Voltage Divider Load Calculator
Calculate output voltage with load, parallel resistance, and divider current.
What the voltage divider load calculator does
The voltage divider load calculator shows what happens to a resistive divider when something is connected to its output. Enter the input voltage, the two divider resistors R1 (top) and R2 (bottom), and the load resistance RL, then press Calculate. The tool reports the unloaded output, the output with the load attached, the parallel combination of R2 and RL, and the current drawn from the source.
The formulas
- Unloaded output: Vout = Vin × R2 / (R1 + R2)
- Effective bottom resistance: Rp = (R2 × RL) / (R2 + RL)
- Loaded output: Vout,load = Vin × Rp / (R1 + Rp)
- Divider current: I = Vin / (R1 + Rp)
The load sits in parallel with R2, so it always lowers the bottom resistance and therefore the output voltage. The only question is by how much.
Worked example with the default values
Vin = 12 V, R1 = 1 kΩ, R2 = 2 kΩ, RL = 10 kΩ.
- Unloaded: 12 × 2000 / 3000 = 8.00 V
- Rp = 2000 × 10000 / 12000 = 1,667 Ω
- Loaded: 12 × 1667 / 2667 = 7.50 V, a 6.3% drop
- Current: 12 / 2667 = 4.5 mA
How much load a divider can tolerate
| RL relative to R2 | Output drop (R1 = R2) | Verdict |
|---|---|---|
| 100 × | 0.5% | Negligible |
| 10 × | 4.8% | Acceptable for indicators, marginal for references |
| 2 × | 20% | Significant; redesign or buffer |
| 1 × | 33% | Divider no longer sets the voltage |
| 0.1 × | 83% | Output collapses |
The common design rule is to make RL at least ten times R2, or equivalently to run the divider at ten times the load current. Loading is worst when R1 and R2 are large; a divider made from two 1 MΩ resistors is disturbed by a 10 MΩ multimeter.
The Thevenin view
A loaded divider behaves like an ideal source of the unloaded voltage in series with an output resistance equal to R1 in parallel with R2. For the example, RTH = 1000 ‖ 2000 = 667 Ω. Any load then forms a second divider with that 667 Ω: 8 V × 10000 / 10667 = 7.50 V, the same answer. This view makes it easy to judge a divider at a glance: the output resistance must be small compared with whatever it feeds.
Fixing a divider that sags
- Lower both resistors by the same ratio. Output stays the same unloaded, RTH falls, but the divider draws more current and dissipates more power; check with the power dissipation calculator.
- Pre-compensate. If the load is fixed, choose R2 so that R2 ‖ RL gives the right ratio. Run the calculator with a few values until the loaded output hits target.
- Add a buffer. An op-amp voltage follower has an input resistance in the gigaohm range and presents a low output resistance to the load.
- Use a regulator or reference when the output must be accurate under changing load.
Where loading matters
ADC inputs on microcontrollers have sampling capacitors that need a source below about 10 kΩ; sensor dividers feeding them should be designed with the ADC’s input as RL. Level shifters between 5 V and 3.3 V logic drive gate capacitance rather than resistance, so their loading is dynamic and set by frequency. Transistor bias networks, covered in the BJT biasing calculator, load the divider with the base input resistance β × RE.
Common mistakes
- Measuring the divider with a cheap meter. A 1 MΩ input meter loads a high-value divider and reads low.
- Entering kilohms as ohms. All four fields expect ohms.
- Ignoring the source’s own resistance. A battery or sensor feeding R1 adds to it.
- Using a divider as a power supply. A load that draws significant current needs a regulator; a divider wastes power and its output varies with load.
Related tools
The voltage divider calculator handles the unloaded case and solving for resistor values. The current divider calculator is the parallel counterpart, and the Ohm’s law calculator and current calculator cover the individual steps.
External references
Frequently asked questions
Why does the output voltage fall when I connect a load?
The load is in parallel with R2, which lowers the bottom resistance of the divider and shifts the ratio toward R1.
How big should the load be to ignore the effect?
At least ten times R2 for errors under about 5%, or 100 times for under 1%.
Can I increase the output by adding a load?
No. A resistive load can only reduce it. Raising the output requires changing R1 or R2.
What is the output resistance of a divider?
R1 in parallel with R2. It is what the load “sees” and determines how much the output sags.
Does the calculator work for AC?
Yes for resistive loads at low frequency. Capacitive or inductive loads need reactance instead of resistance, which changes with frequency.