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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.

更新于 已验证的示例：7

V
更多选项
试一试
Output voltage
V
Output voltage: 3.83673 V
有效数字：6；取最近值，等距时远离零
Division ratio Vout/Vin
0.319728
Current through the divider
0.816327mA

R1 = 10 kΩ over R2 = 4.7 kΩ passes 31.97% of the input: 3.837 V from 12 V, drawing 816.3 µA.

Voltage divider

Vin = 12 VR1 = 10 kΩR2 = 4.7 kΩVout = 3.837 V
计算方法 S

  1. Divider output

    Vout=VinR2R1+R2=12×4,70010,000+4,700=3.83673 VV_{\text{out}} = V_{\text{in}}\frac{R_2}{R_1 + R_2} = 12\times\frac{4{,}700}{10{,}000 + 4{,}700} = 3.83673\ \mathrm{V}

  2. Current through the chain

    I=VinR1+R2=0.000816327 AI = \frac{V_{\text{in}}}{R_1 + R_2} = 0.000816327\ \mathrm{A}

关于Voltage divider and LED resistor calculator

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.

计算示例

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

核验来源：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

核验来源：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

核验来源：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

核验来源：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

常见问题

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.

“Voltage divider and LED resistor calculator”有多准确？

准确性取决于输入值和方法的假设。十进制运算使用50位有效数字，但估算、数值方法和源数据的精度可能较低；显示时的舍入并不能消除这些限制。 已按独立来源核验的计算示例：7。 例如，“12 V with 10 kΩ over 4.7 kΩ”根据Python 3.8 fractions: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA进行核验。

这种方法出自哪里？

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.

关于此计算器

Vout=VinR2R1+R2;R=Vs−nVfIV_{\text{out}} = V_{\text{in}}\frac{R_2}{R_1 + R_2};\qquad R = \frac{V_s - nV_f}{I}

来源

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

已对照来源验证

此计算器包含 7 个已解示例，答案来自独立来源。这些示例会在测试套件中运行，你也可以在此运行验证。

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