Hi!
I am doing a project with WS2811 pixel strings. I am fairly new to Arduino, too. I have never done a project like this before. I do not have the supplies I am going to use yet, as I would like some advice before I buy. I am going to use Arduino Uno R3 (which I do have).
So, my main question is about powering the 150 lights that use 12V I want to utilize. I am not sure if the Arduino will be able to power all 150 lights without blowing its power regulators. I want to plug both the 12V wire and the ground wire of the lights into a separate plug on the transformer. Of course the data wire would be plugged into pin 3 of the Arduino board. Is it ok if it is not plugged into the Arduino? Here is a drawing of my idea.
Another question I have is about the coding for lights. I will use DemoReel100, but I modified it. Specifically, I modified the #define NUM_LEDS from 64 to 150. Is this ok? Will it mess anything up?
Thank you everyone for reading my post! If there is anything that you think I should change or add, please tell me.
Ok, thank you for telling me! The transformer I am going to use (according to specs on Amazon) is an AC to 12V DC. What I meant is if it would be ok if the ground plug was plugged into the transformer instead of the Arduino.
Quoting: "ground plug was plugged into the transformer instead of the Arduino."
What do You mean? All AC connections must be kept on the mains side, no one connected to the DC part.
The DC part of the supply, the negative, must be connected to the controller, negative and positive.
Can You use pen and paper and make a more clear schematics?
That diagram shows you are not powering the LEDs with the Arduino?
It is missing a series resistor ( 220R to 510R) between pin 3 of the Arduino and the data input of the strip and a large capacitor ( at least 470uF) between power and ground located at the very start of the strip.
Are you saying you have two +12V outputs on your regulated supply?
If so are they isolated from each other? If they are then you need to connect the two grounds together.
Are you sure you don't have a +12V and a -12V output?
In order to ask questions, you have to actually provide information. You have not.
You must first provide the most critical piece of information - what "WS2811 pixel string" do you have, or as we gather, intend to buy? Give the actual Web link.
WS2811 LED scripts come in various forms; there are 5 V and 12 V versions. Most 12 V versions as you seem to imply, drive three LEDs from each WS2811 chip, so there may be 150 actual LEDs but only 50 actual "pixels" and NUM_LEDS would therefore be 50. You need to clarify this.
A UNO is a very clumsy version for such projects. I would advise getting a cheap clone Nano.
The Arduino requires 5 V to operate. It is generally inappropriate to attempt to use the "barrel jack" on a UNO or "Vin" to power it with 12 V. Either you have a 5 V power supply or should use a switchmode "buck" regulator to power the Arduino when it is not plugged into a PC. And what other devices do you propose to connect to the Arduino to alter/ control the patterns?
Your diagram is wrong as it indicates that the data line from pin 3 to the LED strip is part of a large open loop of wiring. This would be extremely bad and a troublesome design as Mike has also indicated. The data wire must travel together from Arduino to LED strip with the ground wire, generally as twin cable (and the series resistor Mike mentions should be placed at the data input of the LED strip where the capacitor also is).
So to provide power to the Arduino, it would be most practical to bring the 12 V back from the LED strip together with the data and ground wires, and have the "buck" regulator there to power the Arduino. If you insist on using a UNO, you must disconnect the 5 V from the "5V" pin on the UNO when you connect it to a PC via USB. Not a problem with a Nano.
And by the way, you do not need to delete posts - you can re-think and modify them using the "pencil" icon below your post. Just do not add information to older posts in reply to later questions as it confuses the discussion.
I don't know why they sell these in 12V. They must be very inefficient and get quite hot. Most of the power will turn into heat, not light. As @Paul_B mentioned, in ws2811 led strips, each "pixel" is 3 LEDs, which makes efficient use of the power at 12V. In these strings, each pixel can only be a single led, which only needs a fraction of 12V, so the remaining voltage must be burned of as heat.
I would suggest getting the 5V version of these strings, and a 5V power supply.
In an RGB led, the red element needs around 2V. The green and blue elements need around 3~3.5V. So a 5V supply is enough. The extra voltage, the 1.5V, gets turned into waste heat.
In a 12V ws2811 strip, because there are 3 LEDs per "pixel", the LEDs use 3 X 3.5 = 10.5V, and the extra 1.5V gets turned into heat.
In ws2811 strings, there is only one led per pixel, so 8.5V gets turned into heat. That's two thirds of the total power becoming waste heat!
That doesn’t actually say if the two outputs are independent or not. So if you do go with this one check with your multimeter and the power off that there is continuity between the two -V connectors.
But as a general comment most Amazon electronics sellers haven’t got a clue as to what they are selling.
That is nonsense. They will just have current limiting resistors of higher values. A ws2811 needs a current limiting resistor also at 5v, and for the RGB leds if it is 3 in series, only the Red channels will need limiting (less voltage drop) . With string it is only a single diode so there will be a a current limiting resistor for all 3 channels.
voltage does not need to be burned off at all, just the current needs to limited.
The reason they sell 12v versions is that the voltage drop when using the same cable will be less relatively. Absolute it is the same of course, but if you lose 2 volts from 12 volts you have 10 volts left. Starting at 5 volts you would only have 3v left. Basically you can get away with longer and thinner cables using 12v, which with string can be quite relevant. There will not be any extra heat.
