A divider is a ratio, not a supply
Two resistors in series across a voltage produce a tap between them at Vin × R2 / (R1 + R2). That is exact, it is easy to verify with a meter, and it is where most of the trouble starts, because the formula describes a divider with nothing connected to the tap.
Connect a load and it sits in parallel with R2. The pair R2 and the load together have a lower resistance than R2 alone, so the ratio shifts and the output falls. How far it falls depends entirely on how the load compares with the divider's own output impedance, which is R1 in parallel with R2. A 10 kΩ over 10 kΩ divider from 12 V taps 6 V and has a 5 kΩ output impedance. Hang a 5 kΩ load on it and the output is not 6 V, it is 4 V. Nothing is broken; the circuit is doing what two resistors do.
The working rule is the ten-times rule: the load should be at least ten times the output impedance of the divider for the error to stay near a few percent, and a hundred times if the tap is being used as a reference. If you cannot make that true by choosing resistors, the answer is not different resistors. It is a buffer — an op-amp follower or an emitter follower — so the divider drives a high impedance and the buffer drives the load.
Choosing the pair, not just the ratio
The ratio fixes only the relationship between R1 and R2. You still choose the scale, and that choice is a straight trade.
| Scale of the pair | Standing current | Output impedance | Where it fits |
|---|---|---|---|
| 1 kΩ class | milliamps | hundreds of ohms | Stiff reference, mains-free bench work, noise immunity |
| 10 kΩ class | hundreds of µA | a few kΩ | General purpose, ADC input dividers |
| 100 kΩ class | tens of µA | tens of kΩ | Battery gauges, anything always-on |
| 1 MΩ class | µA | hundreds of kΩ | Only in front of very high impedance inputs; picks up noise |
Low values are stiff and waste current continuously. High values save current and become sensitive to the input bias current of the thing they feed, to leakage across a dirty board, and to capacitive pickup from anything nearby. For a divider feeding a microcontroller ADC, the 10 kΩ to 100 kΩ range is the usual compromise, and adding a small capacitor from the tap to ground gives the sampling capacitor something local to draw from.
Tolerance stacks, and the sign matters
Two 5 percent resistors do not give a 5 percent output. The worst case is R1 at one end of its band and R2 at the other, which for a mid-scale divider works out to roughly the sum of the two tolerances applied to the ratio. That is why precision dividers use 1 percent or 0.1 percent parts, and why matched pairs in one package exist: what matters for a divider is not the absolute accuracy of each resistor but how closely they track each other, including over temperature.
If you need a specific ratio and only have ordinary parts, measure the two resistors before fitting them and pick a pair whose actual ratio is right. Two 5 percent resistors that happen to be 2 percent high together produce an accurate divider, because the errors cancel in the ratio.
Where dividers earn their place
Reading a voltage higher than a microcontroller can accept is the most common honest use. A 12 V battery through 100 kΩ over 33 kΩ presents about 3 V at the tap, safe for a 3.3 V input, and the ADC input is high impedance enough that the loading rule is satisfied. Bias networks, feedback dividers around a regulator and setting a comparator threshold are the other everyday uses. What they have in common is that the tap feeds something that draws almost no current.
What a divider cannot do is power anything. Dropping 12 V to 5 V for a circuit that draws 100 mA through a divider means the divider chain itself must carry far more than 100 mA to stay stiff, dissipating watts to deliver half a watt, and the output moves whenever the load changes. That job belongs to a regulator. Sizing the rails that a regulator feeds is what the power supply rail calculator is for, and the LED resistor calculator covers the other classic two-terminal sizing problem.
Questions people ask
Why does my divider output drop as soon as I connect something?
Because the thing you connected is now in parallel with the bottom resistor, and the pair together is a lower resistance than R2 alone. The output follows the new ratio, which is always lower. The size of the drop depends on the load compared with the divider's output impedance, R1 in parallel with R2. If the load is ten times that impedance the drop is a few percent; if it is equal to it the output falls by a third. Enter the load resistance above and the calculator shows the loaded figure directly rather than the textbook one.
Can I use a voltage divider to power a 5 V circuit from 12 V?
No, and this is the single most common misunderstanding about dividers. A divider holds a ratio, not a voltage. For the output to stay near 5 V while the circuit draws current, the divider chain would have to carry many times that current so the load barely disturbs it, which means burning several watts in two resistors to deliver a fraction of a watt. It also fails the moment the load current changes, so the rail moves with the circuit's own activity. Use a linear regulator for small currents or a switching regulator for larger ones. A divider is for information, not for power.
What resistor values should I pick for the pair?
The ratio is set by the voltages; the scale is your decision. For a general-purpose divider feeding a high impedance input, a total chain resistance in the 10 kΩ to 100 kΩ range is the usual compromise: the standing current is small, and the output impedance is still low enough not to pick up noise. Go lower when you need the tap to be stiff or the environment is electrically noisy, and accept the continuous current. Go higher only when the input it feeds is genuinely high impedance and the standing drain matters, and expect noise pickup and sensitivity to board leakage.
How accurate is a divider made from 5 percent resistors?
Worse than 5 percent in the worst case, because both resistors are uncertain and their errors work in opposite directions on the ratio. For a mid-scale divider the worst-case output error approaches the sum of the two tolerances. The calculator shows this spread from the tolerance field. The practical fixes are to use 1 percent parts, which are barely more expensive, or to measure the two resistors you actually have and pick a pair whose ratio is right, since matched errors cancel.
Do I need a capacitor across the bottom resistor?
For a divider feeding a microcontroller ADC, usually yes. An ADC input takes a short gulp of charge each time it samples, and a high-impedance divider cannot supply it fast enough, which shows up as readings that sit low. A small capacitor from the tap to ground gives that gulp somewhere local to come from. It also filters noise, at the cost of slowing the divider's response, so keep it small if the signal is changing quickly. It does nothing for the loading problem caused by a steady current draw — only lower resistances or a buffer fix that.