The Voltage Divider: The Formula Is Trivial, the Assumption Behind It Is Not

What the calculator computes, and the footnote that matters

The tool solves the divider in three directions, which is more useful than it sounds. In the first mode you give it Vin, R1 and R2 and it returns Vout. In the second you give it the output voltage you want and R2, and it solves for R1 = R2 × (Vin − Vout) / Vout. In the third you give it the output you want and R1, and it solves for R2 = R1 × Vout / (Vin − Vout). The last two are how you actually design a divider, because you start from the voltage you need and one resistor you have decided on. Ask it for 12 V down to 3.3 V with R1 fixed at 10 kΩ and it returns R2 = 3.793 kΩ, which is the number you then round to a real part.

It also reports the current and the power, which is where the honest reading of a divider begins. The default 12 V pair of 10 kΩ resistors draws 0.6 mA and burns 7.2 mW, split evenly because the two resistors are equal. Change the ratio and the split follows the resistances: in the 12 V to 3.3 V case above, 10 kΩ over 3.793 kΩ, the current is 0.87 mA and the 7.57 mW in R1 is nearly three times the 2.87 mW in R2. That asymmetry is worth noticing, because R1 is the resistor that drops the difference and it is always the one running hotter in a divider that steps down hard.

And then there is the footnote, printed under the circuit diagram: the formula shown is the unloaded one and assumes negligible current is drawn from the output. That is not boilerplate, it is the whole boundary of what the tool can tell you. There is no field for a load anywhere in the interface, so every number on screen — the output voltage, the current, both powers — describes a divider with nothing connected to its middle node. Everything below is about what happens when you connect something, and none of it is in the tool.

One exact law: the divider is a source with an internal resistance

Thévenin's theorem says that any network of sources and resistances, looked at from two terminals, behaves exactly like one voltage source in series with one resistance. For a divider the source is the unloaded output — the number the tool gave you — and the resistance is R1 in parallel with R2, seen by shorting the supply in your head and asking what resistance the output terminal sees looking back in. For two 10 kΩ resistors that is 5 kΩ. For 100 kΩ over 22 kΩ it is 18.03 kΩ.

Once you have that, the loaded output is a second divider, this time between the internal resistance and your load, and the whole thing collapses to a single expression: Vout loaded equals Vout unloaded divided by (1 + Rth / RL), where Rth is R1 ∥ R2. That is exact, not an approximation. Check it on the default: 6 V divided by (1 + 5000/100000) is 6 / 1.05 = 5.714 V, which is what the full circuit gives. Check it on a 24 V battery monitor built from 100 kΩ and 22 kΩ feeding a 1 MΩ input: unloaded 4.328 V, Rth 18.03 kΩ, so 4.328 / (1 + 0.01803) = 4.251 V, an error of 1.77 %.

This is also the correction to the rule of thumb. Everyone repeats that the load should be ten times the lower resistor, but the quantity that governs the error is R1 ∥ R2, not R2. The two coincide only when R1 is much larger than R2. When R1 equals R2 the internal resistance is half of R2, and the folk rule turns out to be worth an error of exactly one over (2k + 1) for a load of k times R2 — so ten times R2 gives 4.76 %, twenty times gives 2.44 %, and you need about fifty times R2 before the error drops under one per cent. If you want an accuracy target rather than a superstition, size the load against R1 ∥ R2 and read the error straight off: it is very nearly Rth / RL.

Where a divider belongs, and where it does not

It belongs anywhere the thing reading the voltage draws essentially nothing. The input of a CMOS logic gate, the input of a comparator, the reference pin of a regulator, the gate of a MOSFET, the input of an operational amplifier: all of these are megohms or better, and against a divider with a few kilohms of internal resistance the loading error is a rounding error. Scaling a battery voltage down into the input range of an analogue-to-digital converter is the archetypal good use, and the arithmetic above is exactly how you would check it before trusting the reading.

It does not belong anywhere the load draws real current, and the failure is not gradual — it is a collapse. A divider is not a power supply. It cannot regulate, because the output moves with the load and there is no feedback anywhere to notice; it cannot hold a voltage when the load changes, which every real load does; and its output impedance is fixed by the resistors, so the only way to make it stiff is to make it wasteful. The number that makes this concrete: hang a 1 kΩ load on the tool's 12 V, 10 kΩ default and the output goes from 6 V to 1 V. If your circuit needs a lower voltage rail and can draw current, the answer is a regulator — linear if the current is small and the drop modest, switching if either is large — and never two resistors.

One case sits between the two and catches people out: the input of a microcontroller's analogue-to-digital converter. It is high impedance in the steady state, so the arithmetic above says a divider is fine, but many converters sample onto a small internal capacitor and briefly draw a pulse of current to charge it. If the divider's internal resistance is too high, that capacitor does not finish charging within the sampling window and the reading comes back low — with no warning, and in a way that looks like a calibration error rather than a source impedance problem. Data sheets give a maximum recommended source impedance for exactly this reason. The standard fix costs one component: put a capacitor from the divider's output to ground, so the sampling pulse is supplied by the capacitor rather than through the resistors, which lowers the source impedance at the instant that matters without raising the quiescent current at all.

The quiescent-current trade-off, in five decades

Because the loading error depends on R1 ∥ R2, you can shrink it to nothing by making the resistors small. The cost is that the divider is permanently connected across the supply and draws current whether the circuit is doing anything or not. That current is Vin divided by (R1 + R2), and it never stops. The table below holds the ratio fixed at 12 V down to 6 V and walks the resistance pair from 100 Ω to 1 MΩ, with a 100 kΩ load hanging on the output throughout. At 100 Ω the answer is right to within half a tenth of a per cent and the divider burns 720 mW forever. At 1 MΩ it burns 72 µW and reads 1 V instead of 6.

In mains-powered equipment nobody notices a few milliwatts, so you can afford to be stiff. In anything running from a battery the calculation inverts completely: a divider drawing 0.6 mA continuously flattens a small cell in weeks, and it does so while the device is asleep, which is exactly when a battery-powered design is supposed to be spending nothing. That is why battery monitors use large resistors — the 100 kΩ over 22 kΩ example draws 197 µA and wastes 4.7 mW — and why the really careful ones add a transistor that connects the divider to ground only while a reading is being taken, so the quiescent cost drops to whatever the transistor leaks. If your divider is on a battery and its current is not in your energy budget, it is the first thing to audit.

Two more things worth knowing before you trust the output

The output tracks the input, exactly and forever. A divider has no reference in it: it is a ratio, so a supply that sags by five per cent produces an output that sags by five per cent. That is fine when you are measuring the supply, which is what a battery monitor does, and it is a real problem when you wanted a stable reference for something else. If the thing reading your divider is an analogue-to-digital converter whose own reference is the same supply, the two errors cancel and you get a ratiometric measurement that is better than either part deserves. If the converter has its own precise reference, they do not cancel, and the divider's output inherits every wobble on the rail.

And the tool's two default states do not match, which is worth knowing if you compare the page against a computed example. The interactive divider on the page opens with 12 V and two 10 kΩ resistors and shows 6 V. The registry entry behind the tool — the definition that feeds generated examples elsewhere on the site — carries different defaults, 12 V with 1 kΩ and 2 kΩ, which give 8 V. Both are correct arithmetic for their own inputs; they are simply two different starting points for the same tool, and if you see 8 V quoted somewhere and 6 V on screen, that is the reason and not a bug in either calculation.

Frequently asked questions

Can I use a voltage divider to power a 3.3 V device from 5 V?

No. A device draws current, and the moment it does the divider's output falls. Worse, a device draws different current at different moments — more when a radio transmits, less when it sleeps — so the supply rail would move with the workload, which is the one thing a supply must not do. Even if you sized the divider so hard that the sag were small, you would be dissipating many times the device's own consumption in two resistors, which is why the arithmetic never works out. The answer is a regulator: a linear one if the current is modest and the 1.7 V drop is acceptable as heat, a switching one if the current is large or the efficiency matters. Where a divider is right for a 5 V to 3.3 V job is signalling — bringing a 5 V logic output down to a 3.3 V input that draws microamps — and even there a level-shifter is better if the signal is fast.

Is the ten-times rule of thumb good enough?

It is a floor, not a target, and it is stated against the wrong quantity. The error is governed by R1 in parallel with R2, not by R2 alone, and the two only coincide when R1 is much bigger. For the common case of two equal resistors, a load of ten times R2 gives an error of exactly one over twenty-one, or 4.76 % — enough to matter if you are reading a battery voltage and quoting it to the user, and irrelevant if you are driving a comparator that only cares which side of a threshold you are on. The honest procedure is to pick the accuracy you need, compute R1 ∥ R2, and require the load to be that divided by your error target: for one per cent you need the load to be a hundred times R1 ∥ R2, which for two equal resistors works out at about fifty times R2. Then check the arithmetic with the exact expression rather than the rule.

Why does the tool refuse to solve when I ask for an output above the input?

Because there is no answer to give. A resistive divider can only take a fraction of the input, so the output is strictly between zero and Vin, and asking for anything outside that range makes the algebra produce a negative resistance. In the mode that solves for R2 the formula is R1 × Vout / (Vin − Vout), which goes to infinity as Vout approaches Vin and turns negative beyond it; in the mode that solves for R1 it is R2 × (Vin − Vout) / Vout, which blows up as Vout approaches zero. The tool guards both cases and shows a message rather than a nonsensical component value, which is the right behaviour — a negative resistance is not a part you can order. If you genuinely need an output above the input, you need a boost converter or a charge pump, not a divider.

How do I check the loading error without a load field in the tool?

In two steps, using a second tool. First read the unloaded output from the divider calculator. Then compute R1 in parallel with R2 — the parallel resistor calculator does this directly, and its answer is the divider's internal resistance. Divide that by your load resistance, add one, and divide the unloaded output by the result. Worked on the default: 6 V unloaded, R1 ∥ R2 = 5 kΩ, load 100 kΩ, so 6 divided by 1.05 gives 5.714 V. If you only want the error rather than the voltage, it is almost exactly the internal resistance divided by the load, expressed as a percentage — 5 over 100 is five per cent, and the exact answer was 4.76 %. That approximation is good to a fraction of a per cent whenever the load is more than about ten times the internal resistance, which is the region you should be designing in anyway.

Does a potentiometer behave like a divider?

It is one, with the split point movable. The track between the two end terminals is the total resistance, and the wiper divides it into R1 above and R2 below, so everything in this article applies unchanged — including the loading behaviour, with one extra twist that catches people. The internal resistance R1 ∥ R2 is not constant as you turn the knob: it is zero at both ends, where one of the two is zero, and largest in the middle, where it equals a quarter of the track resistance. So a loaded potentiometer is accurate at the extremes and worst in the centre, which bends the response curve of a control that felt linear when you tested it unloaded. It is also why an audio volume control feeding a low-impedance load misbehaves in the middle of its travel, and why the fix is usually to buffer the wiper rather than to change the pot.