How to wire multiple 2 volt flickering LEDs to work on 12V auto system

I have tried to use the site to find an answer to this, but it seems everyone that has asked questions have different variables than I am dealing with, and I can't ever seem to grasp electrical concepts or formulas. I even tried the automatic calculators, but they show only one resistor, and the articles I find here say one is required for each LED.

I wish to make two flickering LED lit panels to simulate a wood or coal burning type display that will be used in a 12 volt motorcycle system. Each panel will have five individual orange LEDs and five individual yellow LEDs, shining against a small white plastic panel. There will be two panels on one switched circuit.

The LED specs are:

• Size: 5mm
• Animation: Flickering (irregular) like artificial candles
• Source voltage (motorcycle battery): 12V
• LED Forward Voltage (Orange): 1.9V to 2.1V
• LED Forward Voltage (Yellow): 1.8v to 2.2v
• Current per LED (both colors): 20mA

I need to know the best way to wire these, (series or parallel) or and what value of resistor to use, and how many.

• What is the maximum number of LEDs you can afford to lose, i.e. if one in a series chain burns out? Oct 7, 2014 at 20:51
• I don't have time for a full answer right now, but it seems like 5 LEDs in series will give you 10V drop, and the remaining 2V you can use a resistor and Ohms law to get 20mA flowing through it. Sounds like 100 ohms to me! You will only be dissipating 40mW as heat, so any standard 125mW+ rated through-hole resistor will be fine. You just do the chain of 5 LEDs per panel, so the two panels will effectively be in "parallel". You will only need 2 resistors (one for each chain of 5 LEDs) Oct 7, 2014 at 20:53

Five LEDs in series will have a total voltage requirement of about 10V. This is closest to your supply voltage, so you would wire two such series runs in parallel. Each series run should have a current limiting resistor. The following schematic represents one "panel":

simulate this circuit – Schematic created using CircuitLab

Assuming all LEDs have a $V_f$ of 2V, the current-limiting resistor would drop the remaining 2V. Since you want 20 mA, the value of the resistor is easy to calculate using Ohm's law (R = E / I):

$$\frac{2}{0.02} = 100\,\Omega$$

You should determine what actual voltage drop your LEDs have, but this gets you close.

There are other things you may want to do, such as add some overvoltage protection. If you're running this from a motorcycle battery that's not also connected to, say, a motorcycle, you shouldn't have any problems. However the alternator on a motorcycle (if it's like a car) may bring the voltage up to 13.8~14.4 volts.

• Thanks for taking the time to implement my comment haha. Nice work and diagram example too Oct 7, 2014 at 21:31
• @KyranF I think we arrived at the same conclusion at the same time. I could say thanks for implementing this as a comment, as well! :) Oct 7, 2014 at 21:50