Samsung 561C bin S6 voltage help plz.

[Budies 101]

I'm looking to mess around and build my own light.

Ok, so I am admittedly not the best at figuring this stuff out, but I have looked for the information myself. I can't seem to find out the volts per Samsung 561c s6 chips are, and maybe it's because I don't understand.

Basically I'm looking to do 520 chips on a PCB, and I need to find out what driver would do the best job of keeping a middle of the road efficiency attainable or even driving them hard and dimming it down simply for flexibility.

I was thinking of the HLG240H-C1400A. I would think I could dim it around 200watts and be efficient or drive them harder if for some reason I needed the extra watts.

If someone knows, that would help =) Yes I just jump into things that i don't understand, but that's what makes it fun for me =X

Thanks!

 

[CobKits]

you need to decide on the efficiency you want, then the voltage you want and arrange your 520 chips to get whatever you need

samsung has a simulator

 

[Budies 101]

you need to decide on the efficiency you want, then the voltage you want and arrange your 520 chips to get whatever you need

samsung has a simulator

I thought dimming made for better efficiency, was that wrong? Basically use a higher ma and dim it down for better efficiency or turn it all the way up.

I cant find the simulator if you have a link. =)

 

[Budies 101]

Ok, i think I found it, thanks Cobkits

 

[CobKits]

I thought dimming made for better efficiency, was that wrong?

not dimming specifically, but whatever current you run chips at

in other words you can get a driver to run an LED array at 350 mA or you can take a 700 mA driver and dim it to 350 mA and they are equally efficient on the chip side. on the driver side the 350 mA driver is prob more efficient (and def more cost effective) than a driver dimmed to 50% load

 

[Budies 101]

not dimming specifically, but whatever current you run chips at

in other words you can get a driver to run an LED array at 350 mA or you can take a 700 mA driver and dim it to 350 mA and they are equally efficient on the chip side. on the driver side the 350 mA driver is prob more efficient (and def more cost effective) than a driver dimmed to 50% load

Ok, gotcha.

makes sense thanks.

On the calculator for samsung, the IF_Drive is the ma and the VF is voltage? When all is imputed at .14 IF_Drive I'm at 2.7 Volts, per chip? Would I need a driver that can do like 1404 volts? And is the IF_Drive at .14 equal to a 1.4 mA driver?

Confused here if you can help.

 

[CobKits]

not really a single diode expert sorry. its been discussed in the early pages of the quantum thread tho.

but as a start youd put X number of 2.7V diodes in a series string to get whatever voltage you want. that string wants 140 mA so you make 10 strings in parallel and slap it on a 1400 mA driver

 

[Budies 101]

This is what I see.

 

[Randomblame]

They must be installed in series and parallel connections.
For instance,
52 in series (2,8v each) needs ~145v and to get 210w total you need 10 of these strings in parallel.
In this case you need a driver which provides ~ 150vdc and 1.4A. (HLG-240H-C1400(B)
So each string (and LED) gets ~ 140mA!

52x 2,85v = 148,2v x 1,4A(10 strings parallel/140mA per string) = 207,48w, 520 LED's(52x 10),

BTW,
T.S means solder point temperature and should be around 60-70°C at 140mA.
For instance, the H-Series strips have 50°C on pcb level at 80mA.
That why I calculate with only 2,8v...

 

[Budies 101]

They must be installed in series and parallel connections.
For instance,
52 in series (2,8v each) needs ~145v and to get 210w total you need 10 of these strings in parallel.
In this case you need a driver which provides ~ 150vdc and 1.4A. (HLG-240H-C1400(B)
So each string (and LED) gets ~ 140mA!

52x 2,85v = 148,2v x 1,4A(10 strings parallel/140mA per string) = 207,48w, 520 LED's(52x 10),

BTW,
T.S means solder point temperature and should be around 60-70°C at 140mA.
For instance, the H-Series strips have 50°C on pcb level at 80mA.
That why I calculate with only 2,8v...

Great! thx for the help.

I'll build a tester board and see how it goes!

 

[Organic Miner]

Wow that's a lot of soldering.

 

[Budies 101]

Wow that's a lot of soldering.

Haha, no I will have it done on a PCB. =)

 

[bullisok]

Ok, gotcha.

makes sense thanks.

On the calculator for samsung, the IF_Drive is the ma and the VF is voltage? When all is imputed at .14 IF_Drive I'm at 2.7 Volts, per chip? Would I need a driver that can do like 1404 volts? And is the IF_Drive at .14 equal to a 1.4 mA driver?

Confused here if you can help.

When you run things in series Volts add and when running in parallel Amp add. So say each diode is 3.6v max and 350ma max but you want yo run it at 700ma and you want to have 520 diodes. Than means you are going to run 260 diodes and run them in series and do that twice then take those two sets of series and connect them in parallel which will give you a PCB thats rated 936volt and 700ma max with 655w

 

[VegasWinner]

You can use 2.9v for vf. The calculator indicates 2.87v. These numbers are estimated by Samsung's caveat on the page. Good to use 2.9vf as a design voltage.
You will need to establish a design strategy of a series parallel balanced circuit for the driver you plan on using matching vf with a series of diodes in a string and match amps with the rows to establish a desired drive current for the diodes and efficiency desired.
The parallel portion of the equation will establish drive current by using the relationship of rows/driver current = 160mA to 170mA to get the benefit of using the LM561C diode.