Help w/DIY Moonlight

[newsalt]

Can someone help me with my DIY moonlight project? I have attached a picture of what I'm planning to build. I know I need a limiting resistor on each LED, but not sure how many ohms or the watts. Is what I'm showing the best way to do it or is there another way?

Thanks

 

[smaug]

OK this is what you need to do. You will need a resistor on each LED. to figure the value you subtract the led voltage from the supply voltage (12V-3.7V= 8.3V), then you divide the current by the voltage(8.3V/.02A=415 Ohms) These can be any wattage resistors 1/8 or 1/4 whatever you have. Now to figure out the Total current of the system you add the mA toghether so you will have 60mA so you 500mA supply is fine.

 

[smaug]

Oh YOu more than likly wont find a 415 ohm resistor. You just need to go up to the next value that you can find. 420 or 450 what ever as long as it is at least 415. If you get to much the britness will suffer.

 

[sincityreefer]

OR,
you could wire those 3 in series (as opposed to parrallel like you have shown) and make a 11.1 V Load, that way you could use a much smaller resistor around 45 ohms and be more effiecent. I see you have a 12V supply, with the givven equipment thats what i would do. Wiring in series is pretty easy, all you have to do is connect the positive end of the first LED to the + side of the supply (with the resistor in between this part) then the negitive side of the first LED to the positive side of the 2nd. and so on.... I can whip up a little drawing for you if you want to go this way, let me know. HTH
-steve

 

[sincityreefer]

Originally posted by smaug
OK this is what you need to do. You will need a resistor on each LED.

Another bonus of wiring in series you only need 1 reistor between the + side of the supply and the + leg of your first LED. Not that we need to be this concerned with power consumption, but in series the LED's will dissapate 213mW, and a 47 ohm resistor will dissipate 17mW. which is 220mW total. In parrallel. the LED's dissipate 195mW, but the resistors dissipate 147mW EACH, bring the total to 636 mW. The clear advantage in efficency is the series wiring method. But like i said, whith what we are dealing with here, its not really an issue, either way will work awsome. just throwin in my .02 cents.
-steve

 

[smaug]

Yeah series will work fine. The only reason i would use parrallel is to avoid trouble shooting if a bulb were to die but with the given lifespan on LEDs they may last longer than your tank. Post some pics when your done.