Stu; Those are 100mA LEDs not 1000mA. I think they are about the limit of what you can feed LEDs without a heat sink concern. I just ordered a (100) 100mA LEDs and they are in a slightly larger than normal gum-drop size. There would be no way to even heat sink them. They do have kinda large flags on the leads so that may be it for them. The0wn4g3's LEDs have 4 pins, I think, which is how they can reject some heat, though not on that circuit board.
The0wn4g3; You used the perf board upside down. You're supposed to have the copper underneath so when you solder the pins they have the copper to solder to.
That's looking pretty bright!
<a href=showthread.php?s=&postid=14662067#post14662067 target=_blank>Originally posted</a> by lynxvs
No need to be rude,
I was just commenting on this statement:
"But the string with the shorted LED will have a different voltage requirement than the good remaining string. This causes what is termed as "current hogging". The good sting will either go dim or OFF completely while the bad string may have 1200mA running thru it. The remaining LEDS will fail in seconds."
which is false if it's a constant current source. The LEDs will smoke if one string opens however.
Why was that being rude?? You paraphrased my original post into a less accurate statement that seemed to miss the details of mine.
Your statement is correct, if and only if, you're talking about a single string of LEDs. I wasn't. I was talking about the hazards of parallel strings. If one sting changes the other string certainly doesn't continue to have the same current flowing thru it!
Draw a circuit diagram and figure out what I stated. There is nothing incorrect in that statement
whatsoever. If you think it's wrong you must not be understanding it or what current hogging is.
Don't take this wrong I am trying to help you understand this.
Draw a constant current source running two resistors. Make it a 1A source. If each resistor is 10 ohms you will have 5V provided by the constant current source and 0.5A running thru each resistor.
Now change one resistor to an 8 ohm resistor.
The constant current source will see only a 4.44 ohm load and will change its output voltage to:
1A x 4.4 ohms = 4.4V
Now you have 4.4V across a 10 ohm resistor.
4.4 / 10 = 0.44A (it now has less of the original current running thru it)
Looking at the 8 ohm resistor:
4.4 / 8 = 0.55A
The 8 ohm resistor is "hogging" more of the current.
This is VASTLY more pronounced with LEDs and results in the chain reaction failure mode I described.
This is what The0wn4g3 will see happen as soon as one of his LEDs fails since he has ten parallel LEDs all being driven by one constant current source.
The0wn4g3; I would suggest that you add one or more LEDs to the parallel ones you have so you have 11 or 12 sharing the current. Then if one fails the others still won't be over driven and you will get a chance to notice and replace a blown one before they all fry.