across the board, breaks the cancellation and converts a stable ratio into a drifting one.

Self-heating does the same thing on its own. The larger resistor in an unequal divider dissipates most of the power and therefore runs hotter, so the two parts sit at different temperatures even in still air. Keep the dissipation well under the derated rating and the effect stays small; run either part near its limit and the divider will read differently after ten minutes than it did at switch-on.

Questions people ask

Why is my divider output lower than the formula says?
Because something is connected to it. The textbook formula describes a divider with nothing on its output; anything that draws current sits in parallel with R2 and changes the ratio. A 12 V supply through 1 kΩ and 2 kΩ gives 8 V unloaded — hang a 2 kΩ load on it, which is not an unreasonable input impedance, and R2 effectively becomes 1 kΩ, the ratio becomes one to one, and the output collapses to 6 V. Nothing in the circuit looks wrong.
Does it matter whether I use 1k and 2k or 1M and 2M?
Not for the output voltage — only the ratio sets that, so 1k/2k, 10k/20k and 1M/2M all give the same figure. What the absolute values set is the standing current wasted down the chain and how badly the output sags when you load it, and those two pull in opposite directions. Lower values are stiffer and waste more; higher values waste nothing and sag under any load at all.
How stiff does the divider need to be?
Make the 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 per cent or so. Stiffer 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, which removes the problem rather than trading against it.
How accurate is a divider built from 5 per cent resistors?
About 10 per cent on the ratio, not 5, because two parts can land at opposite ends of their tolerance bands and the errors add. One per cent parts give roughly 2 per cent by the same reasoning. It matters most when the divider feeds something that measures: a 10-bit ADC resolves about 0.1 per cent per step, so a 5 per cent divider throws away almost the entire useful precision of the conversion, and the reading is stable, repeatable and wrong by fifty counts with nothing in the firmware to reveal it. Use 1 per cent parts, or calibrate against one known input in software — which is free, and what most production designs actually do.
Can I use a divider to get 5 V from 12 V for a circuit?
No. A divider has no regulation whatsoever: the output moves with the input, with the load, and with temperature. It is for signals — scaling a voltage down for an ADC or a comparator, setting a reference, biasing a transistor base, building a feedback network. Using one to power something that draws real current wastes most of the energy as heat and gives a rail that changes every time the load does. That is what a regulator is for.