How this instrument works
A voltage divider is two resistors wired in series across a supply, with the output taken from the node between them. Because the same current flows through both resistors in a series path, Ohm's law gives that current as Vin ⁄ (R1+R2), and multiplying it back through R2 alone yields the output: Vout = Vin·R2 ⁄ (R1+R2). Kirchhoff's voltage law is what guarantees the two resistor drops add up to the full supply voltage, so nothing beyond series wiring has to be assumed.
The output is a dimensionless ratio, R2 ⁄ (R1+R2), scaled by the supply. That ratio can never exceed 1, so an ideal, unloaded divider can only shrink a voltage, never boost it — a fact that trips up beginners who wire one expecting to step a battery up. The ratio also explains why the same pair, doubled to 2 kΩ and 4 kΩ, gives an identical output: only the proportion between R1 and R2 sets the voltage, not their absolute size.
The formula assumes an ideal load: nothing draws current from the output node besides the divider itself. Connect a real sensor — a thermistor feeding a microcontroller's ADC, say — and that component's input impedance effectively sits in parallel with R2, pulling the true output below the calculated value whenever the load resistance is not much larger than R2. Electronics texts call this the loading effect, and it is why bias networks, including the ones that set a transistor's base voltage, are built from resistors far smaller than whatever they are meant to drive.
- Enter the supply voltage in the Input voltage field — the full voltage across both resistors together.
- Enter the resistor between the supply and the output node in Upper resistor, R1.
- Enter the resistor between the output node and ground in Lower resistor, R2.
- Read the result in Output voltage — this is the voltage that appears across R2.
- Before wiring it, check that anything you plan to connect to the output has an impedance much larger than R2, or the reading will not hold once connected.
Worked example — 12 V through a 1 kΩ / 2 kΩ divider
Take a 12 V supply, a 1 kΩ upper resistor, and a 2 kΩ lower resistor — a genuinely common pairing for scaling a 12 V rail toward a lower logic voltage. The ratio R2 ⁄ (R1+R2) works out to 2000 ⁄ 3000, or 0.6667, and multiplying that by the 12 V supply gives Vout = 12 × 0.6667 = 8.0 V, read straight off the Output voltage field.
The same current flows through both resistors: I = Vin ⁄ (R1+R2) = 12 ⁄ 3000 = 0.004 A, or 4 mA. That current is small enough to ignore in most bias applications, but it still costs power — 0.016 W in R1 and 0.032 W in R2, 0.048 W total — a figure worth checking against a resistor's rated wattage before it goes into a permanent circuit.
Questions
Which resistor is R1 and which is R2?
R1 is the resistor between the supply and the output node; R2 is the resistor between the output node and ground, and Output voltage is always what appears across R2. Swapping the two values does not just relabel the circuit — it changes the ratio, so a 1 kΩ / 2 kΩ pair reading 8.0 V from a 12 V supply reads 4.0 V instead if the two resistor values are exchanged.
Why can the output voltage never exceed the input voltage?
Because R2 ⁄ (R1+R2) is a ratio of a positive resistor to a larger positive sum, so it stays between 0 and 1 for realistic resistor values, and multiplying Vin by a fraction can only shrink it. A plain resistive divider carries no source of extra energy, so stepping a voltage up — turning 12 V into 20 V, say — needs an active circuit such as a boost converter, not two resistors.
Does connecting a load to the output change the reading?
Yes, if the load's resistance is not much larger than R2. The load effectively parallels R2, lowering the resistance on that side of the divider and pulling Vout below the value the plain formula predicts. For the 1 kΩ / 2 kΩ example, adding a 2 kΩ load in parallel with R2 halves R2's effective value to 1 kΩ and drops Vout from 8.0 V to 6.0 V.
How much current flows through the divider, and does it matter?
The same current runs through both resistors: I = Vin ⁄ (R1+R2). For the 12 V, 1 kΩ / 2 kΩ example that is 4 mA, dissipated as roughly 48 mW split between the two resistors. It matters twice over — as continuous draw from whatever supplies Vin, and as heat the resistors must be rated to shed, which is why bias networks favour resistor values in the kΩ–MΩ range rather than tens of ohms.
Can a voltage divider replace a voltage regulator?
Not for a load that draws meaningful or varying current. A regulator actively holds its output steady as load and supply change; a resistive divider's output sags under load exactly as the loading effect describes, and its ratio never adjusts. Dividers suit near-zero-current jobs — setting a reference into a high-impedance ADC or op-amp input — not powering a circuit directly.
What happens if R1 and R2 are equal?
The ratio R2 ⁄ (R1+R2) becomes exactly 0.5, so the output is always half the supply regardless of the absolute resistor value — 6 V from a 12 V input, whether R1 and R2 are both 1 kΩ or both 100 kΩ. Only the proportion between the two resistors sets the voltage; their shared magnitude instead sets how much current the divider draws for a given supply.