Voltage Divider Calculator: Design Resistor Divider Circuits Fast
Calculate output voltage, resistor values, or current for any voltage divider circuit. Includes load effect warnings and practical design tips.
Voltage dividers show up everywhere in electronics — level shifting a 5V signal to 3.3V for a microcontroller, biasing a transistor, setting a reference voltage for a comparator. The math is simple enough to do in your head once you know the formula, but getting resistor values right for a given load takes a bit more work. The CalcHub Voltage Divider Calculator handles both directions: give it the resistors and it tells you the output, or give it the voltages and it suggests resistor values.
The Formula
Vout = Vin × R2 / (R1 + R2)Where R1 is the top resistor (connected to the supply) and R2 is the bottom resistor (connected to ground). Vout is measured across R2.
That's the unloaded version. When you connect something to Vout, the load resistance appears in parallel with R2, which changes the effective bottom resistance and pulls Vout down. This is the part most beginners miss.
Solving in Both Directions
Known resistors, find Vout:- Vin = 12V, R1 = 10kΩ, R2 = 10kΩ
- Vout = 12 × 10k / (10k + 10k) = 6V
- Vin = 5V, Vout = 3.3V — you want to level-shift for a 3.3V GPIO
- The ratio R2/(R1+R2) = 3.3/5 = 0.66
- Pick R2 = 33kΩ, then R1 = (5-3.3)/3.3 × 33k = 17kΩ → use standard 18kΩ
Load Effect: The Thing That Bites You
When you connect a load (like a microcontroller input pulling 100µA, or a relay coil pulling 50mA), the effective R2 drops because your load is in parallel with it.
Effective R2 = R2 × Rload / (R2 + Rload)| R2 | Rload | Effective R2 | Vout Change |
|---|---|---|---|
| 10kΩ | 100kΩ | 9.09kΩ | −9% |
| 10kΩ | 10kΩ | 5kΩ | −33% |
| 1kΩ | 10kΩ | 909Ω | −9% |
| 1kΩ | 1kΩ | 500Ω | −33% |
Choosing Resistor Values
This is where experience matters more than math:
- High impedance loads (op-amp inputs, ADC pins): You can use 100kΩ–1MΩ resistors. Quiescent current is negligible.
- Logic level shifting (5V → 3.3V for MCU pins): 10kΩ–33kΩ is a good range. Low enough to drive most inputs reliably, not so low you waste power.
- Sensor biasing or reference voltages: Keep the divider current at least 10× the expected load current.
- Battery-powered designs: Use high-value resistors (100kΩ+) to minimize drain, but verify your load tolerance.
Practical Example: 5V to 3.3V Level Shifter
Many sensors output 5V signals that would damage 3.3V microcontroller pins. A simple resistor divider works if the signal is slow (not I2C or SPI at high speeds — use a proper level shifter IC for those).
- R1 = 1.8kΩ, R2 = 3.3kΩ gives Vout ≈ 3.3V from 5V
- Or use 2.2kΩ and 3.9kΩ for slightly lower values
Can I use a voltage divider to power a circuit?
Generally no — at least not if that circuit draws any meaningful current. Dividers work well for signal conditioning and biasing, but for powering loads use a voltage regulator. The output voltage of a divider shifts significantly under changing loads.
Why does my measured output differ from the calculated value?
Check the resistor tolerances (±5% resistors can produce a ±10% swing in Vout), verify your input voltage with a multimeter, and consider whether your measurement probe itself is loading the circuit. Oscilloscope probes have 1MΩ input impedance — usually fine, but worth knowing.
What's the current draw of the divider itself?
The quiescent current is simply Vin / (R1 + R2). For 12V across two 10kΩ resistors, that's 12 / 20,000 = 0.6mA constantly flowing through the divider even with no load.
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