The formula
R1 and R2 in series carry one current, I = V/(R1 + R2). The output is taken across R2, so it is that current times R2. Only the ratio of the resistors matters, not their size.
I = V / (R1 + R2) Vout = V · R2 / (R1 + R2)
Where dividers are used
Dividers scale a voltage down to what an input can read: a battery monitor feeding a microcontroller’s ADC, a potentiometer as a volume control, or a sensor such as an LDR or thermistor in place of one resistor, so the output voltage tracks light or temperature (see the Sensors module).
The loading effect
Connect a load across the output and it sits in parallel with R2, lowering the effective R2 and pulling Vout down. Keep the load much larger than R2 (as a rule of thumb, ten times or more) or the divider no longer gives the voltage you designed. This is exactly the problem the Thévenin equivalent solves.
Worked example (the simulator’s default values)
- V = 9 V, R1 = 220 Ω, R2 = 470 Ω.
- I = 9/690 = 13.0 mA; Vout = 9 × 470/690 = 6.13 V; V across R1 = 2.87 V.
- Add a 1 kΩ load: R2 ∥ 1 kΩ = 319.7 Ω, so Vout drops to 9 × 319.7/539.7 = 5.33 V.
Common mistakes
- Putting R1 in the numerator. The output is across R2, so R2 goes on top.
- Ignoring the load. A divider’s output changes as soon as something draws current from it.
- Using huge resistor values to save current, then reading the output with a meter whose input resistance is comparable.
Common questions
What is the voltage divider formula?
Vout = Vin × R2/(R1 + R2), with the output taken across R2. 9 V with 220 Ω and 470 Ω gives 6.13 V.
Why does a voltage divider’s output drop under load?
The load sits in parallel with R2, which lowers the effective bottom resistance and so the ratio. Keep the load at least about ten times R2.