In this guide
Two resistors. The simplest circuit in electronics — and one of the most misused. The voltage divider formula takes ten seconds to learn and years to learn when not to trust. Every semester, student projects fail because a divider that measured perfectly on a multimeter collapses the moment a real load connects, or because a divider used as a "power supply" delivers a voltage that sags with every milliamp drawn.
This guide covers the formula, the loading problem that breaks dividers in practice, the 10× rule of thumb, worked design examples, where dividers are genuinely the right answer, and where you need a regulator, buffer, or level shifter instead.
The formula (and what it assumes)
Vout = Vin × R2 / (R1 + R2)
R1 is the upper resistor (toward Vin), R2 the lower (toward ground), Vout taken at their junction. The formula assumes no current flows out of the junction — the only current is the divider's own, I = Vin/(R1+R2), flowing straight through both resistors to ground.
That assumption is the entire story of this guide. The moment a load draws current from the junction, the formula stops being true.
The loading problem
Connect a load (resistance RL) from the junction to ground, and RL sits in parallel with R2. The effective lower resistance becomes R2‖RL, which is smaller than R2 — so Vout drops. The heavier the load (smaller RL), the worse the error.
Worked example: 5V → 3.3V divider with R1 = 1.7kΩ, R2 = 3.3kΩ. Unloaded: Vout = 5 × 3.3/5.0 = 3.30V. Now connect a sensor that draws the equivalent of a 3.3kΩ load (1mA at 3.3V): R2‖RL = 1.65kΩ, Vout = 5 × 1.65/(1.7+1.65) = 2.46V. Your "3.3V" rail just became 2.46V — a 25% error, and it varies with whatever the sensor is doing.
The general rule: a divider is accurate only when the load draws negligible current compared to the divider's own current. The standard engineering rule of thumb:
The 10× rule: the divider's own current should be at least 10× the maximum load current. Then loading shifts Vout by less than ~10%.
Designing a divider properly: worked example
Goal: scale a 0–5V sensor output to 0–3.3V for an ADC, where the ADC input draws up to 10µA during sampling.
- Load current: 10µA → divider current should be ≥100µA (10× rule).
- Total resistance: R1 + R2 = 5V / 100µA = 50kΩ.
- Ratio: R2/(R1+R2) = 3.3/5 = 0.66 → R2 = 33kΩ, R1 = 17kΩ → nearest standards 33kΩ and 18kΩ.
- Verify: Vout = 5 × 33/51 = 3.24V (close enough; calibrate the last 2% in software).
- Power: P = V²/R = 25/51k ≈ 0.5mW total — negligible.
But wait — check the ADC's input requirements too. Many MCU ADCs want source impedance under ~10kΩ for their sampling capacitor to charge in time. Our divider's Thevenin impedance is R1‖R2 ≈ 11.6kΩ — borderline. For a fast-sampling ADC, drop to R1 = 1.8kΩ, R2 = 3.3kΩ (divider current ~1mA, impedance ~1.2kΩ) and accept the 1mA continuous drain — or buffer with an op-amp follower. This tradeoff (low impedance vs wasted current) is the eternal divider dilemma, covered further in the ADC accuracy guide.
Thevenin equivalent: the one-model summary
Any divider looks, to its load, like a voltage source Vth in series with a resistance Rth:
Vth = Vin × R2/(R1+R2) (the unloaded voltage)
Rth = R1 ‖ R2 (the source impedance)
Every divider question reduces to these two numbers. Vth is what you want; Rth is what ruins it — the load forms a second divider with Rth, so Vout = Vth × RL/(Rth+RL). If Rth ≪ RL, you're fine. If not, you have a problem. Memorize this model and you'll never be surprised by a divider again.
Where dividers are the right answer
| Application | Why a divider works here |
|---|---|
| Scaling a sensor output for an ADC | Load is tiny (µA) and predictable; 10× rule easy to satisfy |
| Setting a comparator threshold | Comparator input draws ~nA; divider is essentially unloaded |
| Feedback network in regulators/op-amps | Designed as part of the circuit; loading is accounted for |
| Pull-up/down with a switch | Actually a divider with a switch — covered in the pull-up guide |
| Reference voltages for high-impedance inputs | Fine, as long as nothing draws real current |
Where dividers fail (use something else)
| Application | Why the divider fails | Use instead |
|---|---|---|
| Powering a circuit ("5V to 3.3V supply") | Load current varies → voltage varies; wastes power as heat | Voltage regulator (LDO or buck) |
| I2C/SPI level shifting | Divider fights the pull-ups; can't handle bidirectional open-drain | Proper level shifter IC |
| Driving an LED | LED's nonlinear I-V curve makes brightness unpredictable | Series current-limiting resistor or driver |
| High-speed signals | Divider + parasitic capacitance = low-pass filter that rounds edges | Active level translator |
| Battery voltage monitoring with high values | 1MΩ dividers look great on paper but ADC leakage and noise dominate | Buffer with op-amp, or use moderate values |
The "divider as power supply" deserves emphasis because it's so tempting: two resistors do make 3.3V from 5V — until the load changes. A microcontroller that draws 5mA sleeping and 200mA transmitting would see its "3.3V" rail swing wildly. Dividers are for signals, never for power.
Practical details
Resistor tolerance matters. With 5% resistors, a nominal 3.30V divider can actually be 3.14–3.47V. For threshold and scaling circuits, use 1% resistors — they cost the same in practice — and calibrate in software where it counts.
Temperature drift is usually negligible for student work (100ppm/°C on both resistors drifts the ratio far less than the absolute values), but keep dividers away from hot regulators if precision matters.
Keep impedances moderate. Below ~1kΩ total, the divider wastes noticeable current. Above ~1MΩ total, PCB leakage, humidity, and the multimeter's own 10MΩ input impedance corrupt the measurement. The 1kΩ–1MΩ window covers nearly all student needs.
Measure with the load connected. A divider measured open-circuit tells you Vth; your circuit cares about Vout under load. Always verify with everything connected and running.
Common mistakes
- Designing the ratio and ignoring the current — the #1 divider failure.
- Using a divider to power a load, then debugging the "unstable" load for hours.
- 5% resistors in a precision threshold circuit, then wondering why the threshold is off.
- Forgetting the ADC's sampling-capacitor requirement — the divider is fine for DC but too high-impedance for the ADC's sampling window.
- Measuring the divider unloaded, declaring it correct, and connecting the load afterward.
- Using a divider for I2C level shifting (it breaks the open-drain protocol).
Quick design checklist
- Load current known (max, not typical) → divider current ≥10× that
- R1, R2 computed; Thevenin impedance R1‖R2 checked against the load's needs
- 1% resistors for anything precision-related
- Total resistance in the 1kΩ–1MΩ sanity window
- Voltage verified with the load connected and running
- If the "load" is a power rail → stop, use a regulator
Where to go from here
- ADC Resolution vs Accuracy — source impedance, sampling, and reference issues when dividers feed ADCs.
- Pull-Up vs Pull-Down Resistors Explained — the divider's close cousin for digital inputs.
- Power Supply Design: Linear vs Switching — what to use when you need real power, not a signal.
- How to Use a Multimeter for Circuit Debugging — measure dividers under load without fooling yourself.
- More electronics fundamentals in the Electronics / E&TC branch hub.