LED Resistor Calculator
Find the value of the current-limiting resistor in series with an LED. You can also see the current that actually flows with a standard (E24) resistor, the power the resistor dissipates and a suggested power rating.
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Results
- Calculated resistance
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- E24 resistor (at or above)
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- Current with that resistor
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- Power in the resistor
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- Suggested rating (≥2×)
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- Power per LED
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The supply voltage is not above the total forward voltage, so the LEDs will not light. Raise the supply voltage or use fewer LEDs in series.
Forward voltage depends on the LED’s color, type, current and temperature. Use the value from the datasheet.
See it as a graph
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
The current-limiting resistor formula
An LED's current rises steeply once the voltage passes its forward voltage Vf, so connecting it straight to a supply leaves the current undefined. A series resistor fixes this: the voltage across it, Vs − n·Vf, sets the current. The required resistance is R = (Vs − n·Vf)/If, where n is the number of LEDs in series. With the defaults of Vs = 5 V, Vf = 2 V, If = 10 mA and one LED, R = (5 − 2) V ÷ 10 mA = 300 Ω. The resistor dissipates 3 V × 10 mA = 30 mW and the LED 2 V × 10 mA = 20 mW.
Typical forward voltages
Vf depends on the LED's color (the semiconductor material that emits the light), its type, the current and the temperature. As a rough guide, red, orange and yellow LEDs are often around 1.8–2.2 V, and blue, white and InGaN green LEDs around 2.8–3.5 V. These are only guides, so check the datasheet for Vf at the current you plan to use. With little headroom, variation in Vf has a large effect on the current. Two 2 V LEDs in series on 5 V with 100 Ω draw 10 mA, for example, but if each Vf is 2.2 V the current falls to 6 mA.
Rounding up to E24, and power rating
The calculated value is rarely one you can buy, so the tool picks the smallest E24 value at or above it. Rounding up keeps the current from exceeding the target. The default 300 Ω happens to be in E24, but with Vf = 1.8 V the result is 3.2 V ÷ 10 mA = 320 Ω, and choosing 330 Ω gives about 9.7 mA. As a guide, choose a power rating at least twice the dissipation; for the default 30 mW, that suggests 1/8 W (125 mW). Resistors can handle less power at high ambient temperature (derating), so leave more margin in hot locations.
LEDs in series or in parallel
To light several LEDs from one resistor, put them in series. The same current then flows through every LED, and less power is lost in the resistor. As the total Vf approaches the supply voltage, however, the current becomes less predictable, and once it exceeds the supply the LEDs do not light. On 5 V with Vf = 2 V, two LEDs (100 Ω) is the limit; three need 6 V. For parallel LEDs, give each LED (or each series string) its own resistor. With one shared resistor, the LED with the lowest Vf takes more of the current, and as it heats up its Vf drops further, which can make the imbalance worse.
Practice analog circuit questions
FAQ
What resistor do I need for an LED on 5 V?
Use R = (5 V − Vf)/If. With Vf = 2 V at 10 mA it is 300 Ω, and with Vf = 3 V (a typical guide for blue or white LEDs) at 10 mA it is 200 Ω; both are E24 values. Actual Vf differs between LEDs, so calculate with the datasheet value and choose the E24 resistor at or above the result.
Can I run an LED without a resistor?
Avoid connecting it directly to a voltage source. A small change in voltage changes an LED's current a lot, so the current can exceed its rating, damaging the LED or shortening its life. If you don't want a resistor, use a constant-current driver or choose an LED with a built-in resistor.
Should I wire multiple LEDs in series or in parallel?
Series is the usual choice when the supply has enough headroom: every LED gets the same current and one resistor is enough. When the total Vf is close to or above the supply voltage, use parallel strings, but give each string its own resistor. Sharing one resistor lets variation in Vf push most of the current through some of the LEDs.