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.
By AdminDrafted with AI writing toolsUpdated Checked examples: 7
R1 = 10 kΩ over R2 = 4.7 kΩ passes 31.97% of the input: 3.837 V from 12 V, drawing 816.3 µA.
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.
An unloaded two-resistor divider gives Vout = Vin × R2/(R1 + R2), with R2 the resistor to ground: 10 kΩ over 4.7 kΩ on 12 V gives 3.84 V. An LED's series resistor is R = (Vs − Vf)/I: a 2 V red LED at 20 mA on 5 V needs (5 − 2)/0.02 = 150 Ω.
R1 runs from the input to the output node, and R2 from the output node to ground. For 12 V with R1 = 10 kΩ and R2 = 4.7 kΩ:
Only the ratio sets the output. Scaled tenfold to 100 kΩ over 47 kΩ, the divider still gives 3.837 V but draws 0.0816 mA. Larger values waste less current; smaller values hold their voltage better under load.
Fix one resistor and solve for the other:
For 3.3 V from 5 V with R1 = 10 kΩ, R2 = 10 × 3.3/1.7 = 19.41 kΩ. That is not a standard value: the nearest E12 parts, 18 kΩ and 22 kΩ, give 3.21 V and 3.44 V, so check the output with the part you will fit.
The target must be below the input voltage. A resistive divider can only scale a voltage down.
Anything connected to the output sits in parallel with R2, so the bottom of the divider becomes R2 × RL/(R2 + RL). Use that value in place of R2 in the divider formula.
Two 10 kΩ resistors on 12 V give 6 V unloaded. A 10 kΩ load makes the bottom 5 kΩ, so the output falls to 4 V and the source current rises to 0.8 mA, split 0.4 mA each through R2 and the load.
A load at least 10 times R2 keeps the drop under about 10%: a 100 kΩ load on the same divider gives 5.71 V. A 100 Ω load on the 10 kΩ/4.7 kΩ divider pulls 3.84 V down to 0.116 V.
That is why dividers feed sensing and reference inputs, while a regulator powers a circuit. OpenStax covers the series and parallel rules behind both calculations.
A blue LED (Vf ≈ 3.2 V) on 9 V at 15 mA leaves VR = 5.8 V. R = 5.8 ÷ 0.015 = 386.67 Ω, rounded up to 390 Ω. The current becomes 14.87 mA and the resistor dissipates 86 mW, so a ¼ W part keeps the 2× margin.
Three red LEDs in series on 12 V at 20 mA leave 12 − 3 × 2.0 = 6 V for the resistor. That gives 300 Ω, rounded up to 330 Ω, for 18.18 mA.
The rows are 10 kΩ over 4.7 kΩ on 12 V (Default), two 10 kΩ resistors on 12 V with a 10 kΩ load (Equal loaded), and one red LED on 5 V at 20 mA (LED default).
| Scenario | Selected result | Source or LED current |
|---|---|---|
| Default | 3.83673 V | 0.816327 mA |
| Equal loaded | 4 V | 0.8 mA |
| LED default | 150 Ω | 20 mA |
Per LED at about 20 mA, the color presets assume red 2.0 V, yellow 2.1 V, green (InGaN) 3.0 V, blue 3.2 V, white 3.2 V and infrared 1.2 V.
Real parts vary by a few tenths of a volt with type, current and temperature, so use the datasheet figure when you have it.
The spread matters most when little voltage is left for the resistor. On 3.3 V, a red LED at 5 mA gets 270 Ω; if its real Vf is 2.2 V, the current falls 15%, from 4.81 to 4.07 mA. On 12 V with 560 Ω, the same shift moves the current 2%.
The resistor turns VR × I into heat. Standard through-hole ratings run ⅛, ¼, ½, 1, 2, 5 and 10 W; pick one at least twice the calculated P = I²R.
Above 10 W, a resistor is the wrong part: a red LED at 1 A from 100 V would burn 96 W in a 100 Ω resistor, which is a job for a constant-current LED driver.
Mounting, airflow and the manufacturer's derating curve also decide how hot a resistor runs.
Combine resistors into one equivalent with the series and parallel calculator, and check current or power with the Ohm's law calculator. The resistor color code calculator gives the color bands for the value you pick.
Checked against: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA
Checked against: R2 = R1·Vout/(Vin − Vout) = 33000/1.7 Ω = 19.411765 kΩ
Checked against: R2‖RL = 5 kΩ, 12 × 5/15 = 4 V (6 V unloaded)
Checked against: (5 − 2.0)/0.02 = 150 Ω (in E12); P = 0.02² × 150 = 0.06 W; 2 × 0.06 ≤ 0.125
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.
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.
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.
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.
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.
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