I'm building a project that involves lighting up 6 separate groups of LEDs for a total of 70 LEDs. The largest group contains 19 LEDs. All I need to know is A) will an Arduino Micro [or potentially LilyPad] push enough current to power 70 LEDs simultaneously, and B) will a single digital pin push enough current to power 19 LEDs simultaneously? I'd test it, but I haven't bought enough LEDs yet ![]()
Each LED can draw up to 20mA, 70 x .02 = 1.4A, you will need an external power supply.
http://www.dipmicro.com/store/DCA-0520
If you want 19 on at one time, then use a 12V supply, connect the LEDs in strings of 4 or 5 with a current limit resister per string, connect the strings in parallel, top LED anode to +12, and control the bottom cathode connection to GND with a Logic Level, Low Rds N-channel MOSFET:
Actually, 19 LEDs, 5 strings = 100mA.
Use TPIC6B595 to sink current from the various strings.
An atlernative to Bob's sound advice:
Use ws2812b leds. Each has its own built-in driver and can display any colour you like. Make a single string of them, but grouped phsically however you want. Use the "neoPixel" library to drive them with a single Arduino pin. You will still need a separate 5V supply for 70 of them!
Paul
Awesome, thanks for the information guys. I'll get to testing.
HI! I have a similar project underway- using a single strand of WS2812 pixels, in 4 separate groups (Left, Right, Front, Back). I'm still learning using the NeoPixel library, but I can figure out how to group them logically. I need the left and right side to be in sync with each other and to be able to control the front and back sides individually. Any example on how to group the LED logically?
Thanks!
Hi, you should really start a new thread for this, but essentially if the leds are in a single string then they arelogically one group. You have divided them into 4 physical groups. Lets say you have 60 leds in 4 equal groups. The groups are addressed by number, i.e. group 1 is leds 0 to 15. Group 2 is leds 16 to 31 etc. and group 4 is leds 45 to 59. That's all there is to it.
Paul