Boneyard Tools

Why a Loaded Voltage Divider Sags

How a load resistor pulls a divider's output below its ideal value, and how to size resistors so the sag stays small.

The ideal divider assumes nothing is attached

The clean formula Vout = Vin x R2 / (R1 + R2) describes a divider whose output pin connects to nothing that draws current. In that state the same current flows through R1 and R2, and the voltage splits in exact proportion to the resistances. This is the number this calculator returns, and it is the right starting point for design. The moment you attach a real circuit to the output, though, that assumption stops holding.

A load resistor rewrites the bottom of the divider

Whatever you connect across the output sits electrically in parallel with R2. Two resistors in parallel always have a lower combined value than either one alone, so the effective bottom resistance shrinks. Because Vout tracks the bottom resistance over the total, a smaller bottom means a smaller output voltage than the ideal formula predicts. You can model this exactly by replacing R2 with the parallel combination of R2 and the load, then reapplying the divider equation.

Stiff dividers versus high-impedance loads

A common rule of thumb is to make the divider current at least ten times the current the load will draw, which keeps R2 much smaller than the load resistance. Such a stiff divider barely moves when the load is connected, but it wastes more power in the resistors as heat. A high-value, power-thrifty divider does the opposite: it sips current yet sags badly under even a modest load. Choosing between the two is a trade between accuracy and efficiency.

When to reach for a buffer instead

If the output must feed something that draws real current, such as an analog input that samples charge or a stage that sinks milliamps, resistor sizing alone may not hold the voltage steady. An op-amp voltage follower placed after the divider presents a very high impedance to the divider and a low impedance to the load, breaking the sag entirely. For power rails, a dedicated regulator is the better answer, since a divider is only meant to set a reference, not to supply current.

Frequently asked questions

How do I include the load in the calculation?

Compute the parallel value of R2 and the load resistance, then use that combined value in place of R2 in Vout = Vin x R2 / (R1 + R2). The result is the true loaded output.

Does a bigger load resistance help?

Yes. The higher the load resistance compared with R2, the closer the parallel combination stays to R2, so the output sags less and the ideal formula stays accurate.