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Thread: Wiring up LED's
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23rd April 2013, 12:57 AM #1Intermediate Member
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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
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23rd April 2013, 07:20 AM #2
I guess that would depend on the voltage requirement of your LEDs
Every day is better than yesterday
Cheers
SAISAY
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23rd April 2013, 08:52 AM #3Intermediate Member
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They are;
- Voltage: DC 3~3.6V
- Current: 15~20mA
Regards,
Kiwiken
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23rd April 2013, 09:11 AM #4
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.Cliff.
If you find a post of mine that is missing a pic that you'd like to see, let me know & I'll see if I can find a copy.
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23rd April 2013, 09:18 AM #5
See this page.
Wiring up multiple LEDs in parallelCliff.
If you find a post of mine that is missing a pic that you'd like to see, let me know & I'll see if I can find a copy.
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23rd April 2013, 01:42 PM #6Intermediate Member
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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
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23rd April 2013, 02:04 PM #7
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
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23rd April 2013, 03:04 PM #8
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... )
- Andy Mc
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23rd April 2013, 03:11 PM #9
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......
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23rd April 2013, 07:32 PM #10
A long time ago in another life...
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
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23rd April 2013, 07:49 PM #11
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.
Those were the droids I was looking for.
https://autoblastgates.com.au
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23rd April 2013, 09:27 PM #12SENIOR MEMBER
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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.
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24th April 2013, 09:04 AM #13Intermediate Member
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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
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24th April 2013, 11:15 AM #14SENIOR MEMBER
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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.