1

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.

Ricardo
  • 6,134
  • 19
  • 52
  • 85
Jason
  • 11
  • 2

2 Answers2

4

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.

Andy aka
  • 434,556
  • 28
  • 351
  • 777
3

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

Community
  • 1
Passerby
  • 72,580
  • 7
  • 90
  • 202
  • Please reread the OP. He is talking about 30 LEDs wired in parallel. – WhatRoughBeast May 03 '14 at 19:47
  • @WhatRoughBeast I knew that. Updated anyway. – Passerby May 03 '14 at 20:11
  • Yes. They are 3 rows of parallel all tied into the same + and - bars on a LED flash light. http://www.dorcy.com/p-407-41-4227-3-aa-24-led-worklight-flashlight.aspx – Jason May 03 '14 at 20:13
  • 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. – Jason May 03 '14 at 20:15
  • 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. – Jason May 03 '14 at 20:24
  • @jason wait, a 3x AA, or 4.5V LED Flashlight, with 24 leds? Typical white leds have a Vf of 3.2~3.4V. Even with 3 Parallel Strings of 8 leds, or 8 Parallel Strings of 3 leds, there is no way for 4.5V to power this. – Passerby May 03 '14 at 20:29
  • @jason it's most likely using a simpler setup of all 24 leds directly tied to the 4.5V, depending on the combined current draw causing the batteries to experience Voltage Droop from their internal ESR, equivalent series resistance. Like cheap dollar store LEDs. It's not a good design, but works for the lifetime of the item (breaks in a year or two). – Passerby May 03 '14 at 20:33
  • @jason also, there is no way they are only drawing 16mA total. Now 160mA total, that's 5mA each, which might be bright enough to be useful. Most 5mm Leds are designed for a typical max usage of 20mA. That said, 8.4V / 160mA = 52Ω resistor. Standard value is 56Ω. It **might** work, but that's what happens when the led array design depends on the intricacies of Diode Forward Voltage Curves and Battery ESR. – Passerby May 03 '14 at 20:37
  • I wired 5 1/4 watt (510 ohm) resistors in parallel to the 30 LED array and it was bright enough. Something doesn't seem correct though. Can it be as simple as just doing what I did? – Jason May 06 '14 at 21:41
  • 510Ω / 5 in parallel = 102Ω. I = V / R. 8.4V / 102Ω = 82mA. 82mA / 30 = ~2.5 mA per led. That seems way too low for a flashlight... @Jason the led array must be wired differently. No way it is 30 directly in parallel. – Passerby May 06 '14 at 21:49