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| R1 | R2 | Actual Vout | Error (%) |
|---|
Given: Vin=12V, R1=6.8kΩ, R2=4.7kΩ
Given: Vin=5V, R1=350Ω, R2=350Ω
Given: Vin=5V, R1=2kΩ, R2=3kΩ
A voltage divider is the simplest and most-used building block in electronics: two resistors in series that split an input voltage into a smaller, predictable fraction. The output is taken from the junction between them and follows Vout = Vin × R2 / (R1 + R2), where R2 is the resistor across which the output is measured. Because the ratio — not the absolute values — sets the output, a 2 kΩ/3 kΩ pair and a 20 kΩ/30 kΩ pair both give the same 60% output.
The formula above assumes nothing is drawing current from the output. As soon as you connect a real load, it sits in parallel with R2, lowers the effective resistance, and pulls the output below the calculated value. The rule of thumb is to make the divider current at least 10× (ideally 100×) the load current, or to buffer the output with an op-amp follower. This is why dividers are perfect for high-impedance inputs (an ADC pin, a comparator, a transistor base bias) but a poor way to power anything.
Pick the ratio for the voltage you need, then scale both resistors together to trade current for noise: smaller resistors waste more power but are less sensitive to stray capacitance and input leakage; larger resistors save power but load more easily and pick up noise. For battery devices, keep the total resistance high (tens of kΩ to MΩ) so the divider itself sips only microamps.
Scaling a battery or supply voltage down so a microcontroller ADC can measure it, setting reference and threshold voltages, biasing transistor and op-amp stages, level-shifting logic signals, and building the resistive half of sensors such as thermistor and photoresistor circuits.
Can a divider supply current to a circuit? Only tiny amounts. For real loads use a proper regulator; a divider is for setting a reference voltage, not delivering power.
Why is my measured output lower than calculated? Almost always the loading effect — whatever you connected is drawing current. Increase the divider current or add a buffer.
The divider formula Vout = Vin × R2/(R1+R2) is exact — but only while nothing is connected to the output. The moment a load is attached it sits in parallel with R2, lowering the effective lower-leg resistance and pulling the output below the calculated value.
The rule of thumb is to make the divider current at least ten times the load current, or equivalently keep the load resistance at least ten times R2. That keeps the error under roughly 10%. For better accuracy, either lower the divider resistances further or buffer the output with an op-amp follower, whose input draws essentially nothing.
A common compromise for an ADC input divider is a total of 10–100 kΩ, which is high enough to be efficient and low enough for a typical ADC sampling capacitor to charge within its acquisition window.
Two ±1% resistors do not necessarily give a ±2% divider. What matters is how the two track each other, and resistors from the same batch and package type track far better than their individual tolerances suggest. This is why matched resistor networks exist: they offer modest absolute accuracy but excellent ratio accuracy and temperature tracking, which is exactly what a divider needs.
A divider cannot regulate. Its output moves with the input and with any load change, and the energy not delivered to the load is dissipated as heat in the resistors. For anything drawing meaningful current, a regulator is the right answer — a divider is for sensing and reference, not for powering.
Vout = Vin × R2/(R1+R2), where R2 is the resistor across which the output is measured. It holds exactly only when no load is connected.
Because the load sits in parallel with R2 and reduces the effective lower-leg resistance. Keep the load resistance at least ten times R2, or buffer the output with an op-amp follower.
For an ADC input, a total of 10–100 kΩ is a common compromise. Lower values resist loading and noise but waste current; higher values save power but are disturbed by input bias current and leakage.
Not necessarily, and usually better. What matters is how the two track each other, and parts from the same batch and package track far more closely than their individual tolerances imply. Matched networks exploit this deliberately.
No. A divider cannot regulate, its output falls as load current rises, and the undelivered energy becomes heat in the resistors. Use a regulator for anything drawing meaningful current.
Buffer the output with an op-amp voltage follower. Its input draws negligible current, so the divider sees effectively no load while the follower drives the real one.