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I am using 3x10 COB LED along with CC driver that is 960 mA, 30-35 V and the connection type is series-parallel. Now the issue is if one of my string fails so will it add up to the remaining LED or no?

I made some calculations too:

  • 1 LED = 30/10 = 3 V
  • 1 LED current rating = 960 mA/10 = 96 mA

So single LED is taking 3 V and 96 mA.

For 1 string:

3x10=30 V, 96x10=960 mA

This is for 1 string, I am extremely confused if the CC driver rating is only for one string or what please help I don't want it to override other strings.

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  • \$\begingroup\$ Please DON'T SHOUT in the question title. All-caps removed. \$\endgroup\$
    – Transistor
    Mar 12, 2022 at 8:46
  • \$\begingroup\$ I read your question twice but I still don’t understand what the issue is. Please draw a schematic of what you have and how it’s connected, along with datasheet links to the parts used. \$\endgroup\$
    – winny
    Mar 12, 2022 at 9:33
  • \$\begingroup\$ The driver rating has nothing to do with the LEDs you connect. "1 LED current rating = 960 mA/10 = 96 mA" - it doesn't make sense to calculate the LED rating from the driver specs, the LED rating is in the LED datasheet. Which is your driver, which is your LED and how are the LEDs connected? (You divide both voltage and current by 10, which also is not correct as you wrote you have 3x10 COBs. \$\endgroup\$
    – Sim Son
    Mar 12, 2022 at 14:12

1 Answer 1

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I am using 3x10 COB LED along with CC driver that is 960 mA, 30-35 V and the connection type is series-parallel.

Assuming you mean that you have 3 parallel strings of 10 LEDs driven by a 960mA driver, then each LED has a forward voltage of 3.0-3.6v and you are driving them at a current of 320mA per string.

Now the issue is if one of my string fails so will it add up to the remaining LED or no?

If one string fails open circuit then the remaining two would get 480mA each.

3x10=30 V, 96x10=960 mA

When elements are in series they all get the same current. It does not divide between them.

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  • \$\begingroup\$ user1850479 thanks a lot for clearing it out for me. so, if remaining strings gets 480mA it would be overdriven and as a result it will burn out soon therefore, in order to overcome i'll be using constant current IC in each string and will be powering it up using constant voltage driver. Does that makes sense to you??? \$\endgroup\$
    – electro_noooobb
    Mar 14, 2022 at 4:31
  • \$\begingroup\$ or i can use LM3466 IC to balance out current in each string even if one of the string fails. \$\endgroup\$
    – electro_noooobb
    Mar 14, 2022 at 11:29
  • \$\begingroup\$ @marwahAbdulGhafoor Using one IC per string is reasonable, especially if you're more worried about reliability. I'm not sure the LM3466 will do that though. A more conventional linear LED driver IC would be a better choice. \$\endgroup\$
    – user1850479
    Mar 14, 2022 at 14:49
  • \$\begingroup\$ Thanks alot for your help and guidance. yeah i checked LM3466 would only balance out current so i am going to be using constant current IC in each string. thanks again :) \$\endgroup\$
    – electro_noooobb
    Mar 15, 2022 at 4:21

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