LED Array Calculator (Series and Parallel Strings)
LED Array Calculator (Series and Parallel Strings)
Turn a supply voltage, an LED forward voltage and a total LED count into a real arrangement: how many LEDs go in each series string, how many strings, the resistor each string needs and its wattage — and how much of your supply the resistors burn rather than the LEDs.
LED array: strings, resistors, efficiency
24 white LEDs (V₁ 3.2 V) at 20 mA from a 12 V supply, leaving at least 15% of the supply across the resistor, E12
Arranging an LED array
R = (Vs − nsVf) ÷ IF, rounded UP to E12 or E24 PR = I²R
Itotal = k·I η = nsVf ÷ Vs
- Vs
- supply voltage
- Vf
- forward voltage of one LED at the current you want
- h
- the share of the supply you insist on leaving across the resistor, as a fraction
- ns
- LEDs per series string — the most that fit while leaving that headroom
- k
- number of parallel strings, each with its own resistor
- η
- efficiency: the LEDs’ share of the supply power. It is exactly the string’s share of the supply voltage, because every part of the string carries the same current
Worked example
24 white LEDs (V₁ 3.2 V) at 20 mA from a 12 V supply, leaving at least 15% of the supply across the resistor, E12
Headroom: 15% of 12 V is 1.80 V, so at most 10.20 V of LEDs per string
LEDs per string: ⌊ 10.20 ÷ 3.2 ⌋ = ⌊ 3.1875 ⌋ = 3; strings: ⌊ 24 ÷ 3 ⌋ = 8, none left over
Each string drops 9.60 V, leaving 2.40 V across its resistor: R = 2.40 ÷ 0.020 = 120.0 Ω, and 120.0 Ω is itself an E12 value, so 120 Ω
Each resistor burns I²R = 0.020² × 120 = 48 mW; 8 of them come to 384 mW, which a ¼ W part handles with room to spare
Supply: 8 × 20 mA = 160 mA at 12 V = 1.92 W; the LEDs take 1.54 W of it, so the efficiency is 9.6 ÷ 12 = 80.0%
Why one resistor must not feed several LEDs in parallel
| Two white LEDs on one 120 Ω resistor from 12 V | V₁ of LED A | V₁ of LED B | What happens |
|---|---|---|---|
| Identical parts | 3.20 V | 3.20 V | Each takes half the current — the case the arithmetic assumes |
| A normal reel spread | 3.10 V | 3.30 V | LED A conducts first and holds the node near 3.10 V; LED B is 0.20 V below its own V₁ and barely lights |
| After A warms up | 3.00 V | 3.30 V | A’s V₁ falls as it heats, so it takes even more of the current, heats further, and can run away |
How many LEDs fit in a string, at 15% headroom
| Supply | Red, 1.85 V | Green (GaP), 2.2 V | White or blue, 3.2 V |
|---|---|---|---|
| 5 V | 2 (2.30) | 1 (1.93) | 1 (1.33) |
| 9 V | 4 (4.14) | 3 (3.48) | 2 (2.39) |
| 12 V | 5 (5.51) | 4 (4.64) | 3 (3.19) |
| 24 V | 11 (11.03) | 9 (9.27) | 6 (6.37) |
| 48 V | 22 (22.05) | 18 (18.55) | 12 (12.75) |
Laying out an LED array
An array is not one big LED problem, it is a small one repeated. Put as many LEDs in series as the supply will carry, give that string its own resistor, then put as many identical strings in parallel as you need. Everything in a series string carries the same current, so one resistor controls the lot; strings in parallel each need their own, because nothing else makes them share.
How many per string. The temptation is to cram in as many as the supply allows. Do not. Whatever voltage the LEDs do not use is what the resistor works with, and the resistor is the only part of the circuit that controls anything. On 12 V with 3.2 V LEDs, 3.750 would fit arithmetically, so three fit and leave 2.40 V and a fourth does not fit at all. Where the arithmetic does allow one more, the headroom rule still stops you: 24 V gives 7.50, so seven LEDs would fit and leave 1.60 V — but that is only 6.7% of the supply, and at that headroom the current is set by the LEDs’ own spread rather than by you, so the page stops at 6 and leaves 4.80 V. The headroom box is where you make that trade: more headroom means a steadier current and a hotter resistor, less means a more efficient array and a current you cannot predict.
What the spread costs. The chart on this page holds the resistor at 120 Ω and sweeps one LED’s forward voltage across a ±0.4 V band, which is about what a reel of 5 mm white LEDs covers. At 3.2 V the string runs at 20.00 mA; at 2.8 V it runs at 30.00 mA and at 3.6 V at 10.00 mA. That is the whole argument for headroom, and the whole argument for a constant-current driver once the array is big enough to matter.
Rounding, and what it does. The exact resistance almost never exists as a part, so the page takes the next E12 or E24 value up. Up, not down: a larger resistor gives slightly less current than you asked for, which is safe, while a smaller one gives more than the LED is rated for. E24 has closer steps if you stock it. Choose a wattage rating of at least twice the dissipation shown, because ratings are quoted at a stated ambient temperature and fall above it.
Where the power goes. Efficiency here means the LEDs’ share of the supply power, and because every part of a string carries the same current it is exactly the string’s share of the supply voltage: 9.6 V of 12 V is 80.0%. That number is usually what changes someone’s design, and the headroom rule caps it: the LEDs can never take more than the share you did not reserve, so 15% headroom puts a ceiling of 85% on the efficiency, and rounding down to a whole number of LEDs costs a little more on top. A switching constant-current driver has no such ceiling and wastes almost nothing. For one LED, use the LED series resistor calculator, which this page does not repeat; for the arithmetic behind the resistor, the Ohm’s law calculator; to read the resistor you already have, the resistor colour code calculator.
Frequently asked questions
How do I work out an LED array’s layout?
Divide the supply by the LED forward voltage, leaving headroom for the resistor, and round down: that is the number per series string. Divide the total LED count by it for the number of strings. 24 white LEDs at 3.2 V on 12 V with 15% headroom gives 3 per string and 8 strings.
Can I use one resistor for LEDs in parallel?
No. LEDs do not share current: the one with the lowest forward voltage takes most of it, warms up, drops further and takes more still. Give each series string its own resistor, or use a driver with one channel per string.
What resistor does each string need?
The voltage the LEDs do not use, divided by the current you want. Three white LEDs on 12 V leave 2.40 V, so 2.40 ÷ 0.020 = 120 Ω, and 120 Ω is the E12 value at or above it.
How efficient is a resistor-fed LED array?
The LEDs’ share of the supply voltage, and nothing more. Three 3.2 V LEDs on 12 V give 80.0%; the other 20.0% is heat in the resistors. It can never beat the headroom you reserved — 15% headroom caps it at 85% — and rounding down to whole LEDs costs a little more. A switching constant-current driver has no such ceiling.
Why leave headroom across the resistor at all?
Because the resistor is the only part that controls the current. With 120 Ω in a three-LED string on 12 V, an LED 0.4 V below typical pushes the current to 30.0 mA and one 0.4 V above drops it to 10.0 mA. With almost no headroom the same spread switches the string between nothing and destruction.
Related calculators
References
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Chapter 1 (Ohm’s law, power in resistors) and §2.2 on driving LEDs from a fixed supply.
- IEC 60063:2015. Preferred number series for resistors and capacitors (the E6, E12, E24, E48, E96 and E192 series). International Electrotechnical Commission.
- Wikipedia, Luminous efficacy — luminous efficacy of a source, lm/W, with the overall luminous efficiency alongside: tungsten incandescent 40 W / 100 W at 230 V, 10.4 and 13.8 lm/W; 100 W at 120 V, 17.5 lm/W; tungsten halogen 100–500 W at 230 V, 16.7–19.8 lm/W; compact fluorescent 9–32 W with ballast, 46–75 lm/W; LED lamp 5–16 W at 230 V, 75–217 lm/W. Retrieved 24 September 2026.