Voltage Divider Calculator

Why a voltage divider "gets it wrong" when you connect something to it

The result Vin × R2 ÷ (R1 + R2) holds with the output open. Any circuit connected to it, such as the input of an analog-to-digital converter or an amplifier, forms an extra path to ground in parallel with R2. Because the equivalent resistance gets smaller, the output falls. A 12 V divider of 1.8 kΩ and 1.3 kΩ gives 5.03 V with no load; with a 100 kΩ load it gives 4.99 V; with 10 kΩ it gives 4.68 V, an error of −7%.

There are two ways out. One is to choose resistors much smaller than the load, but that raises the current wasted all the time: 18 kΩ and 13 kΩ have the same ratio as 1.8 kΩ and 1.3 kΩ, draw ten times less current and suffer far more from the same load: with 10 kΩ, the output falls to 2.87 V. The other is to compute the resistors with the load already in the equation, which is what the "Find the resistors" tab does: with a 10 kΩ load on 12 V, the 1.5 kΩ and 1.2 kΩ pair delivers 5.000 V.

• No load: Vout = Vin × R2 ÷ (R1 + R2)
• With load: R2 is replaced by R2 ∥ RL = R2 × RL ÷ (R2 + RL)
• Output impedance = R1 ∥ R2
• Supply current = Vin ÷ (R1 + R2 ∥ RL)

When not to use a resistive divider

The divider is for scaling down a signal or making a reference with little current. It is not for powering a circuit: the output would collapse as soon as the circuit drew current, and the voltage difference is thrown away as heat in the resistors. For power, use a regulator. With fast signals, the input capacitance of whatever reads the node forms a filter with the divider's output impedance, and ADC readings call for a low source impedance: check the limit in your part's datasheet.

The error the tool shows is the rounding to the standard series. 1% or 5% resistors add their own deviation, and the ratio of two resistors can be off by more than either one alone. At higher voltages, check power too: with 24 V and a 4.7 kΩ and 750 Ω pair the current is 4.4 mA and R1 dissipates about 91 mW.