LED recessed Soffit Light Project using $4 flashlights. - Page 1
Forum section: electronics / projects

author: Mrkirby
author_rank: Newbie
timestamp: April 25, 2013, 12:40:04 am
content: I have more respect for people youtubing videos, because wow I suck at it.My video explanation is here:http://youtu.be/bTB9xMzG7rcFirstly, my intentions were to send this question to Ben Heck from his podcast but I think he is busy like Dave here. That is why I say hi Ben in the beginning.Project Details:I have 12 LED flashlights with 22 LEDs in each.I am using a 12 volt transformer with an unknown amount of amps to power the system in parallel.Total feet like not even 150ft.I am wondering if the resisters included in the flash lights will protect the LEDs from being burnt up using 12volts and the correct amps..(which I don't know the answer too).Secondary question, could I easily add additional lights? What do I need to keep track of to add to the system.Anyway, any help would be great! Thanks

author: Hideki
author_rank: Frequent Contributor
timestamp: April 25, 2013, 02:47:41 am
content: One battery is 1.5 volts. The lights are designed to take 3 of them, so 3 x 1.5 = 4.5 volts total.If you try to run them at 12 volts they will probably light up really bright and die rather quickly. Ideally what you want is a ~4.5 volt source, but it may survive on 5 volts as well.One solution would be to run 3 lights in series. 4.5 * 3 = 13.5 volts, so if you give them 12 they should work fine but perhaps be a bit less bright.You don't show it in any detail, but that's a badly designed LED light if all it has is a single resistor with all the LEDs in connected parallel.The current is unknown. The resistor value is unknown. I guess you don't have a multimeter to measure any of those things?Some rough calculations: If each LED gets 20 mA then 22 of them would be 440 mA. A white led at that current drops around 3.5V, so the remaining 1V is across the resistor. 1V / 0.44A ~= 2.27 ohms. I can't really see the colors in the video, but 2.2 ohms seems unlikely since it would be Red Red Black.Also, a transformer outputs AC, which is not what you want in this case. Check that it is in fact a DC power supply.

author: Mrkirby
author_rank: Newbie
timestamp: April 25, 2013, 03:42:32 am
content: Thank you for the reply!I think that is why I am putting this project in Parallel because I want to use more than just three led flashlights.So working with a parallel system. 440amps. In 3 flashlights....Do I then multiple by 12 flashlights...with 440 amps? (264 LEDs )12 * 440 = 5.2 amps?I tried to understand some other youtube videos and I thought I remember seeing resisters resisting more amps in parallel or doubling it..... Ya... I am not sure.However I will show a new video to showcase the circuit better. If you need. (Let me know)If you tell me what to do to find the current with my manual ranging multimeter I will find a number for you. I have a sperry sp-152aResister colours are: red green (brown/gold). Then silver.Again those colours in that order give me more numbers, I got 25000k. Which doesn't make sense to me.Let me know what you need me to do translate this electrical conundrum.Thanks again!

author: Corporate666
author_rank: Unknown
timestamp: April 25, 2013, 07:35:15 am
content: Some thoughts...1) In the video, you removed 2 LED's. Did you test the light after you did that? If you just pulled the LED's, you may have left an open circuit and the light will no longer function. If you just connected the two leads that went to the LED, you will likely be overdriving some or all of the remaining LED's and cause premature failure.2) An AA/AAA battery will be about 1.5 volts when charged, and drop to more like 1.2 as it discharges. So that light will run on about 3.5 to 4.5 volts. If you put 12 volts into it, you will almost certainly destroy it very quickly considering it just uses a resistor for current limiting. Based on V=IR, the more volts you put in with the value of R being the same, the higher the value of I goes.3) The suggestion about to drive them in strings of three - that does not limit you to only three lights. You can have as many strings as your power supply will support, but each string will be 3 in series. That would give a forward voltage of 4 volts to each light, which would have them running as if they were on not-full-but-not-empty batteries.4) As said above, make SURE your power supply is 12VDC. Many 12V transformers are not rectified, so they take mains voltage and output 12 volts AC, which won't work... well, there's a bit more to it than that, but suffice it to say you want to run these on 12 volts DC.To wire 3 lights in series... if you look at your light, despite having 3 batteries, there will just be 2 wires or connections between that battery pack and the LED board... positive and negative. So you would connect the positive from one LED module to the negative of another. Then connect the positive of that second LED module to the negative of a third. Now you have 3 in a string (in series) with one leftover positive and one negative connection. It will require 10.5 to 13.5 volts to run that string.... which is right in line with what your 12 volt DC power supply will provide.You can have as many strings as your power supply will handle... simply connect the aforementioned positive of a string to the positive of another string - and connect to the positive of your supply. Then do the same with the negatives... and you can have as many strings of 3 as you like.Always do strings of 3 though...any less will cause them to burn out prematurely.The right way to do this is to use an LED power supply that provides constant current, but it sounds like you're looking to do this fast and cheap- so that's likely out of the picture?

author: sleemanj
author_rank: Super Contributor
timestamp: April 25, 2013, 08:14:35 am
content: The lights already had a resistor built into them? Just one resistor in each light?I suggest that you draw the circuit inside one of your flashlights. Also power up one of the flashlights and with your meter on DC volts measure the voltage difference across the two legs of one of the LEDs (put one meter probe on one leg, the other on the other leg), this will tell you the "voltage drop" of your LEDs. You may as well measure the voltage drop of the resistor too using the same method. Finally measure the resistance of the resistor that was built in by turning the light off, putting your meter in resistance (ohms) mode, and measuring across the two legs of the resistor.With this information, the circuit diagram, the voltage drop of an LED, the voltage drop of the resistor, and the resistance of the resistor already in them, you (we) have a picture of exactly what the flashlight requires to operate, and how you can modify it (them) to run off your 12v supply.

author: Mrkirby
author_rank: Newbie
timestamp: April 25, 2013, 11:48:05 am
content: Thanks corporate666!The lite does work fine without the two LEDs that I had removed.I will try and draw out a diagram...then i will close this feed and re-open a new one with more clear questions.Currently all comments suggest to do a series circuit.Can it work in Parallel?As for the transformer. I will acquire the one I need for the job, no matter the volt or amp.

author: Corporate666
author_rank: Unknown
timestamp: April 25, 2013, 05:22:20 pm
content: Quote from: Mrkirby on April 25, 2013, 11:48:05 amIt will not work in parallel unless you get a lower power transformer. If you put 12 volts into that circuit in parallel, way too much current will flow through each device and burn out your LED's very fast. If you change out the resistor to work with 12 volts, then the resistor will be dropping many more volts and will get much hotter - requiring a much larger resistor and making the circuit much less efficient.There is really no reason to run them in parallel vs. series that I can think of... but if you are dead set on doing that, then you would need a DC power supply that puts out about 4 volts, not 12 volts.

author: Mrkirby
author_rank: Newbie
timestamp: April 26, 2013, 01:37:33 am
content: Thanks for the advice. So this is the brutal diagram I worked out. I think I just see the parallel circuit looking like it could work better that is all.Again it would have been sooner but I had a root canal today.... I know excuses excuses...Thanks for the help guys!!Really!James

author: Corporate666
author_rank: Unknown
timestamp: April 26, 2013, 02:41:33 am
content: Quote from: Mrkirby on April 26, 2013, 01:37:33 amYour diagrams are correct - well done.As for the ease of wiring, it's probably easier to do serial OR parallel... combining them requires slightly more thought and planning, but putting some in series lets you use your existing power supply. Also, even if you went out and found a 4 volt power supply, you will need 3 times the amp capacity to run them all in parallel. If each module required, say, 200mA, and you are running 30 modules... that would be 10 strings of three... which would be 2 amps. Not too bad and in the realm of a beefy wall-wart supply. But if you were to run them in parallel, it would be 6 amps, which is getting into the realm of heavy duty "brick" style supplies. It also means heavier gauge wire and more potential for things to go bad if you have a short at some point.