First Project - LED Cube - Page 1
Forum section: electronics / beginners

author: wobblycogs
author_rank: Contributor
timestamp: June 18, 2012, 06:30:44 pm
content: For the last couple of months I've been branching out from my usual hobby of woodworking into electronics. I like the idea of building light into the things I make and LEDs are ideal for that.To get a bit of experience soldering and playing with some safe simple circuits I decided to build an 100 LED cube as described in this Instructablehttp://www.instructables.com/id/STEP-X-STEP-LED-CUBE-NO-Programming/. I've built the four layers of LEDs (25 LEDs in each layer wired in parallel) but now that I come to put the cube together I think I've found a (several actually) mistake in the instructions. The LEDs are 3.5v @ 20mA.Unfortunately the instructions get a bit vague at the point where you put everything together. From the photos it looks like he has wired the cube such that all the LEDs are parallel. This makes some sense because the idea is was to run it off a USB connection which is 5v, if the layers were in series I think it would require 14V. Additionally the instructions specify a 3.3 Ohm 2W resistor but that seems wrong, by the calculations as that would only power a single 25 LED layer (500mA) on a 5V supply.I think what I want is to wire the four layers in parallel and use a 1 Ohm 4W resistor with power being supplied by a 5W "wall wart" that is capable of supplying at least 2000mA. Does that sound right?Follow on question: I have a wall wart that supplies at 5V and up to 3500mA (it originally powered a USB hub), what's the connector on the end of the cable called? It's like a metal cylinder with a hole in the middle where a pin goes. I'd like to get socket for my LED cube.Cheers,hope my questions make some sense

author: codeboy2k
author_rank: Super Contributor
timestamp: June 19, 2012, 04:20:38 am
content: I looked at the pictures, it appears they are in parallel too.the 3.3 ohm resistor will limit the current to about 500mA. What will happen is that each LED will still drop 3.3V but will not get 20mA, instead each LED will get about 5mA. This may be bright enough, depending on the LED.If you want it brighter, and want to give 20mA to each LED, then you can use a single 3.3 ohm resistor in the power line to each level, so each level of 25 LEDs will get 20mA. But that may be too bright. Maybe it's ok at 5mA. You'll have to try it and see. So I suggest wiring it up on your bench a few different ways, before you cube it all together.Quote from: wobblycogs on June 18, 2012, 06:30:44 pmYes, that would work, each LED will get 20ma then. Maybe it's too bright? or use one 3.3 ohm in each leg of 25 LEDs. You still need 2000mA. (2A) total.QuoteThose are called coaxial power connectors. There are many different sizes, 2.1mm, 2.5mm (inside), 5.5mm, 9.1mm (outside), etc. You have to match the plug and jack and pin sizes.http://en.wikipedia.org/wiki/Coaxial_power_connector

author: wobblycogs
author_rank: Contributor
timestamp: June 19, 2012, 07:27:15 am
content: Many thanks codeboy2k I think I've got a better understanding now. Just to make sure I understand could you tell me if this thinking is correct:Assuming we are using a single 3.3 Ohm resistor in series and a 5V supply. The total current that can be supplied to the LEDs is 454mA because (Vsup- Vled)/R = I --> (5 - 3.5)/3.3 = 0.454. So if I wanted to make the cube brighter I would just put in a low value resistor, yes? A 1.5 Ohm would give a total current of 1000mA so (assuming the response to increased current is linear) about a doubling of brightness.The one thing I don't feel I understand now is how do you choose the power / heat dissipation level of the resistor. Why choose 2W not 0.25W or 3W? I realize it's something to do with the amount of current flowing through the resistor (less current = less heat after all) but I'd like to know how to make the choice myself.Thanks again.

author: codeboy2k
author_rank: Super Contributor
timestamp: June 19, 2012, 10:20:19 am
content: Quote from: wobblycogs on June 19, 2012, 07:27:15 amYes, that's right. Ohms law.Quotemore of Ohms law:so 0.5A * 0.5A * 3.3 ohms =0.825 Watts so the designer chose 2W. He could have chose 1W but it would run really hot. thats not a problem, but you would smell the heat, so better to have a little more than you need. I often double the calculated wattage. If you use change it to use a 1.5 ohm resister with 1000mA (1A) , then you have Power = 1A * 1A * 1.5 ohms = 1.5W so I would use at least a 3W resistorand sincethen:and sincethen:Here's a nice summary in a chart form:

author: wobblycogs
author_rank: Contributor
timestamp: June 23, 2012, 07:29:44 pm
content: HI again,I finally got a chance to wire things up tonight and I'm a little concerned that something is wrong.I have 3.3 Ohm 2W resistors and a 5V source (capable of 3.5A) as the design specified.When I wire up a single 25 LED layer with a resistor everything seems to be fine. The resistor gets fairly hot but not so hot that I'm worried the blue smoke will escape. I'd guess it gets to about 60 deg C after a couple of minutes - I can't quite hold it but it's not going to burn me or catch fire. With this set up I measured the current draw at just over 500mA and the voltage across the LEDs as 3.6VI then wired up all four layers (100 LEDs) in parallel again with a single resistor. Everything lit up fine but the resistor got much hotter after about 30 seconds, I'd guess about 80 to 90 deg C - not quite hot enough to burn me instantly. The current draw had increased to around 650mA.Does this sound correct?I'm guessing the increased current draw is because there are more possible current paths when all four layers are soldered in.

author: T4P
author_rank: Super Contributor
timestamp: June 23, 2012, 08:51:41 pm
content: well ... duh ...