# 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.

Versão interativa: https://www.calcopenly.com/pt/science/voltage-divider-led-resistor-calculator
Tema: Calculadoras científicas

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.

## Dados

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

## Resultados

- Output voltage (V) — resultado principal
- 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)

## Fórmula

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

## Exemplos resolvidos

### 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**
- Fonte de verificação: 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**
- Fonte de verificação: 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**
- Fonte de verificação: 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**
- Fonte de verificação: 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**
- Fonte de verificação: 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**
- Fonte de verificação: 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

## Perguntas

### 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.

### Qual é a precisão de “Voltage divider and LED resistor calculator”?

A precisão depende dos dados inseridos e das hipóteses do método. O cálculo decimal usa 50 algarismos significativos, mas estimativas, métodos numéricos e dados de origem podem ter menor precisão; o arredondamento exibido não elimina essas limitações. Exemplos resolvidos verificados com fontes independentes: 7. Por exemplo, “12 V with 10 kΩ over 4.7 kΩ” é verificado com Python 3.8 fractions: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA.

### De onde vem o método?

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.

## Fontes

- 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)