What a voltage divider is for
A voltage divider is two resistors in series across a supply, with the output taken from the junction between them. The output is a fixed fraction of the input, set only by the ratio of the resistances. It is the simplest circuit in electronics and one of the most misused.
The legitimate jobs are all low-current: scaling a sensor or battery voltage down into an analogue-to-digital converter's range, setting a bias point on a transistor base or an op-amp input, providing a reference for a comparator, and reading a potentiometer or a thermistor. In every one of these the thing connected to the output draws almost no current, and that is not a coincidence — it is the condition under which a divider works at all.
The illegitimate job is powering something. A divider has no regulation whatsoever: its output voltage depends on the load, so as soon as the load changes, so does the output. Worse, a divider stiff enough to hold up under a real load has to burn far more current than the load itself. Dropping 12 V to 5 V for a 100 mA device through a divider would mean a standing current of a full ampere and roughly 12 W of heat to deliver 0.5 W of useful power. Use a regulator or a buck converter for that.
Because a series chain is a divider whether you meant it or not, the same arithmetic explains unrelated phenomena: why a long extension cord makes a heater run cool, why a battery's terminal voltage sags, and why a multimeter with too low an input resistance changes the very voltage it is measuring.
The formula, and why loading changes it
Start from the series chain. The same current flows through both resistors, so I = Vin/(R₁+R₂), and the voltage across the lower resistor is that current times its resistance: Vout = Vin·R₂/(R₁+R₂). Everything else on this page follows from that one line and from Kirchhoff's voltage law, which guarantees the two drops sum back to the input.
Notice what the ratio does not depend on: absolute resistance. A 1 kΩ / 2 kΩ divider and a 100 kΩ / 200 kΩ divider produce identical outputs. They differ only in current — the second draws a hundredth as much. That is the knob you turn when a divider dissipates too much: multiply both resistors by the same factor and the output does not move.
Rearranged for design, the formula gives R₂ = R₁·Vout/(Vin − Vout). The denominator makes the limitation explicit: as the target approaches the input, R₂ grows without bound, and at or above the input there is no solution at all. A resistive divider can only step down.
Loading. Connect anything with finite resistance to the tap and it sits in parallel with R₂. The effective lower resistance becomes R₂′ = R₂RL/(R₂+RL), which is always smaller than R₂, so the output always falls. How far it falls is governed by the divider's Thévenin output impedance, R₁ in parallel with R₂. Thévenin's theorem says the divider is exactly equivalent to an ideal source of the unloaded voltage in series with that impedance, so the load and the output impedance form a second divider. A load ten times the output impedance keeps you within about 9% of the unloaded value; a load a hundred times larger keeps you within about 1%.
Worked example: reading a 5 V signal into a 3.3 V ADC input
You want to scale a 5 V logic signal down for a 3.3 V microcontroller input. Choose R1 = 1.8 kΩ and R2 = 3.3 kΩ — both standard E24 values — and work it out by hand.
- Add the resistors. 1,800 + 3,300 = 5,100 Ω.
- Apply the divider formula. Vout = 5 × 3,300 ÷ 5,100 = 16,500 ÷ 5,100 = 3.2353 V. That is 1.96% below 3.3 V, comfortably inside a 3.3 V input's tolerance.
- Standing current. I = 5 ÷ 5,100 = 0.00098039 A = 0.98 mA.
- Power in each resistor. I² = 9.6117 × 10⁻⁷. So R1 takes 9.6117 × 10⁻⁷ × 1,800 = 1.73 mW and R2 takes 9.6117 × 10⁻⁷ × 3,300 = 3.17 mW. Both are trivial for any package; total is 4.90 mW, which matches 5 V × 0.98 mA.
- Output impedance. 1,800 ∥ 3,300 = (1,800 × 3,300) ÷ 5,100 = 5,940,000 ÷ 5,100 = 1,164.7 Ω.
- Now load it. Suppose the thing you connect presents 10 kΩ. Then R₂′ = (3,300 × 10,000) ÷ 13,300 = 2,481.2 Ω.
- Recompute the output. Vout = 5 × 2,481.2 ÷ (1,800 + 2,481.2) = 12,406.0 ÷ 4,281.2 = 2.8978 V.
- Measure the damage. (2.8978 − 3.2353) ÷ 3.2353 = −10.4%. A 10 kΩ load is only 8.6 times the 1,164.7 Ω output impedance, and a 10% error is the price.
The fix is either a much higher load resistance — a typical microcontroller ADC input is in the megohms, which makes the error negligible — or an op-amp voltage follower between the divider and the load. What you must not do is lower R1 and R2 to "stiffen" the divider without checking the current: dividing both by ten fixes the loading but multiplies the standing current by ten.
How to read the result and choose sensible values
Three numbers decide whether a divider is a good design.
The loaded output versus the unloaded output. If they differ noticeably, your load is too heavy for the divider and you need higher resistances, a buffer, or both. Aim for a load at least ten times the output impedance if a 10% error is tolerable, and a hundred times if you need 1%.
The standing current. This flows continuously whether the circuit is doing anything or not. On mains-powered equipment nobody cares about a milliamp; on a coin-cell sensor it is the entire power budget. A 100 kΩ / 100 kΩ divider across 3 V draws 15 µA, which will flatten a 220 mAh CR2032 in about eighteen months on its own. Push the resistances higher and you trade current for noise pickup and sensitivity to leakage.
The dissipation. Both resistors must survive their own I²R. This only becomes a limit at high input voltages: a 400 V bus divided by a 1 MΩ / 10 kΩ pair puts 0.157 W into the upper resistor and, more importantly, nearly 400 V across it — check the resistor's maximum working voltage, which is a separate specification from its power rating and is often only 200 V or 250 V for a small film part. Split the upper leg into several resistors in series to stay within it; the series resistance calculator handles that chain.
As a practical starting point for signal-level dividers, aim for a total series resistance of 10 kΩ to 100 kΩ. That range keeps the current under a milliamp on a 5 V rail, keeps the output impedance low enough for most ADC sampling circuits, and stays far enough below the megohm region that board leakage and humidity do not matter.
Divider ratios for standard E24 resistor pairs
| R₁ | R₂ | Ratio | Out from 5 V | Out from 12 V | Zout |
|---|---|---|---|---|---|
| 10 kΩ | 1.1 kΩ | 0.0991 | 0.495 V | 1.189 V | 991 Ω |
| 10 kΩ | 3.3 kΩ | 0.2481 | 1.241 V | 2.977 V | 2,481 Ω |
| 10 kΩ | 4.7 kΩ | 0.3197 | 1.599 V | 3.837 V | 3,197 Ω |
| 10 kΩ | 5.6 kΩ | 0.3590 | 1.795 V | 4.308 V | 3,590 Ω |
| 10 kΩ | 10 kΩ | 0.5000 | 2.500 V | 6.000 V | 5,000 Ω |
| 1.8 kΩ | 3.3 kΩ | 0.6471 | 3.235 V | 7.765 V | 1,165 Ω |
| 4.7 kΩ | 10 kΩ | 0.6803 | 3.401 V | 8.163 V | 3,197 Ω |
| 10 kΩ | 20 kΩ | 0.6667 | 3.333 V | 8.000 V | 6,667 Ω |
| 10 kΩ | 100 kΩ | 0.9091 | 4.545 V | 10.909 V | 9,091 Ω |
Z<sub>out</sub> is R₁ in parallel with R₂ and is what limits how heavy a load the divider tolerates. Note that 10 kΩ ∥ 4.7 kΩ and 4.7 kΩ ∥ 10 kΩ give the same 3,197 Ω — swapping the two resistors changes the ratio but not the output impedance.
A divider is not a power supply
The most common beginner mistake is using a divider to run a device from a higher rail. It cannot work, for two reasons that compound each other. First, the output collapses under load, because the load is in parallel with R₂. Second, making the divider stiff enough to resist that collapse means running a standing current many times the load current, which turns almost all the energy into heat in R₁.
If you need a lower supply rail, use a linear regulator for small currents where efficiency does not matter, a switching regulator where it does, or a zener shunt reference where you need a rough, cheap clamp. A divider is for information, not for power.
Mistakes and assumptions worth knowing
- Ignoring the load. The unloaded formula is exact only into an open circuit. Always compare the loaded and unloaded outputs before committing to values.
- Forgetting the meter is a load. A 10 MΩ multimeter across a 1 MΩ divider reads low by a measurable amount, and an old 20 kΩ/V analogue meter can be wildly off. Measured disagreement with theory is often the meter.
- Using a divider to level-shift a fast digital signal. The output impedance and the receiver's input capacitance form a low-pass filter. At 1 kΩ output impedance and 20 pF of input capacitance the time constant is 20 ns, which will visibly round a fast edge.
- Overlooking resistor tolerance. Two 5% resistors give an output whose worst-case error is close to 5%, and the errors do not cancel. Use 1% parts where the ratio matters, ideally from the same batch.
- Exceeding the maximum working voltage. On a high-voltage divider the upper resistor may be well within its power rating and still be over its voltage rating. Split it into a series chain.
- Assuming the divider holds when the input is not stiff. A divider hung off a high-impedance source is itself a load on that source, and the two interact.
- Making the resistances too high. Above about a megohm, board leakage, flux residue, humidity and the input bias current of whatever follows all start to matter, and the real output drifts from the calculated one.
Related circuits and when to use them instead
Several close relatives solve the problems a plain divider cannot.
Buffered divider. Follow the tap with an op-amp voltage follower and the load no longer matters — the follower presents a near-infinite input impedance and drives the load from the op-amp's output. This is the standard fix for an ADC front end and costs one cheap part. The op-amp gain calculator covers the amplifier stages that usually follow.
Potentiometer. A pot is an adjustable divider in one package. The same loading arithmetic applies, with R₁ and R₂ being the two halves of the track, and the output impedance is worst at mid-travel where it equals a quarter of the total track resistance.
Current divider. The dual circuit, where a current splits between parallel branches in proportion to conductance — see the current divider calculator and the parallel resistance calculator, which is also where the R₂-parallel-load arithmetic on this page comes from.
Capacitive and inductive dividers. Replace the resistors with capacitors and the same ratio arithmetic applies to reactance instead of resistance, which is how high-voltage probes and some RF matching networks work. Reactance depends on frequency, so start with the capacitive reactance calculator.
For the underlying element-level relationships, the Ohm's law calculator has every rearrangement of V, I, R and P, and the LED resistor calculator handles the very common case where the "lower" element is a diode rather than a resistor and the divider arithmetic does not apply at all.
Key terms
- Thévenin equivalent
- Any two-terminal linear network can be replaced by one ideal voltage source in series with one resistance. For a divider, the source is the unloaded output and the resistance is R₁ in parallel with R₂.
- Output impedance
- The series resistance of the Thévenin equivalent. It determines how much the output sags for a given load current: ΔV = I_load × Z_out.
- Loading error
- The difference between the unloaded and loaded output, expressed as a fraction. Equal to Z_out/(Z_out + R_L) of the unloaded value.
- Standing current
- The current the divider draws continuously through R₁ and R₂ regardless of what the load does. It is pure loss.
- Maximum working voltage
- The highest voltage a resistor may sustain across it, independent of its power rating. Commonly 200 V to 250 V for small surface-mount and axial film parts — check the datasheet.
- E24 series
- The 24 preferred resistor values per decade defined in IEC 60063 for 5% parts: 10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91.