I have 30 LEDs wired in parallel. At 3.4V the circuit draws 16mA. What type of resistor do I need to accommodate this voltage (12V - 3.5V) and current load (16 ma) requirement at a regulated 12V DC power supply? It's for my Jeep.

I'm sorry but the math portion of this I just can't compute.

  • 1
    \$\begingroup\$ 0.16mA or 0.16A? \$\endgroup\$ May 3, 2014 at 17:44
  • \$\begingroup\$ In an automotive setting, you might want to use a constant current source instead of a fixed resistor to account for the rather substantial fluctuations of the supply voltage (10 to 14 V, as @Passerby already mentioned). \$\endgroup\$
    – n.st
    May 4, 2014 at 0:16

2 Answers 2


Wiring LEDs in parallel without a current limit resistor for each LED can mean that all the current is taken by the LED that has the lowest forward voltage. This can mean destruction of that LED, then the next one until they are all dead.

Assuming you do the right thing and use current limiting resistors for them all then the resistor has to drop 8.6 volts at 16mA. This is a resistor value of about 537 ohms.

However, you could wire 3 in series to produce a combined LED voltage of 10.2 volts then use a resistor of 112 ohms. It will be a more efficient way of driving the LEDs . Ten groups of three in series sounds a much better way to me.

  • \$\begingroup\$ From OPs question, it indicates that the array is already wired. \$\endgroup\$
    – Passerby
    May 3, 2014 at 18:59
  • \$\begingroup\$ @passerby yeah and he might have to rethink that potentially. \$\endgroup\$
    – Andy aka
    May 3, 2014 at 19:05

According to Ohm's Law:

$$R = \frac{V - V_f}{I}$$ $$R = \frac{~12V - 3.4V}{16mA}$$ $$R = \frac{8.6V}{16mA}$$ $$R = 537Ω$$

So for a 16mA draw, 3.4V Forward Voltage of the LED, and an Average of 12V, you need a ~537Ω (next value up is 560Ω 10%) for each LED in parallel. Otherwise, you are risking blowing your entire array. Using a single resistor for leds in parallel is bad.

$$P = (V - V_f) \times I$$ $$P = (~12V - 3.4V) \times 16mA$$ $$P = 8.6V \times 16mA$$ $$P = 0.137W$$

So a simple 1/4W resistor will work.

Keep in mind, Automotive Power is typically 12V, but ranges from 10V to 14V or higher. The 560Ω resistor is good for that range, keeping current between 14mA and 20mA depending on actual input voltage.

See LED matrix dimming: How to control the current in a 12V LED lamp for video lighting? for some more info

  • \$\begingroup\$ Please reread the OP. He is talking about 30 LEDs wired in parallel. \$\endgroup\$ May 3, 2014 at 19:47
  • \$\begingroup\$ @WhatRoughBeast I knew that. Updated anyway. \$\endgroup\$
    – Passerby
    May 3, 2014 at 20:11
  • \$\begingroup\$ Yes. They are 3 rows of parallel all tied into the same + and - bars on a LED flash light. dorcy.com/p-407-41-4227-3-aa-24-led-worklight-flashlight.aspx \$\endgroup\$
    – Jason
    May 3, 2014 at 20:13
  • \$\begingroup\$ I'm taking out the LED array PCB and installing it into my jeep as the interior door light behind the factory lens. I just want to add a resistor to it so it will take 13v from the supplied line. \$\endgroup\$
    – Jason
    May 3, 2014 at 20:15
  • \$\begingroup\$ I think it's I = 0.02*n = 0.6 (n = number of LEDs. 30) 30X0.6 = 18 V=9.5 R=V/I = 9.5/.02 = 475ohms? Is that formulation correct? I need one 475ohm resistor? I tried a 510ohm and it was not bright at all. Paralleled 2 510ohm resistors and it went a bit brighter. \$\endgroup\$
    – Jason
    May 3, 2014 at 20:24

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