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Title: LED Moonlights..
Description: Need some help


eL ChiNo LoCo - December 31, 2005 09:02 AM (GMT)
Okay, well I got everything I needed, plugged it in, but no go sad.gif

Stuff that Im using:
6V 300mA Power Supply
6 blue LED's, 3.2-3.4 V, 18,000-20,000 mcd
6 100 ohms resistors
speaker wire 24 gauge

heres a diagram i did to show how i connected it. Also, on the power suppy, I cut off the end of the plug part where youd plug it into like a cellphone or something else (shiny metal tip part), but in this case they were removable, but I dont know which one is positive or negative....on the supply.

user posted image

Cichlas - December 31, 2005 11:36 AM (GMT)
You don't have enough voltage on your circuit to run 6 of those leds in series. You should be looking at 24 volts at least t run them at full brightness i.e. you need to add up the voltage required for each led and then add a bit extra on. The resistors are only there to use up any voltage not used by the leds. In you configuration, you don't have any so they are just adding unneccesary load and stopping it working.

On a general design tip on your config, when wiring them up in series, you only need one resistor in the chain somewhere. No need for one per led. and it only needs to be big enough to consume the unused voltage. Any bigger and it will take power from the leds and they won't work, any less and the leds may burn out.

Hope that helps.

eL ChiNo LoCo - December 31, 2005 08:44 PM (GMT)
Thanks Cichlas, posted on 2 forums, your the first to reply. If I only need one resistor, I dont know how to calculate the ohms since its going to be for all 6, do you?

Cichlas - January 1, 2006 01:13 AM (GMT)
That must be worked out based on how much voltage is left after adding up the led voltages and the current you want to run through the circuit. i.e.

If you had a power supply giving 24V, with the 6 leds each taking 3.2 volts. Your leds would use up 3.2 x 6 = 19.2 volts. That leaves 4.8 volts to get rid of through the resistor. If the current required for the leds was 20 milliAmps then the resistor would be worked out as 4.8 (volts) divided by 0.02 (20mA) = 240 ohms. You also need to make sure the resistors can handle the heat that would be generated, so you need to get ones of a high enough wattage to dissipate that heat. Probably wouldn't be a big issue in this application but you would work it out as 4.8 (volts) x .02 (20mA) = 0.096 watts (a 10th of a watt or there abouts). Get the next size up if no exact match is available.

BTW, I'm not saying you can't use a resistor for each, just that there is no need and one is a much simpler design.

Hope that helps.

eL ChiNo LoCo - January 1, 2006 06:32 AM (GMT)
Thanks, really appreciate the help. Where would I add the resistor, to the positive or negative side? In the begining or end?

Cichlas - January 1, 2006 11:51 AM (GMT)
Doesn't matter.

eL ChiNo LoCo - January 2, 2006 09:16 AM (GMT)
Thanks. Ive been reading around other forums, and someone on this forum http://forums.waterwolves.com/index.php?sh...0&hl=moonlights
Has 10 running a 12V adapter? How is that possible?

The reason Im very confused is because I was following this thread:
http://www.aquaria.info/index.php?name=PNp...ewtopic&t=44381
Could you tell me if the way this person is better or not? Im most likely going to go by what you said, but is this person then incorrect (sorta). He explains in a few post down how its "poor electrician practice" to run diodes in a series with only one resistor.. :blink:

Cichlas - January 2, 2006 02:41 PM (GMT)
Ok, Lets tackle the first question.....

It all depends how you wire them up. If you do them all in series, like you had, then the supply voltage has to be high enough to cater for all of the led voltages added together. If you wire them in parallel, say one in each parallel leg, then the total voltage for the the whole thing will be just 3.2 volts (1 x led voltage). In the series circuit the current in total is 20mA. In a parallel circuit it will be 20mA per leg. so if you had 6 leds (1 in each leg = 6 legs) your total current for the circuit would now be 120mA (20mA x 6 legs). It is like a trade off, you can use a lower voltage but you need a higher current. Depending on what you are doing, it may be cheaper to get a higher voltage supply than a higher current one. So you need to make sure your power supply can handle the current. If you deviate from your current design and make a parallel circuit design, you can run what you want. In your case you would wire each one in parallel on its own leg because two leds would still go over what your power supply voltage is. You therefore need a resistor to drop 2.8 volts in each leg. It maybe that this is what you intended to do in the first place but just wired them wrong. Not sure. The total current for your circuit would be 6 x 20mA = 120mA. Your power supply can do 300mA so you have no worries there.

2nd question.......

His reference to "Bad Practice" is to do with wiring them in parallel, not series. In series, one resistor is the way to go as long as your resistor can handle the power (the watts I was talking about in my previous reply). If it can't, then you may add extra resistors in there to share the power handling as long as the total resistance of them is the same. For parallel circuits, you should have one in each leg. That isn't a matter of bad practice, that is just how it should be done for the circuit to work efficiently and properly.

OK, having said all that, there are draw backs to parallel circuits to consider. Each led must be very close or identical in their behaviour. If not, it can effect how the whole circuit works and drag the rest of the circuit down. having bought all your leds from the same place and most likely from the same batch, they hopefully should be close enough not to cause any problems. If you later decided to add some more leds, it could have unexpected results. i.e. one led might be much brighter than the rest and the rest dimmer than before.


Hope that make sense and helps with your questions.

eL ChiNo LoCo - January 3, 2006 10:20 AM (GMT)
Thank you very much Cichlas, youve been the most help, if not all the help Ive gotten for this as my first DIY for my tank. Ill see if I can post pictures, but I doubt my crappy camera can take pictures in low light. For some reason, 6 lights isnt real bright, compared to http://www.kaotica.com/frag/diy/moonlight/ this guys 5 leds, running on 6V.

Cichlas - January 3, 2006 01:36 PM (GMT)
It is difficult to really tell from a photograph how bright the lights actually are. You can fiddle with the camera settings and the picture after the shot was taken to make things look different. He may also have brighter leds than you have. Glad you have it all sorted, a pic would be great :)

Vip - January 13, 2006 08:49 AM (GMT)
Damn Cichlas, I didnt know you knew you lighting so well.

Cichlas - January 14, 2006 01:56 AM (GMT)
QUOTE (Vip @ Jan 13 2006, 09:49 AM)
Damn Cichlas, I didnt know you knew you lighting so well.

lol.... I don't really. I was an electronics engineer on Radars and airfield comms stuff in the Royal Air Force for 11 years before leaving and taking up programming computers. Sometimes, albeit not very often, something crops up that I can use the old trade for :)




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