DIY multi LED grow light

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bob0816

bob0816

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I want to built an LED grow light from scratch but I'm not sure what to look for when wiring in series and parallel on the same board.

I want it to have different types of LEDs with different wavelength, like the Spider Farmer SF4000 or the SSK-272-RR-V2 for example, so I have to deal with different voltages and currents needed.

What do I have to look for when wiring series (of LEDs) in parallel when it comes to the forward voltage U_f (V) and forward current I_f (A) of the loads?

I tryed to find a schematic but didn't so I tryed to look for the paths in the LED light PCB. From the SSK-272-RR-V2 I could see a lot of them in the pictures found online and so I made a scetch. Here the simplistic schematic as I guessed it:
Unbenannt

Installed are:
256 Nichia 757, NFSW757H-V1 [U_f: 2.84 V, I_f: 65 mA, I_max: 180 mA] warm whites (black squares) and
16 Cree XP-G3, XPGDPR-L1-0000-00F01 [U_f: 1.99 V, I_f: 350 mA, I_max: 1.5 A] 660 nm "photo reds" (red squares)

Then I tryed to calculate and see how they did it.
The whites are wired in series of four, the reds parallel in four.
When I look at one quart of the board and would try to drive it, my thoughts are:
- in parallel wiring, every row sees the same voltage U, so to run them, I have to put at least 11.36 V on the circuit because of the series of whites adding up (4x 2.84 V = 11.36 V to deliver U_f)
- that means that the reds will get 11.36 V in parallel too, the U_f of this part is just 1.99 V as of parallel wiring but the I_f is 1.4 A (4x 350 mA)
My question now is (if I am right so far is another one): where does more energy travel? The way of lower current needed, or the way of lower voltage needed?
I tryed to find the way of least resistance (literally) and calculated R=U/I:

series of 4 whites: 11.36 V, 65 mA -> R=174.77 Ohm
parallel of 4 reds: 1.99 V, 1.4 A -> R=1.42 Ohm

so in my mind there would be much more current flow through the red LEDs in parallel, maybe with a CC driver set to 1.5 A on a safe level (because it could take 4x I_max: 1.5 A=6 A), but the white LEDs would run way under there capacity.


What do I have to take in account when planning such a circuit regarding different V_f and I_f values?

Where am I wrong? I think have miscalculated the resistances (1/R=1/R_1+1/R_2+1/R_n)
 
bob0816

bob0816

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The schematic in the thread start is wrong.

Actual SSK-1-272-RR-V2 schematic
So it would be 4 "quarters" in parallel,
each quart
would be:
- a series of: 8*(4 Nichia in parallel), 4 Cree in series and again 8*(4 Cree parallel)
Nichia 2.84 V, 65 mA
Cree 1.99 V, 360 mA

so it would need 8*2.84 V + 4*1.99 V + 8*2.84 V
= 53.72 V to run

and 4*65 mA
= 260 mA tu run on optimal current.

driven by hypothetical 53.72 V and 260 mA, the Nichia would run optimal, the Cree on 0.72 of optimum, correct?

Just for the purpose of understanding:
Could I wire, lets say, 2 Samsung LH351H Blue (450 nm) 3 V, 350 mA in parallel to the quart with a resistor infront?
I would calculate:

53.72 - 2*3 V = 47.72 V to be resisted
47.72 V/260 mA = 183.53 Ω (R=U/I)

so I would take a 180 Ω resistor like the Y1496183R000F9W, voltage drop would be 46.8 V (180 Ω*260 mA=46.8 V) and the voltage behind the resistor would be 6.92 V (53.72 V-46.8 V=6.92 V),

What would that mean for the two 3 V blue LEDs? What current would they get?
In my mind is: A resistor restricts the flow of current, but reduces the voltage and the current has the same value behind the resistor as before.
 
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