📉 Voltage Divider Calculator

Last updated: June 16, 2026

Voltage Divider Calculator

Two-resistor resistive divider — Vout = Vin × R2 / (R1 + R2)

Vin R1 Vout R2 V+
V
Ω
Ω

* Assumes ideal resistors with no load on Vout. Actual output will be lower when a load is connected.

What Is a Voltage Divider and Why Should You Care?

If you have spent any time tinkering with microcontrollers, sensors, or analog circuits, you have almost certainly run into a situation where the voltage coming out of one component is too high for the thing you want to connect it to. Maybe you have a 12 V sensor output and a 3.3 V ADC pin on your ESP32. Maybe you need a stable 2.5 V reference from a 5 V rail. This is exactly where the humble voltage divider earns its keep.

A voltage divider is one of the simplest circuits in electronics — just two resistors in series, with the output tapped from the junction between them. No capacitors, no inductors, no active components. Yet understanding it deeply, beyond just memorizing the formula, will save you from a surprising number of real-world headaches.

The Formula, Explained Like a Human Being

The output voltage of a two-resistor divider is:

Vout = Vin × R2 / (R1 + R2)

Where R1 is the top resistor (connected from Vin to the midpoint) and R2 is the bottom resistor (connected from the midpoint to ground). The output is taken at that midpoint.

Think of it this way: current flows through R1 and R2 in series, so the same current passes through both. Voltage drops proportionally to resistance — R2 gets a fraction of the total voltage equal to its share of the total resistance. If R1 and R2 are equal, you get exactly half of Vin. If R2 is much smaller than R1, you get a small fraction. If R2 is much larger, Vout approaches Vin.

This proportional relationship is what makes voltage dividers so useful and also what makes them tricky. The moment you connect a load to Vout, that load appears in parallel with R2, effectively reducing R2's value — and pulling your output voltage down. We will come back to this.

Finding R2 for a Target Output Voltage

The inverse problem is just as common as the forward one: you know the Vin you have, you know the Vout you need, and you want to find what resistor values to use. Rearranging the formula gives you:

R2 = R1 × Vout / (Vin − Vout)

You pick R1 freely, then calculate R2. But how do you pick R1? This is where engineering judgment comes in. The trade-off is between power consumption and load sensitivity:

  • High resistance values (100 kΩ+ for R1): Low quiescent current, low power waste. Great for battery-powered designs. But the output is very sensitive to any load — even a few kilohms of load resistance will shift Vout noticeably.
  • Low resistance values (1 kΩ or less for R1): High quiescent current, more power wasted as heat. But the output holds steady even with moderate loads connected.

A practical rule of thumb: make the divider current at least 10× larger than the expected load current. This keeps the output stable enough for most applications.

The Load Effect — The Part Everyone Forgets

This is the gotcha that trips up beginners more than anything else. When you connect a load (say, an ADC input, a transistor base, or another circuit) to the Vout node, that load has some impedance. Let us call it R_load. It now sits in parallel with R2, and the effective bottom resistance becomes:

R2_eff = (R2 × R_load) / (R2 + R_load)

If R_load is much larger than R2 (say, 1 MΩ load with 10 kΩ R2), the effect is negligible — R2_eff is barely different from R2. But if R_load is comparable to or smaller than R2, Vout will drop significantly from your calculated value.

This is why ADC inputs with high input impedance (like those on STM32 or ESP32, which are typically in the megaohm range) work fine with dividers using 10–100 kΩ resistors, while driving a 1 kΩ load from a 10 kΩ divider will give you a nasty surprise.

Power Dissipation — It Matters More Than You Think

Every resistor dissipates power as heat: P = I² × R, or equivalently P = V² / R. In a voltage divider, the quiescent current flows through both R1 and R2 continuously, even when nothing is connected to Vout. In a 5 V system with R1 = R2 = 100 Ω, you are dissipating 125 mW — which adds up fast in a battery device and can get warm enough to matter in a compact enclosure.

Standard resistors come in ¼ W (250 mW) and ⅛ W (125 mW) ratings. Push a resistor beyond its rated wattage and you risk drift, failure, or fire. Always check power dissipation, especially when using low-value resistors with higher voltages. The calculator above shows you power in R1 and R2 separately — use that number to verify your chosen resistors are within spec.

Practical Example: 5 V to 3.3 V for an ADC Input

Suppose you have a sensor outputting 0–5 V and you want to feed it into a 3.3 V-max ADC on a microcontroller. You need to scale the full 5 V range down to 3.3 V. Using the formula:

R2 = R1 × 3.3 / (5 − 3.3) = R1 × 3.3 / 1.7 ≈ R1 × 1.94

If you choose R1 = 10 kΩ, then R2 ≈ 19.4 kΩ. The nearest standard E24 value is 20 kΩ, giving you Vout = 5 × 20 / (10 + 20) = 3.33 V — close enough for most 3.3 V systems with a bit of margin. The quiescent current would be about 0.17 mA, well within comfortable bounds for a USB or regulated supply.

Choosing Standard Resistor Values

Real resistors come in standard series — E12, E24, E48, E96 — with increasing granularity. For most voltage divider work, E24 (24 values per decade) gives you enough precision to get within 2–5% of your target. If you need tighter tolerance, use E96 or E192 values with 1% resistors. For critical reference voltages, you may want a precision resistor divider module or a dedicated reference IC instead.

When the calculated R2 does not match any standard value exactly, you have two options: pick the nearest standard value and accept a small Vout error, or combine two standard resistors in series or parallel to hit a closer value. The calculator gives you the exact mathematical R2 — it is then up to you to map that to available parts.

When NOT to Use a Resistive Divider

Voltage dividers are not regulators. The output voltage shifts with load current, with supply voltage variations, and with temperature. For powering logic or anything that draws significant current, use a proper linear regulator (like an LDO) or a switching regulator. A divider is best reserved for signal-level attenuation, biasing high-impedance inputs, setting reference levels for comparators, and creating mid-rail bias points in op-amp circuits.

Also, never use a plain resistive divider to "reduce" voltage for a battery charger input, a motor driver, or anything that demands a stable low-impedance supply. You will end up with a circuit that works on the bench and fails in the field when current demand changes.

Getting the Most Out of This Calculator

Use the "Find Vout" mode when you already have resistors in hand and want to know what output voltage they produce. The results include quiescent current and per-resistor power dissipation — always check those numbers before committing to a design. Use the "Find R2" mode when you have a target voltage and want to know what bottom resistor value you need for a given R1. The verified Vout readout confirms the math is self-consistent.

One last tip: if you are designing for a production board rather than a one-off prototype, build in a bit of tolerance margin. Use 1% resistors instead of 5%, and aim for a Vout that is slightly below your absolute maximum (for example, 3.2 V instead of 3.3 V) to absorb component variation and supply fluctuations without ever exceeding the safe input range of your next stage.

FAQ

What is the voltage divider formula?
The output voltage is Vout = Vin × R2 / (R1 + R2), where R1 is the top resistor connected from the input to the midpoint, and R2 is the bottom resistor from the midpoint to ground. Vout is measured at that midpoint.
Can I use a voltage divider to power a component?
Generally no. A resistive voltage divider is not a voltage regulator — its output voltage drops as load current increases. It is suitable for attenuating signals into high-impedance inputs (like ADC pins) but not for supplying power to circuits that draw meaningful current. For that, use an LDO or switching regulator.
Why does my calculated Vout differ from what I measure on the bench?
The most common cause is load effect. Anything connected to Vout appears in parallel with R2, reducing the effective resistance and pulling the output voltage down. Use resistor values that are much smaller than the load impedance — a good rule is to make the divider current at least 10× the expected load current — to keep the error small.
How do I choose R1 when using the 'Find R2' mode?
Pick R1 based on your power budget and load requirements. Higher values (10 kΩ–100 kΩ) waste less current but are more sensitive to load. Lower values (1 kΩ or below) are more stable under load but dissipate more power. For feeding into high-impedance ADC inputs, 10 kΩ for R1 is a common and sensible starting point.
What resistor values should I actually buy — the calculator gives me an odd number?
The calculator gives you the exact mathematical value. In practice you need to round to the nearest standard resistor value in the E12, E24, or E96 series. For most applications, E24 with 1% tolerance resistors is sufficient. Plug the standard values back into the 'Find Vout' mode to see the actual output you will get.
Does resistor tolerance affect the output significantly?
Yes, especially with 5% resistors. If both R1 and R2 are off in opposite directions (R1 high and R2 low, for example), the errors compound. For a divider that needs to stay within ±1% of a target, use 1% tolerance resistors. For precision reference applications, use 0.1% resistors or a dedicated precision reference IC.