October 8, 2012 AT 7:02 am

ASK AN EDUCATOR – “How can I interface an LED array to my EL Inverter?”

Jeffery asks:

hello. i have a question. i have a 6v mini el inverter which has a built in switch and is powered by 2 coin cell batteries. This inverter will be powering 2 seperate small squares of el sheet. But I also need it to power 3 rows of 4 LED’s.

How would I integrate the LED’s and connect them to my El Inverter? is it possible?

Ahh, there is nothing better then a little hardware hackery. For this post, I am going to assume that you are an “ebay special” 6v 2xCR2032 EL Inverter.

After inspecting the image, it appears that the circuit uses a standard slide-type switch for applying power to the EL circuitry. Assuming that this circuit switches power at the switch, which would turn the 6V on and off to the circuit, is where you would connect the + side of your LED array. The schematic below illustrates a rough diagram of the potential circuit:

Now, about connecting those LEDs. You have two main options for configuring them, series and parallel. Alternatively you could be fancy and use a switching driver, instead we are going to focus on the basics.

For these solutions, I am going to use LEDs with a Vf (forward voltage) of 1.5V and a Vi (forward current) of 15mA. You will need to know the proper values of your LEDs, these just happen to be at or below most.

In order to supply the appropriate power to your LEDs, you will need to use a series resistor. This “dropping resistor” can be calculated with the following formula:

R = (Vs – Vl) / I

Vs = 6V

Vl = Vf = 1.5V

I = Vi = 0.015A

R = (6V – 1.5V) / 0.015A = 300Ohms

Now that we know the value of the dropping resistor for one LED, we need to determine it for an array. The first two options are configured in series and the second two are in parallel.

Option 1 is the simplest solution as it only requires 4 resistors. Hypothetically, they could be connected to the battery without a resistor as (1.5V * 4 = 6V). This solution can result in an uneven illumination of the LEDs and if one burns out, the rest of the series string goes with it. The benefit here is a low component count and a circuit that only consumes 60mA.

Option 2 breaks the string of 4 LEDs into 2 series strings. By increasing the number of series strings, the chances for one string burning out goes down and you have proper circuit protection with the presence of the 200Ohm resistors. The disadvantage lies with an overall circuit current going up to 120mA, which would result in a shorter battery life.

Option 3 is configured in parallel and every LED has an appropriate dropping resistor. This method ensures an even illumination of the LEDs and if one burns out, the others are not effected. The major disadvantage to this configuration is the current draw, which rings in at a whopping 180mA.

Option 4 is a derivative of option 3, where the parallel LEDs are broken into pairs with 1 dropping resistor on each pair. This method can yield unpredictable results if the forward voltages are not matched. The result can be an over-illumination of the LED that can cause permanent damage or total failure.

Each of these options illustrate a potential method for interfacing 12 LED’s with your inverter. Personally I would chose one of the first two options due to their low component count and lower current draws. You might want to experiment with all 4 just to witness their differences. Before you begin your hacking, make sure that you REMOVE THE BATTERIES and turn the switch on for a few seconds to ensure the caps have discharged through the EL sheet and not YOUR BODY!

I hope this has helped answer your question! Good luck with you circuit and have fun with your LEDs!

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“Ask an Educator” questions are answered by Adam Kemp, a high school teacher who has been teaching courses in Energy Systems, Systems Engineering, Robotics and Prototyping since 2005.


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1 Comment

  1. Neat :D

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