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LED Array Calculator (Series/Parallel)

Compute the resistor each string needs and the total current when arranging multiple LEDs in series and parallel.

Resistor per string
300 Ω
nearest std: 300 Ω
Total current
40 mA
2 × 20 mA
Power per resistor
120 mW

Each series string needs its own resistor; parallel strings share the supply. Keep string voltage a couple of volts below the supply for stable current.

Disclaimer: This calculator is provided for general informational and educational purposes only, on an “as is” basis and without any warranty of accuracy or fitness for a particular purpose. Results may contain errors — always verify independently before relying on them in real designs. PartAndStock accepts no liability for any loss or damage arising from use of this tool, including when embedded on third-party sites.

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Step-by-step: How to Wire an LED Array

How to use

  1. 1Enter the supply voltage (Vs).
  2. 2Enter one LED's forward voltage (Vf) and current (If).
  3. 3Enter the series LEDs per string (N) and the number of parallel strings (M); the resistor and total current are computed.

How it works

Enter the supply voltage, LED forward voltage and current to find the per-string resistor and total current for a series/parallel LED array.

String voltage = N·Vf · Rs = (Vs − N·Vf) / If · Total current = M·If (N: series, M: parallel strings)

Arranging LEDs in series and parallel

LEDs are driven by current, not voltage, so every series string needs its own current-limiting resistor. In series the LED voltages add (N·Vf) and a single resistor limits the whole string: Rs = (Vs − N·Vf)/If. To share the same current across several strings you wire them in parallel — each string gets its own resistor so manufacturing spread doesn't split the current unevenly.

Why each string needs its own resistor

With parallel LEDs sharing one resistor, the LED with the lowest Vf draws more current, heats up, drops its Vf further and hogs the current (thermal runaway). Give each series string its own resistor to prevent this. Keep the string voltage (N·Vf) at least a couple of volts below the supply, or current gets unstable as the supply or Vf drifts.

Worked examples

  • Vs=12 V, Vf=2 V, If=20 mA, N=3, M=2 → string voltage 6 V, Rs = 6/0.02 = 300 Ω/string
  • Same circuit: total current = 2·20 = 40 mA
  • Vs=5 V, Vf=3.2 V (white), N=1 → Rs = 1.8/0.02 = 90 Ω

Typical LED Forward Voltages (Vf)

ColorVf (approx.)
Red1.8–2.2 V
Yellow/Orange2.0–2.2 V
Green2.0–3.0 V
Blue/White3.0–3.4 V
IR1.2–1.5 V

Frequently Asked Questions

How do I calculate the LED array resistor?+
For each series string, Rs = (Vs − N·Vf)/If. E.g. 12 V supply, 3 series LEDs (Vf=2 V), 20 mA → Rs = (12−6)/0.02 = 300 Ω. Each parallel string must have its own resistor.
Why can't I parallel LEDs without resistors?+
On a shared resistor, manufacturing spread makes the lowest-Vf LED draw more current, heat up and hog it. A separate resistor per series string balances the current.
How many LEDs can I put in series?+
The total string voltage (N·Vf) must stay a few volts below the supply so the resistor has headroom to drop. E.g. with 12 V and 2 V LEDs, up to ~5 in series is safe.
Should I wire them in series or parallel?+
Series shares the same current, so it's more efficient (one resistor, less loss) but needs a higher supply. If the supply is low, split the LEDs into parallel strings, each with its own resistor.
What power rating should the resistor be?+
Power per string is P = (Vs − N·Vf)·If. E.g. a 6 V drop at 20 mA → 120 mW; a 1/4 W (250 mW) resistor gives safe margin.

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