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Old 04-01-2004, 11:00 PM   #1
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HELP! DIY moonlights not working!

OK, I tried to build my own moonlights...got it all put together, plugged it in......and nothing happened.

Here's what I did:

I've got a DC9V 500mA AC adapter....I connected a 30 ohm resistor to the positive side (the one without a colored stripe). I then connected the positive end of five 4.5V 30 mA LEDs to the resistor with wires, and the negative end of the LEDs with wires to the negative side of the AC adapter.

Is there something I'm doing wrong? If not, how can I track down the problem?
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Old 04-01-2004, 11:40 PM   #2
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Use a VOM (volt ohm meter) to check the LED, remember, they are nothing more than a diode that emits light so current should only flow in one direction. You could rig up a battery to make sure the LEDs are ok as well. I think a C or AA is close to 4V. The formula for resistance, voltage, current eludes me atm. You could also try running just a pair in parallel. Also, make sure your getting 9V DC from the supply and that the polarity is not reversed..GL.
-Paul C.
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Old 04-02-2004, 01:08 AM   #3
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I can't remember, but it sounds like your numbers are all good. Did you test the first led in the chain before you soldered up the rest? Try switching the leads from the adapter.
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Old 04-02-2004, 02:09 AM   #4
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Here's the basic formula's for ohm's law.
v = i x r
r = v/i
i = v/r

i just have some basic questions..
how are the led's arranged ? are they in series or parallel ?
from the sound of what you did, you might have them in series. if they are in series then you don't have enough voltage present to lite up the leds or even get them into saturation (a fancy term, that means they turn on, or lite up). in series voltage changes per component, and current is deadlocked, so what that means is you will need to have 4v for every led you're using. however if you arrange the circuit in parallel, you will have all the voltage you require, in parallel voltage is deadlocked in the circuit, while current changes per componet.

depending on how they are hooked up in the circuit, one led might have been doa, or burned up on assembly (led's can only take a few seconds of extreme heat, from a soldering iron before they melt internally), the best way to figure it out is use a 2 AA batteries, and try out each led. i am assuming though you have blue leds, probably in the 4000mcd range, they go into saturation around 3.5v and top out at 4.3v so you need to get your supply voltage in the range, or pretty close to it, anything manufactured has a tolerance, and are never face value or follow the theoretical values. (resistors are a great example, the 4th color band is for tolerance, gold = +/- 5% ; silver = +/- 10% ; no 4th color band = +/- 20%)

so here's what i suggest you do,
1. make sure your dc transformer works
2. make sure you have the dc transformer hooked up right, if it happens to be a recycled dc transformer (from a cd player for example) the wire that has a white band running down the length of the cord, is the neutral portion (or ground).
3. if the circuit is still a no go, rig up 2 AA batteries and try out each led, like pc suggested, if they are fresh akl. batteries you should be able to squeeze by with 2, if they been used, you might want bump it up 3 AA batteries (1.5 x 3 = 4.5v)
4. if all the led's lite up with the battery, rebuild the circuit, then get access to a voltmeter (or multimeter) and run through the circuit, checking the voltage from the transformer then to the rest of the circuit. if can you spring for it.. HD has multimeters for 20, and walmart has them for 10-15. both stores have great return policies..

i am electronics major, so if you need any help let me know, i can run you through the basics, and give you some ideas how to build this circuit, provide wiring diagrams, ect ect. just let me know.

bry
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Old 04-02-2004, 04:20 PM   #5
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What wattage resistor whould I use? I've got a 1/4 watt 30 ohm one on there now.
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Old 04-02-2004, 05:06 PM   #6
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well the less commonly used formula from ohm's law..
P = watts = V x I
i am sorry for not catching this last night, i didn't do any math for this problem but your resistor pick is wrong.

let's do some math!

dc source = 9v 500mA
led forward voltage requirement = 3.7 - 4.2v, 20ma (Max)

So 9v(total) ‚Äď 4v(led requirement) = 5volts
R= 5volts / 20mA (LED max Amperage) = 250 Ohms
P= v x I = 5 volts x 20ma = 100mW
so.. 1/4 = 250mW so 1/4 watt is exactly what you want for this circuit, however you will need to replace the resistors you have, with 250ohms. how exactly did you figure out, you would want to use 30ohms ?

anyway since the leds were getting alot more than 4 volts and more than 20mA of current, the led's you have are probably toast (5v / 30ohm = 166mA, most led's can only take 20mA before they burn up). so it looks like you will need to replace them, if i may suggest a vendor.. go to www.lsdiodes.com they have blue led's @ 4000mcd for 45cents a piece, with 2dollar shipping (untill you spend more than 20bucks). if you want some more help, or wiring diagrams, or more explantions plz email ijedi@hotmail.com

just for reference, led's, diodes, some types of capacitors, are polarity specific. they will only work in one fashion, the longer leg on an led, is the anode (positive side) the shorter leg is the cathode (negitive). the cathode leg needs to be put on the ground (netural) wire of the power source. with out getting completely technical, if you don't install the led's in the correct fashion they can be destroyed. resistors though have no polarity, and can be installed in any fashion.

hope that helps..
bry
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Old 04-03-2004, 01:20 AM   #7
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Turns out the polarity of the adaptor was just backwards. Stupid noobie electronics mistake. My 30 ohm resistor seems fine and everything's working now.
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Old 04-03-2004, 01:34 AM   #8
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Well, after plugging it in and letting it run for about 5 minutes, the resistor gets so hot it nearly burned me. Any idea why and what I can do about it?


edit:
I think I have an idea....
I used a 30 watt resister because I'm using 1 resistor for all 5 LEDs...if it were 1 resistor for each LED it would be 150-180 ohms.

I did some more math of my own and discovered that with the multiple-resister setup I needed 1/4 watt resisters.....but with my single-resistor setup, I need a resistor that's ~ 1.125 watts.

So, I believe my problem is solved. I'll head to the electronics shop tomorrow morning and grab a different 30 ohm resister and see how it works. 8)
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Old 04-03-2004, 03:04 AM   #9
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Quote:
Originally Posted by JProx
P= v x I = 5 volts x 20ma = 100mW
so.. 1/4 = 250mW so 1/4 watt is exactly what you want for this circuit,
heat = power = wattage. if the resistors you have are hot enough to burn you, or start to change color, then you don't have a high enough wattage. there is only 2 fixes for that problem
1) replace the power source with a lower voltage (extreme i know..)
2) replace the resistors you have, with ones with a larger wattage rating. Again use the formula for power p = V x I then grab a resistor with a rating above what you calculated. (if wattage calced. is 270mW then get 1/2 watt resistor, or safer yet 1watt)

and sorry.. i was a bit confused on how you wired the circuit (parallel or series) and over looked some keywords. 5v / 30ohm = 166mA / 5 # of leds in circuit (i am guessing parallel now) = 33mA, so you're practically within their max rating, personally though i would use a higher ohm rating, just to reduce the current and get under the max value (anything over max either burns it up in a few seconds/minutes of operation, or shortens life span of the component, led's are put into saturation by voltage, not current so lowering the current will not reduce their lighting output)

using a current limiting resistor (1 resistor in series with the load) is about the most efficient way to build the circuit, and the best way to build an led driver, when all the led's are the same. however power generated becomes a problem (as you learned) since radio shack sells resistors in a 5pack for 99cents.. it doesn't hurt to give each led their own resistor, just to avoid the heat/power problems.

personally though, i have built my moonlight with a resistor on each led, just to make the circuit a bit safer, and less prone to break, resistors can be killed running at hi temps for extended periods of time. instead of just breaking like an led, or an IC, or the toys we get at x-mas, a resistor can start to leak. Leaking (nothing to do with liquid or releasing anything) as in losing resistance, running at hi temps for a long time, can make the resistor burn up internally, lowering it's resistance thus letting more current and voltage flow into the led, and eventually killing the led. how likely is that to happen though? with a higher watt resistor, unlikely to happen. but just thought i would pass on extra knowledge and get on tangent.

bry
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Old 04-03-2004, 04:24 PM   #10
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Well, I put in a 1 watt resistor and it's better, but it's still mightly hot. Plenty hot to burn me in 1-2 seconds. I'm trying leaving it plugged in for a while to see if the resistor starts to melt, but I'm starting to see the wisdom of using one resistor per LED. Would that make this much less of an issue?
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