What the LED Series Resistor Calculator does
This calculator gives the series resistor an LED needs: the exact value from the supply voltage, the forward voltage and the current you want, then the nearest preferred value you can actually buy, the current that value really produces, and the power each part has to get rid of.
An LED is not a resistor. Its current rises very steeply with voltage, so it must be fed from something that limits current. For an indicator, a resistor is that something; the only question is which one, and what the LED's current really becomes once you round to a value that exists.
How to use it
- Enter the supply voltage. Use the real one - a 5 V USB rail measured at 4.8 V changes the answer.
- Enter the LED's forward voltage at the current you intend, taken from the data sheet rather than from memory. It varies by colour and by part.
- Enter the current you want. Modern indicator LEDs are bright at 2-5 mA; 20 mA is a legacy default that mostly wastes power.
- Set how many LEDs are in series. They share one resistor and one current; their forward voltages add.
- Read the exact value, then the preferred value above it, and check the actual current and the resistor's dissipation before choosing a part.
Reading the results
The exact value almost never exists as a part. Rounding up gives slightly less current than asked for, which is the safe direction; rounding down gives more. This page defaults to the next preferred value at or above the exact one.
Resistor dissipation decides the physical part. A quarter-watt resistor should not be asked to dissipate more than about half its rating in a closed box, which is why the calculator suggests a rating of twice the calculated power.
Efficiency is the share of supply power that reaches the LEDs rather than the resistor. Low headroom looks efficient but makes the current unstable; high headroom is stable and wasteful. That trade-off is the reason constant-current drivers exist.
Worked example: three amber LEDs on a 12 V supply
Three LEDs at 2.1 V each drop 6.3 V, leaving 5.7 V for the resistor. At 20 mA the exact value is 5.7 / 0.02 = 285 ohms, and the resistor dissipates 0.02^2 x 285 = 0.114 W.
The nearest E24 value at or above 285 is 300 ohms, which gives 5.7 / 300 = 19.0 mA - about 5% below the target, which no one will see. A 330 ohm part, the nearest E12 value, gives 17.3 mA instead.
Dissipation at 300 ohms is 0.108 W, so a 1/4 W resistor is the right part: running it at 43% of rating leaves proper margin. Of the 0.24 W drawn from the supply, 0.126 W reaches the LEDs - 52%, which is what happens when half the supply voltage ends up across the resistor.
Formulas and scoring rules
- Series resistor
R = (Vsupply - n x Vf) / In is the number of LEDs in series; their forward voltages add.- Current with a real resistor
I = (Vsupply - n x Vf) / RThis is what the LED actually gets once you round to a stock value.- Resistor power
P = I^2 x R = (Vsupply - n x Vf) x IChoose a part rated at roughly twice this, so it runs at about half its rating.- LED power
P = n x Vf x I- Efficiency
eta = LED power / total power = n x Vf / VsupplyIndependent of the current: it is just the share of supply voltage the LEDs take.
When a resistor is the wrong answer
A series resistor works because the voltage across it is large enough to swamp the LED's variation. When the headroom is small - a white LED at 3.2 V on a 3.3 V rail, say - a 0.1 V change anywhere moves the current by a large fraction, and two LEDs from the same reel will not match.
The same applies to power LEDs, whose forward voltage falls as the junction heats: the current rises, which heats it further. For anything above a few tens of milliamps, or wherever brightness has to be consistent, use a constant-current driver. A resistor is for indicators.
Series or parallel for several LEDs
In series, every LED carries exactly the same current, so they match. The cost is supply voltage: the forward voltages add, and you need headroom on top.
In parallel with one shared resistor, the LED with the lowest forward voltage takes most of the current and is brightest, and the mismatch gets worse as it warms. If you must run LEDs in parallel, give each branch its own resistor - and treat each branch as a separate calculation here.
Limitations: what the result does not prove
- Forward voltage is not a constant. It varies between parts of the same type, falls as the junction warms and rises with current; data sheets give a range, and this calculator uses the single figure you enter.
- The result is an indicative design figure, not a compliance statement. Check the LED's absolute maximum current and the resistor's voltage and power ratings against their data sheets.
- It assumes a steady DC supply. Pulsed or PWM drive, multiplexed displays and mains-derived supplies all need their own analysis, including peak current limits.
- Nothing here says anything about brightness. Luminous intensity is not proportional to current, and two LEDs at the same current can differ in output by a factor of two.
Privacy: where your data goes
Everything you paste, type or drop is processed in this browser tab. It is not uploaded, logged, stored or sent to analytics. Session recording and tag-manager scripts are switched off on this page.
Standards and sources
- IEC 60063 - Preferred number series for resistors and capacitors (E series) - checked 19 Sep 2026
- Texas Instruments - LED drive circuits and current limiting (Op Amps for Everyone appendix)
- Nichia - LED data sheet showing forward voltage spread and derating
Frequently asked questions
What resistor do I need for an LED on 5 V?
For a red LED at about 2 V running at 10 mA, (5 - 2) / 0.01 = 300 ohms, so fit 330 ohms. For a white or blue LED at about 3.2 V at 10 mA it is 180 ohms. Always use the forward voltage from the data sheet for your part rather than a colour rule of thumb.
What happens if I use a bigger resistor than calculated?
Less current, so a dimmer LED and a cooler resistor. That is harmless, and often desirable: most modern indicator LEDs are uncomfortably bright at 20 mA and perfectly visible at 2-5 mA. Going smaller than calculated is the risky direction.
Can I connect an LED without a resistor?
Only if something else limits the current - a constant-current driver, a current-limited port, or a microcontroller pin with a defined source impedance, which is a fragile assumption. Without a limit the LED draws whatever the supply can give and fails, sometimes instantly.
How many LEDs can I put in series?
As many as the supply voltage allows, with headroom left over for the resistor. On 12 V with 2.1 V amber LEDs, three leaves 5.7 V and four leaves 3.6 V - both work, but four is more sensitive to supply variation. The calculator refuses arrangements that need more voltage than you have.
Should the resistor go before or after the LED?
Electrically it makes no difference: the same current flows through both, so the resistor limits it either way. Layout, switching arrangements and whether you are switching the high or low side are what decide in practice.
What power rating does the resistor need?
At least twice the dissipation shown here, so that it runs at about half its rating. For typical indicator currents that is a 1/8 W or 1/4 W part; the calculation only starts to demand more when the headroom voltage and the current are both large.
Why is my LED dimmer than another one at the same current?
Luminous efficiency varies enormously between parts, colours and manufacturers, and it is not proportional to current. Matching current matches current, not brightness - for that you need parts from the same bin, or per-channel adjustment.
Does this work for an LED strip?
Not directly. Most 12 V strips already have resistors built in for each segment and are designed to be driven from 12 V without anything extra; adding a series resistor just dims them and wastes power. Addressable strips have a constant-current driver in each package.
Last reviewed by the A2Z.Tools team against the sources listed above.
