Introduction: Convince Yourself to Just Use a 12V-to-AC-line Inverter for LED Light Strings Instead of Rewiring Them for 12V.
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.
