Lesson 2.1 · Module 2: Reactive Components in Time & Frequency

Voltage Dividers & the Loading Effect

Two resistors that turn 5 V into a smaller voltage. This is how a microcontroller measures a voltage that is too big for its input.

The live, interactive simulation for this lesson opens here when JavaScript is on.

Key ideas

A voltage divider splits the voltage in proportion to the two resistors.
Whatever you connect to the middle point (the load) draws current and pulls the voltage down — the loading effect.
Bigger resistors waste less current, but the output is “weaker” and more easily pulled down.

Try this in the simulation

Make R1 and R2 equal: the output is exactly half the input.
Click “ADC sample” to connect and disconnect the measuring input, and watch U_out shift.
Make both resistors very large (1 MΩ) and see how much worse the loading gets.

How does a voltage divider work, and what is the loading effect?

A voltage divider uses two resistors in series to make a smaller voltage from a larger one: the output is the input times R2 ÷ (R1 + R2). The formula only holds if nothing draws current from the middle point. A load connected there sits in parallel with R2, lowers its resistance and pulls the output down. That is the loading effect, and it is why a divider feeding a microcontroller’s analogue input needs sensible resistor values.

What you’ll learn

How to calculate the output voltage of a two-resistor divider.
What the loading effect is and how a load pulls the output down.
Why very large resistor values make the output weak and easy to disturb.
How to size a divider that feeds an ADC input.

The formula, letter by letter

\htmlClass{mv mv-v_out}{\htmlStyle{font-size:1.11em}{\htmlData{sym=U_out}{U_{\mathrm{out}}}}} = \htmlClass{mv mv-v_in}{\htmlStyle{font-size:1.00em}{\htmlData{sym=U_in}{U_{\mathrm{in}}}}}\cdot\frac{\htmlClass{mv mv-r2}{\htmlStyle{font-size:1.06em}{\htmlData{sym=R_2}{R_2}}}\parallel \htmlClass{mv mv-rl}{\htmlStyle{font-size:1.13em}{\htmlData{sym=R_in}{R_{\mathrm{in}}}}}}{\htmlClass{mv mv-r1}{\htmlStyle{font-size:1.06em}{\htmlData{sym=R_1}{R_1}}} + \htmlData{sym=R_2}{R_2}\parallel \htmlData{sym=R_in}{R_{\mathrm{in}}}} = \htmlData{funit=V,v=2.380952380952381}{\text{2.38 V}}\;\;(\text{ideal } \htmlData{funit=V,v=2.5}{\text{2.50 V}})

Shown with this lesson’s starting values. In the simulation the numbers change as you move the sliders.

$U_{\mathrm{out}}$: Output voltage (V · volt)
$U_{\mathrm{in}}$: Input voltage (V · volt)
$R_2$: Resistance of R2 (Ω · ohm) : How strongly something opposes current.
$R_{\mathrm{in}}$: Input resistance (Ω · ohm)
$R_1$: Resistance of R1 (Ω · ohm) : How strongly something opposes current.

Common questions

What is the voltage divider formula?

U_out = U_in × R2 ÷ (R1 + R2), where R1 is the resistor on the supply side and R2 is the one connected to ground. Equal resistors give exactly half of the input voltage.

Why does my divider read the wrong voltage when I connect something to it?

The thing you connect draws current and acts like another resistor in parallel with R2. That lowers the bottom resistance, so the output drops. Use smaller divider resistors, or put a buffer in between.

Can I use a voltage divider to power a circuit?

Only a tiny one. The output changes as soon as the load changes, and the resistors waste power. A voltage regulator is the right tool for powering things.