LEDs in series and parallel: how many per string, one resistor per string, and the maths
One LED needs one resistor. Twenty LEDs do not need twenty resistors, nor should they share one: the answer is strings. LEDs in series share a single current and add their forward voltages, so a 12 V supply runs five red LEDs in a row through one resistor; strings then go in parallel, each with its own resistor, until all the LEDs are placed. This calculator works out how many LEDs fit a string with enough voltage left for the resistor to regulate, splits the array into the fewest even strings, sizes each string's resistor (rounded up to E24) with its power, and totals the current and the efficiency. It also lets you force a shorter string and compare every possible split.
How to use the LED array calculator
- Type the supply voltage and how many LEDs you want to light.
- Pick the LED colour in the menu row (it fills in a typical forward voltage; change it to the data-sheet figure) and set the current, 20 mA for a standard indicator.
- The array is split automatically. Force a number of LEDs per string, or change the volts kept for the resistor (1 V by default). The drawing shows every string lit; the Other splits tab compares each possible arrangement.
Series: sharing a current
R = (Vs − n Vf) / I nmax = ⌊(Vs − Vmin) / Vf⌋
LEDs in a chain pass the same current, so they are equally bright, and their forward voltages add. The resistor takes what is left of the supply: for five 2 V LEDs on 12 V that is 2 V, and at 20 mA the resistor is 100 Ω. The chain can hold as many LEDs as the supply has volts for, less a margin for the resistor. Beyond that the resistor cannot do its job, and a string with no resistor at all is a short circuit waiting for the supply to drift.
Parallel: one resistor per string
When the LEDs do not all fit one string, make more strings and connect them across the supply in parallel. Each string must have its own resistor. LEDs in parallel behind a shared resistor do not share the current: forward voltages differ by tenths of a volt even within a batch, and the LED with the lowest Vf takes most of the current, overheats, and when it fails the rest take its share. With a resistor per string the strings are independent, and a failure affects only its own string. The calculator splits the LEDs as evenly as it can (strings differ by at most one LED) and gives each size its own resistor.
The resistor's margin
The resistor regulates: if the supply rises or the LEDs' Vf falls, the extra voltage lands on the resistor and the current rises only in proportion to its share. With 2 V across the resistor, a 0.2 V change in the string is a 10 % change in current; with 0.5 V it is 40 %. Keep at least 1 V, 2 V for a loose supply, and more for LEDs with a wide Vf spread such as white and blue. Rounding the resistor up to a standard value lowers the current a little, which is the safe direction.
Efficiency
The share of the supply's power that reaches the LEDs is the string voltage over the supply voltage: five 2 V LEDs on 12 V put 10 of every 12 watts into light, 83 %; one LED per string on 12 V puts in 2 of 12, 17 %, the rest heating resistors. So fill the strings, or choose a supply a volt or two above the string. For many LEDs or high-power ones a constant-current driver replaces the resistors and holds the current whatever the voltage.
Your array, step by step
- Per string: (12 V − 1 V) / 2 V → up to 5; 10 LEDs make 2 × 5.
- 5-LED string (× 2): R = 2 V / 20 mA = 100 Ω → 100 Ω, 20 mA, 40 mW.
- Totals: 40 mA from the supply, 480 mW in, 400 mW to the LEDs, efficiency 83%.
Worked example: 10 red LEDs from 12 V
Red LEDs drop about 2 V and are happy at 20 mA. Keeping 1 V for the resistor, (12 − 1) / 2 = 5.5, so five LEDs fit a string, and ten LEDs make two strings of five. Each string's LEDs use 10 V, leaving 2 V for the resistor: R = 2 V / 20 mA = 100 Ω, which is a standard value, so each string gets a 100 Ω resistor dissipating 40 mW, and runs at 20 mA. The supply delivers 40 mA and 480 mW, of which 400 mW lights the LEDs: 83% efficient. Had there been twelve LEDs, the split would be three strings of four with 200 Ω each (4 V across the resistor, 67%); with white LEDs at 3.2 V only three fit a 12 V string.
Questions
Can I put LEDs in parallel with one resistor?
Do not. Their forward voltages differ, the lowest takes the current, and they fail one after another. One resistor per string, always.
Can I mix colours in a string?
Yes: they share the current, so add up their different forward voltages for the string. Mixed colours at the same current will not look equally bright, though.
Why not use the whole supply voltage for LEDs?
With nothing across the resistor there is no regulation: a small rise in supply or fall in Vf sends the current through the roof. Keep a volt or two.
When should I use a constant-current driver instead?
For high-power LEDs (350 mA and up), long strings from mains-derived supplies, or anywhere the resistor would waste watts. A driver holds the current directly and needs no margin.