What is the voltage divider formula?

Vout = Vin × R2 / (R1 + R2). The output voltage is set by the ratio of R2 to the total resistance. To design for a target Vout, choose R2 first then calculate R1 = R2 × (Vin/Vout − 1).

How do I choose resistor values for a voltage divider?

The divider current (Vin / (R1+R2)) should be at least 10× the load current to maintain regulation. Higher divider current improves accuracy but increases power loss. For signal-level dividers, 10–100kΩ total resistance is common.

Can I use a voltage divider as a power supply?

No. A voltage divider is only suitable for light loads where the load current is small compared to the divider current. For power supply applications, use a linear regulator or switching converter instead.

Voltage Divider Calculator

How the Voltage Divider Calculator Works

A resistive voltage divider is two resistors in series between a supply voltage and ground. The junction between them produces an output voltage proportional to the resistance ratio. It is one of the most fundamental circuits in electronics, used for biasing, ADC input scaling, feedback networks, and level shifting.

Core Formula

Vout = Vin × R2 / (R1 + R2)

R1 is the top resistor (between Vin and Vout). R2 is the bottom resistor (between Vout and GND). Rearranging for design: R1 = R2 × (Vin/Vout − 1).

Choosing Resistor Values

The calculator uses a design heuristic: the divider current (I_divider = Vin / (R1+R2)) is set to at least 10× the specified load current. This ensures the load does not significantly disturb the output voltage. The auto-calculated resistance pair satisfies this constraint while keeping power dissipation low.

Signal-level inputs (ADC, comparator): 10–100 kΩ total — load current is negligible
Biasing BJT base: 1–10 kΩ — base current drives the lower bound
Power monitoring sense divider: 100 kΩ–1 MΩ — minimize quiescent loss

Power Dissipation

P_R1 = I_divider² × R1 · P_R2 = I_divider² × R2

Total power lost in the divider is Vin × I_divider. For battery-powered designs this can be significant — a 10 kΩ divider on a 3.3V rail draws 0.33 mA and wastes 1.1 mW continuously. On a 2000 mAh cell this alone drains 250 μAh per hour.

Design Example: 5V → 3.3V, 1 mA load

Target: Vout = 3.3V, Vin = 5V, ratio = 3.3/5 = 0.66
Divider current heuristic: 10 × 1 mA = 10 mA → total R = 5V / 10 mA = 500 Ω
R2 = 500 × 0.66 = 330 Ω, R1 = 500 − 330 = 170 Ω
P_total = 5V × 10 mA = 50 mW
Note: For lower power, use 33 kΩ + 17 kΩ (same ratio, 10× less current)

Frequently Asked Questions

Why does my output voltage drop under load?
The load resistor appears in parallel with R2, reducing the effective bottom resistance and lowering Vout. The larger your load current relative to the divider current, the worse the regulation. If you need stable voltage under varying load, use a voltage regulator, not a divider.

What is a Wheatstone bridge?
A Wheatstone bridge is two voltage dividers sharing the same supply, with the output taken as the differential voltage between their midpoints. It is used for precision resistance measurement and sensor interfacing (strain gauges, RTDs, thermistors) where small resistance changes must be detected.

Can I use voltage dividers for high-voltage measurement?
Yes. High-voltage probes and power meter front ends use voltage dividers with resistors rated for the operating voltage. Use resistors with appropriate voltage ratings and keep the divider ratio large enough that the measured node stays within the ADC input range. Resistor tolerances directly set measurement accuracy.