Choose an LED resistor: current, power and tolerance

Choose the current you need first

Suppose an indicator runs from 9 V at about 15 mA. This example assumes a 2.1 V LED drop. That value must come from the real component’s data sheet and operating conditions; color alone does not determine an exact value.

The resistor drops the voltage left between the supply and LED. After finding the ideal value, three decisions remain: choose an available value, recalculate current and check dissipation. A number in ohms does not answer all three.

From a calculated 460 Ω to an available 470 Ω

470 Ω is an E12 value and slightly reduces current from the target. Using the suggested ×2 margin, look for a power rating of at least 0.2026 W. A 0.25 W resistor exceeds that nominal calculation; its operating temperature and heat derating still depend on the manufacturer.

Check the worst-case corner too

For this exercise only, assume the supply can rise to 9.6 V, the LED drop can fall to 1.9 V and resistor tolerance is ±5%. Maximum current combines high supply, low LED drop and low resistance.

The nominal choice no longer meets a strict 15 mA limit. With these hypothetical bounds, nominal R must be at least 7.7 / (0.015 × 0.95) ≈ 540.35 Ω; the next E12 is 560 Ω. Recheck current and power with that value.

Vf is held fixed within each calculation. Its current–temperature curve is not modeled, nor does the model turn a microcontroller output into an ideal 9 V supply. These examples do not describe a particular commercial LED or its maximum permitted current.

Check a decision, not just a division

1. Enter 9 V, 2.1 V and 15 mA in the tool. Compare ideal and commercial values.
2. To analyze a resistor you already have, use I = (Vs − Vf)/R with fixed R. The LED calculator proposes a different resistor when inputs change.
3. Explain what tolerance changes and which data you need before building.