Sorta new to circuitry and all but I have a quick question. I bought 10x UV Leds and figured out the resistor I need(47 Ohms). What Im doing is putting like 5-6 of them embedded in the clear MoBo tray I made. Now if I wire them in series after the resistor will each one get decreasign less bright? Also, If I do parallel after the resistor to them, will all of them be less bright then the potential but the same brightness?
The missing number in your problem statement is the supply voltage. I'm also a little skeptical about your resistor value unless you're talking about a series combination of 3 with a 12V supply. I would expect UV LEDs being a InGaN type to drop around 3.7V @ 20 mA. If you were to use 12V supply, you could stack three in series with a 47 ohm resistor. R = (12 - 11.1) / 0.02 = 45 ohms (round up to 47) You would want to run parallel strings, each with its own resistor.
To answer what you actually asked (not to say that linear wasn't helpful): No. They'll each draw the same amount of current in either configuration (if in series you used mismatched components, then they'd recieve differing amounts of power, which could be a bad thing). If you used the right amount of resistance, then the LEDs will be at their maximum safe brightness (overpowering LEDs continuously = smell of smoke, bright flash, potential explosion like what happened to cheese).
LOL. So itd be cool either way is what im hearing? I believe if I just pull it from the 5v then the 47ohm resistor would be enough to keep my LEDs from frying. Then if I just go series connections then all the LEDs will be same brightness. Thanks for help guys.
Well, what I edited out of my original post is: The cumulative forward voltage can't exceed the supply voltage. If your LEDs drop 3.7V @ 20mA, then you can not stack them in series with a 5V supply. They won't get dimmer, they just won't turn on. I don't know the exact turn-on voltage, but I expect it's above 2.5V. Now if they were red, those drop more like 2.0V 2 20 mA, and two in series from 5V supply would work, But three in series would not. So that's why I said,
Oh I sorta get what you're saying now. I was thinking of getting like 5-6 on one circuit so with the resistor that wouldnt happen correct? I thought that I had to have 3.6v running through the circuit AFTER the resistor and everything after that would run at the remaining voltage. If I wanted 5-6 or even more LED's on one circuit, how would I go about accomplishing that?
Leds are funny things, the resistor isn't there to knock the voltage down, it's there to set the current through the led. If the supply is 5v and the led 3.7v, the spare 1.3v across a 68R resistor will produce 19mA current through both resistor and led. To run 5-6 of these leds you've a number of choices: 1. From a 5v supply, 5-6 parallel strings, each consisting of 1 led and a 68R resistor 2. From a 12v supply, strings consisting of 2 leds in series with a 220R resistor, each string connected in parallel 3. From a 12v supply, strings of 3 leds in series with a 47R resistor, each string connected in parallel. (For 5 lights, you can wire a string of 2 in parallel with a string of 3, but each string will need its own resistor). But your 10 leds, 5v and 47R is a
So what I get is that I should take the 12v supply(yellow) to 1 resistor then from there take 2 wires split up with 3 LEDs on each? Sorry for noobness. I just want to make surebefore I screw anything up. Thanks for the help Itll look like this?: __(led)__(led)__(led) / \ ---12v-----(resistor)/ ---(grnd) \ / \__(led)__(led)__(led)/
