LessonThe voltage divider · 17 min

The loading effect - and why a divider is not a power supply

The divider formula assumes nothing else draws current from the middle point. Connect something that does - a load - and the output voltage drops. Understanding this loading effect is what separates a divider that works from one that mysteriously reads low.

A divider looks like a weak battery

Seen from its output, any divider behaves exactly like a battery of its unloaded voltage in series with a resistance R1 ∥ R2 (this is Thévenin's theorem, which you will use a lot later):

For the example: V_th = 6 V, R_th = 5 kΩ, so V_loaded = 6 × 10 / 15 = 4 V - the same answer, and now you can see why. The load must be much bigger than R_th for the divider to hold its voltage:

InteractiveTry it: load a divider

Supply

VR1 (top)

ΩR2 (bottom)

ΩLoad

Ω

V_out with no load-V

Output resistance R1 ∥ R2-Ω

V_out with the load-V

Drop-%

Divider current (no load)-mA

Make R1 and R2 ten times smaller: the drop shrinks, but the divider's current grows ten times.

Your meter is a load too

A digital multimeter on volts has about 10 MΩ inside. On a 1 MΩ / 1 MΩ divider across 10 V (R_th = 500 kΩ) it reads 5 × 10 / 10.5 = 4.76 V, not 5 V. When a reading of a high-resistance circuit looks a little low, suspect the meter's own loading.

Choosing the values

Why a divider is not a power supply

A 5 V motor drawing 100 mA looks like 50 Ω. Make "5 V" from 12 V with 140 Ω / 100 Ω and connect it: R2 ∥ 50 Ω = 33 Ω, so the motor gets 12 × 33 / 173 = 2.3 V. It barely turns, the voltage swings as its current changes, and R1 burns 0.7 W as heat. Any load that takes real current - a motor, an LED strip, a module, a phone - needs a regulator (a 7805 or LM317, or a switching buck converter), never a divider.