DIY multi LED grow light

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 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)

SSK-1-272-RR-V2 as basic for the schematic: EDIT: actual schematic: Nichia 757 data sheet: http://www.nichia.co.jp/en/product/led_product_data.html?type='NF2W757H-F1'
Cree XP G3 data sheet: https://www.cree.com/led-components/media/documents/dsXPG3_1.pdf

I want blue, red and white on one board and also thought about installing resistors, like done in the Solskin:

So who says the Cree needs to run its rated current?

Also, what are you going to mount them on, wire them up with? Making your own backplane?

printer said:

Spoiler: quote

So who says the Cree needs to run its rated current?

Also, what are you going to mount them on, wire them up with? Making your own backplane?

Click to expand...

That's one of my questions. Will it all run all loads on the same current according to the lowest/highest current-need in the circuit, if available from the driver?
All lines in parallel will get the same current. The voltage is the same on all lines. The current is split by all lines equally. Will it run through all loads in line at the same value or will there be differences and what determens the value? the overall resistance of the circuit?

Yes, custom PCB.

The schematic is wrong, edited the actual one in the post above

So by the new SSK-1-272-RR-V2 schematic 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?
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.

A voltage applied to a bunch of parallel loads will be the same across all the loads. The current in each branch will be different if the loads will be different. In a series string there is only one current value, voltage across each component can vary due to their resistance.

Diodes can be thought of as an ideal diode with no resistance (say it needs 1.5V to switch on) and a resistor in series. Once the diode is switched on the resistance determines how much current goes through the device. Just mentioning this as just thinking Leds as a resistor is not correct. Once the required voltage is provided then the excess voltage is used up across the device's internal resistance. Hope that this does not muddy the waters too much.

Just looked at your new schematic (actually it is a layout diagram, not a schematic). So you have 4 x 0.65 mA = 0.26A. That is in series with the 0.35mA devices. So what is the total current in the string and the voltage, given the information you have?

I think you made a few errors on the original board your looking at as far as connections go.

(Only drew part of the board, but sure you see how it works.)
Its basically the 2 power rails on top and bottom and then traces go up or down depending on polarity needed. Then the 'reds' run from top to bottom in series with the rest being series of 8's and 8 of those paralleled together. Sets of 8 are easy to work with mainly due to 8 @ 3.x VDC equals 24 VDC +/- and then its multiplied from there. So for your 6 diodes in series it would be 18-ish volts needed...hard to find an inexpensive driver @ that value I would think.
I'd say in this setup without any resistors used the voltage and amperage for 8 'red' diodes is fairly equal to 8x8 'white' diodes. So if the reds use say 400mA then the whites would use 50mA.
I trying to remember from some tests I did awhile back, but I think thats the basics...lol

PS...you don't need the extra paralleled traces between the series diodes.
Hope that helped in some way...lol
Best of luck and happy growin'

printer said:

The current in each branch will be different if the loads will be different.

Click to expand...

I don't really understand what that means. Can the current be divided asymetrically and what would that mean for loads and "electrical burden" on the loads? It's different from voltage in series, where every load takes just it's needed part, right?

Diodes can be thought of as an ideal diode with no resistance (say it needs 1.5V to switch on) and a resistor in series. Once the diode is switched on the resistance determines how much current goes through the device.

Click to expand...

So powered by a 5 V source, to put 20 mA on the 1.5 V diode the resistance would have to be 5 V/20 mA=250 Ω?

Once the required voltage is provided then the excess voltage is used up across the device's internal resistance.

Click to expand...

What is meant by excess voltage and being used up across the device's internal resistance,
the 3.5 V (5 V-1.5 V) that are used up in the resistor, or voltage that exceeds the 5 V of the device, heating it up? I meant

So you have 4 x 0.65 mA = 0.26A. That is in series with the 0.35mA devices. So what is the total current in the string and the voltage, given the information you have?

Click to expand...

Since the current is all the same in one series string, it depends on something I'm not sure about, either the driver output divided by the number of strings, or the strings resistance, or what each load takes (but that I think is least likely, since they have a wide band of current-finctioning, which is why the current determines their performance/brightness). Here I'm with the same puzzlement as with the first quote

For the quart the total voltage is 53.72 V, current would be determined by the CC drivers output, 260 mA by the smaller load, 360 mA by the bigger load,
the total board considered (just one PCB board = four quarts in parallel) is 53.72 V, 1.04 A / 1.4 A if calculated correctly, meaning it would run at 55.87 W (53.72 V*1.04 A) or 75.21 W (53.72 V*1.4 A). The full SSK-1-272-RR-V2 (eight quarts) is given as 135 W, I_max=2.8 A (so simple current calculation succeeded, 2.8 A meaning running the board on Cree optimum and Nichia on roughly 1.4 the optimum.
I still ask my self, how current and voltage flow in a parallel with different loads, for example, if I have 3 V source and power 4*3 V LED in parallel with resistors before them and they have differences in V (dV) due to normal manufacturing variability: 3.1 V, 3.0 V, 3.05 V, 2.9 V, what consequences would the dV have electrically? Will they just get all 3 V and the connected current and so the 2.9 V would nominally "use up" faster?
And what if they are, hypothetically, all 3 V, but one is 20 mA, two 50 mA and one 65 mA, what would that mean in regards to: flow of current, and then, resistor choice? I think the smaller the needed current, the bigger the resistor because current would flow equally destributed through the circuit and then we just have to reduce it to the wanted lvl...?

And that would mean, that the current in the 4quarts board can be anywhere from (I_f_string)*(n_strings) up to (I_max_string)*(n_strings)

(n_strings)=total number of strings
(I_f_min)=lowest I_f on all strings (I_f: forward current)
(I_max_string)=I_max of one string

Mak'er Grow said:

Spoiler: quote

I think you made a few errors on the original board your looking at as far as connections go.
View attachment 4698863
(Only drew part of the board, but sure you see how it works.)
Its basically the 2 power rails on top and bottom and then traces go up or down depending on polarity needed. Then the 'reds' run from top to bottom in series with the rest being series of 8's and 8 of those paralleled together. Sets of 8 are easy to work with mainly due to 8 @ 3.x VDC equals 24 VDC +/- and then its multiplied from there. So for your 6 diodes in series it would be 18-ish volts needed...hard to find an inexpensive driver @ that value I would think.
I'd say in this setup without any resistors used the voltage and amperage for 8 'red' diodes is fairly equal to 8x8 'white' diodes. So if the reds use say 400mA then the whites would use 50mA.
I trying to remember from some tests I did awhile back, but I think thats the basics...lol

PS...you don't need the extra paralleled traces between the series diodes.
Hope that helped in some way...lol
Best of luck and happy growin'

Click to expand...

Yeah, saw it better when I inverted the color of the board picture and played around a little with brightness and contrast there's a new one in the second post, unfortunately I can't edit the thread-post.
Thanks mate!

bob0816 said:

I don't really understand what that means. Can the current be divided asymetrically and what would that mean for loads and "electrical burden" on the loads? It's different from voltage in series, where every load takes just it's needed part, right?


So powered by a 5 V source, to put 20 mA on the 1.5 V diode the resistance would have to be 5 V/20 mA=250 Ω?


What is meant by excess voltage and being used up across the device's internal resistance,
the 3.5 V (5 V-1.5 V) that are used up in the resistor, or voltage that exceeds the 5 V of the device, heating it up?


Since the current is all the same in one series string, it depends on something I'm not sure about, either the driver output divided by the number of strings, or the strings resistance, or what each load takes (but that I think is least likely, since they have a wide band of current-finctioning, which is why the current determines their performance/brightness). Here I'm with the same puzzlement as with the first quote

Click to expand...

Lots of math errors in this reply to @bob0816 to start with.
Should it not be like "0.65A" aka 650mA and not "0.65mA" and then multiplied by 4 doesn't equal "0.26mA" it equals 2.6A.

Beyond that if memory serves me right...if you reduce the voltage then you need more amperage to turn on the LED...and the reverse comes into play as well...so theres lots of variations of the 2 can be used, but you do still need to be aware of the minimum requirements. For example if the LED uses 3.0 VDC @ 100mA up to say 3.5 VDC@ 250mA then you could turn it on with with any value between those, but NEVER go above the max or the LED with fry...most boards I've tested seem run about 50-75% of the max in hopes to save or lengthen the life of the LED.

Mak'er Grow said:

For example if the LED uses 3.0 VDC @ 100mA up to say 3.5 VDC@ 250mA

Click to expand...

How tight are U and I tied together? (getting kinda personal all of a sudden) Is it an electrical/physical constant conecting them,
or a material, load-specific constant (maybe temperature related?) ?

bob0816 said:

How tight are U and I tied together? (getting kinda personal all of a sudden)

Click to expand...

No idea what your referring to.
For the other part, yes, it has to do with brightness and therefore heat created.
You could light it dimmly at say 3 VDC @ 100ma and it will brighten as you add voltage (constant current driver basics)...then if you start with constant voltage of say 3 VDC and increase amperage the LED will also get brighter...up to the max values, but as said before the heat also increases with brightness and the life span also shortens when pushed to the limits...sorry for repeating myself...lol

I maent, how close/tight are the voltage (V) (or U) and amperage (I) connected, because you mentioned them in two sets, implying the connectedness (and I know they are), but was thinking constant current / constant voltage and how things would go different between raising them together for the component and the driver, for a deeper understanding.
I don't think it has the same consequences for component and circuit although it has the same visible effects.
I know I can read it out the current-voltage voltage-temperature diagrams. I would like to know the advantages and disadvantages and the explanation, I want to know everything (grow light LED) circuit before I start.

edit: that's actually a fantastic point, I will check the needed voltage of all components to flow estimated current at which all could work efficiantly and long lasting and back check the total voltage again. Driving with a CC+CV driver, setting the current and then going up to calculated voltage with room under and above

bob0816 said:

I maent, how close/tight are the voltage (V) (or U) and amperage (I) connected, because you mentioned them in two sets, implying the connectedness (and I know they are), but was thinking constant current / constant voltage and how things would go different between raising them together for the component and the driver, for a deeper understanding.
I don't think it has the same consequences for component and circuit although it has the same visible effects.
I know I can read it out the current-voltage voltage-temperature diagrams. I would like to know the advantages and disadvantages and the explanation, I want to know everything (grow light LED) circuit before I start.

edit: that's actually a fantastic point, I will check the needed voltage of all components to flow estimated current at which all could work efficiantly and long lasting and back check the total voltage again. Driving with a CC+CV driver, setting the current and then going up to calculated voltage with room under and above

Click to expand...

I think you got it now...lol
Using different LEDs is why there is no standard/constant answer...each set of LEDs will probably be slightly different then another...some use/like more voltage and some use/like more current.

I am repotting plants so I do not have time to cover all the points above. Just things to think about. Led characteristics, the resistance is not constant until the curve generally makes a straight line plot. At the lower voltages the resistance increases, otherwise the straight line section would intersect the Voltage axis. the UV curve would intersect at 3V.



256 Nichia [U_f: 2.84 V, I_f: 65 mA,]
16 Cree [U_f: 1.99 V, I_f: 350 mA, I_max: 1.5 A]

8 x 0.065 mA = 0.52A 8 x 2.84V = 22.72V

8 Cree - 1.99V x 8 = 15.92V

You can not have a voltage rail supplying 22.72V and 15.92V at the same time. The yellow diodes add up to 22.72V and the red add up to 15.92V.



You can have different currents through parallel branches across the same voltage. You can not have the battery feed two different voltages at the same time.

printer said:

You can not have a voltage rail supplying 22.72V and 15.92V at the same time. The yellow diodes add up to 22.72V and the red add up to 15.92V.

Click to expand...

I said nothing about different voltage or anything of the sort.
If you re-read my post I say the picture/schematic he was using wasn't right compared to the board he posted.
Also you can have different voltages from 1 rail...just step the higher value down to secondary value with resistor(s) or regulator(s).

Mak'er Grow said:

I said nothing about different voltage or anything of the sort.
If you re-read my post I say the picture/schematic he was using wasn't right compared to the board he posted.
Also you can have different voltages from 1 rail...just step the higher value down to secondary value with resistor(s) or regulator(s).

Click to expand...

The boards do not look to have a regulator, quite a difference in voltage for resistors. And that only creates heat and lowers the over all efficiency of the board. As your board is pictured it will not work. No resistors or regulators are installed as drawn.

In parallel wiring, the voltage is split evenly on all lines.
The whites are 2.84 V, the reds are 1.99 V.
In this circuit diagram there are three lines in parallel, 2*(8 series of 8* 2.84 V diodes (8*2.84 V in series=22.72 V) in parallel =22.72 V_f per each one of that two lines) and 1* series of 8*1.99 V= 15.92 V,
so leaving 6.8 V (22.72 V-15.92 V) to be resisted in that line to drive the two 22.72 / to overcome the minimum forward voltage V_f and not put an excess of 6.8 V on the series reds. The amperage given by the driver/source would be divided by three, 1/3 flowing through each line.
(how can I subscript? superscript via ^ (^2 -> ²))
In this circuit diagram there are three lines in series, 2*(4 series of 8* 2.84 V diodes =)22.72 V and one series of of 4* 1.99 V =7.96 V, adding up to 53,4 V
(2*22.72 V+7.96 V), the amperage given by the driver/source would be the same through all diodes.

Here's no place for insults and personal dispute! Stick to facts (@ all)

printer said:

You can have different currents through parallel branches across the same voltage. You can not have the battery feed two different voltages at the same time.

Click to expand...

So a resistor is "taking up" voltage and thereby reducing current?

I think about either wiring in parallel with resistors like seen in the diagram above or in this diagram or in series as in this diagram
advantages and disadvantages:

series:
- advantages:
--- voltage adds up and can be neglected, all diodes will be run on the same current
--- -> no need for resistors

- disadvantages:
--- current dictated by lowest individual I_f_optimum, running all other types below optimum
--- -> no mix of high and mid power LEDs (high power LEDs will have a higher photon flux but therefore generate more heat to be dissipated)

parallel:
- advantages:
--- current splits up equally over all lines and can be adapted to needs, voltage is the same on all lines
--- -> high and mid power LEDs can be run in the same circuit

- disadvantages:
--- need for resistors to even out/adapt current as needed by individual diode type per line
--- -> efficiency loss via resistors

(list will be updated over time)

Characteristics I want to see in my setup:
- UV A and B (will be seperat because of high prices an inefficiency so far known to me, most likely a reptile florescence light, added in late flower)
- blue 430 nm and 450 nm (photosynthetically active, inhibits cell expantion/stretch (very small grow area of 40 cm*20 cm))
- white (high green proportion, photosynthetically active, high leave penetration)
- red 640 nm and 660 nm (highly photosynthetically active, stimulates cell expantion/stretch)
- far red 730 nm (highest leave penetration, induces/shortens flowering, Emerson Effect; in switchable extension of basic circuit or driven seperately)
(- IR 840 nm probably not to not get overly complex, just mentioned for complementation)

some interesting videos on the topics of wavelengths and benefits for growing:

due to parts going out of stock before I was finished planning I had to give up a higher performing-less piece setup I tryed to design:

λ, part number, V_f, I_f, I_opt, I_max
451 nm, 350 mA I_opt
WW, 700 mA I_opt
WW, 350 mA I_opt
640 nm, 350 mA I_opt
660nm, 350 mA I_opt
730nm, 350 mA I_opt
(will be edited, all work went down the drain when I hit some key by accident )

with current knowledge I see, that the 700 mA WW (warm white), which I selected because of their higher PPF (photoactive photon flux) value of 228 lm in higher piece number over the 110 lm 350 mA, would have lead to wiring the 350 mA diodes in parallel of 2 in series with the 700 mA diodes.

bob0816 said:

Can the current be divided asymetrically and what would that mean for loads and "electrical burden" on the loads? It's different from voltage in series, where every load takes just it's needed part, right?

Click to expand...

Yes you can have different currents flowing in different paths across a voltage source. Where voltages divide up across the parts in a series circuit a parallel circuit can have a greater amount of current through a lower value resistor (say 10 ohm) than a higher value (20 ohm). The voltage across them will be the same but the current through the 10 ohm will be twice as much.

So powered by a 5 V source, to put 20 mA on the 1.5 V diode the resistance would have to be 5 V/20 mA=250 Ω?

Click to expand...

No, there is a voltage drop in the diode that needs to be overcome, then the diode starts conducting. The above graph using the Red curve is roughly 1.5V. The 1.5V has to be supplied before the current starts to flow, after that the additional voltage across the diode's internal resistance determines the current.

Taking the slope of the line we can determine the internal resistance of the diode. To make it easy I will take 2V, 40mA and 2.5V, 100 mA. The voltage difference is 0.5V, the current 0.06A. So the Led's resistance is 0.5V / 0.06A = 8.3 ohms. Now if you put this Led across 5V the diode drop will cause the voltage across the internal resistance to be, 5V - 1.5V -= 3.5V. Now we find the current, 3.5V / 8.3 ohms = 0.42 A. While I have the calculator out the power through the diode, 0.42 x 0.42 x 8.3 = 1.4W

What is meant by excess voltage and being used up across the device's internal resistance,
the 3.5 V (5 V-1.5 V) that are used up in the resistor, or voltage that exceeds the 5 V of the device, heating it up? I meant

Click to expand...

Excess voltage (could not think of a better term at the time) is the voltage above the internal diode voltage drop.

Since the current is all the same in one series string, it depends on something I'm not sure about, either the driver output divided by the number of strings, or the strings resistance, or what each load takes (but that I think is least likely, since they have a wide band of current-finctioning, which is why the current determines their performance/brightness). Here I'm with the same puzzlement as with the first quote

For the quart the total voltage is 53.72 V, current would be determined by the CC drivers output, 260 mA by the smaller load, 360 mA by the bigger load,
the total board considered (just one PCB board = four quarts in parallel) is 53.72 V, 1.04 A / 1.4 A if calculated correctly, meaning it would run at 55.87 W (53.72 V*1.04 A) or 75.21 W (53.72 V*1.4 A). The full SSK-1-272-RR-V2 (eight quarts) is given as 135 W, I_max=2.8 A (so simple current calculation succeeded, 2.8 A meaning running the board on Cree optimum and Nichia on roughly 1.4 the optimum.

Click to expand...

I still ask my self, how current and voltage flow in a parallel with different loads, for example, if I have 3 V source and power 4*3 V LED in parallel with resistors before them and they have differences in V (dV) due to normal manufacturing variability: 3.1 V, 3.0 V, 3.05 V, 2.9 V, what consequences would the dV have electrically? Will they just get all 3 V and the connected current and so the 2.9 V would nominally "use up" faster?
And what if they are, hypothetically, all 3 V, but one is 20 mA, two 50 mA and one 65 mA, what would that mean in regards to: flow of current, and then, resistor choice? I think the smaller the needed current, the bigger the resistor because current would flow equally destributed through the circuit and then we just have to reduce it to the wanted lvl...?

Click to expand...

3.1 V, 3.0 V, 3.05 V, 2.9 V, As I showed with the graph, there is a diode drop and a resistance drop in a Led (actually any diode). Just to make things simple let us assume the internal resistance of each of these Leds are the same and it is just the diode drop that is different. And let's say that once the diode drop voltage is supplied it only takes 0.1 V to get 20 mA through the Led. So a quick approximation, say the first diode needs 3.0 V to cover the diode voltage drop. That means 0.1V / 20 mA = 5 ohm. The math for the others is close enough, we will call all of them 5 ohms.

And say we turn up the power supply until it supplies 3.1 V. That means the first diode is passing 20 mA. The second Led has a diode drop of (3.0 - 0.1 V = 2.9 V) so it has 3.1 V - 2.9 V = 0.2 V across its 5 ohm resistance. 0.2 V / 5 ohms = 40 mA.

The third one, 3.1 V - 2.95 V = 0.15 V. 0.15 V / 5 ohms = 30 mA

The fourth, 3.1 V - 2.8 V = 0.3 V. 0.3 V / 5 ohms = 60 mA.

The power supply would be supplying 20 + 40 +30 + 60 = 150 mA.

So it is not good to put parallel Leds across a constant voltage power supply without balancing resistors inline with them. These resistors are much bigger than the internal resistance and it balances out the currents. It does come at a cost of added voltage and power wasted in them. But that is life.

Now what happens if you have a constant current power supply and you have it set for 80 mA? I will not do the math, just to say the 2.9 V and 3.0 V Led will get most of the current and the 3.05 V 3.1V Leds will get a fraction of the current and will barely turn on if at all.

And that would mean, that the current in the 4quarts board can be anywhere from (I_f_string)*(n_strings) up to (I_max_string)*(n_strings)

Click to expand...

They usually balance out but yes.

(n_strings)=total number of strings
(I_f_min)=lowest I_f on all strings (I_f: forward current)
(I_max_string)=I_max of one string
[/QUOTE]

bob0816 said:

In parallel wiring, the voltage is split evenly on all lines.
The whites are 2.84 V, the reds are 1.99 V.
In this circuit diagram

Spoiler: parallel wiring

View attachment 4699485

there are three lines in parallel, 2*(8 series of 8* 2.84 V diodes (8*2.84 V in series=22.72 V) in parallel =22.72 V_f per each one of that two lines) and 1* series of 8*1.99 V= 15.92 V,
so leaving 6.8 V (22.72 V-15.92 V) to be resisted in that line to drive the two 22.72 / to overcome the minimum forward voltage V_f and not put an excess of 6.8 V on the series reds. The amperage given by the driver/source would be divided by three, 1/3 flowing through each line.
(how can I subscript? superscript via ^ (^2 -> ²))
In this circuit diagram

Spoiler: series wiring

View attachment 4699486

there are three lines in series, 2*(4 series of 8* 2.84 V diodes =)22.72 V and one series of of 4* 1.99 V =7.96 V, adding up to 53,4 V
(2*22.72 V+7.96 V), the amperage given by the driver/source would be the same through all diodes.

Here's no place for insults and personal dispute! Stick to facts (@ all)

Click to expand...

Here's no place for insults and personal dispute! Stick to facts (@ all)

Click to expand...

Who has been insulting the other?

And it seems you get the series or parallel thing. Since I did not see any additional parts I assumed the board to be as you have drawn it up with the black strings in series with the red. The operating voltage of the board will tell how the thing is designed. I assume you have a reasonably close diagram. If it were the three circuits in parallel then a regulator or resistor would be needed for the red squares. And rather than wasting the power why not put in a couple more leds to use up the difference in voltage between 22.7 V and 15.9 V?