Voltage Divider Calculator

What a voltage divider does

Two resistors in series across a supply carry the same current, and the supply voltage splits between them in proportion to their resistance. The output is taken across the lower resistor, the one connected to ground, and it is always a fraction of the input. Nothing is amplified and nothing is regulated: the output follows the input, scaled down.

The rule follows from Ohm's law. Current through the pair is Vin divided by the total resistance, and the voltage across the lower resistor is that current times R2. Combining the two steps gives the standard form, output equals Vin times R2 divided by the sum of R1 and R2. The William and Mary analog electronics notes derive it the same way and name it the most useful thing two resistors can do.

The lower resistor is not always the output side. If the output is taken across R1 instead, the fraction becomes R1 over the total, and the same numbers produce the complement of the result. Getting this backwards is the first thing to check when a measured output is not what a calculation predicted.

The unloaded formula, with a worked example

Take 12 V, a 10 k resistor on top and a 4.7 k resistor on the bottom. The total is 14.7 k, so the output is 12 times 4.7 divided by 14.7, which is 3.8367 V. The ratio is 0.3197, a little under one third, and the output tracks the input with that fixed ratio as long as nothing else draws current from the tap.

The current is worth computing at the same time, because it decides how stiff the divider is. It is 12 divided by 14,700, or 0.8163 mA. That current flows through both resistors and through the supply, whether or not anything is connected to the output. The power in each part is the current squared times its resistance: 6.66 mW in the 10 k and 3.13 mW in the 4.7 k, for 9.79 mW total from the supply.

The output resistance of the divider is the two resistors in parallel, 10 k with 4.7 k, which comes to 3.197 k. That number is the Thevenin resistance of the network, and it is what any load has to work against. The Bucknell voltage divider notes make the same point: the divider behaves like a source of 3.8367 V behind 3.197 k of resistance.

Why a load pulls the output down

A load connected from the output to ground sits in parallel with R2, so the lower leg becomes the parallel combination of the two. That combined resistance is always smaller than R2 alone, which lowers the fraction and therefore the output. The effect is small for a light load and large for a heavy one, and the crossover is set by how the load compares with R2.

With the 12 V, 10 k and 4.7 k divider, a 10 k load makes the lower leg 4.7 k in parallel with 10 k, which is 3.197 k. The output then is 12 times 3.197 divided by 13.197, or 2.9072 V. That is 0.9295 V below the unloaded figure, a drop of 24.2 percent, and it happens with a load only about twice the value of R2.

The practical rule is to keep the load at least ten times R2 if a few percent of droop is acceptable. At 47 k, which is ten times 4.7 k, the output loses 6.37 percent. A load of 100 times R2, here 470 k, would keep the loss under 1 percent, but a resistor that large makes the divider easy to disturb with a meter or a stray capacitance.

Choosing resistor values for a target output

Working backwards is a ratio problem. The desired output over the input gives the fraction, and any pair of resistors with that ratio will do. A 5 V signal read by a 3.3 V input needs a fraction of 0.66, so R2 should be twice R1: 10 k and 20 k gives 3.333 V, which leaves 0.033 V of headroom under the 3.3 V rail.

The ratio fixes the voltage, and the absolute values fix the current. A 10 k and 20 k pair draws 0.1667 mA from a 5 V source and presents an output resistance of 6.667 k, while a 100 and 200 pair gives the same 3.333 V with 16.67 mA and 66.67 ohms of output resistance. The first wastes almost nothing and is easy for a load to disturb, the second is stiff and burns 83 mW doing nothing useful.

Standard values rarely land exactly on the target. The 12 V to 3.3 V row uses 22 k and 8.2 k from the E12 series and gives 3.2583 V, which is inside the usual 3.0 to 3.6 V logic window. When a precise ratio matters, either use 1 percent parts or put a trimmer in place of R2 and set it with a meter.

Current, power and the cost of small resistors

The same ratio can be built from almost any resistance, and the choice comes down to current. Halving both resistors doubles the current and quadruples the power in each part, since power goes with the square of the current. The voltage output stays identical, which is why a divider that reads correctly can still run hot.

With 12 V across 100 and 47 ohms, the total is 147 ohms and the current is 81.63 mA. The 100 ohm part dissipates 0.666 W and the 47 ohm part 0.313 W, for 0.980 W of heat from a circuit whose only job is to produce a 3.837 V reference. Quarter watt resistors would fail in that circuit, and even half watt parts would run near their limit.

The 10 k and 4.7 k version of the same divider burns 9.79 mW, which is 100 times less, and works with ordinary eighth watt parts. Small resistors are worth using when the load is heavy or when the divider feeds something noisy, because a low output resistance rejects interference. Large resistors are worth using when the supply is a battery and the divider runs continuously.

One more check is the resistor voltage rating. A 12 V divider is far below the 200 V or so that a common through hole part tolerates, but a divider across a 400 V rail is not, and the top resistor then needs to be split into two parts in series.

Tolerance, temperature and real parts

The formula assumes the resistors are exactly their marked values. Real parts are not, and the error in the ratio is larger than the error in either resistor when the ratio is far from one half. For the 10 k and 4.7 k pair, 1 percent parts can push the output to 3.8891 V at one extreme and 3.7847 V at the other, which is plus or minus 1.4 percent around 3.8367 V.

The worst case happens when one resistor is high and the other is low, so a designer who adds the two tolerances and stops there will overestimate the error in some circuits and underestimate it in others. The safe method is to evaluate the formula twice, once with each resistor at the end of its range that hurts the output.

Temperature adds a second drift. A common metal film resistor moves about 50 to 100 parts per million per degree Celsius, so a 40 degree rise changes a 10 k part by 0.2 to 0.4 percent. That is small next to a 1 percent tolerance but not next to a 0.1 percent part, and it matters for a divider used as a reference rather than as a signal scaler.

Where dividers are used, and where they are the wrong tool

Dividers appear wherever a voltage needs scaling down for something that draws almost no current. A battery monitor on a microcontroller uses a pair to bring 12 V or 24 V into a 3.3 V analog input, and a thermistor circuit uses one fixed resistor against a sensor whose resistance changes with temperature. A level shifter for a 5 V signal into a 3.3 V input is the same idea with the roles of the resistors fixed by the two logic levels.

They are the wrong tool for power. A divider cannot supply a load without its output falling, and any current it delivers has to come through R1. Feeding a 3.3 V rail from a 12 V supply with a divider would need a very small R1, which turns the divider into a heater. A linear regulator or a buck converter does that job with efficiency above 80 percent, while a divider is at best 32 percent efficient at the ratio in the example.

They are also the wrong tool for a high impedance source. A divider built from 1 M resistors feeding an analog input with a 10 k input resistance will read low by the same parallel rule as the loaded table. The voltage drop calculator covers the related problem of loss along a wire, which is a divider formed by the wire resistance and the load.

Checking a result against Ohm's law

Every number on this page can be confirmed with two rules. The current through the series pair is the input voltage divided by the sum of the resistors, and the voltage across either resistor is that current times its resistance. For 12 V with 10 k and 4.7 k, the current is 0.8163 mA, the drop across the 10 k is 8.163 V and the drop across the 4.7 k is 3.837 V, and the two drops add back to 12 V.

That sum is the check worth doing by hand: the two resistor voltages must add to the input. If they do not, the output was probably taken across the wrong resistor, or a load was included in one part of the calculation and left out of the other. The Ohm's law calculator handles the single resistor step, and the Ohm's law reference page covers the wheel of rearrangements.

Once the current is known, power follows from current times voltage, and the amps to watts calculator converts that current to watts for a supply of any voltage. For a divider that feeds a real load, add the load current to the divider current before sizing the supply, since both come from the same source.

A final sanity check is the output range. A two resistor divider with positive resistors always produces an output between 0 and the input, never above it and never negative. A result outside that window means a sign error, a load entered as a negative resistance, or a resistor left at zero.