PART number is L-53F3C , according to your suggestions and datasheet

Source voltage 5

diode forward voltage 1.26

diode forward current (mA) 40

number of LEDs in your array 8

Thanks for the clarification on the P/N.

I know you won't be using these LEDs, but here's how to check the LED calculator results:

1. Solve for resistor voltage drop (single resistor)

W = V * A (wattage formula)

W/A = V

108.8mW/40mA = 2.72v

2. Solve for current draw through the resistor. (ensures voltage in step 1 is right)

E = I * R (ohm's law)

E/R = I

2.72v/68 ohms = 40mA

3. Solve for voltage drop across each LED

(Source voltage - resistor voltage)/2 = voltage drop across each LED

(5v-2.72v)/2 = 1.14v per LED

4. Run that 1.14v LED voltage drop through the chart on pg 3 of the datasheet and you get about 8mA.

The calculator provided a VERY safe current limiting resistor value that will make your LEDs VERY dim.

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https://ledcalculator.net/#p=5&v=1.26&c=40&n=2&o=wUsing the same values as above, this calculator returns a resistor value of 62 ohms, 5% or better tolerance, 1/8W (min.) or 1/4W (better heat dissipation) for 2 LEDs and one resistor per row.

1. Solve for resistor voltage drop (single resistor)

W = V * A (wattage formula)

W/A = V

99.2mW/40mA = 2.48v

2. Solve for current draw through the resistor. (ensures voltage in step 1 is right)

E = I * R (ohm's law)

E/R = I

2.48v/62 ohms = 40mA

3. Solve for voltage drop across each LED

(Source voltage - resistor voltage)/2 = voltage drop across each LED

(5v-2.48v)/2 = 1.26v per LED

4. Run that 1.26v LED voltage drop through the chart on pg 3 of the datasheet and you get about 40mA.

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You're working on a very steep part of the curve so even a small change in resistor value results in a very large change in current.

- Might be worth the effort and expense to use a variable resistor.

Another good LED calculator site:

http://ledcalc.com/Good luck finding suitable LEDs.

Scott