Voltage Divider Calculator
Solve any one of Vin, R1, R2 or Vout — with current, power per resistor, real E12/E24 parts and the loaded output.
Supply across the pair. Volts, so 3.3 not 3300.
470, 4k7, 4.7k and 1M all parse.
Vout is taken across this one.
Voltage at the tap with nothing connected to it.
Optional. What the tap actually drives — the input resistance of an ADC, a gate, a meter.
26.36kΩ
R1 — solved from the other three
IEC 60063 preferred values. The error is against the ideal Vout above and ignores tolerance, which on E24 parts is ±5% per resistor.
Vout = Vin × R2 / (R1 + R2) · I = Vin / (R1 + R2) · P = I² × R · Rth = R1 ∥ R2
Ideal resistors, DC, and a stiff supply with no source impedance. Tolerance is not modelled: a pair of ±5% parts moves Vout by up to about ±10% on its own, on top of whatever the E-series rounding costs. The unloaded formula assumes nothing draws current from the tap — enter a load resistance to see what one does.
How it works
Enter any three of Vin, R1 and R2 (the top and bottom resistors) and Vout, and leave the fourth blank. The blank one is computed from the other three.
Resistances accept a multiplier letter, so 470, 4k7, 4.7k and 1M all parse. R1 sits between the supply and the tap; R2 sits between the tap and ground, and Vout is measured across R2.
Pick the package rating you are actually using — 1/8, 1/4, 1/2 or 1 W. The tool works out the current through the pair and the power in each leg, and warns before either resistor is over its rating.
Pick E12 or E24 and read the nearest preferred value for anything computed, together with the Vout you would really get from parts you can order and the error that introduces.
If something is going to be connected to the tap, enter its resistance in the load field. The output is recomputed through the Thévenin equivalent, which is usually the difference between a divider that works and one that does not.
The ratio, and the three things it does not tell you
The unloaded output is the supply scaled by R2 as a fraction of the pair. That equation is exact and it is also the least interesting thing about a divider, because it holds for 10 Ω / 10 Ω and 10 MΩ / 10 MΩ alike, and those two behave nothing like each other. The same current flows through both resistors, so the pair dissipates Vin squared over the total resistance — which is how a divider that is arithmetically perfect destroys a quarter-watt part. And because the tap has a source resistance of R1 in parallel with R2, anything connected to it forms a second divider with the first: that is the Thévenin equivalent, and it is why the answer changes the moment you attach a load.
Vout = Vin × R2 / (R1 + R2) · I = Vin / (R1 + R2) · P = I² × R · Rth = R1 ∥ R2 · Vloaded = Vout × RL / (Rth + RL)
Use cases
Scaling a voltage into an ADC input
The commonest use: reading a 12 V rail on a 3.3 V microcontroller. Enter Vin 12, Vout 3.3, and one resistor, and the other falls out. Then check the source resistance — most ADC sample-and-hold inputs want to see roughly 10 kΩ or less to charge the sampling capacitor in time, so a 1 MΩ divider that looks efficient on paper gives readings that drift with sample rate.
Why your divider gets hot
Power is the failure nobody calculates first. A 10 kΩ / 10 kΩ divider across 230 V passes 11.5 mA and dissipates about 1.3 W in each resistor — five times what a quarter-watt part can take. Enter the real Vin and set the package rating, and the tool says so before the part discolours.
Rounding to a resistor you can actually buy
A ratio almost never lands on a stock value. If the maths asks for 26.36 kΩ, E24 offers 27 kΩ and E12 offers 27 kΩ as well; the tool substitutes the preferred value, re-solves, and shows the Vout you would really measure and the percentage error. That is the number worth designing against, not the ideal one.
What a load does to the output
The unloaded formula assumes nothing draws current from the tap, which is only true for a high-impedance input. Attach a load comparable to R2 and the output sags: a load equal to Rth halves it. A useful rule is that a load ten times the Thévenin resistance costs about 9%, so if you need better than a per cent, aim for a hundred times.
Level-shifting a 5 V signal to 3.3 V
A divider works for a slow, one-way logic line and fails for a fast one, because the source resistance and the receiver capacitance form an RC filter that rounds the edges. Keep the pair low enough that the time constant is small against the bit period — and note that a divider cannot shift 3.3 V up to 5 V, which needs an active part.
Setting a feedback ratio on a regulator
Adjustable regulators set their output with a divider on the feedback pin, and the datasheet gives a reference voltage rather than a Vout. Enter that reference as Vout and the target rail as Vin to get the ratio, then check the feedback pin bias current against the divider current — a divider drawing less current than the pin leaks will not hold the rail where you set it.
Tolerance is on top of everything here
Two ±5% E24 resistors can move the output by roughly ±10% between them, and that error stacks on whatever the preferred-value rounding already cost. If the ratio matters more than that, the answer is 1% E96 parts or a matched network rather than a tighter calculation — the arithmetic here is exact and the parts are not.
When a divider is the wrong answer entirely
A divider is a resistor in series with your load. It cannot supply meaningful current, its output moves with the supply, and its dissipation scales with Vin squared. If the tap has to drive anything, or the rail has to stay put while the load changes, the part you want is a regulator or a buffer — a divider only measures and attenuates.
Questions
- Which resistor is R1 and which is R2?
- R1 is the top one, between the supply and the tap. R2 is the bottom one, between the tap and ground, and the output is measured across it. Swapping them inverts the ratio, so a divider that gives almost all of the supply instead of almost none is usually this.
- Can I type 4k7 instead of 4700?
- Yes. The resistance fields accept 470, 4k7, 4.7k, 1M and similar, which is how values are written on schematics and in parts lists. The voltage fields are plain numbers in volts, so enter 3.3 rather than 3300.
- What are E12 and E24?
- IEC 60063 preferred value series. E12 is twelve values per decade, spaced for ±10% parts; E24 is twenty-four, spaced for ±5%. They repeat every decade, so the 47 in the table covers 4.7 Ω, 47 Ω, 470 Ω, 4.7 kΩ and so on. Tighter ±1% work uses E96, which this tool does not offer because at that tolerance you are usually choosing a ratio rather than rounding to one.
- Why does it warn me at half the rating rather than at the rating?
- Because a resistor at its rated power is at its maximum permitted temperature, where its value has already drifted and its life is shortest. Derating to about half is ordinary practice and costs nothing but a larger package, so the warning fires early enough to be useful rather than after the part has failed.
- Does it account for resistor tolerance?
- No, and that is deliberate. The solver is exact for the values you give it. Tolerance is a separate, larger error — two ±5% parts can shift the output by roughly ±10% on their own — so it is stated as a caveat rather than folded into a single number that would imply more precision than exists.
- Why does entering all four values sometimes report a disagreement?
- Because three of the four determine the fourth exactly. If you fill in every field, the tool checks the set instead of silently overwriting one of your numbers, and tells you when they cannot all be true at once. Clear whichever field you want recomputed.
- Does this work for AC?
- Only as an approximation. The maths assumes ideal resistors, DC, and a supply with no source impedance of its own. At AC, stray capacitance and inductance, the source impedance and any reactance in the load all matter, and a purely resistive model stops describing the circuit.
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