LED Series and Parallel Array Calculator
Plan an array of identical LEDs on one supply: how many in each string, how many strings, and the resistor for each string.
- LEDs per string = floor((12 − 0.5) ÷ 3.2) = 3
- R = (Vs − n × Vf) ÷ If = (12 − 3 × 3.2) ÷ 20 mA = 120 Ω
- Rounded up to the next E24 value so the current stays at or below the target.
LED Array Design
R = (Vs − n × Vf) ÷ If
PR = (Vs − n × Vf)² ÷ R
Every string needs its own resistor. Never put LEDs directly in parallel with one shared resistor, because the LED with the lowest Vf takes most of the current.
Worked example
Ten white LEDs (3.2 V, 20 mA) on 12 V: 3 per string, so 3 strings of 3 plus 1 single LED. Each 3 LED string needs (12 − 9.6) ÷ 0.02 = 120 Ω; the single LED needs 440 Ω, so fit 470 Ω. Total current is 80 mA.
Frequently Asked Questions
Why does each LED string need its own resistor?
LED forward voltages differ slightly. Without separate resistors one string hogs the current and fails first.
How many LEDs can I put in series on 12 V?
With white LEDs at about 3.2 V, three per string, leaving about 2.4 V for the resistor.
Should I round the resistor up or down?
Up, so the current stays at or below the LED rating.
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