Calculators
Voltage divider calculator
By Bulan Sarkar
Check a divider you already have, with or without a load on the output, or give the voltage you want and get the best pair of standard resistors. Both modes show the current drawn and the power in each resistor.
The input resistance of whatever the output drives. Leave empty for no load.
- Output
- 3.243 V
- 27.03 % of the input
- Worst case at ±1 %
- 3.196 V to 3.291 V
- Output resistance
- 7.3 kΩ
- R1 ∥ R2: what the load sees
- Current from the input
- 324 µA
- Power in R1 / R2
- 2.84 mW / 1.05 mW
- ⅛ W parts are enough
The formula, and what the load does to it
Vout = Vin × R2 / (R1 + R2)
The same current flows through both resistors, so each one takes a share of the input voltage in proportion to its resistance. The output is the voltage across R2. The voltage divider lesson derives it and works through where dividers turn up.
That formula assumes nothing draws current from the output. Anything you connect does, and electrically it sits in parallel with R2, so the bottom arm becomes R2 ∥ RL and the output drops. How far it drops depends on the divider's output resistance, R1 ∥ R2, against the load. Thévenin's theorem is the tool for this: the divider looks like a source of Vout behind R1 ∥ R2. A load 100 times that resistance costs about 1 %; a load equal to it halves the output. Meter loading is the same effect showing up on your multimeter.
The worst-case range assumes both resistors sit at opposite ends of their tolerance, R1 high and R2 low, then the reverse. When R1 and R2 are equal, 1 % parts give a 1 % spread on the output. The pair finder tries each standard value of R1, solves for the R2 that would give the exact output (allowing for the load when there is one), and tests the standard values either side of it. It keeps pairs whose total lands within about a factor of three of the total you asked for.
Worked examples
5 V down to 3.3 V from E24 parts
A 5 V signal needs to reach a 3.3 V input. The ratio is 3.3 / 5 = 0.66, so R2 has to be about 1.94 times R1. Asked for E24 values totalling around 30 kΩ, the pair finder returns R1 = 4.7 kΩ and R2 = 9.1 kΩ: 5 V × 9.1 / 13.8 = 3.297 V, which is -0.088 % from the target.
The pair draws 362 µA and has an output resistance of 3.1 kΩ. With 1 % resistors the output can land anywhere from 3.275 V to 3.319 V, so resistor tolerance matters far more here than the 0.09 % the E24 rounding costs.
Halving 12 V into a 100 kΩ input
Two 10 kΩ resistors split 12 V into 6 V with nothing connected. Connect an input with 100 kΩ of resistance and the bottom arm becomes 10 kΩ ∥ 100 kΩ = 9.09 kΩ. The output falls to 5.71 V, -4.76 %. The divider's output resistance is 4.76 kΩ, and the load is only 20 times that.
Drop both resistors to 1 kΩ and the same load costs much less: the output is 5.97 V (-0.5 %). The price is current. The 1 kΩ pair draws 6.03 mA against 629 µA, and each resistor dissipates about 36 mW. When neither the error nor the current is acceptable, put a buffer (an op-amp follower) between the divider and the load.
Choosing the resistance scale
The ratio fixes the output; the size of the resistors fixes everything else. Low values make a stiff divider that shrugs off a load but wastes current and warms up. High values sip current, which matters on a battery, but a load or even the input current of an ADC pulls them around. A few kilohms to a few tens of kilohms suits most signal work. For battery monitoring people go to hundreds of kilohms and add a small capacitor across R2 for the ADC to draw from. The power dissipation lesson covers the heat side.