Learn · Circuit fundamentals
Voltage Dividers Explained: How They Work and When to Use Them
A voltage divider uses two resistors to produce a fraction of an input voltage. The idea is simple, but the result depends on resistor ratio, current, and what you connect to the output.
This guide explains the circuit from first principles, then shows why a real load can change the voltage you expected.
Need the result first? Use the Voltage Divider Calculator
What is a voltage divider?
A basic voltage divider is two resistors in series between an input voltage and ground. The upper resistor is R1; the lower resistor is R2. Vout is measured at the node between them, relative to ground.
Because the same series current flows through R1 and R2 in the unloaded case, each resistor takes a share of the input voltage. The lower resistor's share is the output.
The voltage divider formula
Vout = Vin × R2 / (R1 + R2)Here, Vin is the source voltage, R1 is the resistor from Vin to the output node, and R2 is the resistor from the output node to ground. The ratio R2 / (R1 + R2) determines what fraction of Vin appears at Vout.
R1 and R2 must be in the positions described above. Swapping them changes the result because it changes which resistor is connected to ground.
Why the formula works
Start with the total resistance. In a simple series path, the resistances add:
Rtotal = R1 + R2I = Vin / (R1 + R2)That current flows through R2. Since the output voltage is the voltage across the lower resistor, Ohm's law gives:
Vout = I × R2Substitute the current expression into the voltage expression:
Vout = (Vin / (R1 + R2)) × R2Vout = Vin × R2 / (R1 + R2)This is why the divider is a ratio, not a fixed voltage source. Change either resistor, and the fraction changes.
Worked example
Use the same starting values as the calculator:
Rtotal = 10 kΩ + 20 kΩ = 30 kΩI = 5 V / 30 kΩ ≈ 166.7 µAVout = 166.7 µA × 20 kΩ ≈ 3.33 VThe same result comes directly from the divider formula: 5 V × 20 / 30 ≈ 3.33 V. The current-based calculation is a useful cross-check because it makes the voltage drop across R2 explicit.
What happens when resistor values change?
- R2 increasesMore of the total resistance is below the output node, so Vout tends to increase.
- R1 increasesMore of the input voltage is dropped above the output node, so Vout tends to decrease.
- R1 equals R2The ratio is one half, so Vout = Vin / 2.
If both resistor values are multiplied by the same factor, the ideal divider ratio stays the same. However, the current changes, and so do source impedance and sensitivity to a connected load. For example, 10 kΩ / 20 kΩ and 100 kΩ / 200 kΩ have the same ideal ratio, but they will not behave identically with the same input impedance attached.
Loading effect
The unloaded formula assumes nothing significant is drawing current from Vout. In a real circuit, a load resistance RL may be connected from Vout to ground. RL is then in parallel with R2, so the effective lower-leg resistance becomes:
Req = R2 || RL = (R2 × RL) / (R2 + RL)Vout_loaded = Vin × Req / (R1 + Req)Using Vin = 5 V, R1 = 10 kΩ, R2 = 20 kΩ, and RL = 20 kΩ:
Req = 20 kΩ || 20 kΩ = 10 kΩVout_loaded = 5 V × 10 / (10 + 10) = 2.5 VThe ideal unloaded value was about 3.33 V, but the connected load pulls the output down to 2.5 V. If RL is comparable to R2, the effective lower-leg resistance decreases substantially. That lowers Vout.
When voltage dividers are useful
Voltage dividers can create bias or reference signals, scale a sensor signal, or reduce a measurable voltage for an input circuit when the receiving input is sufficiently high impedance. They can also be useful for logic or signal sensing when the divider's source impedance and the input behavior are understood.
For ADC scaling, check the input impedance and sampling behavior of the receiving circuit. The input may draw current or briefly demand charge, changing the voltage from the ideal unloaded value.
When not to use one
A voltage divider is generally not a regulated power supply. If a load draws meaningful or changing current, its voltage will change the divider's effective resistance and therefore change Vout. Use a regulation approach intended for powering that load instead of treating a resistor divider as a universal 5 V to 3.3 V converter.
Real-world considerations
- Resistor tolerance means the actual ratio can differ from the nominal ratio.
- Each resistor dissipates power; verify that the selected component has appropriate rating and margin.
- Source impedance and load/input impedance affect the voltage seen at Vout.
- ADC sampling behavior can introduce a dynamic load, especially when source resistance is high.
- Higher resistance reduces divider current but can make the node more sensitive to leakage, noise, and input loading.
Common mistakes
- Using R1 and R2 in the wrong positions in the formula.
- Assuming the no-load voltage remains unchanged after attaching a load.
- Treating a divider as a power regulator.
- Ignoring resistor power dissipation or wasting current with unnecessarily low values.
- Choosing very high resistance without checking input impedance and noise behavior.
Calculate → Understand → Apply
Try it yourself
Change R1 and R2 to see how the ratio changes. Then enable RL and observe the loaded Vout, loading error, current, and power.
Open the Voltage Divider Calculator