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Voltage Divider Calculator and Formula

14 min

This interactive voltage divider calculator allows you to obtain the output voltage (VoutV_{out}) of a passive voltage divider circuit immediately. Below, we explain the physical and mathematical theory behind this fundamental circuit.


What is a Voltage Divider?

A voltage divider (or potential divider) is a simple passive linear circuit that reduces an input voltage (VinV_{in}) to a fraction of it as output voltage (VoutV_{out}).

It typically consists of two resistors in series (R1R_1 and R2R_2) connected to an input power supply. Voltage dividers are widely used in electronics to:

  • Adapt signal levels for microcontrollers (such as ADC pins on Arduino or ESP32)
  • Bias active components like BJT transistors and operational amplifiers
  • Create stable reference voltages
  • Interface 5V sensors with 3.3V microcontrollers

Voltage Divider Formula

The physical principle of a voltage divider is based on Ohm's Law and Kirchhoff's voltage laws. Since the same current flows through both resistors in series, the general formula is:

Vout=Vin×R2R1+R2V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}

Where:

  • VoutV_{out} — Output voltage at the midpoint between the two resistors (V)
  • VinV_{in} — Input supply voltage (V)
  • R1R_1 — Top resistor (connected to VinV_{in}), in ohms (Ω)
  • R2R_2 — Bottom resistor (connected to GND), in ohms (Ω)

Current through the divider:

I=VinR1+R2I = \frac{V_{in}}{R_1 + R_2}

Solving for R1 given a target Vout:

R1=R2×VinVoutVoutR_1 = R_2 \times \frac{V_{in} - V_{out}}{V_{out}}


Step-by-Step Example: 5V Sensor to 3.3V ADC

A common real-world problem: you have an ultrasonic sensor (HC-SR04) that outputs 5V signals, but you want to read it on an ESP32 ADC pin that is limited to 3.3V.

Given: Vin=5VV_{in} = 5\,\text{V}, Vout=3.3VV_{out} = 3.3\,\text{V}, choose R2=20kΩR_2 = 20\,\text{k}\Omega.

Find R1:

R1=20,000×53.33.3=20,000×0.515=10,303ΩR_1 = 20{,}000 \times \frac{5 - 3.3}{3.3} = 20{,}000 \times 0.515 = 10{,}303\,\Omega

Use the nearest standard value: 10 kΩ (E12 series).

Verify:

Vout=5×20,00010,000+20,000=5×0.667=3.33VV_{out} = 5 \times \frac{20{,}000}{10{,}000 + 20{,}000} = 5 \times 0.667 = 3.33\,\text{V}

Current through the divider: I=5V/30kΩ=167μAI = 5\,\text{V} / 30\,\text{k}\Omega = 167\,\mu\text{A} — negligible.


Common Voltage Divider Pairs Quick Reference

VinV_{in}VoutV_{out}R1R_1R2R_2
5 V3.3 V10 kΩ20 kΩ
5 V2.5 V10 kΩ10 kΩ
12 V5 V14 kΩ (15 kΩ std)10 kΩ
12 V3.3 V26 kΩ (27 kΩ std)10 kΩ
9 V3.3 V17 kΩ (18 kΩ std)10 kΩ
3.3 V1.65 V10 kΩ10 kΩ

Always verify by plugging your chosen standard values back into the formula, since the nearest E12/E24 resistor may shift VoutV_{out} by a few percent.


Loading Effect

When you connect a load resistance RLR_L in parallel with R2R_2, the effective bottom resistance becomes:

Req=R2×RLR2+RLR_{eq} = \frac{R_2 \times R_L}{R_2 + R_L}

And the actual VoutV_{out} drops to:

Vout,loaded=Vin×ReqR1+ReqV_{out,\text{loaded}} = V_{in} \times \frac{R_{eq}}{R_1 + R_{eq}}

Rule of thumb: To minimize loading error to less than 10%, choose RL10×R2R_L \geq 10 \times R_2. For a microcontroller ADC pin (typically 100 kΩ–1 MΩ input impedance), a divider using 10 kΩ–20 kΩ resistors experiences negligible loading.


Power Dissipation

The power dissipated by the divider resistors as heat:

Ptotal=Vin2R1+R2P_{total} = \frac{V_{in}^2}{R_1 + R_2}

For a 5V–to–3.3V divider with 10 kΩ + 20 kΩ:

P=2530,000833μWP = \frac{25}{30{,}000} \approx 833\,\mu\text{W}

This is tiny — standard 1/8W resistors are more than adequate. However, for 12V systems with low-value resistors (e.g., 1 kΩ + 2 kΩ), power reaches 48 mW — still fine for 1/4W resistors, but worth checking.


Frequently Asked Questions

How does load resistance affect the voltage divider output?

When a load (RLR_L) is connected in parallel with R2R_2, it reduces the effective bottom resistance and pulls VoutV_{out} lower than the unloaded value. To keep the output stable, the divider's source resistance (R1R2R_1 \parallel R_2) must be much smaller than the load — ideally 10× or more. If the load impedance varies widely, consider a buffered divider using an op-amp voltage follower.

Can I use a voltage divider to power a motor or LED strip?

No. A resistive voltage divider is only suitable as a signal-level shifter or reference bias, not as a power supply. The output voltage collapses under load, and the power wasted in the resistors scales with current. Use a switching regulator (buck converter) or LDO regulator to step down voltage for power delivery.

What is the difference between a voltage divider and a Zener regulator?

A voltage divider produces an output that varies with load and input voltage. A Zener diode placed across R2R_2 clamps VoutV_{out} to a fixed reference voltage as long as the current through the circuit exceeds the Zener's minimum holding current. Zener regulators are still inefficient under heavy loads; for precision references, use a dedicated voltage reference IC (e.g., LM4040, TL431).

Why do I get a different voltage than the formula predicts?

The three most common causes: (1) the load draws enough current to change the divider ratio (loading effect), (2) you are measuring with a cheap multimeter whose internal resistance (~10 MΩ) participates in the divider at very high source impedances, or (3) the actual resistor values differ from nominal due to tolerance — check values with a multimeter.

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