Voltage divider

Find the two resistors you can actually buy, or calculate what comes out of the ones you have, load included.

Type values the way they're written on a schematic: 4.7k, 2k2, 10k.

Input voltageVVoltage I wantVResistor seriesHow much the divider may draw

Load connected to VoutΩOptional. The input resistance of whatever you connect (an ADC, a transistor, an LED with its resistor…). If you set it, the design accounts for it.

Pairs you can actually buy

R1	R2	Vout (unloaded)	Error	Worst case ±5 %	Current
240 kΩ	91 kΩ	3.3 V	-0.0 %	3.07 V – 3.54 V	36.3 µA
180 kΩ	68 kΩ	3.29 V	-0.3 %	3.06 V – 3.53 V	48.4 µA
47 kΩ	18 kΩ	3.32 V	+0.7 %	3.09 V – 3.57 V	185 µA
200 kΩ	75 kΩ	3.27 V	-0.8 %	3.04 V – 3.52 V	43.6 µA
39 kΩ	15 kΩ	3.33 V	+1.0 %	3.1 V – 3.58 V	222 µA
150 kΩ	56 kΩ	3.26 V	-1.1 %	3.03 V – 3.51 V	58.3 µA

Vout range if both resistors sit at the edge of their ±5 %.

How it works and where it breaks

Two resistors in series split the voltage in proportion to their values. The output is taken at the midpoint:

V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2}

What matters is the ratio between R1 and R2, not their absolute values: 1 kΩ and 2 kΩ give the same output as 100 kΩ and 200 kΩ. What changes from one pair to another is how much current the divider wastes and how well it stands up to a load.

Why it sags under load

The formula assumes nothing draws current from Vout. The moment you connect something, its input resistance $R_L$ sits in parallel with R2 and the divider is no longer the one you designed:

V_{out} = V_{in}\,\frac{R_2 \parallel R_L}{R_1 + R_2 \parallel R_L} \qquad\text{where}\qquad R_2 \parallel R_L = \frac{R_2\,R_L}{R_2 + R_L}

Bench rule: if the load is at least ten times R2, the drop stays under 10 %; a hundred times, under 1 %. A microcontroller ADC pin is usually hundreds of kΩ or more, so a divider with a 10 kΩ R2 drives it fine. An LED, a transistor or a relay draws real current: there the divider does not work as a supply.

Real resistors and tolerance

Resistors come in standard series. E12 (±10 %) is 12 values per decade: 1, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, 8.2. E24 (±5 %) adds one between each pair and is what any shop stocks. E96 (±1 %) is for when the voltage has to be exact.

Tolerance shows up at the output. With two ±5 % resistors the worst case moves Vout by about ±5 % too; with ±1 % it stays within ±1 %. That is why the Design mode shows the worst-case range: if your circuit cannot tolerate it, change series before changing values.

How much current to spend

The divider draws $I = V_{in} / (R_1 + R_2)$ all the time, load or not. Large resistors (hundreds of kΩ) draw microamps, ideal for sensing a battery, but are more sensitive to load and noise. Small resistors (a few kΩ) give a stiffer output at the cost of milliamps and warm resistors. Each one dissipates $P = I^2 R$ ; a 1/4 W part is almost always plenty, and the calculator tells you when it is not.

The typical case: reading a battery with an ADC

A 12 V battery and an ADC that takes 3.3 V. Ask for 3.3 V from 12 V in the E24 series with low draw: you get pairs like 270 kΩ and 100 kΩ, giving 3.24 V and drawing 32 µA. Leave margin (keep the output a bit below the ADC maximum) because a charged battery rises to 12.6 V or more, and resistor tolerance adds up.