Voltage divider output:
formula and worked example

What a resistor divider actually does

A voltage divider is two resistors in series across a supply, with the output taken from the point where they join. It exists to produce a lower voltage than the one you have: scaling a battery voltage down so a microcontroller pin can read it, setting a bias point for a transistor, or creating a reference level for a comparator.

The useful property is that the output depends on the ratio of the resistors, not on their absolute values. A 2 kΩ and 1 kΩ pair gives the same output as 20 kΩ and 10 kΩ. That freedom lets you choose the absolute size to suit how much current you can afford to waste and how much noise the circuit must tolerate.

• Vin: the voltage across the whole chain, in volts.
• R upper: the resistor between Vin and the output tap, in ohms.
• R lower: the resistor between the output tap and ground, in ohms.

The formula and why it works

The reasoning is short. The two resistors carry the same current, which is Vin divided by their sum. The output is simply the voltage that current creates across the lower resistor, so Vout equals that current times R lower. Rearranged, this gives the expression above.

Use any consistent unit for the resistors, as the units cancel in the ratio. Ohms, kilohms and megohms all work as long as both resistors use the same one.

Worked example: 12 V across 2 kΩ and 1 kΩ

Take a 12 V supply with a 2,000 Ω upper resistor and a 1,000 Ω lower resistor. The lower resistor is one part in three of the total, so you can predict the answer before doing any arithmetic: about one third of 12 V.

1. 1Add the resistors: 2,000 + 1,000 = 3,000 Ω.
2. 2Find the fraction below the tap: 1,000 ÷ 3,000 = 0.3333.
3. 3Multiply by the supply: 12 × 0.3333 = 4 V.
4. 4Sanity-check with current: 12 V ÷ 3,000 Ω = 4 mA, and 4 mA through 1,000 Ω drops 4 V.

The upper resistor drops the remaining 8 V, which is why the two drops add back to the 12 V supply.

Reading the result: ratio, current and loading

A divider answer is only half the design. Two other numbers matter: the current the chain wastes, and how much the output sags once something is connected to it. The current here is 4 mA, which is fine for a bench circuit but heavy for a battery device that must last months.

Loading shows the effect clearly. Connect a 10,000 Ω load from the output to ground and it sits in parallel with the 1,000 Ω resistor, giving about 909 Ω. The output drops to roughly 3.75 V. A 1,000 Ω load halves the lower leg to 500 Ω and the output falls to 2.4 V, a 40% error.

• Load at least ten times the lower resistor: output stays within roughly 10% of the ideal value, which is a common design convention.
• Load comparable to the lower resistor: the output is badly wrong and the divider is the wrong tool.
• Load that varies: use a buffer or a regulator rather than a plain divider.

How each input moves the output

Because the output is a fixed fraction of Vin, it moves in exact proportion to the supply. Raise Vin by 10% and the output rises by 10%, as the table shows. That is useful when you are measuring a battery through a divider, since the reading tracks the battery, and a nuisance when you wanted a stable reference.

Changing the resistors is different. Raising R lower from 1,000 to 1,100 Ω lifts the output from 4 V to about 4.26 V, a 6.5% rise for a 10% change, because R lower also appears in the denominator and partly cancels itself. Raising R upper by the same amount pushes the output down to 3.75 V.

Where divider calculations go wrong

• Ignoring the load: the formula describes an unloaded tap, and an ADC input, op-amp bias path or LED all change the answer.
• Swapping the resistors: the output resistor goes in the numerator, so exchanging upper and lower gives the complement (8 V instead of 4 V in this example).
• Forgetting tolerance: two 5% resistors can shift the ratio by several percent, which matters when the result feeds a threshold.
• Using it to supply power: a divider is a signal-level tool, not a power supply, because its output collapses as current is drawn.

A quick design check for a battery monitor

Suppose a 3.3 V logic input must read a battery that charges up to 14.4 V. The divider ratio must stay below 3.3 ÷ 14.4, about 0.229. A 39 kΩ upper and 10 kΩ lower resistor give a ratio of 10 ÷ 49 = 0.204, so the pin sees about 2.94 V at full charge, with some margin for tolerance.

The chain draws 14.4 V ÷ 49 kΩ, about 0.29 mA. That is small enough for a mains-powered monitor and still worth a second thought for a battery that sits unused for weeks. Larger resistors reduce the drain but make the reading more sensitive to the input's own leakage and to noise.