That annoying Christmas LED light string question.

1. Dec 9, 2013

Terranova

Hello, my name is Mike, and I have just enough electrical engineering knowledge to be dangerous. This season I have inherited a programmable LED Christmas light set up from my father-in-law who is no longer with us to help guide me with his EEngeneering knowledge.

Here's the issue.
I have multiple strings of Red, Green, Blue and White LED Christmas lights that I would like to shorten to "custom" lengths in order to fit "precisely" around windows, gables etc.
(here's the little knowledge is dangerous thing)
I know that it is possible to remove x LEDs and put a resistor in line to drop enough voltage to not fry the rest of the LEDs with to much current. I've read several things on several forums, none of which have been thorough enough process wise or "dumbed down" enough for me to understand exactly what I need to do in order to figure out how to calculate the size resistors I will need.

Im having trouble synthesizing everything I've read into knowing what I need to do for my specific situation.
Yes OHMs law, led resistor calculators etc.

I don't know the voltage drop(?) for each color because I don't have a spec sheet for the LEDs.

Here's what I do know (insert snarky comment here)

Each strand is roughly 24' long with approximately 4" between each LED.
120v AC
They flicker, so I assume they're not rectified.
I do not have a true spec sheet for the LEDs, per say, but I do have partial info from manufacturer lit stating that the string is rated at 4.8 Watts (0.04 amps)
There are 70 M5 LEDs per string.
The Greem strings have no resistors in line that I can find.
The R/B/W strings each have three resistors in line. One in the middle, and one at each. end between the plug and first LED, dividing the strings into what I assume (according other info I've read on the net) are two circuits of 35 LEDs.
There are (on the R B W) two wires from the plug to the first resistor. From there the LEDs are wired in series and the two other wires run uninterrupted to the next resistor.
The LEDs are not able to be removed from their sockets

I have a multimeter and Im not afraid to destroy a string or two to find the info I need. I've already destroyed a white string trying to get to the inline resistor. That failed miserably.

How do I go about figuring out how much ( heres where I'm confused) voltage or current (?) each color LED uses so I can multiply that figure to come up with a resistor?

I'm apologize if my tone is a little frantic or frazzled but this has been vexing me for over a week.

Thanks,
Mike T.

Last edited: Dec 9, 2013
2. Dec 9, 2013

the_emi_guy

Why not do it empirically. Measure the voltage across and current through the LED you intend to remove. Ohms law gives the needed resistance, no room for error.

3. Dec 9, 2013

Terranova

Could you walk me through that process?
My father in laws multi meter is more complex than I'm used to.
The Ω quadrant on the dial has settings for
Continuity
4000M
40M
4000K
400k
40k
4k and
400 with the sound waves.

In fumbling around with it, I've plugged in the red LED string, set the dial to 4000K (the first place that I didn't get the OL reading) and put the probes into the wire on either side of the LED where I striped the insulation away. With the red probe closest to the plug I get OL. When I switch the leads I get a reading of -972 to -966. Why negative?

4. Dec 9, 2013

Terranova

When I set the multi meter to 4 (in the V quadrant) I get a reading of .252 to .244 (I understand there's supposed to be some variation there).

5. Dec 9, 2013

Terranova

You just helped me put together that I needed to measure BOTH resistance and voltage. Up till now I was just trying to measure voltage and not under standing the reading the MMeter was giving me. So if V =I*R then R = V/I. So .252/970 = .000259

6. Dec 9, 2013

Terranova

Now I went back and tried to measure the resistance on different settings on the MMeter.
At 40M I got -9.67
At 4000K. -967
At 400K. -101.2. (?!?!)
At 40K. -12.62
That's not just a shift in decimals that I was expecting, is it?
This is why I'm so confused.

7. Dec 9, 2013

the_emi_guy

You can't measure the resistance of the LED. You can measure V and I, then calculate R from ohms law.

Once you cut an LED out of the chain (I think that is what you said you wanted to do), you will have access to wires on both sides of the LED. Re-insert the LED by (twisting the wires together) and measure V across it. Next untwist and disconnect one side of the LED so that you can insert ammeter in series then measure I through the LED.

8. Dec 9, 2013

Terranova

Ok, I think I did what you said.

So with .252 voltage (that seems low compared to everything else I've read) from before and a measurement of 12.25 miliamps? (multimeter setting is 40m/10A). So R = V / I.
So .252/.01225 = 20.57. That doesn't seem right either.

9. Dec 9, 2013

Crazymechanic

if the leds are big enough , why wont you just take one full string which is made for a certain voltage , say 110v mains and has a built in rectifier , so if there is a rectifier inside the string then all you need to know is the mains voltage , say 110v take that x(times) 1.414, you get about 155v DC , because leds run on dc.

Now count how many leds you have , and watch are they all in series or are they say 5 paralell then those in series with the next 5 paralell ones or otherwise.
if you will count the amount of leds you have and the way they are connected you could then easily obtain the necessary voltage for each one.

Once you would know that you could understand how much volts you need if you cuit the string shorter.

But basically I think searching in google for led voltage drop , you could find alot of tables of different color leds and their respective voltages and miliamps.
once you know that you need to check the way your leds are wired up , in chunks like 5which are paralell and then in series to the next five ones or differently and from that you could as I said tell the required voltages.

10. Dec 10, 2013

pantaz

Make sure the power to the circuit is off before measuring resistance.