Case 1
- Supply voltage
- 5V
- LED forward voltage
- 2V
- Target LED current
- 0.02A
150.00Ω
Open with these valuesLED Resistor Calculator
150.00Ω
Result: 150.00 ΩThe resistor swallows whatever voltage the LED leaves over: R = (Vs − Vf) ÷ I. A red LED at 2 V on a 5 V rail at 20 mA needs 150 Ω. Enter the current in amperes — 20 mA is 0.02 — and round the answer up to the next value you actually own.
Worked examples
Case 2
- Supply voltage
- 12V
- LED forward voltage
- 2V
- Target LED current
- 0.02A
500.00Ω
Open with these valuesCase 3
- Supply voltage
- 5V
- LED forward voltage
- 3.2V
- Target LED current
- 0.02A
90.00Ω
Open with these valuesHow it's calculated
R = (Vs − Vf) ÷ I
- StepEnter the supply voltage feeding the LED and resistor in series.
- StepEnter the LED forward voltage from its datasheet — about 2 V red, 3.2 V blue.
- StepEnter the current in amperes: 20 mA is 0.02, 5 mA is 0.005.
- ResultRound the answer up to the next standard value you have.
What this number means
The answer is the ideal value, not a part
690 Ω is arithmetic, not a stock value: E24 jumps from 680 straight to 750. Round up, never down — 6.9 V across 680 Ω passes 10.1 mA instead of the 10 mA you asked for, while 750 Ω gives 9.2 mA.
The resistor has to shed (Vs − Vf) × I watts
The 150 Ω example sheds (5 − 2) × 0.02 = 0.06 W, and the 690 Ω case (9 − 2.1) × 0.01 = 0.069 W. An ordinary quarter-watt part covers both; leave a margin above the calculated figure.
The supply must sit above the forward voltage
The resistor can only drop what the LED leaves over. At or below the forward voltage there is nothing left, the formula returns zero or a negative value, and the LED stays dark.
Commonly misread
690 Ω does not exist, so I fit the 680 Ω next to it.
680 Ω is the next value down and lets more current through: 6.9 V across it gives 10.1 mA. Go up to 750 Ω instead, which gives 9.2 mA.
The resistor dissipates Vs × I, so 5 V × 0.02 A = 0.1 W.
It only sees what the LED leaves over, so the figure is (5 − 2) × 0.02 = 0.06 W.
I want 20 mA, so I type 20 into the current field.
The field is in amperes: 20 mA is 0.02. Entering 20 yields 0.15 Ω and a dead LED.
Reference table
| Supply, forward, current | What it is | Resistor |
|---|---|---|
| 3.3, 1.8, 0.005 | Red LED on a 3.3 V rail at 5 mA | 300 |
| 5, 2, 0.001 | The same red LED at 1 mA — dim but visible | 3000 |
| 5, 2, 0.02 | Red LED on 5 V USB at 20 mA | 150 |
| 5, 3.2, 0.02 | Blue or white LED on 5 V at 20 mA | 90 |
| 9, 2.1, 0.01 | Red LED on a 9 V block at 10 mA | 690 |
| 12, 2, 0.02 | Red LED on a 12 V rail at 20 mA | 500 |
Questions
How do I calculate an LED resistor?
Subtract the LED forward voltage from the supply voltage, then divide by the current you want: R = (Vs − Vf) ÷ I. A red LED at 2 V on a 5 V supply at 20 mA needs (5 − 2) ÷ 0.02 = 150 Ω.
What is a current-limiting resistor?
It is a resistor in series with the LED that holds the current at a safe level. An LED on its own draws far too much current and burns out almost immediately, so the resistor takes the leftover voltage.
What current should I use?
Most indicator LEDs are rated around 20 mA, which is 0.02 A here, and many look fine at 5 to 15 mA. Check the datasheet for the maximum continuous forward current and stay below it.
Why must the supply exceed the forward voltage?
The resistor can only drop what the LED leaves over. If the supply equals or is below the forward voltage there is nothing left, the formula gives zero or a negative value, and the LED stays dark.
What power rating does the resistor need?
It dissipates (Vs − Vf) × I watts. The 150 Ω example above sheds (5 − 2) × 0.02 = 0.06 W, so an ordinary quarter-watt part is plenty; always leave a margin above the calculated figure.
Does this round to a standard resistor value?
No — the answer is the exact arithmetic value, so 690 Ω rather than the nearest E24 value of 680 Ω. Round up yourself: a higher resistance means slightly less current and a slightly dimmer but safer LED.
Sources and last check
Information, not professional advice.