Hello

I did not find any posts about this one (maybe I could be wrong)/

I have a power supply 30V @ 5A which can be current or voltage regulated from Meanwell, and I have 2 led modules of 100W each, each module needs 2.7A @ 31v, obviously if I connect them in parallel I could only obtain 2.5A for each which is ok, my question is how I can protect them in case one of them blows (rare case) because let's say one of them blows or die the other one will receive 2.5A but current tends to increase and the led only can support until 3A absolute maximum and then it will die eventually (in case is not noticed before this occurs).

So how I can protect each other from overcurrent?, I was thinking on resistors using ohm's law but probably the resistors will turn too hot due the power handling. Capacitor?
Any advice on this? Thank you beforehand.

Greetings

 

You could add an inline fuse for each LED module.

Something like this:
http://www.safco1.com/Images2007/Large-In-Line-Fuseholder.jpg

Check the fuse datasheet for current derating with temperature and check out the blowing times. Some fuses I used recently will hold 110% of the rated current for many hours.

 

The led modules are purely leds, or do they have additional circuitry? Do you have the datasheet?

 

Have you read this?

LEDs, 555s, Flashers, and Light Chasers

My suggestion is two separate circuits, if one goes out the other will not notice. They can share power supplies parallel to each other.

 

Thank you all for your kind replies.

Ghar: even though is the cheapest and easy way to protect the leds I wouldn't want to stay in darkness if one of them blows (if that is the case I would use ultrafast blow fuses) thank you though.

Norfindel: these leds comes with a copper plate and the serie-parallel arrangement inside of a small area (1x1"), only leds no other circuitry.

Bill_Marsden: I apologize for not understanding which circuitry do you refer, the ones you already wrote? or any other circuitry in particular?

I have read in other sites about capacitors for protection like this one:
http://www.ledsmagazine.com/features/3/5/2

even though is used for AC voltages.
I know that a capacitor would act as a resistor in an AC supply due its characteristics but what about in a DC? would work in my power supply that has PWM?

As I said before I would use resistors but this ones will turn too hot to act as a current protection.

My power supply is PWM and is constant current and constant voltage, (150W)30V @ 5A, if I use 2 led modules in parallel if one of them blows it will lower the current for a few minutes but it will increase until it blows or damage the led itself.
Any ideas?

Thank you again for your replies.

 

The link has everything you ask about. How to bias LEDs, PWM, and a bit more. You might read it a bit.

Thing about electronics, it always involves reading if you want to learn.

 

Seems to me like you need to get the datasheet.
Strange that it's a series/paralell combination, i wonder how they do it with the different Vf.
Anyways, leds are better driven with a constant current. Find what current the module requires, set your supply to that current, and it will light as it's supposed to do.
If you connect both modules in paralell directly, the one with lower Vf will conduct much more current, and there's going to be trouble. If you know what's the higher and lower limit on the Vf, you could calculate what series resistor you need to improve current sharing. If it's small enough, there won't be huge losses.
I also noticed your modules need 31v, but your supply can deliver up to 30v. You're at least 1v short.

 

Here is the link of the led modules and the power supply as well, I hope you can understand better than me explaining this.

http://www.edison-opto.com.tw/Datasheet/EdiStar/EdiStar%20Series_Eng_v9.pdf

http://www.meanwell.com/search/clg-150/clg-150-spec.pdf

This last one power supply by meanwell can adjust the voltage from 26 to 32 volts at 5A, I'm telling you this because it seems that theorically can distribute 2.5A at each led module (this to prevent damage) but is not and I got a proof of one module that it blows and I was sure that the heatsink is very big enough (I have readings of 36°C on the substrate which is 55° maximum) to handle the heat and that it has a very good thermic compound.

Anyways any help here would be appreciated.

Thank you everyone.

 

From the datasheet, it seems like the 100W modules have an If of 3A. You would need to adjust the power supply for 3A constant current, then connect one module to it.

The Vf can range from 30v to 36v. That looks like too much for putting resistances for current sharing, as they would have a heavy power dissipation. The modules just aren't supposed to be used like that, you need to supply them with a constant current source, and anything but a switching supply would waste too much power. It's a 100W device, after all.

It seems to need a quite big heatsink. That's important.

Do you have the CLG-150-36-A supply?

 

Show us the LEDs you are actually planing on using please?

 

不建议这种接法,首先2路并联均流不好做,其次输出电流不足,影响发光效果。最好用双输出电路。现在市场上已经有此类电源

Translates to: (Via GOOGLE)

This access method is not recommended, first of all 2-way flow of good to do both in parallel, followed by the output current, which can affect glow. Is best two-output circuit. Now the market has such power!

And, I personally, am glad the market has such power!

 

Norfindel:Yes, I do have the power supply and is constant current adjustable, ideally if I adjusted to the full amount of current (5A) then I would have "theorically" 2.5A in each module, but in the field this won't happen , as I said before a couple resistors for this would be too much heat to dissipate in order to protect them.

Bill_Marsden: the link is included above in my last reply, is the first one of the links. These modules are the ones I'm using.

Well, I appreciate all your help for this guys I will probably use the fuses that Ghar suggested before even though if one of them blows i could have a few minutes to grab a lantern or flashlight before the other blows, according to the datasheet this should not happen but in real life did happen. I was make sure that the modules received less than 2.5A each and they did but for some reason current increases and then blows, also I was make sure that the modules has enough silicon thermo-compund and that they had a nice bound with the heatsink. Well maybe I need to decrease a little bit more the current output and then try again.

I will do that and then I will tell you how did go.

Meantime any other ideas would be appreciated.
Thank you.

 

Hi retched:

Did you mean 2 power supplies, one for each led?
I want to save some money and use 2 L.E.D. with one power supply.

What power supply do you recommend?
Also could you explain what two output circuit you refer?

Thank you for your reply.

 

If you connect those modules in parallel, they will blow, as they won't have the same Vf, so the current sharing won't be 2.5A in each module, but almost 5A in one module. When that module blows, then the other will carry 5A and will also burn. You need a constant current source per led module.

 

Actually two low ohmage resistors (high wattage) might work, as they would tend to balance the currents. The power supply voltage to do this would be unreasonable though, over 30V or so.

 

Hello guys, I'm pretty sure that I left a reply 1 day ago but it does not appear here as did with the other replies, I don't know what happend there.

Anyways, yes I'm agree to put one power supply for each led, that would be the best solution. You are right Bill Marsden I test the current on each led with 2 amp-meters and one of them reads 2.2A and the other reads 2.95A, I was playing with a resistor (0.2ohms 5W) in the one that reads 2.95A and when I turn the power supply the reads on the ampmeters were 2.51A and the other reads 2.49A, it seems that compensates one with the resistor and the temperature on the resistor barely pass the 50°C which attached to the heatsink on the coolest side with a thermo compound would do the trick, don't you think?
I think I will put a fuse as an extra protection just in case.

Well, I think is not the best option but is the cheapest. I will let this to run for a while and see what happens, I will let you know what's going on.
In mean time should I close this thread or live it open? or what can I do?

Please let me know

Thank you all for your comments and support, I'm grateful with your help.

 

Leave it open. If you have further information, just make a new post at the end of the thread.

Consider using a thermal fuse or breaker that is physically attached to the heat sink. That way if it exceeds the fuse/breaker's temperature, it will turn off the LED before it burns up, and allow it to cool. Another alternative might be to use a fan to cool the LED's heat sink.