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Still figuring out some fundamentals here, so forgive me if this is a newbie question, if I'm totally missing the mark, or if my math is incorrect (hoping to learn, even if this is all wrong).

I have a constant current power supply (12-20V 350mA) driving a burned out LED chip (6 SMD LEDs), and I have a replacement LED chip (300mA, 6 SMD LEDs.)

Would it be correct, or even more importantly, safe (AKA no fire,) to use a parallel resistor to divide the 350mA from the driver, pulling 50mA to the parallel resistor and leaving 300mA for the replacement LED chip?

I've attached a rough wiring diagram:

  • The yellow box of LEDs is the replacement chip - I don't have a specific voltage for it from a data sheet, so I'm assuming 3.2V/LED for the resistor math.
  • The 8Ω series resistor is intended to bring the voltage of the replacement LED chip circuit up into range of the power supply.
  • The 240Ω parallel resistor is creating the current division.
  • Both resistors are 1 watt resistors.

Obviously, in regards to the safety aspect of the question, the power supply is wired into 110V AC, so I don't want to create any safety issue that ultimately backs up the chain and involves that in any way.

proposed current divider wiring diagram

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    \$\begingroup\$ Does this answer your question? Why exactly can't a single resistor be used for many parallel LEDs? \$\endgroup\$
    – Long Pham
    Commented Jun 14, 2021 at 8:49
  • \$\begingroup\$ A much better and safe option would be to use two LED chips in parallel so that each LED chip gets 350mA /2 = 175 mA. Yes, that's a lot less current but the LEDs will get less warm and last a lot longer. Also efficiency will be better so you will get more light. Adding resistors only wastes power and generates more heat. Your suggested 240 ohm resistor will also waste a lot of power. So lower the current by using two LED chips or use a driver that outputs 300 mA or less. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 14, 2021 at 8:57
  • \$\begingroup\$ Your 240 ohm resistor will do what you want it to but as Long Pham says, you should split your 8 ohm into two 16 ohm ones, one for each string. Best case you’ll be to get a power supply which could run everything in series with no resistors in the circuit. \$\endgroup\$
    – winny
    Commented Jun 14, 2021 at 8:57
  • \$\begingroup\$ @Long Pham Thanks, but not really. My question was more about the current dividing resistor, not the current limiting resistor for the LEDs. That said, your link does provide good additional info. The LEDs in my diagram are not assembled by me - that parallel LED chip comes pre-assembled (everything in yellow in the diagram). I understand the parallel problem described in the link, but now it confuses me how any of these widely available LED chips don't all suffer from this problem... or maybe that's why they're prone to burning out... \$\endgroup\$
    – b. insler
    Commented Jun 14, 2021 at 8:58
  • \$\begingroup\$ @winny thanks for clarifying Long Pham's point. Makes more sense now why the 8Ω resistor was called out. And agreed on getting a 300mA driver. There are actually a number of lighting fixtures to repair, so I was curious if there was a way to not replace the driver in every one. \$\endgroup\$
    – b. insler
    Commented Jun 14, 2021 at 9:04

1 Answer 1

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Would it be correct, or even more importantly, safe (AKA no fire,) to use a parallel resistor to divide the 350mA from the driver, pulling 50mA to the parallel resistor and leaving 300mA for the replacement LED chip?

Both resistors are 1 watt resistors.

The resistor values are correct, but their power ratings are marginal (the 240 Ω resistor will dissipate ~0.6 W and the 8 Ω resistor ~0.72 W). I recommend using resistors rated for at least 2 W, preferably metal film or wirewound rather than carbon (which decreases resistance when it gets hot).

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