Free calculator

Voltage Divider Calculator

A voltage divider uses two resistors in series to make a smaller voltage. The output is the input times R2 ÷ (R1 + R2). Connecting something to the output changes that answer, so this calculator also shows the loaded output and the error the load causes.

Result

Output voltage

3.837 V

Current from the source: 816.3 µA
Power from the source: 9.796 mW
Output resistance (R1 and R2 in parallel)a load much bigger than this barely disturbs the output: 3.197 kΩ

The formula

Output with nothing connected

\htmlData{sym=U_out}{U_{\mathrm{out}}} = \htmlData{sym=U_in}{U_{\mathrm{in}}}\,\dfrac{\htmlData{sym=R_2}{R_2}}{\htmlData{sym=R_1}{R_1} + \htmlData{sym=R_2}{R_2}}

With a load across R2

\htmlData{sym=R_p}{R_{\mathrm{p}}} = \dfrac{\htmlData{sym=R_2}{R_2}\,\htmlData{sym=R_L}{R_{\mathrm{L}}}}{\htmlData{sym=R_2}{R_2} + \htmlData{sym=R_L}{R_{\mathrm{L}}}},\quad \htmlData{sym=U_out}{U_{\mathrm{out}}} = \htmlData{sym=U_in}{U_{\mathrm{in}}}\,\dfrac{\htmlData{sym=R_p}{R_{\mathrm{p}}}}{\htmlData{sym=R_1}{R_1} + \htmlData{sym=R_p}{R_{\mathrm{p}}}}

R1 for a wanted output

\htmlData{sym=R_1}{R_1} = \htmlData{sym=R_2}{R_2}\,\dfrac{\htmlData{sym=U_in}{U_{\mathrm{in}}} - \htmlData{sym=U_out}{U_{\mathrm{out}}}}{\htmlData{sym=U_out}{U_{\mathrm{out}}}}

What each letter means

$U_{\mathrm{out}}$: Output voltage (V · volt)
$U_{\mathrm{in}}$: Input voltage (V · volt)
$R_2$: Resistance of R2 (Ω · ohm) : How strongly something opposes current.
$R_1$: Resistance of R1 (Ω · ohm) : How strongly something opposes current.
$R_{\mathrm{p}}$: Parallel resistance (Ω · ohm) : What two resistors side by side add up to. Always less than the smaller one.
$R_{\mathrm{L}}$: Load resistance (Ω · ohm)

How it works

The same current flows through both resistors, so the input voltage is shared between them in proportion to their sizes. R2, the bottom resistor, gets the share R2 ÷ (R1 + R2). Equal resistors give exactly half.

Anything you connect to the output sits in parallel with R2 and pulls the output down. A load that is ten times bigger than R2 lowers the output by less than 10 %; a load of 100 times lowers it by less than 1 %. That is why a voltage divider is good for feeding a signal to a high-resistance input, but poor for powering something.

To design a divider, choose R2 to set how much current you are willing to waste (a few kΩ to a few tens of kΩ is common), then work out R1. In the reverse mode the calculator tries every pair of E12 values from 100 Ω to 1 MΩ, or every E12 value for R1 if you fix R2, and shows the error you get with real parts.

Worked example

12 V across R1 = 10 kΩ (top) and R2 = 4.7 kΩ (bottom). U_out = 12 V × 4.7 kΩ ÷ (10 kΩ + 4.7 kΩ) = 3.837 V.
Now connect a 10 kΩ load across R2. It sits in parallel with R2, giving 3.197 kΩ. The output falls to 12 V × 3.197 kΩ ÷ (10 kΩ + 3.197 kΩ) = 2.907 V, which is 24.2 % lower than 3.837 V.
Reverse: you want 5 V from 12 V and choose R2 = 10 kΩ. R1 = 10 kΩ × (12 V − 5 V) ÷ 5 V = 14 kΩ. The nearest E12 value that lands closest is 15 kΩ, which gives 4.8 V (−4.0 % from the target).

Common questions

Can I power a device from a voltage divider?

Usually not. A device that draws current acts as a load in parallel with R2, and the output sags as its current changes. Use a voltage regulator to power things, and keep dividers for signals and measurements, where the input takes almost no current.

How big should the resistors be?

Big enough not to waste current, small enough that the load does not disturb them. A good rule is to make the load at least 10 times bigger than R2. For a 3.3 V microcontroller input or an ADC, values from 1 kΩ to 100 kΩ are typical.

Why does the output change when I connect my meter or ADC?

Because the meter or ADC has its own input resistance, and it is in parallel with R2. If that resistance is not much bigger than R2, the output drops, and the loading error row shows by how much.

Try it live

More calculators

All calculators · All lessons · How components work