Voltage Divider Calculator (with Load)

Background.

Two resistors in series across a supply, with the output taken from the joint between them: that is a voltage divider, and it is the most-used circuit in electronics after the wire. It scales a voltage down by a fixed ratio, it costs almost nothing, and it appears in feedback networks, sense inputs, bias chains, level shifters and reference points on nearly every board ever made. The formula is one line — the output is the input times R2 over R1 plus R2 — and if that were the whole story this page would not need to exist.

It is not the whole story, because that formula describes a divider with nothing connected to it. The moment you attach something to the tap, the load sits in parallel with R2, the effective lower resistance falls, and the output falls with it. This is the single most common way a divider design goes wrong, and it goes wrong quietly: the tap measures correctly with a high-impedance meter, then reads low as soon as the real circuit is connected. So this calculator asks for the load as a first-class input and reports the loaded output beside the ideal one, along with the divider's own output impedance and a verdict on whether the droop is acceptable.

The circuit follows from the series rule. Because R1 and R2 form a single path, the same current flows through both, and by Ohm's law that current is the input voltage divided by the total resistance. The output is the voltage across R2 alone, which is that current times R2 — giving Vout = Vin x R2 / (R1 + R2). Everything else on this page comes from the same two facts. The quiescent current is Vin over R1 plus R2. The power wasted is that current times the input voltage. And the output impedance — the number that decides everything about loading — is R1 in parallel with R2, which is what you see looking back into the tap with the supply shorted.

That output impedance is where the interesting design tension lives. The ratio of the resistors sets the voltage, and nothing else about them matters to it: 10 kΩ over 4.7 kΩ and 10 MΩ over 4.7 MΩ produce the same output. But they behave nothing alike. The megohm pair has an output impedance of 3.2 MΩ and will be dragged around by any load, by leakage on a dirty board, and even by a modest multimeter. The kilohm pair has an impedance of 3.2 kΩ and holds up far better — at the cost of drawing a thousand times as much quiescent current, which never stops flowing and, on a battery design, is often the largest drain in a sleeping circuit. Choosing divider values is choosing where to sit between those two failures, and both numbers are printed here so you can see the trade rather than guess at it.

The second mode designs rather than checks. Give it the input voltage, the upper resistor and the output you want, and it returns the lower resistor that produces it: R2 = R1 x Vout / (Vin − Vout). Two things are worth knowing about that result. It will almost never be a stock value, so pick the nearest E24 or E96 part and run the first mode again to see what you actually get. And it goes only downwards — a resistive divider cannot produce more than its input, so a target at or above the supply is refused rather than returned as a negative resistance.

A few limits, stated here rather than buried further down. The supply is assumed stiff: a real source has its own output impedance which adds to R1, and a divider hung off a weak reference will not behave as calculated. This is a DC and low-frequency model with no parasitic capacitance, so nothing here describes a compensated scope probe or a divider feeding a fast edge. Resistor tolerance is not modelled either — two ±5 % parts can put the ratio out by nearly ten percent in the worst case, which is why precision dividers use ±0.1 % parts or a trimmer. And a divider is never a regulator: the droop figure on this page moves with the load, so a design trimmed to compensate for one load current will be wrong at every other one. If the load varies, buffer the tap with an op-amp follower or use a real reference.

What is voltage divider calculator?

A voltage divider is two impedances in series across a source, with the output taken from the node between them. In its resistive form it produces an output that is a fixed fraction of the input, set purely by the ratio of the two resistors.

The derivation is two applications of Ohm's law. R1 and R2 form a series path, so the same current I flows through both, and I = Vin / (R1 + R2). The output is the voltage developed across R2, which is I x R2. Substituting gives Vout = Vin x R2 / (R1 + R2). Nothing is approximated, and the result is exact for an unloaded divider driven by an ideal source.

The divider's Thevenin equivalent is the more useful way to think about it once anything is connected. Looking back into the tap with the source replaced by a short, R1 and R2 appear in parallel, so the divider behaves as an ideal source of Vout behind a series resistance of R1 x R2 / (R1 + R2). A load RL then forms a second divider with that Thevenin resistance, which is exactly why the output droops and by how much.

Dividers appear everywhere: setting the feedback ratio of a switching regulator, scaling a battery voltage into an ADC's input range, biasing a transistor base, generating a reference for a comparator, forming the gain-setting network of an op-amp, and — with one resistor replaced by a sensor — turning a thermistor, photoresistor or potentiometer into a readable voltage. The rules on this page apply to all of them; only the values change.

How to use this calculator.

1. 01Pick the mode. Use the first if you already have two resistors and want to know the output; use the second if you know the output you need and want the lower resistor that produces it.
2. 02Enter the input voltage — the supply across the whole divider, not the output you are aiming for.
3. 03Enter R1, the upper resistor between the supply and the tap, then R2, the lower resistor between the tap and ground. The output is the voltage across R2. Getting these the wrong way round is the most common mistake with this circuit and it produces a perfectly plausible wrong number.
4. 04Enter the load resistance at the tap. If nothing is connected, leave it at 0 and you get the open-circuit output. If something is connected — an ADC pin, a transistor base, a meter — put its input resistance in, because that is the number that decides whether the design works.
5. 05Read the output impedance. This is R1 in parallel with R2 and it is the figure to design against: aim for a load at least ten times higher, and preferably a hundred.
6. 06Read the loaded output and the droop beside the ideal output. The loaded figure is what your circuit will actually see.
7. 07Check the quiescent current if the design runs from a battery. It flows continuously whether the circuit is doing anything or not, and on a sleeping device it is frequently the dominant drain.
8. 08If the droop is too large, scale both resistors down together. That keeps the ratio and the output voltage identical while cutting the output impedance in proportion — at the cost of more quiescent current. If you cannot afford either, buffer the tap with an op-amp follower.
9. 09In solve mode, take the returned R2 to the nearest stock value and re-run the first mode with it. The result will not be exactly your target, and it is better to know by how much than to assume.

The formula.

The divider relation comes from applying Ohm's law twice to a series pair. Because R1 and R2 sit in a single path with no branch between them, the same current flows through both: I = Vin / (R1 + R2), which is the series rule from OpenStax Equation 10.2 combined with V = IR. The output is the voltage across R2 alone, so Vout = I x R2 = Vin x R2 / (R1 + R2). Notice that only the ratio appears — multiply both resistors by a thousand and the output is unchanged.

Rearranging for the design case gives R2 = R1 x Vout / (Vin − Vout). The denominator makes the constraint explicit: as the target approaches the input voltage it goes to zero and R2 goes to infinity, and past it the result turns negative. That is why a target at or above the input is refused rather than returned — a resistive divider steps down and only down. Test the boundary yourself by asking for 11.9999 V from a 12 V supply: the answer is a real, finite, absurdly large resistor.

The quiescent current is that same I, and the power the divider burns is Vin x I, all of it heat. On a 12 V rail with 10 k and 4.7 k that is about 10 milliwatts, which no resistor minds. On a 400 V rail with the same ratio it would be 400 x (400/14,700) = 10.9 watts, which is why high-voltage dividers use megohm values and why their maximum operating voltage — 200 V for a Vishay MRS16, 500 V for a 2512 chip — matters as much as their wattage.

The output impedance is the divider's Thevenin resistance: short the supply in your head and R1 and R2 appear in parallel between the tap and ground, giving R_th = R1 x R2 / (R1 + R2). This is the whole story of loading. A load RL at the tap forms a second divider with R_th, so the loaded output is Vout x RL / (R_th + RL). The calculator computes it the equivalent way — by replacing R2 with R2 in parallel with RL and re-running the divider formula — which gives the identical answer and keeps the arithmetic in one place.

The droop percentage is (unloaded − loaded) / unloaded x 100, and the verdict beside it sorts that figure into four bands: under 1 % negligible, 1 to 5 % small but real, 5 to 20 % significant, and 20 % or more severe enough that the divider is no longer acting as a voltage source. Those bands are decided on the unrounded droop, not on the displayed one, so a value that prints as exactly 5.00 % is classified by what it actually is.

Every step is exact decimal arithmetic at forty significant digits, rounded once at the return boundary to ten decimal places. That matters here because the divider ratio is almost always a non-terminating decimal — 4700/14700 has no finite expansion — and accumulating it in binary floating point loses digits at every stage.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax (Rice University), "University Physics Volume 2", §10.2 "Resistors in Series and Parallel". Equation 10.2 gives the series rule RS = R1 + R2 + … = Σ Ri and the common-current constraint the divider relation is derived from; Equation 10.3 gives the parallel rule used for the loaded case and states that the parallel equivalent "is less than the smallest of the individual resistances". Retrieved and read 2026-07-29.
2. [2]"Series and parallel circuits", reference article, retrieved and read 2026-07-29. Independent statement of both rules used here: R = Σ Ri in series, R = (Σ 1/Ri)⁻¹ in parallel with the two-resistor form R = R1R2/(R1 + R2) — the expression this page uses for the divider's Thevenin output impedance and for the loaded lower arm. Agrees with OpenStax term for term.
3. [3]Vishay BCcomponents, "MRS16, MRS25 Professional Thin Film Leaded Resistors", document 28724, revision 07-Mar-16. Read directly from the PDF on 2026-07-29. Source of the maximum-operating-voltage figures quoted in the high-voltage-divider answers: Umax AC/DC of 200 V (MRS16) and 350 V (MRS25), alongside rated dissipation P70 of 0.4 W and 0.6 W and a peak permissible film temperature of 155 °C — a voltage limit that is specified separately from, and often binds before, the power limit.
4. [4]Vishay, "D/CRCW e3 Standard Thick Film Chip Resistors", document 20035, revision 14-Apr-2026. Read directly from the PDF on 2026-07-29. Gives the surface-mount maximum operating voltages used on this page — 75 V (0402 and 0603), 150 V (0805), 200 V (1206, 1210, 1218), 400 V (2010), 500 V (2512) — together with rated dissipation quoted as P70 and a 155 °C film limit.
5. [5]IEC 60063:2015, "Preferred number series for resistors and capacitors", third edition, 27 March 2015. Defines the E24 (±5 %) and E96 (±1 %) value sets referred to when turning a solved R2 into a resistor you can actually buy. Retrieved 2026-07-29 via the reference summary of the series values and their standard tolerances.
6. [6]Vishay Draloric / Beyschlag / BCcomponents, "Resistor Color Card", document VMN-MS6212-1501 (2015). Read directly from the PDF on 2026-07-29. Prints the E24 and E96 preferred-value tables under the heading "According to IEC 60063" — the source used to check that a solved R2 such as 3,793 Ω has a nearest stock value of 3.9 kΩ in E24 and 3.83 kΩ in E96.