What the Voltage Divider Calculator does
This calculator gives a resistive divider's output voltage, the current it draws, the power each resistor dissipates and - the part most calculators leave out - how far a real load pulls the output down. It also searches the E12, E24 and E96 series for the resistor pairs that come closest to the ratio you want.
A divider is the simplest circuit in electronics and still the one most often misapplied, because the textbook formula assumes nothing is connected to the output. The moment something is, the answer changes.
How to use it
- Enter the input voltage and the two resistors. R1 is the top one, between the input and the junction; R2 is between the junction and ground, and the output is taken across it.
- If the output feeds anything with a finite input resistance - an ADC, a meter, a transistor base - enter that as the load and compare the loaded and unloaded figures.
- Check the output impedance. It is R1 in parallel with R2, and it is what every downstream circuit actually sees.
- Open the pair finder, enter the output voltage you want and pick a series; the closest pairs are listed with their error and their current.
- Watch the dissipation figure on anything battery powered: a divider draws its current continuously, whether the circuit is doing anything or not.
Reading the results
The loaded output is the real one. If the load resistance is not at least ten times R2, the divider is no longer a divider - it is a three-resistor network, and its output moves with the load.
Accuracy comes from matching, not from absolute value. Two 1% resistors give roughly 2% worst-case error in the ratio; if one drifts with temperature and the other does not, the error grows with it.
There is a direct trade-off between accuracy and current. Low resistances make the divider stiff and immune to loading and noise, and burn power continuously; high resistances save power and pick up interference.
Worked example: reading a 24 V rail with a 3.3 V microcontroller
47 kOhm over 10 kOhm gives 24 x 10 / 57 = 4.21 V, which is too high for a 3.3 V ADC - so this divider needs a different ratio, and the pair finder is the quick way to one. The current is 24 / 57,000 = 0.42 mA and the whole chain dissipates about 10 mW.
The output impedance is 47 k in parallel with 10 k = 8.25 kOhm. That is at the edge of what a typical successive-approximation ADC likes: its sampling capacitor has to charge through that resistance in the acquisition window, and a small capacitor at the ADC pin is the usual fix.
Load the same divider with a 100 kOhm meter and R2 effectively becomes 9.09 kOhm, so the output falls to 3.92 V - a 7% error introduced entirely by the act of measuring it. Lower both resistors by a factor of ten and the same meter causes only 0.8%.
Formulas and scoring rules
- Output, unloaded
Vout = Vin x R2 / (R1 + R2)- Output, loaded
Vout = Vin x R2' / (R1 + R2'), where R2' = R2 x RL / (R2 + RL)The load sits in parallel with R2, lowering it.- Chain current
I = Vin / (R1 + R2)- Dissipation
P1 = I^2 R1, P2 = Vout^2 / R2- Output impedance
Rth = R1 x R2 / (R1 + R2)The Thevenin resistance seen looking back into the output.
Why a divider cannot supply power
A divider's output impedance is R1 in parallel with R2, which for any sensible resistor values is hundreds or thousands of ohms. Draw a milliamp from it and the output falls by volts. That is why a divider is never a substitute for a regulator, however tempting the arithmetic looks.
If you need a lower voltage that can supply current, use a regulator, or buffer the divider with a unity-gain op-amp. The buffer keeps the divider's ratio and gives it an output impedance of a fraction of an ohm.
Choosing resistor values, not just the ratio
Infinitely many pairs give the same ratio, and the choice between them is about current. Start from what the output feeds: make the divider current at least ten times any current the load draws, and make R2 at most a tenth of the load resistance.
Then check the other end. A divider across a 400 V rail made of 4.7 MOhm and 100 kOhm draws almost nothing, but leakage across a dirty board, and the input bias current of whatever reads it, become comparable with the divider current - at which point the reading is not trustworthy. High-voltage dividers also need resistors rated for the voltage across them, which often means several in series.
Limitations: what the result does not prove
- Everything here is DC and resistive. Stray and input capacitance turns a high-impedance divider into a low-pass filter, which matters for fast signals and for compensated probe dividers.
- Resistor tolerance and temperature coefficient are not modelled; the output is calculated from the exact values you enter.
- The result is an indicative design figure and not a compliance statement. High-voltage dividers in particular need parts rated for the working voltage and creepage distance, which is a safety question for a qualified engineer.
- It assumes a stiff source. If the input itself has significant impedance, the divider loads that in turn and the real output is lower still.
Privacy: where your data goes
Everything you paste, type or drop is processed in this browser tab. It is not uploaded, logged, stored or sent to analytics. Session recording and tag-manager scripts are switched off on this page.
Standards and sources
- IEC 60063 - Preferred number series for resistors and capacitors (E series) - checked 19 Sep 2026
- Texas Instruments - Op Amps for Everyone (source impedance and buffering)
- Analog Devices - Driving an ADC input and source impedance requirements
Frequently asked questions
How do I calculate a voltage divider?
Vout = Vin x R2 / (R1 + R2), where R2 is the resistor between the output and ground. Two equal resistors halve the voltage. The formula assumes nothing is drawing current from the output, which is why the loaded figure above usually matters more.
Why is my divider's output lower than calculated?
Almost always loading. Whatever the output feeds sits in parallel with R2 and lowers it. Enter the load resistance above to see the real figure; if the shift is too large, scale both resistors down or buffer the output with an op-amp.
What resistor values should I use for a voltage divider?
Pick the ratio first, then the scale. Make the divider current at least ten times what the load draws, and keep the output impedance below about 10 kOhm for an ADC. Lower values are more accurate and less noisy; higher values save power.
Can I use a voltage divider to power a circuit?
No. A divider has a high output impedance, so any current drawn collapses the output. Use a regulator for power. A divider is for producing a reference or scaling a signal for something that draws almost no current.
How accurate is a divider made from 1% resistors?
About 2% in the worst case, since both resistors can be off in opposite directions. Matched pairs in one package do far better because they drift together, which is why precision instrumentation uses resistor networks rather than two separate parts.
What is the output impedance of a voltage divider?
R1 in parallel with R2. For 47 k and 10 k that is 8.25 kOhm. It is the figure that decides whether the next stage loads the divider, how quickly an ADC can sample it, and how much noise the node picks up.
Can a voltage divider scale an AC signal?
Yes, but stray capacitance across the resistors makes the ratio frequency dependent. Oscilloscope probes solve this with a deliberate compensating capacitor across the top resistor, which is what the trimmer on a x10 probe adjusts.
How do I get a voltage that is not available from standard resistor pairs?
Use the pair finder above with E96 values, which get within a fraction of a per cent of almost any ratio. If you need better than that, use a precision reference and an amplifier, not a divider - a divider's accuracy is capped by its resistors' tolerance.
Last reviewed by the A2Z.Tools team against the sources listed above.
