# Voltage divider and LED resistor calculator

> Voltage divider output, or R1 or R2 for a target voltage; the current-limiting resistor for LEDs with its nearest E12 value and power rating.

Interactive version: https://www.calcopenly.com/science/voltage-divider-led-resistor-calculator
Subject: Science calculators

A voltage divider of two resistors gives Vout = Vin × R2/(R1 + R2). The calculator solves for the output, or for R1 or R2 when you know the output you want, and shows how a load across R2 pulls the output down. An LED needs a series resistor R = (Vs − n·Vf)/I, where Vf is the forward voltage of each of n LEDs in series and I is the current you want.

The divider default, 10 kΩ over 4.7 kΩ on 12 V, gives 3.84 V and draws 0.82 mA. A red LED (Vf ≈ 2.0 V) at 20 mA on 5 V needs exactly 150 Ω, an E12 value, and the resistor dissipates 60 mW, so a ⅛ W part covers it with a 2× margin.

Exact values are rounded up to the next E12 value (IEC 60063), which keeps the LED current at or below target. Forward voltages are typical figures; datasheets vary by a few tenths of a volt, and the current is most sensitive when little voltage is left across the resistor.

## Inputs

- **Circuit** (options: Voltage divider, LED resistor)
- **Solve for** (options: Output voltage, R2, R1)
- **Input voltage**
- **Wanted output voltage**
- **R1 (top)**
- **R1 unit** (options: mΩ, Ω, kΩ, MΩ)
- **R2 (bottom)**
- **R2 unit** (options: mΩ, Ω, kΩ, MΩ)
- **Load resistance across R2**
- **Load unit** (options: mΩ, Ω, kΩ, MΩ)
- **Supply voltage**
- **LED colour** (options: Red, Yellow, Green (InGaN), Blue, White, Infrared, Enter forward voltage)
- **Forward voltage per LED**
- **LED current**
- **Current unit** (options: µA, mA, A, kA)
- **LEDs in series**

## Results

- Output voltage (V) — main result
- R1 (kΩ)
- R2 (kΩ)
- Division ratio Vout/Vin
- Current through the divider (mA)
- Output without the load (V)
- Exact resistor (Ω)
- Nearest E12 at or above (Ω)
- LED current with the E12 resistor (mA)
- Power in the resistor (W)
- Resistor rating to buy
- Power in the LEDs (W)

## Formula

$$
V_{\text{out}} = V_{\text{in}}\frac{R_2}{R_1 + R_2};\qquad R = \frac{V_s - nV_f}{I}
$$

## Worked examples

### 12 V with 10 kΩ over 4.7 kΩ

- Circuit: Voltage divider
- Solve for: Output voltage
- Input voltage: 12 V
- R1 (top): 10
- R1 unit: kΩ
- R2 (bottom): 4.7
- R2 unit: kΩ
- **Output voltage: 3.83674 V**
- **Current through the divider: 0.816327 mA**
- Checked against: Python 3.8 fractions: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA

### R2 for 3.3 V from 5 V with R1 = 10 kΩ

- Circuit: Voltage divider
- Solve for: R2
- Input voltage: 5 V
- Wanted output voltage: 3.3 V
- R1 (top): 10
- R1 unit: kΩ
- R2 unit: kΩ
- **R2: 19.4118 kΩ**
- Checked against: Python 3.8 fractions: R2 = R1·Vout/(Vin − Vout) = 33000/1.7 Ω = 19.411765 kΩ

### Equal divider loaded by 10 kΩ

- Circuit: Voltage divider
- Solve for: Output voltage
- Input voltage: 12 V
- R1 (top): 10
- R1 unit: kΩ
- R2 (bottom): 10
- R2 unit: kΩ
- Load resistance across R2: 10
- Load unit: kΩ
- **Output voltage: 4 V**
- **Output without the load: 6 V**
- Checked against: Python 3.8 fractions: R2‖RL = 5 kΩ, 12 × 5/15 = 4 V (6 V unloaded)

### Red LED on 5 V at 20 mA

- Circuit: LED resistor
- Supply voltage: 5 V
- LED colour: Red
- LED current: 20
- Current unit: mA
- LEDs in series: 1
- **Exact resistor: 150 Ω**
- **Nearest E12 at or above: 150 Ω**
- **Power in the resistor: 0.06 W**
- **Resistor rating to buy: ⅛ W**
- Checked against: Python 3.8 fractions: (5 − 2.0)/0.02 = 150 Ω (in E12); P = 0.02² × 150 = 0.06 W; 2 × 0.06 ≤ 0.125

### Three white LEDs on 12 V

- Circuit: LED resistor
- Supply voltage: 12 V
- LED colour: White
- LED current: 20
- Current unit: mA
- LEDs in series: 3
- **Exact resistor: 120 Ω**
- **Nearest E12 at or above: 120 Ω**
- **Power in the LEDs: 0.192 W**
- Checked against: Python 3.8 fractions: (12 − 3 × 3.2)/0.02 = 120 Ω; LED power 9.6 × 0.02

### Blue LED on 9 V at 15 mA rounds up to 390 Ω

- Circuit: LED resistor
- Supply voltage: 9 V
- LED colour: Blue
- LED current: 15
- Current unit: mA
- LEDs in series: 1
- **Exact resistor: 386.667 Ω**
- **Nearest E12 at or above: 390 Ω**
- **LED current with the E12 resistor: 14.8718 mA**
- **Power in the resistor: 0.086256 W**
- **Resistor rating to buy: ¼ W**
- Checked against: Python 3.8 fractions: 5.8/0.015 = 386.67 Ω → E12 390; I = 5.8/390; P = 5.8²/390 = 0.0862564 W; 2P > ⅛ W

## Questions

### What resistor do I need for an LED on 5 V?

For one red LED at 20 mA, 150 Ω: (5 − 2.0 V)/0.020 A. A blue or white LED with a 3.2 V forward drop needs (5 − 3.2)/0.020 = 90 Ω, rounded up to 100 Ω in the E12 series, which gives 18 mA. Many indicator LEDs are bright enough at 5–10 mA, which doubles to quadruples the resistance.

### What resistor do I need for an LED on 12 V?

For a single red LED at 20 mA, (12 − 2.0)/0.020 = 500 Ω, rounded up to 560 Ω (E12). That gives 17.9 mA and 0.18 W in the resistor, so use a ½ W part to keep a 2× margin. Three white LEDs in series instead (3 × 3.2 V = 9.6 V) need only 120 Ω, and the resistor wastes 48 mW.

### How do you calculate a voltage divider?

Vout = Vin × R2/(R1 + R2), where R2 is the resistor between the output and ground. 10 kΩ over 4.7 kΩ on 12 V gives 12 × 4.7/14.7 = 3.84 V. For a target output, fix one resistor and solve for the other: R2 = R1 × Vout/(Vin − Vout), so 3.3 V from 5 V with R1 = 10 kΩ needs R2 = 19.4 kΩ.

### Why does a voltage divider's output drop under load?

The load sits in parallel with R2 and lowers the bottom resistance. Two 10 kΩ resistors on 12 V give 6 V unloaded, but a 10 kΩ load makes the bottom 5 kΩ and the output falls to 4 V. Keep the load at least 10 times R2 to stay within about 10%, or buffer the output; dividers suit reference and sensing inputs, not powering circuits.

### What wattage resistor do I need for an LED?

Work out P = I²R and choose a standard rating at least twice that. A 150 Ω resistor passing 20 mA dissipates 0.02² × 150 = 0.06 W, so a ⅛ W (0.125 W) resistor meets the 2× margin. With one red LED on 12 V the resistor dissipates 0.18 W and needs ½ W. Common through-hole ratings are ⅛, ¼, ½, 1 and 2 W.

### How accurate is the voltage divider and LED resistor calculator?

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 7 worked examples whose answers come from independent sources; for example, “12 V with 10 kΩ over 4.7 kΩ” is checked against Python 3.8 fractions: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA.

### Where does the method come from?

Horowitz & Hill, The Art of Electronics (3rd ed.), §1.2.3 Voltage dividers and §2.1 (LED current limiting); IEC 60063:2015 — Preferred number series (E12); OpenStax University Physics Volume 2, §10.2 Resistors in series and parallel.

## Sources

- Horowitz & Hill, The Art of Electronics (3rd ed.), §1.2.3 Voltage dividers and §2.1 (LED current limiting)
- IEC 60063:2015 — Preferred number series (E12)
- [OpenStax University Physics Volume 2, §10.2 Resistors in series and parallel](https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel)