Voltage Divider Calculator
Calculate the output voltage of a resistive voltage divider, or solve for Vin, R1 or R2. Enter your values, add an optional load resistance, or find the closest standard E-series resistor pair for a target ratio.
Solve Your Resistive Voltage Divider
Voltage Divider Formula
Where:
- \(V_{out}\) — voltage at the divider tap (V)
- \(V_{in}\) — supply voltage across the whole divider (V)
- \(R_1\) — resistor between \(V_{in}\) and the tap (Ω)
- \(R_2\) — resistor between the tap and ground (Ω)
The formula states that the tap voltage is the fraction of \(V_{in}\) proportional to the lower resistor's share of the total resistance. Rearranged, it also solves the other unknowns: \(V_{in} = V_{out}\frac{R_1 + R_2}{R_2}\), \(R_1 = R_2\left(\frac{V_{in}}{V_{out}} - 1\right)\) and \(R_2 = R_1\frac{V_{out}}{V_{in} - V_{out}}\).
Voltage Divider – Calculation Example
Example: a 12 V supply feeds a divider with R1 = 10 kΩ on top and R2 = 8.2 kΩ on the bottom (the calculator's default values). What is the output voltage?
\[ V_{out} = 12 \cdot \frac{8200}{10000 + 8200} = 12 \cdot \frac{8200}{18200} = 5.407 \text{ V} \]
The divider ratio is \( \frac{8200}{18200} = 45.05\% \). The current through the chain is:
\[ I = \frac{V_{in}}{R_1 + R_2} = \frac{12}{18200} = 0.659 \text{ mA} \]
so R1 dissipates \( I^2 R_1 = 4.35 \text{ mW} \) and R2 dissipates \( I^2 R_2 = 3.56 \text{ mW} \).
Result: Vout = 5.41 V at 45.05% of Vin, with 0.66 mA flowing through the divider.
How to Use the Calculator
- In Solve for, choose the unknown: Vout from Vin, R1, R2, Vin from Vout, R1, R2, R1 from Vin, Vout, R2, R2 from Vin, Vout, R1, or Standard resistor pair (E-series). The calculator shows only the fields needed for your choice.
- Enter the known voltages in Input voltage (Vin) and Output voltage (Vout), selecting V or mV in the unit dropdown.
- Enter the known resistances in Resistor R1 (top) and Resistor R2 (bottom), choosing Ω, kΩ or MΩ. R1 sits between Vin and the tap; R2 sits between the tap and ground.
- Optionally set a Load resistance (RL) wired in parallel with R2 — for example an ADC input or bias network. Leave it at 0 for an unloaded divider.
- In the Standard resistor pair mode, enter the Target ratio (Vout/Vin) (e.g. 0.45 for 45% of the supply) and pick an E-series under Resistor series: E12 (±10%), E24 (±5%) or E96 (±1%).
- Read the results: output voltage, divider ratio, divider current, power dissipated in each resistor, the loaded-output analysis, and a live schematic of your divider.
What Is a Voltage Divider?
A voltage divider is two resistors in series across a supply. Because the same current flows through both, the supply voltage is split between them in proportion to their resistances, and the junction between them — the tap — sits at an intermediate voltage. It is one of the most common building blocks in electronics, used to scale a sensor signal into an ADC input range, set a transistor's bias point, shift the level of a signal, or read a potentiometer's position.
The output is always a fixed fraction of the input, never more than the input: \(V_{out}/V_{in} = R_2/(R_1 + R_2)\) can only take values between 0 and 1. If you need a voltage higher than what a divider can give, or a stiff regulated supply, consider a regulator or a Zener diode circuit with its series resistor instead.
The choice of absolute resistor values is a trade-off. Large resistors waste little current, but the divider's output impedance (roughly \(R_1 \parallel R_2\)) becomes high and the tap is easily pulled down by whatever it feeds. Small resistors make a stiffer tap but burn more current in the chain — the calculator's Divider Current and Power in R1/R2 fields let you check that the current budget and the resistors' power ratings are respected. A common starting point is a total chain resistance of 10 kΩ to 100 kΩ.
When a load \(R_L\) is connected across R2, it forms a parallel combination \(R_2 \parallel R_L\) that is smaller than R2 alone, so the tap voltage sags below the unloaded value. The calculator computes this loaded output; a widely used rule of thumb is to keep \(R_L\) at least ten times \(R_2\), which limits the sag to roughly 10%.
Assumptions, Limitations and Practical Tips
- DC or low-frequency only. The formula treats the resistors as ideal. At high frequencies, stray capacitance and inductance matter — see the RC time constant calculator for the bandwidth of a divider feeding a capacitive load, or the skin depth calculator for conductor behavior at high frequencies.
- No current may be drawn from the tap. The unloaded formula assumes \(R_L = \infty\). Any real load in parallel with R2 lowers the output; always check the loaded result before trusting the tap voltage.
- The resistors load the source. The divider draws \(I = V_{in}/(R_1 + R_2)\) continuously from the supply. In battery-powered devices, this quiescent current may dominate the power budget.
- Tolerance shifts the ratio. Two ±5% resistors (E24) can move \(V_{out}\) by roughly twice that, so don't rely on a divider for precision without measuring or using E96 (±1%) parts. The standard-pair finder reports each candidate pair's ratio error so you can see the nominal accuracy.
- A divider is not a regulator. Every ripple or drift on \(V_{in}\) appears at the tap, scaled by the same ratio. If the input moves, so does the output.
- Use it for sensing, not powering. Dividers are ideal as reference taps for high-impedance inputs like op-amp followers, comparators and ADCs. For driving real loads, follow the tap with a buffer or use a separate supply sized with the Ohm's law calculator.
- Alternative topologies. If you need a ratiometric measurement that rejects resistance drift, a Wheatstone bridge is the classic solution.
Reference: Solving Each Unknown
| Unknown | Rearranged formula | What you enter |
|---|---|---|
| Vout | \(V_{in} \cdot \frac{R_2}{R_1 + R_2}\) | Vin, R1, R2 |
| Vin | \(V_{out} \cdot \frac{R_1 + R_2}{R_2}\) | Vout, R1, R2 |
| R1 | \(R_2 \left(\frac{V_{in}}{V_{out}} - 1\right)\) | Vin, Vout, R2 |
| R2 | \(R_1 \cdot \frac{V_{out}}{V_{in} - V_{out}}\) | Vin, Vout, R1 |
| Standard pair | Closest \(R_2/(R_1 + R_2)\) from the chosen E-series | Vin, target ratio, series |
Frequently Asked Questions
How do I choose the actual resistor values, not just the ratio?
Any pair with the right ratio gives the same Vout, so pick the total resistance for your current budget. A 10 kΩ to 100 kΩ chain is a typical compromise: small enough to stay stiff against load, large enough to avoid wasting current. On batteries, going higher (e.g. 1 MΩ) saves power, but the tap becomes sensitive to noise and leakage. The calculator's divider current and power fields help you verify the choice.
Can a voltage divider step a voltage up or power a circuit?
No. The ratio \(R_2/(R_1 + R_2)\) is always between 0 and 1, so Vout can never exceed Vin. A divider is also not a power source: it has a source impedance of roughly \(R_1 \parallel R_2\), so any current drawn collapses the tap voltage. Use it only for signals feeding high-impedance inputs, or follow it with an op-amp buffer. To actually derive a supply rail, use a linear regulator, buck converter or Zener circuit.
Why is my measured output voltage lower than the calculated value?
The most common cause is load current: whatever the tap feeds (ADC input, microcontroller pin, downstream resistor) acts as a resistance in parallel with R2 and pulls the voltage down. Enter that equivalent resistance as RL in the calculator to see the real loaded output. Other usual suspects are resistor tolerance (±5% parts add up to roughly ±10% error in the ratio) and a Vin that is not actually what you think it is — measure it.
How do I read the divider ratio as a percentage?
Multiply \(R_2/(R_1 + R_2)\) by 100. For example, R1 = 10 kΩ and R2 = 8.2 kΩ give a ratio of 0.45, meaning the tap sits at 45% of the supply. Equal resistors always give 50%. If you are working with a 10-bit ADC, interpret the reading directly in ratiometric terms: code × divider ratio × full-scale counts gives the input fraction.
Is a potentiometer a voltage divider?
Yes. A pot is a single resistive track with a wiper that slides along it — mechanically equivalent to R1 on one side and R2 on the other, with the wiper as the tap. Turning the knob changes the ratio continuously from 0% to 100% of the applied voltage. The same load rule applies: keep the circuit reading the wiper high-impedance, or the output won't track the knob position linearly.
What resistor tolerance do I need for an accurate output?
The worst-case ratio error is roughly twice the individual resistor tolerance, so ±5% (E24) parts can be off by about ±10%. Use ±1% (E96) parts for roughly ±2% accuracy, or ±0.1% precision resistors if you need better. For critical applications, measure the actual resistors with a multimeter or calibrate the divider in software. The E-series pair finder shows each candidate's nominal ratio error.
Do I need to convert kΩ and V before using the formulas?
No. The formula is a pure ratio, so it works with any consistent resistance units — 10 kΩ and 8.2 kΩ can be entered as 10000 and 8200 ohms or 10 and 8.2, the result is identical. Only be consistent: don't mix 10 (meaning kΩ) with 500 (meaning Ω). Voltages must share one unit, though — the calculator lets you pick V or mV per field and converts automatically. In this calculator, just select the right unit in the dropdown next to each field.