Author Options:

i Need help with leds? Answered

the question is in the picture


DON'T parallel them like that !
What supply do you want to run them from ?
This LED driver I've attached will let you control the current.

You could run this off one supply like a laptop brick, just put all the LEDs in series.

Copy of maths.jpgtrimmable current source.JPG

why cant i parallel them like that what happens or what? wont there just be like 10.2 forward voltage and current (mA) would raise to 40 mA?

LEDs are NOT electrically identical - you might be lucky, but in my experience, you probably won't.

If your LEDs were identical, then a parallel pair would carry identical currents, and that current would be stable with temperature and current.

What will happen though, with real LEDs is that one will pull slightly more current, so it will get warmer, and its resistance will decrease, so it will pull slighly more current, and get warmer....

This is called "runaway", and it leads to burnout, if you're not careful, especially if you have decided to run the LEDs at their maximum current.

so all LEDS have varied specifications and the specificatons that the seller gives you is just an estimate to what it would be? by the way all the leds i have are all white.

You've got them in a strange series / parallel arrangement, so  the Vf for  the whole thing is going to be that of 3 in series which is 9.6V min, 10.2V.
The crucial thing is your supply voltage - you're going to need 12V or more.
If 12V, the resistor value will be the supply voltage minus the min. forward volt drop, all divided by the current which is 40mA (which should share itself equally between the paths, as long as the diodes are pretty much identical.  If they're from the same batch they will be.
So the calculation is (12 - 9.6) / 0.04 which is 60 ohms.  Nearest value to this is 62 ohms.
If you're using higher than 12V, just substitute in the calculation accordingly.

I've seen too many LED projects where LEDs have been paralleled to trust the process at all.

I wouldn't do it that way myself, but I'm guessing he's already got that assembly and the LEDs are pretty much identical if they're from the same batch.

I've just read that Vf drops with heating - I always thought it increased - So current hogging would be distinct possibility.

I've seen hogging happening in real time with some of the cheap LED camping lights. Pretty, explosive, brief and smelly.

yea all the leds are white and have the same specs. would this be a good resistor even though it says 1 watt(http://www.frys.com/product/997666?site=sr:SEARCH:MAIN_RSLT_PG)? Im also aware that it is 68 i just couldnt find a resistor that is 62 on frys. and well i found another one that is 1/2 watts
which one should i be better off with cause i dont want the resistor to turn real hot

Diablo,do you have the option of wiring these LEDs in a different way?
If so, you should go with Steve's suggestion.

well i can but i already put it into a reflector and glued it but i would make another one im open for suggestions, so please tell me how am i suppose to wire them for it to be more efficient then the one i made

Put them all in series. Run them from a laptop supply and the circuit I've given you.

This is the thing with the light im doing, I'm trying to run it from a portable rechargeable battery that i had bought that has 12v and has a current of 6500mA, so i could put it into a water bottle then place that in a bicycle water bottle cage and run wires to the lights so i could have my own made bike light. I already made two: one for the front and one for the back . so with the laptop supply it wont really work for me. i hope you get the idea to what im trying to make. so all i wanted to know what type of resistor i need for this application and/or if i could use a driver instead to make the flow of current a bit safer to the LEDS.

One question if i put them in series wont all 6 have a voltage of 20.4, thus needing to have that much voltage in a battery or would that driver schematic you showed me would work to power all those Leds in series because if i was to put them in parallel then the current of the leds would increase to like 120mA, am i correct?

Ah, so you need to put them in two parallel chains then. That needs two resistors, or two of the circuits I showed you.

I think, because these are very low power LEDs, I'd go with the resistors.

120 Ohms, 1/8Watt will be fine

wont the battery burn off the resistor because it is 1/8 watt.

i was researching on how to choose the watt in a resistor.this is the page from where i saw it from(http://wiki.answers.com/Q/How_do_you_determine_a_resistor%27s_wattage).so this is what i understood from what i had read: i divide 12v into 68ohms(i got 68 from the calculator in http://led.linear1.org/led.wiz(these are the values i inputted:12 source voltage, 10.2 for the diode forward voltage 40 for the diode forward current and i counted the whole 6 l.e.d's as 1 whole l.e.d) which got me 0.18mA(rounded). from 0.18mA i multiplied 12v which got me 2.12(rounded).so from the last number u got i think its the type of watt resistor i would need meaning i would get the 68ohm resistor with 2 watts. is this right or am i wrong

No, the resistor dissipation is 120 x 0.020^2

Your maths is wrong, but its very hard to decide, because you don't use punctuation.

Resistor = (V supply - VLed)/ Iled

Power in resistor = Iled^2 * Resistor.

How do you get 40mA as the diode forward current, when they are in series, and the maximum they can accept is 20mA ?

i had read before that when you parallel LEDS the Current increases and when you put them in series the voltage increases.So with my setup there is 2 parallel of 3 LEDs in series, thus making it 10.2V and the current 40mA, isn't that correct or what i had read is totally wrong

You have two strings, carrying 20 mA a piece. The battery supplies 40mA, the current limiting resistors feed 20mA.

yea 20 mA for each string and those two strings are paralleled together doesn't the mA increases to 40mA?

That's right, from the point of view of the battery, but what sets the resistor currents (remember, there is ONE resistor PER STRING), is the 20mA in the string, forget the 40mA total.


That means that the resistor dissipation is ALSO controlled by the 20mA, not 40.


so i cant put one resistor for 2 strings that are paralleled together?

No, it puts you back where you were to start with ! Each string needs a resistor.

o damn. is there any way i could just use one resistor instead of two because i cant really go back and fix it. the leds are already glued to the reflector and i just have a + and - wire that is soldered to both ends of the strips of leds that are soldered together as you see in the above picture?

Try it with one resistor, but be prepared for the LEDs to die. 60 Ohms should do you.

I'm sorry I am kind of lost here. Plus im a noob at this and hoping to understand it. I appreciate your help in trying to make me understand.

Diablo,you haven't told us the application for this or the voltage you want to drive them from.  You mention a reflector so it may be a headlight type thing.  Does it need to be portable?
Steve's is the corret way to do this as it strictly controls the current, but to keep the component count down, I'd just split the top and bottom rows (from your diagram) to give you two chains of 3 LEDs each, and use a 120R resistor in series with each chain - any of the resistor types you've mentioned above will do.  That's assuming a 12V supply.  You'd have to break the LEDs down more (3 chains of 2) if you're using less.

ok so a 62 ohms resistor but which one should it be 1/4 watts, 1/2 watts, or 1 watt?