Science & Engineering
Voltage Divider Calculator - Solve Vout, R1, R2, or Vin
Calculate Vout, Vin, R1, or R2 in a resistor voltage divider. Leave one field blank to solve for it.
Leave exactly one field blank - it will be solved from the other three.
Ratio (Vout/Vin)
0.3333
Divider Current
1.0000 mA
Voltage divider formula
Vout = Vin × R2 / (R1 + R2)
The two resistors form a series circuit. The output is taken across R2. Current through the divider: I = Vin / (R1 + R2).
Common uses
- Level shifting - reduce a 5 V signal to 3.3 V for a microcontroller
- Biasing transistors
- Reference voltages for ADC inputs
- Volume controls (potentiometer)
Circuit diagram
A voltage divider consists of two resistors in series between the supply voltage and ground. The output is measured at the junction between R1 and R2:
Vin
│
[R1]
│
├───► Vout
[R2]
│
GND R1 is the top resistor (between Vin and Vout). R2 is the bottom resistor (between Vout and ground). To get a higher Vout, decrease R1 or increase R2. The formula Vout = Vin × R2 / (R1 + R2) confirms that Vout is always less than Vin (for positive resistances).
Load effect warning
When you connect a load (RL) across R2, it is in parallel with R2. The effective lower resistance is R2 ∥ RL = (R2 × RL) / (R2 + RL), which is always less than R2. This lowers Vout. For a stable Vout, choose R1 and R2 such that the divider current is at least 10× the load current, making the load’s effect negligible.
Resistor selection guide
Standard resistor values follow the E12 and E24 series, which contain 12 or 24 values per decade. When designing a voltage divider, choose a ratio R1/R2 matching your target and round to the nearest standard values.
Example: 5 V source to 3.3 V output requires Vout/Vin = 0.66, so R2/R1 ≈ 0.515. From the E24 series: R1 = 15 kΩ, R2 = 27 kΩ gives Vout = 5 × 27/(15+27) = 3.21 V (within 3.3% of target - acceptable for most digital logic level conversion).
Power dissipation calculation
Total power dissipated by the resistor network is:
P = Vin² / (R1 + R2) For battery-powered circuits, use large resistor values (100 kΩ+) to keep quiescent current low. For a 5 V source and R1+R2 = 200 kΩ: P = 25/200,000 = 0.125 mW - negligible even for coin-cell batteries.
Zener diode comparison
A Zener diode shunt regulator maintains a more stable output voltage under varying load conditions than a resistive divider:
- Stability: Zener ≪ resistive divider under variable load.
- Cost: a Zener diode + one resistor is often inexpensive.
- Current draw: Zener regulators waste current as heat; not ideal for ultra-low-power designs.
- Recommendation: use a resistive divider for signal conditioning and level shifting; use a Zener or IC regulator for powering loads.