Voltage Divider Calculator
Solve for output voltage or either resistor in a two-resistor divider, with optional load to show the real-world loading effect.
- Vout = Vin × R2 / (R1 + R2)
- R2 = R1 × Vout / (Vin − Vout)
- R1 = R2 × (Vin − Vout) / Vout
- Rth = R1 ∥ R2 = (R1 × R2) / (R1 + R2)
How It Works
- 1
Pick what you want to solve for
Choose Find V_out when you have both resistors and the supply voltage. Switch to Find R1 or Find R2 when you know the target output and need to size one of the resistors.
- 2
Enter the known values
Type V_in, then the two knowns for your chosen mode. Units are volts and ohms. The output updates live and flags impossible inputs such as V_out greater than V_in.
- 3
Add a load to check the real output
Toggle the load resistor and enter its value in ohms. The result panel shows the loaded V_out alongside the unloaded Thévenin voltage and output impedance, so you can see at a glance whether your divider is stiff enough.
Voltage Dividers: The Two-Resistor Rule That Sets Reference Voltages
Connect a 9-volt battery across two 1 kΩ resistors in series. Measure at the junction. You read 4.5 volts. That is a voltage divider. The math traces back to Kirchhoff's 1845 voltage law combined with Ohm's 1827 result. Apply them in series and the ratio V_out/V_in equals R2/(R1+R2). No new physics, just arithmetic. The formula is exact on paper and wrong in most real circuits. Connect any load that draws current (an ADC pin, a transistor base, an LED) and the load sits in parallel with R2. The effective divider shifts. The standard engineering fix is the 10× rule: pick your divider resistance at least ten times smaller than the load impedance. Ignore the rule and a 5 V to 3.3 V level shifter that measures right with a multimeter will clip edges on a 400 kHz I²C bus, because the signal line acts as capacitance and a stiff divider cannot drive it fast enough. Voltage dividers are used where small currents are acceptable: thermistor and RTD readouts, transistor bias networks, ADC reference taps, volume potentiometers, and high-voltage oscilloscope probes with continuous bleeder currents. They are not power supplies. A divider that lights a 20 mA LED from 12 V wastes 200 mW continuously and changes brightness every time the battery sags. Linear regulators and switching supplies exist for that job. Texas Instruments still ships the 10× rule as footnote 1 on the SN74HCS72 datasheet, almost two hundred years after Ohm.
Common pitfalls
Using a divider as a power supply. A 12 V to 5 V divider feeding a 20 mA LED wastes 140 mW in the divider and another 140 mW the moment load current rises; output voltage sags by volts. Use a linear regulator (7805, LM317) or a buck converter instead.
Ignoring the 10x rule. Pick the divider resistance at least ten times smaller than the downstream input impedance. A 100 kΩ / 100 kΩ divider feeding a 1 MΩ ADC input reads about 5% low; the same divider feeding a 10 kΩ op-amp input reads about 50% low.
Forgetting bandwidth. Source impedance of an unbuffered divider combines with line and input capacitance to form a low-pass filter. A 10 kΩ Thévenin impedance plus 20 pF trace capacitance gives a 796 kHz corner; a 100 kΩ divider drops that to 80 kHz, which clips I2C edges on a 400 kHz bus.
Cascading two dividers without a buffer between them. Each stage loads the one before it, and the final ratio is neither the product of the individual ratios nor anything simple to hand-calculate. Buffer with an op-amp follower when you need accurate cascaded references.
Frequently Asked Questions
What is a voltage divider used for?
A voltage divider produces a smaller, fixed fraction of an input voltage. Engineers use it to set reference voltages for analog-to-digital converters, bias transistor bases, interface sensors like thermistors and RTDs, attenuate signals for oscilloscope probes, and shift logic levels between 5 V and 3.3 V systems. It is not a power supply — use a linear or switching regulator when the load draws meaningful current.
Why does connecting a load change the output voltage?
Any load you attach sits in parallel with R2. The effective R2 drops, the ratio R2/(R1+R2) shifts, and V_out falls below the ideal unloaded value. This is called the loading effect. The standard engineering remedy is the 10× rule: pick the divider so its output impedance R_th = R1 ∥ R2 is at least ten times smaller than the load impedance. Our calculator shows both the loaded V_out and the unloaded Thévenin values so you can see the error directly.
How do I choose R1 and R2 values?
Start from the ratio V_out/V_in = R2/(R1+R2). That ratio determines the division, not the absolute resistance. The absolute values trade off two things: low resistance stiffens the divider against loading but wastes more power, while high resistance saves power but is more sensitive to load and stray capacitance. Typical microcontroller bias and ADC dividers sit between 1 kΩ and 100 kΩ. High-voltage probes run megohms with bleeder currents in the low milliamp range.
Can I use a voltage divider to power a circuit?
No. A divider is a series-pass arrangement that continuously dissipates power in both resistors, and its output voltage collapses under significant load. A 20 mA load pulled through a 12 V divider wastes hundreds of milliwatts and sags every time the source voltage dips. Use a linear regulator (LM317, LM7805, LDO) or a switching converter for any current above a few microamps.
Does the formula work for AC signals?
The resistive formula works for DC and low-frequency AC. At higher frequencies, stray and parasitic capacitance across R1 or R2 starts to matter, turning the divider into a frequency-dependent filter. Compensated oscilloscope probes solve this with a deliberate parallel capacitor on each resistor, matched to the scope input impedance — this is what the probe compensation adjustment trims.
What is the Thévenin equivalent of a voltage divider?
Looking into the divider tap with V_in applied, the source behaves like an ideal voltage of V_th = V_in × R2/(R1+R2) in series with a resistance of R_th = R1 ∥ R2 = R1·R2/(R1+R2). Any load you attach sees that equivalent source. This is why the 10× rule uses R_th, not R1 or R2 individually, to judge whether the divider is stiff enough.
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