Which LED driver can I use? Samsung SL-B8R5C9H2AWW


Currently, I am using 6 SL-B8R5C9H2AWW with an HLG-320H-42B. It works perfectly, but I want to add 6 to 12 more in 1, 2 or 3 different circuits.

I would like to know what alternative to the HLG-320H-42B is available to me.

The dimmer function is not a priority and I install everything in a protective box.

I had thought of LRS-350-48 to connect 6.

Do you have a good alternative to suggest to me?



When looking for an LED driver, the most common is to find the exact constant current, and the voltage will adjust depending on the number of LEDs of the load. Every one I found for 1.2A, will only power one strip of the SL-B8R5C9H2AWW-ND. The best option I found is 1866-2374-ND, which will be a constant 3.6A (3 strips in series), and the voltage will adjust to 43.1V. This should work, but you would need an LED driver for every set of 3 LED strips.

Firstly, thanks for the quick response.

I would like to add UVA LEDs but there is no strip. I was thinking of going for VAOL-5EUV8T4. If I understand your explanation correctly, I need to have a DC current of 3v and I add the number of amps (30mA) by the number of LEDs?

With the power supply LDA10F-3
I can power 66 LEDs (VAOL-5EUV8T4)

What would be your power supply (Supply Modules/Wall connect) to power 100 LEDs from 120V?

Last thing, do you have power supply (Supply Modules) for my Samsung L2 strip?

Thanks a lot for the help!

HI KKirouac,

There are a couple ways you can do this.

LEDs are generally constant current driven devices, you need to regulate current somehow.

One way to do this is to run them in parallel with a LED Series Current Limiting Resistor. If you’re running them in parallel you need a resistor for every LED in series and then wire up all the LEDs in Parallel.

You’d want to give an LED like this something like 5V to leave some amount of voltage to be able to drop across your resistor.

For the formula to calculate the resistor value is

( Source Voltage - LED Forward Voltage) / Current
For a 5V source it would be

( 5V - 3V ) / 0.03 (for 30milliAmps)
2 / 0.03 = 66.66 Ohms

We have an LED Series Resistor Calculator for the math on this available here.

Otherwise you can run LEDs in series, ten 3V LEDs in series would be 30V then you’d need one resistor for all of them. So then your LED resistor formula would be something like

35V source voltage - 30V (10 LEDs in Series ) = 5V

5V / 0.03A = 166.66 Ohm resistor.

Hi KKirouac,

Another option besides using resistors is to use an in-line current regulator, such as one of these. You simply place them in series with a string of LEDs and supply a bit more voltage (a minimum of roughly 3V - 5V, depending on the particular part) than the LEDs require, and they regulate the current through themselves and the LEDs.


These cost a bit more than resistors, but they allow for more consistent current regulation over varying voltages and temperature ranges, for which resistors cannot compensate. They come in both surface mount and through-hole versions.