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Voltage divider calculator

Two resistors in series split a voltage. The output, taken across the bottom resistor, is a fixed fraction of the input set by their ratio. Find Vout from your resistors, or pick a Vout and get the resistor that gives it.

Across both resistors.

4.7k = 4,700 Ω.

Vout is the voltage across this one.

Whatever you connect to Vout. Leave empty for none.

Output voltage, Vout = Vin × R2 ÷ (R1 + R2)

4 V
Ratio, R2 ÷ (R1 + R2)
0.333
Current through R1
400 µA
Power from the supply
4.8 mW

The formula

Vout = Vin × R2 over R1 + R2

The same current flows through both resistors, so each takes a share of the voltage in proportion to its resistance. To get a Vout you want, rearrange it for R2:

R2 = R1 × Vout over Vin − Vout

A worked example

To read a 12 V battery on a 5 V Arduino input, scale it down by 3. With R1 = 20 kΩ and R2 = 10 kΩ:

Vout = 12 × 10k over 20k + 10k = 4 V

Only 12 ÷ 30 kΩ = 0.4 mA flows through the pair, so it barely drains the battery.

Free lesson

Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Voltage dividers” is lesson 7 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

Or start from the very first lesson, no account needed

The ideas behind it

Questions people ask

What is the voltage divider formula?

Vout = Vin × R2 ÷ (R1 + R2), where R1 is the top resistor (to Vin) and R2 is the bottom one (to ground). Vout is the voltage across R2.

Why is my divider's output lower than the formula says?

Whatever you connect to the output sits in parallel with R2 and lowers it. Keep the load at least ten times bigger than R2, or buffer the output with an op-amp. Put the load into the calculator to see the drop.

What resistor values should I use?

The ratio sets Vout; the size sets the current. 10 kΩ-range resistors suit most sensors and Arduino inputs: a few hundred microamps flow, and a high-impedance input barely loads it.

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