# Powering an LED matrix

I am building an led matrix having 75 SMD 2835 60mA leds. I am connecting 3 in series with a resistance and then connecting the 25 series parallely to a 12V power source supplying 3A@12V. This means that 120mA will be divided among the 25 series, will this set up burn up the leds. Also what value of the resistance will i need to connect to the 3 leds in series?

• It would help to know the exact models of LEDs, as we don't know the specific forward voltage. – SomeoneSomewhereSupportsMonica May 13 '18 at 9:43
• Forward voltage of an led is 3.2-3.3v – Kangkan Kakaty May 13 '18 at 9:44

If forward voltage is 3V DC, the drop across 3 diodes is 3*3V=9V. The remaining voltage is the battery voltage V_battery-9V. The resistance needed (for each series connection of 3 diodes) is (from Ohm's law) : R=(V_battery - 9V)/60mA. E.g. 50 Ohm.

I am building an led matrix having 75 SMD 2835 60mA leds

I am going to assume you mean you want to light up approximately 75 white LEDs with a test current of about 60 mA.

To get the resistance value use an online calculator such as Hobby Hour LED Series Resistor Calculator Forward voltage of an led is 3.2-3.3v

This is not true. The IV curve and temperature determines Vf. Today typical Vf for a 60mA white LED is 2.75-2.85V.

a 12V power source supplying 3A@12V

Is there any particular reason you are using 12V?

A 48v or 54v constant current would be better than using resistors. Recommended: Mean Well HLG-40H-54B

In the 60 mA area I would use a Samsung LM561C (2.75V @ 65 mA) with an efficacy of 212 lm/W.

Using the 561C's Vfmax 3.0V you can drive strings of 18 LEDs using 54V LED driver. I would drive 18 sets of four parallel LEDs wired in series.

To drive these at 65 mA. You will need 4 x 65 mA = 260 mA.
72 x 37 lm @ 65mA = 2,664 lumens

With a \$25 HLG-40H-54B you can run them up to 750mA (≈8,000 lumens) without exceeding the max current (200 mA) for a 561C. Current is easily adjusted with a variable or fixed resistance.