Size the current-limiting resistor for one LED or a string of LEDs: this LED series resistor calculator rounds up to a standard value and shows the current it really gives, the resistor’s power and a chip size. Pick a color for typical datasheet forward voltages or enter your own.
Each string gets its own resistor.
Optional: transistor saturation or output-pin drop in series.
E24 resistor (at or above the exact value)
160 Ω
Preset forward voltages are typical values for one 5 mm LED family. Your LED’s datasheet wins: color, chip and package all change Vf.
Nearby E24 values
| Resistor | LED current | vs target | Resistor power |
|---|---|---|---|
| 130 Ω | 23.5 mA | +17.3 % | 71.6 mW |
| 150 Ω | 20.3 mA | +1.7 % | 62 mW |
| 160 Ω | 19.1 mA | -4.7 % | 58.1 mW |
| 180 Ω | 16.9 mA | -15.3 % | 51.7 mW |
| 200 Ω | 15.3 mA | -23.8 % | 46.5 mW |
Use R = (Vs − Vf) ÷ I. For a red LED with a typical forward voltage of 1.95 V on a 5 V supply at 20 mA, R = (5 − 1.95) ÷ 0.02 = 152.5 Ω. Round up to the next standard value, 160 Ω, which gives 19.1 mA and dissipates 58.1 mW in the resistor.
An LED holds a roughly fixed voltage across itself once it conducts, so the resistor has to drop the rest of the supply. Its value sets the current.
Once you round R to a standard value, the current it actually gives and the power in each part follow from the same voltages:
The share of the supply power that reaches the LEDs is n × Vf ÷ Vs, so a long string on a supply just above its forward voltage is efficient, and a single LED on 24 V mostly heats the resistor. With several strings in parallel, each one has its own resistor and the supply current is the sum of the string currents.
R = (5 − 1.95) ÷ 0.02 = 152.5 Ω. The next E24 value is 160 Ω, which gives 19.1 mA. The resistor drops 3.05 V and dissipates 58.1 mW, so with a 2× margin the smallest chip is 0805 (2012 metric), 1/8 W. If the LED is at its 2.5 V datasheet maximum, the current is 15.6 mA.
The note drives a lamp at 42 mA from 5 V through a transistor with 0.1 V saturation, where Vf = 1.95 V. R = (5 − 0.1 − 1.95) ÷ 0.042 = 70.24 Ω. The note uses 70 Ω, which dissipates 123 mW at 42 mA, and specifies a 1/4 W part for the 2× margin. Rounding up to E24 instead gives 75 Ω, 39.3 mA and 116 mW.
Three white LEDs at 3.3 V typical leave 2.1 V for the resistor, giving 110 Ω. But the datasheet maximum is 4 V, and three such parts need the full 12 V, so the current can fall to zero. With two LEDs per string, R = 270 Ω gives 20 mA typical and still 14.8 mA with maximum-Vf parts.
One LED per resistor, E24 value at or above the exact result, using the typical forward voltages of the Kingbright WP7113 5 mm family. * marks values where a maximum-Vf part would get less than half the current.
| LED (typ. Vf) | 3.3 V | 5 V | 9 V | 12 V | 24 V |
|---|---|---|---|---|---|
| Red (AlGaInP, 630 nm), 1.95 V | 68 Ω | 160 Ω | 360 Ω | 510 Ω | 1.2 kΩ |
| Yellow (AlGaInP, 590 nm), 2.00 V | 68 Ω | 150 Ω | 360 Ω | 510 Ω | 1.1 kΩ |
| Yellow-green (GaP, 568 nm), 2.20 V | 56 Ω | 150 Ω | 360 Ω | 510 Ω | 1.1 kΩ |
| Green (InGaN, 525 nm), 3.20 V | 5.1 Ω * | 91 Ω | 300 Ω | 470 Ω | 1.1 kΩ |
| Blue (InGaN, 465 nm), 3.30 V | no headroom | 91 Ω | 300 Ω | 470 Ω | 1.1 kΩ |
| White (InGaN), 3.30 V | no headroom | 91 Ω | 300 Ω | 470 Ω | 1.1 kΩ |
| Infrared (940 nm), 1.20 V | 110 Ω | 200 Ω | 390 Ω | 560 Ω | 1.2 kΩ |
The higher the supply, the more power the resistor burns. Sizes are the smallest Yageo RC_L chip with at least twice the dissipation as its 70 °C rating.
| Supply | Resistor (E24) | LED current | Resistor power | Chip size | Efficiency |
|---|---|---|---|---|---|
| 3.3 V | 68 Ω | 19.9 mA | 26.8 mW | 0402, 1/16 W | 59 % |
| 5 V | 160 Ω | 19.1 mA | 58.1 mW | 0805, 1/8 W | 39 % |
| 9 V | 360 Ω | 19.6 mA | 138 mW | 1210, 1/2 W | 22 % |
| 12 V | 510 Ω | 19.7 mA | 198 mW | 1210, 1/2 W | 16 % |
| 24 V | 1.2 kΩ | 18.4 mA | 405 mW | 2512, 1 W | 8 % |
The presets in this LED calculator use the typical and maximum forward voltage at 20 mA and 25 °C, and the absolute maximum DC current, from each Kingbright datasheet.
| LED | Part | Vf typ. | Vf max. | Max DC current |
|---|---|---|---|---|
| Red (AlGaInP, 630 nm) | WP7113SURCK | 1.95 V | 2.5 V | 30 mA |
| Yellow (AlGaInP, 590 nm) | WP7113SYD | 2 V | 2.5 V | 30 mA |
| Yellow-green (GaP, 568 nm) | WP7113SGD | 2.2 V | 2.5 V | 25 mA |
| Green (InGaN, 525 nm) | WP7113ZGC/G | 3.2 V | 4 V | 30 mA |
| Blue (InGaN, 465 nm) | WP7113QBC/D | 3.3 V | 4 V | 30 mA |
| White (InGaN) | WP7113QWC/D | 3.3 V | 4 V | 30 mA |
| Infrared (940 nm) | WP7113F3C | 1.2 V | 1.6 V | 50 mA |
The formula treats Vf as a constant, but it is not: it rises with current, falls as the LED warms (−1.9 mV/°C for the Kingbright red part, −3.0 mV/°C for the white and blue ones), and varies from part to part, which is why datasheets give a typical and a maximum value. When the resistor drops only a volt or two, those changes move the current a lot; when it drops several volts, they hardly matter. A microcontroller pin is not an ideal source either: its output voltage sags as it sources current (enter that loss as the driver drop), and the pin has its own current limit. Do not put LEDs in parallel on one resistor. For high-power LEDs or long strings near the supply voltage, a constant-current driver holds the current regardless of Vf and wastes far less power than a resistor.
R = (Vs − Vf) ÷ I: subtract the LED's forward voltage from the supply and divide by the current you want. A red LED (1.95 V typical) on 5 V at 20 mA needs (5 − 1.95) ÷ 0.02 = 152.5 Ω; round up to the next standard value, 160 Ω.
The series resistor that sets the LED current. An LED holds a roughly fixed voltage across itself once it conducts, so the resistor drops the rest of the supply. The current-limiting resistor formula follows from that: R = (Vs − Vf) ÷ I.
At 20 mA, 160 Ω for a typical red LED and 91 Ω for a typical white or blue one (E24 values at or above the exact result). Check the forward voltage in your LED's datasheet, because it sets the answer.
68 Ω for a red LED at 20 mA. A white or blue LED with a typical forward voltage of 3.3 V leaves nothing for a resistor on 3.3 V, so run it from 5 V, at a lower current where its forward voltage is lower (read the datasheet curve), or with a constant-current driver.
510 Ω for one red LED at 20 mA. It dissipates 198 mW, so with a 2× margin use a 1210 (3225 metric), 1/2 W resistor. Putting three or four red LEDs in series on 12 V wastes much less power in the resistor.
LEDs in series can, because the same current flows through all of them. LEDs in parallel should not: their forward voltages differ from part to part, so the one with the lowest Vf takes most of the current. Give each parallel string its own resistor.
Subtract every forward voltage: R = (Vs − Vdrop − n × Vf) ÷ I, with n in the “LEDs in series per string” field. Two white LEDs (3.3 V typical) on 12 V at 20 mA need 270 Ω, which gives 20 mA.
A larger resistor gives a slightly lower current, which keeps the LED at or under the current you chose and under its maximum rating. Light output falls roughly in proportion to current, so 19 mA instead of 20 mA is about 5 % dimmer.
P = I² × R. Broadcom's LED application note AN 1005 rates the resistor at twice its dissipation. A red LED on 5 V with 160 Ω dissipates 58.1 mW in the resistor, so the smallest chip with that margin is 0805 (2012 metric), 1/8 W.
The forward voltage varies between parts and with temperature. The Kingbright WP7113SURCK red LED is 1.95 V typical but up to 2.5 V, which on 5 V with 160 Ω lowers the current from 19.1 mA to 15.6 mA. The more voltage the resistor drops, the less these variations matter.
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