Faulty Led grow light, bypassing leds.

Thread Starter

Yami

Joined Jan 18, 2016
343
I have got a LED grow light which has got 2 LEDS which has gone open and resulted in one half of the light not turning on.
The whole light consists of 60 x 10W leds. 30 leds are grouped together and powered from one power supply.
First of all I couldn't find any 10W leds locally and it seems that its hard to come by online as well. Only leds I could source locally are 1W to 3W which presumably I couldn't interchange.
There are diodes paralleled with the Leds, I'm not sure what they really are but looks they are 3V Zeners. The diodes which were paralleled with the two dead Leds also were burnt.
I was thinking of bypassing the two empty slots with jumper wires, would this be a good idea? I have seen that done on backlights of LED tvs. I am not too concerned with having two brightness levels. Instead of jumper wires should I use resistors? If so what value should I use. The voltage output of the power supply is 95V (I know these controller would be constant current sources).
Thanks for the help in advance.
P.S: Could you please advise on soldering/desoldering these leds. The whole body seems to be soldered along with the two legs. I am able to desolder the two legs however when trying to remove the body it the whole LED disintegrates leaving the metallic part soldered. I tried with my hot air station and had no luck the plastic cover disintegrates and to take the metallic part I had to tap with a screw driver. (Refer to photo 2)
 

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Delta Prime

Joined Nov 15, 2019
1,014
You said half of the LEDs emit light.
Two rows of 30 LEDs. It could be cross series connected allowing the LEDs to light on each have cycle of the AC power applied. Could you take another photo of the support circuitry? Anything that is connected to these LEDs. Do not apply power! do not turn on the unit!
Do you know how to use a multimeter or DMM?
 

Thread Starter

Yami

Joined Jan 18, 2016
343
Thanks for the replies guys, to test out I bypassed the two dead LEDs and took some measurements. On the 'faulty' side the controller output voltage is lowered to 84V compared to 95V which is on the functioning side. There are two LED controllers in the unit each connected to 30 LEDS which makes 60 LEDs in total. The current draw remains to be same for both sides which is 550mA. I don't think these are 10W LEDs I am guessing it's some kind of marketing BS.
 

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Thread Starter

Yami

Joined Jan 18, 2016
343
Those don't look like any 10 watt LEDs I have seen, and doing the math they are more likely to be 3 watt.

I have used this supplier and I trust them, you might be able to get what you need from them.

relg6796 | eBay Stores
Thanks @ElectricSpidey - I also figured they cannot be 10W hehehe, here is the math I did could you please confirm that if this is right.
Vf measured across LED = 3.1V
Current from source = 560mA
P = VI; 3x560x10^-3 = 1.68W
So there is 1.68W of power dissipation at each LED. So I could use the 3W;3V LEDS available locally as a replacement right?
The total power consumed for the 30 LED string would be =1.68W*30 = 5.04W right?
I am slightly confused with the Watt ratings of a LED. When its says '3W LED' does it mean that its the highest power it could dissipate or does it mean that it'll draw that much current from the supply for it to dissipate 3W?(hope I made sense). What would happen if a LED strip is made up of different Watt rating LEDs but all have got the same Vf?

Appreciate the help.
 

BobTPH

Joined Jun 5, 2013
3,315
If the current stays the same when the two LEDs are bypassed, it is a constant current driver and everything should be fine

However, the fact that two went open indicates it is not a robust device from the start.

Bob
 

ElectricSpidey

Joined Dec 2, 2017
1,655
My calculations are

3.1 x .560 = 1.736 watts
1.736 x 30 = 52.08 watts

And yes, the wattage rating of an LED is its maximum not what it will dissipate.
 
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