My plan was simple. I wanted to cut up a wall-powered LED light string into pieces then rewire it to run off 12 volts. The alternative was to use a power inverter, but we all know they are terribly inefficient, right? *Right?* Or are they?

## Step 1: Figure Out the Voltages of Each LED Color

I was all set so I set to figuring out how to split up the string. I ran a 9V battery through a 470 ohm resistor to clip leads (limiting current to no more than 20mA or so). I clipped a volt meter between the 9V negative and the resistor. Without anything inline, it naturally read 9 volts. Then I popped out one of the LED's and put it in parallel to the voltmeter. I flipped it around so it would light up, and then read the meter. The first one was blue and it read 3.0 volts -- that's the voltage drop of the LED. The others are as follows:

Blue: 3.0V

Green: 3.2V

Orange: 2.0V

Red: 5.2V *

Yellow: 2.0V

Blue: 3.0V

Green: 3.2V

Orange: 2.0V

Red: 5.2V *

Yellow: 2.0V

- Note that the red surprised me at 5 volts ... I was expecting more like 2 volts.

## Step 2: Figure Out How to Split Up the String.

The string I have is 60 LED's long. I wanted to minimize the amount of time I spent on the project so I figured I would just take them in order and add a current-limiting resistor to each mini-string that would drop the 12-volt input to whatever is needed by the LED's. The original string had a sequence that went green, blue, red, orange, yellow.

And from the last step, the voltages for each LED were:

Blue: 3.0V

Green: 3.2V

Orange: 2.0V

Red: 5.2V

Yellow: 2.0V

So now we start at green (3.2V) and add orange (2.0V for 5.2V total) then red (5.2V for 11.4V) and that's it because adding yellow (2.0V) pushes the total to 13.4V which is more than the 12V input voltage. Here's a chart of what happens:

This works out quite well because now the sequence is once again back to green where we started! Now it's a matter of figuring out the resistors. For instance, in the first string, there's 0.6 more volts to reach 12V so that's what the resistor will have to drop. Using Ohm's law, that's 0.6V / 30mA = 0.6V / 0.03A = 20 ohms. The rest of the resistors are as follows:

So there's 60 LED's total and the three sequences contain a total of 10 LED's each so that's 6 sets of sequences. Or 18 sequences -- each that need to be soldered up.

Ugh ... am I even on the right track?

And from the last step, the voltages for each LED were:

Blue: 3.0V

Green: 3.2V

Orange: 2.0V

Red: 5.2V

Yellow: 2.0V

So now we start at green (3.2V) and add orange (2.0V for 5.2V total) then red (5.2V for 11.4V) and that's it because adding yellow (2.0V) pushes the total to 13.4V which is more than the 12V input voltage. Here's a chart of what happens:

Color Voltage Total Green 3.2 3.2 Blue 3 6.2 Red 5.2 11.4 Orange 2 2 Yellow 2 4 Green 3.2 7.2 Blue 3 10.2 Red 5.2 5.2 Orange 2 7.2 Yellow 2 9.2

This works out quite well because now the sequence is once again back to green where we started! Now it's a matter of figuring out the resistors. For instance, in the first string, there's 0.6 more volts to reach 12V so that's what the resistor will have to drop. Using Ohm's law, that's 0.6V / 30mA = 0.6V / 0.03A = 20 ohms. The rest of the resistors are as follows:

Sequence Voltage For 12V Resistor G-B-R 11.4V 0.6V 20 ohms O-Y-G-B 10.2V 1.8V 60 ohms R-O-Y 9.2V 2.8V 93 ohms

So there's 60 LED's total and the three sequences contain a total of 10 LED's each so that's 6 sets of sequences. Or 18 sequences -- each that need to be soldered up.

Ugh ... am I even on the right track?

## Step 3: Is It Really Worth It?

I also happen to have a 12V inverter to convert to line-current. Will that really waste the battery more than this?

Remember the sequences?:

Consider this spin: each of the 18 sequences of LED's will use 30mA of current for a total of 540mA or 0.54 amps. Note also that in the first sequence, 11.4V goes to light and 0.6V to waste heat out the resistor. Again at 30mA, that's 0.342 watts and 0.018 watts, respectively. If you do the math for the whole string, it's 5.54 watts of light and 0.936 watts of heat for an efficiency of 5.54 / (5.54+0.936) = 86%. That's in the ballpark of a cheap inverter.

So I connected up the inverter and found it drew 0.380mA at 12.34 volts which is 4.69 watts. Now the string is actually rated at 0.046 amps at 120 volts or 5.52 watts, wired without any large limiting resistors as best I could see (and it's very close to 30mA I calculated above). Anyway, this makes the actual efficiency of the inverter ( 4.69 watts / 5.52 watts ) = 85%.

I guess I could gain 1 whole percentage point of efficiency by going with wiring it by hand. In the end, though, it's probably not worth it.

Remember the sequences?:

Sequence Voltage For 12V Resistor G-B-R 11.4V 0.6V 20 ohms O-Y-G-B 10.2V 1.8V 60 ohms R-O-Y 9.2V 2.8V 93 ohms

Consider this spin: each of the 18 sequences of LED's will use 30mA of current for a total of 540mA or 0.54 amps. Note also that in the first sequence, 11.4V goes to light and 0.6V to waste heat out the resistor. Again at 30mA, that's 0.342 watts and 0.018 watts, respectively. If you do the math for the whole string, it's 5.54 watts of light and 0.936 watts of heat for an efficiency of 5.54 / (5.54+0.936) = 86%. That's in the ballpark of a cheap inverter.

So I connected up the inverter and found it drew 0.380mA at 12.34 volts which is 4.69 watts. Now the string is actually rated at 0.046 amps at 120 volts or 5.52 watts, wired without any large limiting resistors as best I could see (and it's very close to 30mA I calculated above). Anyway, this makes the actual efficiency of the inverter ( 4.69 watts / 5.52 watts ) = 85%.

I guess I could gain 1 whole percentage point of efficiency by going with wiring it by hand. In the end, though, it's probably not worth it.

## 9 Discussions

4 years ago on Step 3

Great Article. One thing where this would be worth it, is if your going to use a dimmer pack to adjust the intensity. It's cheaper to find a 12v dimmer pack, than it is a 120v dimmer pack, especially if your looking to dim multiple channels. And I'm not sure how the inverter would handle a dimmer on the 12v side.

7 years ago on Introduction

Very informative and easy to understand, thanks for this. It demonstrates the old adage, we learn more from our failures than our mistakes!

I've often wondered if it would be possible to have one DC power source, say a solar array with battery backups, and run all household lighting (using LEDs) from that.

11 years ago on Introduction

do you have any idea what i would need to run 40 led lights from AC power

Reply 11 years ago on Introduction

Not off hand. It also depends on how you want to hook them up. If they are all the same voltage, I would divide them into lengths for a standard transformer from the wall (i.e. if 8 LED's make 12 volts, then get a 12V DC transformer), and make strings of that length with a low-value current limit resistor (like 10 ohms in case it's a rock-steady 12.1V that would blow the LED's). Wire all the strings in parallel, hook them to the transformer and presto.

If the LED's are different voltages, divide them up into strings that total to the same voltage and do the same thing. Like if you have blue, green, red and the voltages are 3.1, 3.0, and 2.0 volts, then wire them up blue-green-red-blue-green-red and get a (3.1+3+2)*2=16.2 V transformer. Something like that anyway.

Reply 11 years ago on Introduction

thankyou this is my first shot at making circutry and i am not very keen on circutry in general however i know i need to learn to make the artwork i want to create. so I am looking at wireing 10 decently bright white led lights together. what would help me out is to know what type of led lights to buy and also what resistors and tranformer to get. i think that this basic of a project will help me to understand circutry. i have also already ordered a circut that someone else has already put together i guess what i will do is take it apart and put it back together in my own form. i am also playing with some led throwies

Reply 11 years ago on Introduction

I'm sure there are Instructables on wiring up LED's somewhere, but it's a bit too much to try and explain here (and -- no offense -- more effort than I want to put in to your project). Start experimenting a bit with batteries to get a feel for how to wire up LED's then once you get the hang of that, the transformer solution might be easier to figure out.

Reply 11 years ago on Introduction

thank you for your help I think i will be able to understand your first comment soon It seems to be all about adding up the leds until the curent drop equals the amount given in the transformer if 8 LED's make 12 volts, then get a 12V DC transformer in this statement would i need any resistors

Reply 11 years ago on Introduction

do you know of some Instructables that would be good starting points also I misspelled circuitry

11 years ago on Introduction

good instructable, i think you are worrying a little too much about being exact. Your 60 lights are assuming about 2v per led with a 120v AC voltage source. I would hook up the LEDs without the inline stand resisters first and then add a current limiting resister if you really need it at the source.