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I bought this Led Driver(HLG-480H-48A) for my Strips that use around 43V 9A. I've never used a LED driver before and have got a few questions about them:

  1. The output current adjustment range (according to datasheet) is from 5-10A. Does that mean that the driver "pushes" 5A minimum through the circuit?

  2. If I connect a LED strip that uses same voltage but only 3A what will happen?

  3. I want to connect 3 strips(total I=9A), all in parallel to the same driver. What will happen if one of them has a short-circuit/fault? Will the other ones get damaged as well or should I independently fuse them with a fast acting 3A fuse?

Thank you for your time.

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1) The driver "pushes" as much current through the load (the LEDs) as what you have set, if you set it to 5 A that current will be 5 A.

However the driver only supports a voltage of around 48 V (actually that's adjustable between 40.8 V and 50.4 V). So with the LEDs connected and the current set to 5 A, the voltage across the LEDs must be less than 48 V.

Suppose you connected LEDs that need 60 V when 5 A is flowing through them. This driver cannot support that, you would get the maximum voltage of 48 V and the current would have a very low value (how much depends on the LEDs).

2) The LEDs are rated for 3 A then you must use a driver that outputs 3 A or less. This driver outputs 5 A, which is much more than 3 A. The LEDs will be overdriven, get too hot and will die very soon. Just don't do this. Underdriving (using a lower current than the LEDs need) is always OK, overdriving is not.

3) Connecting 3 x 3 A LED strips for 9 A and using a 9 A driver will work but is risky.

When one LED fails open: the other strips get 9 A / 2 = 4.5 A and are overdriven. They will soon fail as well.

When one LED fails short: then the other LEDs in that string will take most of the current and will be overdriven. This will make that strip fail sooner. When one of those LEDs then fails open, see above what happens.

It is really a much better idea to use 3 separate 3 A drivers.

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