LED resistor calculator
Resistor value for an LED or a series string, rounded up to a standard value, with the real current and the power rating the resistor needs.
- Next E24 value up
- 300 Ω Gives 9.67 mA.
- LED current with that resistor
- 9.67 mA
- Resistor dissipation
- 28 mW
- Minimum rating at 50% margin
- 56.1 mW
- Next common rating
- 1/16 W Below the part’s derating temperature. Check its curve above that.
- Power in the LEDs
- 20.3 mW
- Share of power reaching the LEDs
- 42 %
How this is calculated
R = (Vs − n · Vf) / I R_std = next E-series value at or above R I_actual = (Vs − n · Vf) / R_std P = I_actual² · R_std Minimum rating = P / (rated power ratio × design margin)
- Forward voltage
- Take Vf from the LED datasheet at your operating current. It varies between parts; more voltage across the resistor makes the current less sensitive to that spread.
- Rounding
- The resistor is rounded up, so the current never exceeds what you asked for.
- Power rating
- Required rating is dissipation divided by the design margin, assuming the resistor runs below its derating knee (100% rated power ratio). Above that temperature, multiply the margin by the ratio from the part’s derating curve. 50% is a typical margin.
- OpenStax, University Physics Volume 2 (Ling, Sanny, Moebs), sections 9.4 Ohm's Law and 9.5 Electrical Energy and Power2016
- KOA Corporation, Technical Note: Procedure for power derating based on terminal part temperature, KDST_TN-007v0102
- IEC 60063, Preferred number series for resistors and capacitorsEdition 3.0, 2015-03-27
Questions
With a 2.1 V LED at 10 mA, (5 − 2.1) / 0.01 = 290 Ω. The next E24 value up is 300 Ω, which gives 9.67 mA and dissipates 28 mW.
Keep the total forward voltage well below the supply. Three 3.0 V LEDs on 12 V leave 3 V for the resistor: 150 Ω at 20 mA. With little headroom, small changes in Vf swing the current widely.
Dissipation divided by your design margin. 0.06 W at a 50% margin needs at least 0.12 W, so a 1/8 W part. Above the resistor’s derating temperature, allow more.
Not reliably. Forward voltages differ between parts, so the LED with the lowest Vf takes most of the current. Give each LED or string its own resistor.