Wiring up LED's

I'm making a dust shoe for my router and I'm putting a ring of LED's around the spindle.
I have 12 LED's wired in parallel, what I need to know is what is required for me to hook them up to
my 48 volt power supply.

Regards,

Kiwiken

I guess that would depend on the voltage requirement of your LEDs

They are;

- Voltage: DC 3~3.6V
- Current: 15~20mA

Regards,

Kiwiken

I would put a resistor in series with each led.

Try one with a 2.4K resistor in series & measure the drop across the resistor.

See this page.
Wiring up multiple LEDs in parallel

I have areadly soldered my LED's together and was hoping
to only put something on the positive wire going to the LED's.

Regards,

Kiwiken

13 leds at 3.6V each would be 47 v if done in series (+ve of one to -ve of the next etc).
Better to put 14 in series to be between the 3 and 3.6 volt.
no pull down resistors needed.
Otherwise you'll need 1 to pull the 48 volt down to 3.6V

Those LEDS with those specs need a resistor of 220Ω(±30), power rated to suit. The page Cliff linked to explains it well enough.

You say you've already soldered 'em up in parallel... personally, if I couldn't 'undo what's done' I'd go for two 500Ω resistors, one to each end of the "+ve loop." So in effect they're in series.

But if I could start again, I'd definitely wire one per LED. If physical room was an issue, I'd consider 1 resistor per group of 2 or 3 LEDs. (Then again, my 'tronics skills are even more rustic than my woodwork... :rolleyes:)

Get the resistor value wrong, and you'll either get a dim light for a very long time, or a really bright light for a very short time indeed......:doh:

I was involved with building a large LED clock (ie each number segment was made up of lots of individual round LEDs). We initially tried a single resistor in series with a bunch of LEDs in parallel. Due to the manufacturing differences, the LEDs lit at different rates. It looked quite poor, for a clock display. Agree with most above. Use a resistor for each LED. We cured our problem in the end by using parallel LED driver chips, which were available then. I suspect you probably don't want that option.

Regards
SWK

Cliff is spot on. You need a resistor in series with each LED. 2.235k ohms 2.2k is a common size and would be close enough. This will give you a Vf of 3.3V and a current of 20mA per LED.

I've done strings of LEDs for a few things, and, IMO, by far the best option, if you can, is series strings. As kinda mentioned, if you had done this with your 12, you'd only need a single 250R resistor with very little current going through it and thus little power dissipated (100mW).

As it is, you've got a 48v supply and LEDs that require 3.6v@ 12 x 20mA = 240mA which will need a 180R resistor. The catch here is that you'll need a 10W resistor, and will be getting very hot, and wasting a lot of power, and as mentioned, LEDs in parallel often don't behave well.

As I haven't enough room to put a resistor on each LED and don't want something that's going to
get hot, it looks like I will make up a ring with the LED's in series.
The only reason I was go the parallel way was in case one failed.

So what do I have to ask for when I go in to the local leading edge store to get one as they
are as dumb as dog crap. I would have more luck asking someone at McDonald's and getting the right part.

Regards,

Kiwiken

Well, if you wire them in series, that's 12 LEDs each with 3.6V drop across them, or 43.2V total drop across the string. For a 48V supply and a LED current of 20mA, the resistor required is (48V - 43.2V)/0.02 = 240R.

You can't buy a 240R resistor, only a 220 or 270. If you use a 220 the LED current will be 22mA, if you use a 270 it'll be 17mA, either will be fine.

As mentioned, power dissipation will be low - for the 220R it'll be 0.02^2*220 = 0.1W, so a cheap 1/4W carbon resistor will be fine.

So go get a 220R or 270R 1/4W carbon resistor.