Voltage Divider Calculator
Enter Vin, R1, R2 and an optional load to get the loaded output, Thevenin equivalent and power.
Step-by-step
- Enter valid values to begin.
Ideal source and ideal resistors assumed: no tolerance, temperature coefficient or source impedance. The load RL sits in parallel with R2. In Solve mode R2 is calculated so the loaded Vout (including RL if entered) equals the target, then rounded to the chosen series; the R2 and mode-irrelevant fields are ignored. The 0.125 W check assumes 1/4 W resistors at 50% derating, which is practice, not a standard.
Voltage Divider: Why the Load Changes Your Answer
Core Engineering Principles
A divider of two resistors gives Vout = Vin × R2 / (R1 + R2), and every textbook stops there. The catch is that the formula assumes nothing is connected to the output. The moment you attach a load, that load sits in parallel with R2, the lower leg gets smaller, and the output sags. The cleanest way to think about it is through Thevenin: seen from the output, the divider is a source of Vth = Vin × R2/(R1 + R2) with a series resistance Rth = R1 ‖ R2. Any load draws current through that Rth, so the output drops by exactly that amount.
From that picture the rule falls out. If the load is at least ten times Rth, the output falls by roughly 10% or less, and a hundred times keeps you near 1%. The cure is low resistances, but low resistances burn power all the time. On a battery product we choose the highest values the load impedance and the noise floor allow, and we buffer the tap with an op-amp when the load is unknown.
Rth = R1 ‖ R2 • Solve: R2 = R1 × Vout / (Vin − Vout)
NEC & Standard References
IEC 60063 sets the E12, E24 and E96 preferred values you pick R2 from. IEC 60115-1 is the generic specification for fixed resistors; it covers rated power, tolerance, temperature coefficient and the derating curve, all of which move your ratio. IEEE Std 100 gives the formal definitions of the terms used here, such as Thevenin equivalent circuit and source impedance. The 10× load rule and 50% power derating are practice, not standards.1. Unloaded Vout = 12 × 3.3 / 13.3 = 2.977 V (ratio 0.2481).
2. R2 ‖ RL = 3.3 × 100 / 103.3 = 3.195 kΩ, so loaded Vout = 12 × 3.195 / 13.195 = 2.905 V, which is −2.42%.
3. Rth = 10 ‖ 3.3 = 2.481 kΩ; ten times is 24.8 kΩ, and the 100 kΩ load clears it.
4. Divider current = 12 / 13.195 kΩ = 0.9095 mA. PR1 = 8.27 mW, PR2 = 2.56 mW, total = 10.9 mW.
5. To get exactly 5 V from the same R1 with no load, R2 = 10 k × 5 / 7 = 7.143 kΩ. E24 gives 7.5 kΩ (5.14 V); E96 gives 7.15 kΩ (5.003 V).
- Don’t use a divider as a power supply. It has no regulation and a high output impedance; any real current drags it down. Use a regulator.
- Check the high-voltage case. Mains-sense and battery-monitor dividers must handle the full rail across R1, with resistor voltage ratings and creepage respected. Use several resistors in series for hundreds of volts.
- Input leakage matters at high impedance. A nanoamp of op-amp bias current through 1 MΩ is a millivolt of error.
- Protect the tap. If R1 shorts, the full Vin arrives at a pin rated for a few volts; add a clamp diode or Zener.