Voltage Divider Calculator
A voltage divider is two resistors that split an input voltage into a smaller one — the workhorse of sensor circuits and reference voltages. Enter Vin, R1 and R2 to get Vout, or switch to design mode to find the R2 that produces the exact output you need.
Output voltage (Vout)
5 V
Divider ratio 0.5 · current 5 m A
- Vout
- 5 V
- R2
- 1 k Ω
- Ratio
- 0.5
- Current
- 5 m A
Vout = Vin × R2 ÷ (R1 + R2). Design mode: R2 = R1 × Vout ÷ (Vin − Vout). All values assume an unloaded output.
How it works
- Choose mode: Calculate Vout, or Design (find R2 for a target Vout).
- Calculate mode: enter Vin, R1 and R2 — Vout, ratio and current appear instantly.
- Design mode: enter Vin, R1 and the Vout you need — the calculator returns the required R2.
- The divider ratio (Vout/Vin) is shown — the same value you'll use in analysis problems.
- Circuit current is included for load checks.
The formula
Output voltage
Vout = Vin × R2 ÷ (R1 + R2)
The output is the fraction of R2 over the total resistance — equal resistors split the voltage in half.
Design (solve R2)
R2 = R1 × Vout ÷ (Vin − Vout)
Rearranged for the “I need 3.3 V from 5 V” problem. The target must be between 0 and Vin.
Worked example
Worked example
- Vin = 10 V, R1 = 1 kΩ, R2 = 1 kΩ
- Vout = 10 × 1000 ÷ 2000 = 5 V
- Design: Vin = 10 V, R1 = 10 kΩ, target 3.3 V → R2 = 10000 × 3.3 ÷ 6.7
Calculate: Vout = 5 V (ratio 0.5). Design: R2 ≈ 4.93 kΩ — use a standard 4.7 kΩ or 5.1 kΩ value and verify with the calculate mode.
Common mistakes
- Loading the output — a divider only behaves as calculated when the load current is much smaller than the divider current. Check the current value this calculator provides.
- Choosing resistor values that waste power — a 1 kΩ divider across 12 V draws 12 mA continuously.
- Forgetting the target must be below Vin — a divider can only divide down, never up.
Frequently asked questions
What is a voltage divider used for?
Creating reference voltages (e.g. 3.3 V from 5 V for sensors), level shifting, and biasing in amplifier circuits.
What resistor values should I use?
High values waste less power but are more affected by load. As a rule of thumb, keep the divider current at least 10× the load current.
Does this work for AC?
For resistive dividers at low frequency, yes — the same formula applies to instantaneous values.
Why is my measured Vout lower than calculated?
Almost always loading: the circuit you're feeding draws current through the divider. Add a buffer (op-amp) or use much smaller resistor values.
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