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Old 04-04-2004, 12:05 AM   #21
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lol well radioshaft wants 4bucks an led, lsdiodes is like 400times cheaper than radioshack, you will get you're leds from them next week, mid week i will bet. they are pretty fast shippers. i know your joking (at least i hope you are) but that entire sub fixture with a reflector is totally not neccessary. leds push close to 80% of the light through their tip, which means only a small % is actually escaping from the backside, so going through the pain of installing a reflector to gain 10% extra light (led's are not 100% effecient, they are pretty close though) is pretty much a waste

what i did for my 12gal eclipse tank, was grab some scrap 3/4(width) plywood, cut to fit (and boy was that a P.I.T.A) polyurthaned it, then drilled 5 holes, counter sinked them so the leds would fit inside the wood, the soldered wires to the legs, filled the top section of the countersink with aquarium sealent (to prevent water from entering there) ran the wire through the holes, installed the resistors on the top of the wood, soldered all the wires together, then ran them out of the tank, and to the dc adapter. it's really simple and i will take some pics if you want.

i was just doing some math, and i believe your 130ohm resistor choice is bit off.
9v source - 4v required voltage = 5v / 20mA = 250ohms, so you should pick something between 250 - 300 ohms, to be on the safer side.
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Old 04-04-2004, 08:36 AM   #22
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I looked at those led's and they seem to be the same ones I ordered from lebos.com. Thanks
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Old 04-04-2004, 01:07 PM   #23
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I have a question on this. Tripoli was saying (on the phone with me anyway) that his circuit didn't work unless he had the resistor on the positive side. Should that matter? I didn't think it would matter one way or the other since it would still change the wattage going through the LED.
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Old 04-04-2004, 02:48 PM   #24
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it will work with the resistors on either side of the led, they *should* be on the negitive side, but will generate the same results if they are on the positive (or annode) of the led. as a matter of fact i have another led project in my prototype board (rigged up to a few different ic's) that have the resistors on the positive leg of the led.

the resistors don't change the wattage, what they do change is the voltage and current running through the led. wattage is a product of voltage and current (p= v x i) and it's used to measure how effecient a circuit is (Pin / Pout = Eff /100 = xx %) and how much heat it's generating.
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Old 04-04-2004, 03:19 PM   #25
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Quote:
what they do change is the voltage and current running through the led
But actually current runs from negitive to positive so you are correct in stating it should be on the "negitive side".
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Old 04-04-2004, 08:48 PM   #26
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well..
you're right, current does run from the negitive / netural portion to the positive side of a power supply. however an ideal power supply generates very little (or theoritcally zero) amps and the circuit is responsible for generating the current it needs.

added info: a battery though, does generate a little bit of current, since there is a very small resistance value inside the battery, which is why a battery loses it's charge after a while.

in a simple current limiting / voltage divider circuit having a resistor on the positive or negitive side of an led, will generate the exact same results. it should be wired so the resistor is on the negitive led, but really it doesn't matter in this application.
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Old 04-04-2004, 09:42 PM   #27
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Okay, I have a new question. Tripoli was saying that he put two of the LEDs in series as a test. He was using a 9VDC power supply and the LEDs are 4.5V (typical, not max). Shouldn't that work without a resistor? He said that it blew both the LEDs. I tried it at home, and got the same result - they lit for about half an hour and then blew.
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Old 04-04-2004, 09:49 PM   #28
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A new 9v battery has actually 9.8 v to begin with. I know that with resistors if you series them you cut the resistance in half but I don't think it's the same for voltage. In other words you still had 9 vdc going through the led.

The way to check this is place your power supply on the led, and measure voltage with them in series. I think you will find that voltage remains the same and current is what changes.
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Old 04-04-2004, 11:44 PM   #29
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2cents you're a little confused on the subject..

in series voltage changes (each voltage drop is added together) and current stays the same. in parallel voltage stays the same and current changes. if 2 blue led's are in series with 9volt source, they will blow, why? because blue led's operate in a range of voltage, from 3.8 to 4.2volts. 9volts / 2 = 4.5volts, about 300mV over their max value. the greater the difference over their max value the faster they will blow.

resistors in series add up (R1 + R2 + R3 ect = Re ) in parallel they divide
[(1/r1) +(1/r2) + (1/r3) = 1/Re] Re = equivalent resistance. a little reminder i always thought was helpful, if 2 resistors in parallel are the same value divide the value by 2, and if they are all different the result of the formula will be less than the lowest resistor you have.
an example:
r1 = 1k
r2= 2.5k
r3=5k
1/1k + 1/2.5k + 1/5k = .0016
1/.0016 = 625ohms

so i still feel the best way to build a moonlight is wire the led's in parallel, so you don't need more voltage (in parallel volage is dead locked, it's the same for all components in the paralleled branch) in series you will need to have ~4volts for every led you have in the circuit (5leds x 4v = 20volts). so parallel is the best possible way to build this circuit, it's just so much more efficient. and i have been thinking about doing this for a while.. but perhaps i should just write up a post on how to build moonlights, and the theroy behind electronics... didn't think there was enough of a need though before.

hope that helps.
bry
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Old 04-04-2004, 11:55 PM   #30
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Got my series and parallel mixed up again. I'm always doing that. yea that's right.

I got a "moonlight" instruction book if you want it. pm me with your email addy. (no hotmails & yahoo) the file is 2.5 meg.
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