Voltage Divider Calculator
The real output of a resistor divider including the load it drives, plus standard E24 pairs for a target voltage and the output impedance.
Pick standard resistors for a target output
A divider is not a power supply
The formula Vout = Vin × R2 / (R1 + R2) describes a divider with nothing attached to it. Connect a load and it sits in parallel with R2, which lowers the bottom leg and drops the output below what the formula promised. Two 10 kΩ resistors on 12 V read a tidy 6 V on a meter and collapse to 4 V the moment you hang a 10 kΩ load on them. That is why this calculator asks for the load rather than pretending it does not exist.
- Aim for ten times the load current. If the chain carries at least ten times what the load draws, the output barely moves as the load varies. Below that, the output is a suggestion.
- Output impedance is R1 in parallel with R2. That single number tells you how badly a load will drag it. 10 kΩ over 10 kΩ has a 5 kΩ output impedance, which is enormous by supply standards.
- Ratio and stiffness are separate choices. Scaling both resistors by the same factor leaves the voltage identical while changing current draw and stiffness proportionally. Pick the ratio for the voltage, then the scale for the load.
- They burn power continuously. A divider draws current whether or not anything uses the output, which matters on a battery. A 1 MΩ chain across a coin cell is a different design from a 1 kΩ chain across a bench supply.
- For a fixed reference, use a reference. Dividers drift with temperature and with the input, which is why voltage references and regulators are separate parts. A divider is for scaling a signal, not for setting a rail.
How to use
- Enter the input voltage and both resistor values.
- Add the load resistance across R2 — this is the input most calculators skip.
- Compare the loaded output against the unloaded one to see how far the load drags it.
- Or set a target voltage below and get standard E24 pairs that hit it.
Frequently asked questions
Why does my divider read lower than the formula says?
Because whatever you connected is in parallel with R2. The textbook formula describes a divider with nothing attached, and attaching a load lowers the effective bottom leg, which lowers the output. Two 10 kilohm resistors on 12 V measure a tidy 6 V on a meter and collapse to 4 V the moment a 10 kilohm load hangs on them. That is not a fault, it is the circuit doing what dividers do.
How stiff does a divider need to be?
The usual rule is that the chain should carry at least ten times the current the load draws. At that ratio the output barely moves as the load varies. Below it, the output voltage becomes a function of whatever the load happens to be doing, which is rarely what anyone wants. Getting there means lowering both resistors by the same factor — the ratio, and so the voltage, is unchanged.
What is the output impedance and why does it matter?
It is R1 in parallel with R2, and it is the resistance your load sees looking back into the divider. A 10 kilohm over 10 kilohm divider has a 5 kilohm output impedance, which is enormous compared with any real supply. That single number predicts how badly a load will drag the output, which is why it is shown here alongside the voltage.
Can I use a divider to power something?
Almost never. A divider has no regulation — its output moves with the input, with temperature, and with anything the load does. It also burns current continuously whether or not the output is being used. Dividers are for scaling a signal down to something a meter or an analogue input can read. For a supply rail you want a regulator, and for a fixed reference you want a voltage reference part.
How do I pick the resistor values?
Pick the ratio first, because that sets the voltage. Then pick the overall scale, because that sets the current draw and the stiffness. These are genuinely independent choices: 1 kilohm over 1 kilohm and 1 megohm over 1 megohm both halve the input, but one is stiff and wasteful while the other sips current and is dragged by anything you connect.
Why are the suggested pairs never exact?
Because resistors are made in defined value sets, and the exact ratio you want usually falls between them. This tool searches E24 pairs and reports how far each lands from the target, so you can see whether a 0.1 percent miss matters for what you are building. It also keeps the chain near the resistance you asked for, since 10 ohm and 10 megohm pairs share ratios but not usefulness.
Does the load have to be a resistor?
It has to be expressible as one for this calculation. An op-amp or microcontroller analogue input is close enough — quote its input impedance and the numbers hold. Anything reactive or non-linear, like a transistor base or a switching load, changes with frequency and operating point, so treat the figure here as a starting estimate rather than an answer.
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