Parallel LED Strings
parallel LED strings each with their own resistor share the same current. Each string works like the series calculator. Adding parallel strings increases total current and LED count, with enough left over for the current-limiting resistor. Each string is independent — if one LED fails, only its string goes dark.
R — Current-limiting resistor. Drops the excess voltage and sets the LED current.
Vf — Forward voltage per LED. Red ~2.0 V, Green ~2.2 V, Blue/White ~3.0–3.5 V.
I — Current through the entire series string. Same for all LEDs. Typically 20 mA for standard LEDs.
LED Parallel Calculator
When you need more LEDs than one series string can hold, you run multiple strings in parallel. Each string has its own current-limiting resistor and operates independently. If one string fails, the others keep working. This calculator finds the resistor value per string, total current draw, total power, efficiency, and the number of resistors needed.
How Parallel LED Strings Work
Each parallel string is a self-contained series circuit — LEDs in series with a resistor, all connected across the same supply voltage. The strings share the supply but carry independent currents. This is exactly how LED strip lights work: segments of 3 LEDs + 1 resistor repeated along the strip, all connected in parallel across 12V or 24V. For single-string calculations, use the LED Series Calculator.
Why Each String Needs Its Own Resistor
You cannot connect LEDs directly in parallel without individual resistors. LEDs have a steep V-I curve — a tiny difference in forward voltage between two LEDs causes a large difference in current. The LED with the lowest Vf hogs most of the current, overheats, its Vf drops further, it takes even more current, and it burns out. Then the next-lowest takes over and the cascade continues. One resistor per string forces equal current distribution regardless of Vf variations.
The Formulas
R = (Vs − ns × Vf) / I
Parallel totals:
Itotal = Istring × np
Ptotal = Vs × Itotal
LEDs total = ns × np
Resistors needed = np (one per string)
12V / 3s × 4p — 12 Red LEDs
12V supply, 3 red LEDs (Vf = 2.0V) per string, 4 parallel strings, 20 mA each.
4 strings × 20 mA = 80 mA total
Ptotal = 12 × 0.080 = 960 mW
Presistors = 4 × 6.0 × 0.020 = 480 mW
Efficiency = 50%
Need: 4 × 300 Ω resistors + 12 LEDs
480 mW wasted in resistors. For indicator LEDs at 20 mA this is fine. For high-power lighting, the waste adds up — 50 parallel strings would waste 6 W in resistors alone. For the energy cost of running LED arrays, see the LED Power Calculator.
5V / 1s × 10p — 10 White LEDs
5V supply, 1 white LED (Vf = 3.3V) per string, 10 parallel strings.
10 strings × 18.7 mA = 187 mA total
Ptotal = 5 × 0.187 = 935 mW
Efficiency = 66%
Need: 10 × 91 Ω resistors + 10 LEDs
Each string has only 1 LED because 2 × 3.3V = 6.6V exceeds 5V. This is the worst case for efficiency — 34% wasted. A 12V supply with 3 per string would be 83% efficient for the same 10 LEDs (4 strings of 3, with one extra). For high-efficiency solutions, consider an LED driver IC — the LED Driver Calculator sizes boost and buck-boost topologies.
24V / 6s × 3p — 18 Blue LEDs
24V supply, 6 blue LEDs (Vf = 3.2V) per string, 3 parallel strings.
3 strings × 20 mA = 60 mA total
Ptotal = 24 × 0.060 = 1.44 W
Efficiency = 80%
Need: 3 × 240 Ω resistors + 18 LEDs
80% efficient. 24V systems allow long strings that use most of the supply voltage, leaving little for the resistor. This is the standard approach for architectural and signage LED lighting.
Series vs Parallel — When to Use Which
Use parallel strings when you need more LEDs than one string can hold, or when you need fault tolerance (one failure only kills one string). Each string needs its own resistor. Total current = strings × current per string.
LED strip lights use series-parallel: segments of 3 LEDs + 1 resistor (series), repeated in parallel along the strip. Cut marks are between parallel segments.
Power Supply Sizing
Total current = current per string × number of strings. Size the supply with at least 20% margin. For 10 strings at 20 mA each: 200 mA minimum, choose a 250 mA supply. Total power = Vs × Itotal. For the underlying V = IR relationship, see Ohm’s Law.
Frequently Asked Questions
Last updated: March 2026