⚡ Voltage Divider Calculator

A voltage divider is the simplest way to step a voltage down: two resistors in series across a supply, with the reduced voltage tapped at the point between them. It is the classic trick for reading a sensor, biasing a transistor, or feeding a reference voltage to a comparator — and it sits alongside the parallel resistance and LED resistor patterns as a core electronics building block.

The output depends only on the ratio of the two resistors, not their absolute values. Doubling both resistors keeps the same output voltage but halves the current — handy when you want to waste less power.

In practice the divider is a swiss-army knife. It scales a 0–12 V sensor signal down to the 0–3.3 V range a microcontroller can read, it builds a fixed reference for a comparator, and — with a potentiometer standing in for R2 — it becomes a continuously adjustable knob. The same ratio maths reappears whenever you bias a transistor base or set the gain of an op-amp, so learning it once pays off across the whole electronics toolkit.

For a 5 V supply, the table shows what output you get for common resistor ratios:

The formula is a direct consequence of Ohm's law. The same current flows through both resistors, so the voltage across each is proportional to its resistance. The output is the fraction of the supply that appears across R2:

V_out = V_in × R₂ / (R₁ + R₂)

The current through the chain is I = V_in / (R₁ + R₂), and the power burned in each resistor is P = I²R. That last figure is the one that decides the wattage rating you need — a divider built from 10 kΩ resistors barely warms up, while one made from 10 Ω resistors can get toasty.

As a derived check, the ratio V_out / V_in equals R₂ / (R₁ + R₂). If you know the fraction you want, you can solve for the resistor ratio directly. And there is a practical rule of thumb: whatever you connect to the output should have an input resistance at least ten times R2, otherwise the load drags the output voltage down (this is the divider's "loading" error).

You can predict that loading exactly. The divider, seen from its output, is just a Thevenin source: an ideal voltage V_out in series with a resistance of R₁ ∥ R₂. Connect a load R_L across it and the real output becomes V_out × R_L / (R_L + (R₁ ∥ R₂)). When R_L is huge the correction vanishes; when R_L approaches R₁ ∥ R₂ the output collapses to half — so the "ten times" rule is really just a 10% error budget, and you can tighten it by raising both resistor values (at the cost of more source impedance).

In microcontroller circuits a voltage divider is often used to read an analog sensor whose output range exceeds the ADC's reference pin: a pair of resistors scales the sensor voltage down to a safe level so the internal analog-to-digital converter can measure it. Keep the divider current small enough that it does not load the sensor, but large enough that the ADC's input impedance does not distort the reading.

1. Enter the input (supply) voltage Vin.
2. Enter the top resistor R1 and the bottom resistor R2.
3. Press Calculate to get the output voltage at the junction.
4. Read the chain current and the power dissipated in each resistor.
5. If you are driving a load, keep its resistance at least 10x R2 so the output stays accurate.

Why does my divider output drop when I connect a load?

The load sits in parallel with R2, lowering the effective bottom resistance and therefore the output voltage. A high-impedance load (op-amp input, MCU ADC) barely loads it; a speaker or motor will collapse it. Use a buffer (voltage follower) for heavy loads.

Do the resistor values matter, or just the ratio?

The ratio sets the voltage. The absolute values set the current and power: larger resistors draw less current (better for batteries) but make the divider more sensitive to load.

Can I use a divider to power a device?

Not directly. A divider is for signals and references, not for delivering power — most of the energy is wasted as heat in the resistors, and the voltage sags under load. Use a regulator or buck converter instead.

What if I swap R1 and R2?

You get the complementary voltage: the output becomes Vin x R1/(R1+R2). Swapping the resistors simply moves the tap to the other side of the ratio.

How do I pick resistor wattages?

Compute P = I²R for each resistor (the calculator shows this) and choose a rating at least twice that. Most signal dividers dissipate well under a milliwatt, so ¼ W parts are plenty.

For low-power designs, use large resistors (e.g. 10 kΩ – 100 kΩ) so the divider draws microamps and hardly touches your battery — pair this thinking with our battery life calculator when sizing a supply.

Standard E12 values make most ratios easy: 10 kΩ over 10 kΩ gives exactly half, and 22 kΩ over 10 kΩ gives about 0.31 of the input.

If the output must stay steady under a varying load, follow the divider with an op-amp voltage follower rather than lowering the resistors, which would waste more power.

For adjustable dividers, replace R2 with a potentiometer — the wiper becomes your V_out tap and you get a continuous range instead of fixed steps.

Feeding a microcontroller ADC? Make sure the divider's source resistance (R₁ ∥ R₂) is well under the ADC's recommended input impedance, or the sample-and-hold capacitor will not charge fully between conversions and you will read a sagging, wrong voltage.

Thermal drift is the quiet enemy of precision dividers: a 100 ppm/°C resistor pair that is not matched will let the ratio wander as the board warms. For reference duties, use a resistor network where both parts share one substrate and track each other, or skip the divider for a dedicated voltage-reference IC.

When the divider feeds a high-impedance input, also keep the chain current well above that input's bias current — picoamps for modern CMOS, microamps for older parts. If the chain current drops below the bias current, the bias current itself distorts the tap, which is another reason not to push the resistors so large they save only microwatts.

A concrete sizing example ties it together: to read a 0–12 V battery with a 3.3 V ADC you need V_out/V_in = 3.3/12 ≈ 0.275, which from the ratio table means R₂/R₁ ≈ 0.38. A 10 kΩ top resistor with a 3.8 kΩ bottom (use the nearest E12 value, 3.9 kΩ) lands within a percent. Always leave headroom so a freshly charged cell cannot push the tap past the ADC's rail.