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Electrical

Voltage Divider Calculator

Use the voltage divider calculator to find output voltage or solve for the source voltage and either resistor when the other three values are known.

Output voltage (Vout)
6.00 V
  • Output across R2V 6 · 50%
  • Drop across R1V 6 · 50%
The source voltage divides between the two resistors in proportion to their resistance. The output is the share that falls across R2.

The output voltage is the fraction of the source voltage dropped across R2. The reverse modes rearrange the same voltage-divider relationship to solve for the missing circuit value.

Written by
MPhil in Mathematics, Founder of Mathovia
Published

Mathematics verification: Mathematical formulas, calculations, worked examples and solutions are reviewed for accuracy before publication. How we check our work

A voltage divider uses two resistors to turn a higher voltage into a lower one. It is useful for sensor signals, reference voltages, and measuring a voltage that is too high for an input pin. The calculator can find the output voltage, source voltage, R1, or R2 when the other three values are known.

Source voltage (Vs)
        │
        R1
        │──── Output voltage (Vout)
        R2
        │
       GND
Voltage Divider Calculator formula
The Voltage Divider formula

How to Calculate Voltage Divider Output and Resistor Values (step by step)

Step One: Identify the resistor positions

R1 is the resistor between the source and the output node. R2 is the resistor between the output node and ground. The output voltage is measured across R2.

R1 = resistor connected to Vs
R2 = resistor connected to GND

For the default values, the source is 12 V, R1 is 10000 Ω, and R2 is 10000 Ω.

Step Two: Calculate the output voltage

Use the voltage divider relationship to calculate the voltage at the junction between the resistors.

Output voltage = Source voltage × R2 ÷ (R1 + R2)
Vout = 12 × 10000 ÷ (10000 + 10000)
Vout = 6 V

When R1 and R2 have equal resistance, the output is half of the source voltage.

Step Three: Solve for the source voltage

If the output voltage and both resistors are known, rearrange the formula to find the source.

Source voltage = Output voltage × (R1 + R2) ÷ R2
Vs = 6 × (10000 + 10000) ÷ 10000
Vs = 12 V

Step Four: Solve for R1

To find the upper resistor, use the source-to-output ratio and multiply it by R2.

R1 = R2 × (Source voltage ÷ Output voltage − 1)
R1 = 10000 × (12 ÷ 6 − 1)
R1 = 10000 Ω

Step Five: Solve for R2

To find the lower resistor, multiply R1 by the output voltage and divide by the difference between source and output voltage.

R2 = R1 × Output voltage ÷ (Source voltage − Output voltage)
R2 = 10000 × 6 ÷ (12 − 6)
R2 = 10000 Ω
Voltage Divider Calculator worked example diagram
A visual read of the worked example above — the same numbers the calculator returns for its default inputs.

The formula, and why it works

The same current flows through both resistors in series, so the voltage splits between them in proportion to their resistance. gets the fraction of the total, and that’s the output. Equal resistors split the source in half; making larger pushes the output toward , making it smaller pulls the output toward ground.

The divider current — which sets how much power the resistors waste and how stiff the output is — is:

Common output ratios

For a 5 V source, unloaded:

Target VoutR2 ÷ (R1 + R2)Example pair
1.65 V0.33R1 = 6.8 kΩ, R2 = 3.3 kΩ
2.50 V0.50R1 = 10 kΩ, R2 = 10 kΩ
3.30 V0.66R1 = 10 kΩ, R2 = 20 kΩ
4.00 V0.80R1 = 2.5 kΩ, R2 = 10 kΩ

Any pair in the right ratio gives the same unloaded output; the absolute values decide the current and the loading sensitivity.

The loading effect

This calculator assumes nothing draws current from the output. In a real circuit, whatever you connect there — an ADC pin, a transistor base, a comparator input — sits in parallel with . The effective lower resistance becomes , which is smaller than , so the real output is lower than calculated. Rule of thumb: keep the load resistance at least ten times , or buffer the node with an op-amp voltage follower so the divider itself is barely loaded.

Practical uses

  • Scaling a voltage to read it. A 24 V rail divided down to under 3.3 V so a microcontroller ADC can measure it.
  • Setting a reference. A fixed fraction of a supply as a comparator threshold.
  • Reading a resistive sensor. A thermistor or photoresistor as one half of the divider, with a fixed resistor as the other, turns resistance change into a voltage the ADC can sample.
  • Level-shifting a logic signal from 5 V down to 3.3 V (for slow signals only — the resistors and stray capacitance form a low-pass filter).

Common mistakes

  • Ignoring the load. The single biggest source of “why is my output low?”
  • Resistors far too large, making the node a high-impedance antenna for noise and bias current.
  • Setting Vs equal to Vout when solving for a resistor, which divides by zero.
  • Expecting the divider to supply current. It sets a voltage; it can’t power anything.

For the single-resistor case, the volts to amps calculator and amps to volts cover plain Ohm’s law.

About the formula: The output voltage is the fraction of the source voltage dropped across R2. The reverse modes rearrange the same voltage-divider relationship to solve for the missing circuit value.

Frequently Asked Questions

What is the voltage divider formula?+

Vout = Vs × R2 ÷ (R1 + R2). R1 is the resistor between the source and the output node; R2 is between the output node and ground. The output is the fraction of the source that R2 claims.

What resistors give me 3.3 V from 5 V?+

You need R2 ÷ (R1 + R2) = 3.3 ÷ 5 = 0.66. Any pair in that ratio works — for example R1 = 1.7 kΩ and R2 = 3.3 kΩ, or R1 = 10 kΩ and R2 = 20 kΩ. Use the smallest values that keep the divider current acceptable.

Can I mix kilohms and megohms?+

Yes. The calculator normalizes both resistors to ohms before computing, so R1 can be in kΩ and R2 in MΩ.

Does a connected load change the output?+

Yes, and it's the number-one reason a real divider reads low. A load on the output sits in parallel with R2, lowering the effective R2 and pulling Vout down. Keep the load resistance at least 10× R2, or buffer the output with an op-amp follower.

How do I choose the resistor values, not just the ratio?+

Trade current against loading. Small resistors (1 kΩ range) waste more current but shrug off load and noise. Large resistors (100 kΩ+) sip current but are sensitive to loading, stray capacitance, and input bias current. For a sensor into a microcontroller ADC, tens of kΩ is a common middle ground.

Can a voltage divider power a load?+

Not really. It can set a reference or scale a signal into a high-impedance input, but it can't supply meaningful current — the output sags as soon as current is drawn. Use a regulator for that.

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