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PostPosted: Thu Mar 14, 2019 10:22 pm 
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First time poster here so would just like to say hey to everybody.

Also first time doing any DIY LED projects and would like to check that i have selected the right cob and driver and that they will work together.

Driver: https://www.digikey.co.uk/product-detail/en/mean-well-usa-inc/ELG-200-C1400/1866-1522-ND/7702977
Voltage - Output 71 ~ 143V
Current - Output (Max) 1.4A

LED cob: https://www.mouser.co.uk/datasheet/2/90/ds-CMA3090-1507477.pdf
Maximum drive current: 3600 mA (48 V), 2400 mA (72 V)
Forward voltage options: 48‑V class, 72‑V class

It seems ok to me as both the voltage and currents match up but i'm worried about blowing up my LEDs before i can even see them light up. Also could anyone with more experience doing LED projects than me let me know how much heat i would expect to be getting from a set up like this and if i would get away with using an old cpu cooler without a fan to cool it?
Thanks like i said at the top this is my first project so sorry if i've made any stupid mistakes.


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PostPosted: Thu Mar 14, 2019 11:07 pm 
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:hello2: Hi, not sure how to put this...........

Why do you say the current and voltage are matched when they are clearly not?
Is it you do not understand current or are you just guessing?

I look at it that if I tell you your error, you will not learn. But if I point you in the right direction you will learn from your error. That is why I have not said what the error is.

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PostPosted: Fri Mar 15, 2019 10:20 pm 
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Hey someone-else thanks for replying to my post. :)
I was under the impression that as long as my driver could out put more voltage than the forward voltage of the LED that it would light. But for the current i've chosen one that is about half of the max current of the LED so that it doesn't blow and this should make it more efficient right?
Maybe using the worded matched wasn't the best choice.


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PostPosted: Sat Mar 16, 2019 12:12 am 
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The LEDs will draw as much current as they require (Unless you limit it) your chosen powersupply can not supply enough current and will not last that long.

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PostPosted: Sat Mar 23, 2019 7:48 pm 
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You are correct the LED is a current device, and current is the important bit not voltage, however in the main to reduce heat we control the output of a power supply by switching it on/off so the average current is correct.

One university project I did was to try to get more output out of the LED by flashing it off/on very quickly, however we found this was not as easy as we expected, the main problem was how to measure the output, we realised to our eye the light seemed brighter but the testing instrument was recording no change, so we could not evaluate our results.

The other was the damage caused to the LED, we found once damaged it did not recover, and the output reduced but did not actually fail and we could not without blowing many more LED's to work out how long it takes to damage the LED, so for a LED able to take 1 mA you can likely supply 2 mA for 1 mS and zero for 1 mS so average of 1 mA without a problem, but 2 mA for 1 second and zero for 1 second would likely blow it.

So your looking at the speed at which the LED chip can get rid of the heat which will other wise destroy it, so you can have two chips with the same spec, one will work with a 50 mS pulse the other would work with a 100 mS pulse and the longer the pulse the less heat is produced within the power supply.

I only trained to level 5, I had to work out heat sink sizes, but not involving time, it is possible of course to use capacitors to absorb and release energy so it is not a simple on/off but the net result is it it too complex to work out. At least by me.

The LED seems to be two different units, 48 and 72 volt and it is likely the 72 volt versions would work, however 2400 mA is required and the supply is 1400 mA so it would seem unlikely.


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