How does a voltage divider work?
Two resistors in series across a supply produce an output at their junction of Vin × R2 ÷ (R1 + R2). The same current flows through both, so the voltage splits in proportion to resistance. With equal resistors you get half the supply; with R2 nine times R1 you get 90%. It is the simplest way to scale a voltage down, and one of the first circuits anyone learns.
Why does my divider output sag under load?
Because connecting a load places a third resistance in parallel with R2, lowering the effective resistance and therefore the output. A divider only holds its ratio when the load draws negligible current compared with the divider itself. The rule of thumb is that the load should be at least ten times R2 — ideally a hundred. This is the single most common divider mistake.
Can I use a voltage divider to power something?
No, and it is worth being blunt about it. A divider is a signal-level reference, not a power supply. Anything that draws real current — a motor, an LED at full brightness, a microcontroller — will pull the output down and waste the difference as heat in the resistors. Use a regulator or a buck converter for power. Dividers are for reading a voltage, not supplying one.
How do I choose resistor values?
Trade quiescent current against loading. Small resistors hold the ratio better under load but waste more current continuously — a 1 kΩ pair across 12 V draws 6 mA all the time. Large resistors save power but are more easily disturbed by the load and by input bias currents. For an ADC input, tens of kilohms is typical; for battery-powered designs, hundreds.
What is a common real use for a divider?
Scaling a voltage into a measurable range. A microcontroller ADC reading 0–3.3 V cannot see a 12 V battery directly, so a divider brings it into range and the firmware multiplies back up. Dividers also set reference voltages, bias transistor bases, and combined with a thermistor or LDR form the classic sensor circuit where the changing resistance changes the output voltage.
How much power do the resistors dissipate?
Current is Vin ÷ (R1 + R2), and each resistor dissipates I²R. A 12 V supply with two 1 kΩ resistors passes 6 mA and dissipates about 36 mW in each — comfortable for a standard quarter-watt part. Halve the resistances and the power quadruples. The calculator shows both resistors' dissipation so you can check the parts are adequately rated before building.