Voltage Divider Calculator
Two resistors in series split a voltage in proportion to their values. It is one of the most-used circuits in electronics, and one of the easiest to misapply.
Formula
Vout = Vin × R2 ÷ (R1 + R2)
- R2 is the resistor across which the output is measured
- Current through both = Vin ÷ (R1 + R2)
12 V with R1 = 1 kΩ and R2 = 2 kΩ: 12 × 2/3 = 8 V
The Formula
Vout = Vin × R2 ÷ (R1 + R2)
Where R2 is the resistor between the output and ground.
With 9 V in, R1 = 10 kΩ and R2 = 5 kΩ: 9 × 5,000 ÷ 15,000 = 3 V.
Only the ratio matters for the voltage. 10 k and 5 k give the same output as 100 k and 50 k — but not the same current draw.
Why Ratio Is Not the Whole Story
The resistor values set how much current the divider wastes continuously:
I = Vin ÷ (R1 + R2)
At 9 V with 15 kΩ total that is 0.6 mA, drawn whenever the circuit is powered. Lower values waste more; much higher values make the divider fragile against loading.
The Loading Trap
A divider only gives the calculated voltage when almost nothing is connected to its output. Attach a load and it appears in parallel with R2, pulling the output below the predicted figure.
The usual rule of thumb is to make the load impedance at least ten times R2. This is why a divider is fine for feeding a high-impedance input such as a microcontroller ADC, and unsuitable as a power supply for anything that draws real current — use a regulator for that.
Quick Reference
| Vin | R1 | R2 | Vout |
|---|---|---|---|
| 12 V | 1 kΩ | 1 kΩ | 6.00 V |
| 12 V | 1 kΩ | 2 kΩ | 8.00 V |
| 5 V | 10 kΩ | 10 kΩ | 2.50 V |
| 9 V | 4.7 kΩ | 10 kΩ | 6.12 V |
Frequently Asked Questions
- How do I calculate a voltage divider?
- Multiply the input voltage by R2 divided by the sum of both resistors. With 9 V, 10 kΩ and 5 kΩ, the output is 3 V.
- Can I use a voltage divider to power a component?
- Generally no. Drawing current loads the divider and drops the output below the calculated value. Use a voltage regulator for anything drawing meaningful current.
- What resistor values should I pick?
- The ratio sets the voltage; the absolute values set the wasted current. Values in the kilohm range suit most low-power circuits, with the load at least ten times R2.
- Why is my measured output lower than calculated?
- Almost always loading. Whatever you connected sits in parallel with R2 and lowers the effective resistance, pulling the output down.