Voltage divider calculator
Enter Vin and the two resistors — get the output voltage and the current the string burns. The design section solves R2 for a target output, and the load check shows what a real next stage does to it. More bench math lives on the electronics hub.
0 V
unloaded output voltage
| Quiescent current through the string | – |
| Output as share of Vin | – |
Unloaded values — anything connected to the output changes them. Check the sag in the load section below before trusting the number.
Design: find R2 for a target output
| Ideal R2 | – |
| Nearest E24 (5%) value | – |
| Actual Vout with the E24 part | – |
A plain divider can only divide down — the target must sit between 0 V and Vin. The suggestion snaps to the nearest E24 value; a 1% E96 part gets closer when the exact voltage matters.
Load check
Uses the Vin, R1 and R2 from the calculator above. The load sits in parallel with R2 — any finite load pulls the output down. Leave the field at 0 for the unloaded answer.
| Loaded output voltage | – |
| Effective bottom leg (R2 ‖ load) | – |
Enter a load to check the sag
Why the output sags under load
The unloaded divider is one formula: Vout = Vin × R2 ÷ (R1 + R2), because the same current runs through both resistors. Take 12 V and a 10k top resistor; for a 5 V sense point the bottom resistor works out to 7.14k, so a 7.15k E96 part gives 5.003 V while the string draws 0.7 mA. That number is exact only while nothing else is connected.
Whatever you hang on the output sits in parallel with R2. Connect a 10k load to that 12-to-5 V divider and the bottom leg becomes 7.15k ‖ 10k = 4.17k; the output falls to 3.53 V, a 29% sag, and any threshold you were sensing is gone. The 10x rule keeps this honest: make the load at least ten times R2 — 71.5k here — and the sag drops to about 5%; a hundred times holds it under 1%. If the next stage is low-impedance, buffer the node with an op-amp or shrink both resistors and accept the extra quiescent current.
Dividers are for sensing and references, never for powering anything; the LED resistor calculator sizes a series dropper properly, and the parallel resistor calculator does the R2 ‖ load math above. For the value behind the color bands, use the resistor color code tool.
Frequently asked questions
What is the voltage divider formula?
Vout = Vin × R2 ÷ (R1 + R2), where R1 is the top resistor tied to Vin and R2 the bottom one tied to ground. The same current flows through both resistors, so the output is simply the fraction of the total resistance sitting below the sense point. 12 V across 10k and 7.15k gives 12 × 7.15 ÷ 17.15 = 5.0 V.
Can I power an LED or a module from a voltage divider?
No. The moment a load connects it sits in parallel with R2 and drags the voltage down — a 10k load on a 12-to-5 V divider pulls 5.0 V to 3.5 V, and the voltage now moves every time the load current changes. Use the LED resistor calculator for an LED, or a proper voltage regulator for a module; dividers are for sensing and reference nodes only.
How does the input impedance of the next stage affect the divider?
It appears in parallel with R2, exactly like the load in the check above. A microcontroller ADC input around 100k on a divider built from 10k resistors already costs several percent of the reading; a 1 MΩ meter input on the same divider is harmless. When the next stage draws real current, buffer the node with an op-amp voltage follower or scale both resistors down.
When is a divider enough, and when do I need a voltage regulator?
A divider is enough when the load is high-impedance and the exact voltage is not critical: ADC sense points, transistor bias, reference taps. It is the wrong part when the current varies or the voltage must hold — a regulator keeps 5 V at 5 V from microamps up to its rated current, while a divider output moves with both the load and the supply.
What is the 10x rule for voltage dividers?
Make the load impedance at least ten times R2 — equivalently, let the divider string carry ten times the current the load takes. Loading then shifts the output by only a few percent. Push the ratio to a hundred and the error drops under 1%, at the cost of more quiescent current and more noise pickup on the high-impedance node.