I need someone who knows about LEDs, resistors and voltages

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Hi everyone

Originally I simply wanted a nice bright red light in my case to light everything up.

I thought of putting cathodes in there, but most cathodes turn out to be orange.

I thought of putting EL wire in there, but it wasn't bright enough.

So I decided to wire my own LEDs.

Ok so I did some pretty stupid stuff, I ignored everything I learned at school and tried to use my own dumb ass way of doing things. I hooked up 32 LEDs in parallel, connected it to a 6v battery, it was fine, a bit dim (the battery was dying) Then I proceeded to hooking it up to the 12v rail in my PSU, and magic smoke came out in a shape of a middle finger pointing at me.

So in order to prevent this from happening again, I'm asking the pros (you guys).

I want to connect 32 LEDs in parallel, I should probably connect it to the 5v rail this time.

Here are my LED specs:

14000 mcd
Size: 5mm
Lens Color : Water Clear
Forward Voltage (V) : 1.8~2.0
Forward Current (mA):18
wavelength (nm):620-625-630

Here are the materials I have available to me:
x200 LEDs (14000mcd)
x200 100ohm resistors
x200 470 ohm resistors

Would Having a 470ohm resistor connected to each LED that's hooked up to the 12v rail be better then
having a 100ohm resistor connected to each LED hooked up to the 5v rail?

Well, you need the voltage drop across the resistor to be roughly 3V at 18mA if you want to hook it up to the 5V rail, and roughly 10V at 18mA if you want to hook it to the 12V rail. This makes the resistor value easy to calculate. V=IR, so for a 3V drop at 18mA, 3=0.018*R, therefore R = 167 ohms. 100 ohms will probably work, but you'll be pushing significantly more than the rated current. Ideally, you'd get some 150 or 180 ohm resistors instead (both are standard values).

Alternatively, to hook them up to the 12 volt rail, you need a 10V drop at 18mA, so 10 = 0.018*R, so R = 556 ohms. Once again, your resistors will probably work, but they are a bit small, and will result in a higher-than-rated current flow through the LEDs. 560 ohms is a standard value, and would work well for this.

If you do want to use the resistors you already have, I would use a 470 with each LED and hook it up to the 12V rail. That should give you around 21mA of current through each LED, which should work fine.
a c 144 ) Power supply

If those are the only two resistor values available, try a series string of 4 LED's with two 470 ohm resistors in parallel connected to the 12 volt PSU output.
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That works out pretty nicely actually - that would give 17mA through each LED (assuming a 2V forward drop), with the added bonus of using fewer resistors (only one per two LEDs).
a b ) Power supply

Actually, allowing 2.0 v forward drop per LED, putting 6 of them in series with NO resistor fed from the 12 VDC supply would get you a load supposed to handle the 12 V, and using about 18 mA current. Set up five of these strings and connect them all in parallel across the 12 VDC supply, and you've used 30 LED's (not 32) and no resistors to do the job. So you'd have light output reduced by 2/32 of your max, or just over 6%, probably so small a difference you would not notice. Current consumption would be 18 mA per string, or 5 * 18 = 90 mA.

Another option is a series connection of 7 (not 6) LED's, which allows for a forward drop of 1.71 VDC per LED, slightly under-driving each. This reduces light output slightly, although probably not enough to notice. You could do this on two of your five series structures, thus using up all 32 of your LED's. Or, buy three more and make all five series strings this way. The 5 x 7-LED structure would consume slightly less power than 5 x 6 LED's because the current through 7 LED's in series will be a little lower.

In doing series connections of LED's, the two potential problems arise from malfunctions. What happens if one LED shorts out? Then you have only 5 LED's with a 12 VDC drop, or 2.4 VDC per LED, over-driving them. They would probably survive, but have a shortened life if left that way. (The 7-LED series string option above solves this by making the drop become 2.0 VDC after one shorts out.) Or, if one goes open, ALL the LED's in that line go dead and you will notice it quickly and have to hunt for the open one (and find a replacement).

Any series string of LED's will have these issues. The only way to avoid is to connect each LED with its own 560 ohm resistor in series, Then put all those 32 sets across the 12 VDC supply in parallel. That should consume 18 * 32 = 576 mA, or 6.4 times as much as the set of five 6-LED series strings. The 486 mA extra current calculates out to 5.8 extra Watts power consumption. The extra power consumed all will be converted to heat (by the resistors) inside your case that needs to be removed, but that's not a lot. The resistors, at 560 ohms each, will be dropping 10 VDC, and that means they will dissipate 0.18 Watts of heat. A ¼ Watt rating of the resistors is sufficient for them.

ah thats great!

Thank cjl & jsc.

@ jsc

Just to confirm what your saying I drew a crappy little MS Paint drawing of how it would look like lol

http://i25.tinypic.com/206e8if.png

but would running 17mA of current through a 18mA rated LED reduce its brightness?
I got 14000mcd LEDs so they are really bright. Should Illuminate my case nicely.

2143477,1,662614 I hooked up 32 LEDs in parallel, connected it to a 6v battery, it was fine, a bit dim (the battery was dying) Then I proceeded to hooking it up to the 12v rail in my PSU, and magic smoke came out in a shape of a middle finger pointing at me.
[/quotemsg said:

if 32LED in parallelworks fine in 6v

im thinking
32 LED in parallel so your drop against each LED is 6v and LED work fine.
so if you add 2 LED in series, and connect it to 12V, your drop will again be 6V against each LED.

however, with 32LED, you can get 16 pair of LED, and so you can connect 16LED in parallel
but to get 32LED in parallel, you can just use 64LED in total, thereby, make it glow brighter

Quote:
but would running 17mA of current through a 18mA rated LED reduce its brightness?

18mA is the max rating of your LED, it will not reduce brightness, and even if it does, you wont see it
a b ) Power supply

Watch out! You got away with putting 32 LED's in parallel on a 6 VDC battery that, in your estimation, was dying. So you don't know what voltage it really was supplying to those LED's. But in general, putting 6 VDC into a LED designed for 2 VDC is over-driving it badly, and I would expect that to burn out the LED. So do NOT design for 6 volts into a 2-volt LED.

Yes, running at slightly lower voltage so that the LED current is 17 mA instead of 18 will produce less light. But you'd have to be making VERY close comparisons to see such a small change in light output.

@ Paperdoc

Thanks alot man!!

I never knew LEDs work like that (forgot actually lol).

So If I hook up 32 LEDs in parallel to a 5v rail, each LED gets 5v.
But, If I hoot up two LEDs in series and 16 of those sets in parallel, each LED gets 2.5v.

To lower the voltage each LED gets to exactly 2v I need 56ohm resistors.
However I dont have that =/

Im thinking .5v is insignificant and I can get away with it.
I just don't want them to burn out, I don't mind if their life span is reduced from 100,000 hours to 75,000 hours.

Ok was stupid of me.

I came to a final solution (Exactly what Paperdoc said in his first post) and it looks great

Hooked up 6 LEDs in series and Have five sets of those hooked up in parallel (30 LEDs in total) which is all connected to the 12v rail so each LED gets exactly 2v

Thank you for all your responses, very informative.

That's not the greatest way to go though - generally, you don't want to hook up LEDs directly to power supplies without a resistor, because the I-V curve for a diode is strongly nonlinear. A resistor in series with an LED acts as somewhat of a current limiting device, reducing the sensitivity of the LEDs to small voltage changes.

Basically, you almost always want to run LEDs with resistors, rather than by themselves.
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