0

I'm really new to hobby electronics, just got a raspberry pi and trying to read different data sheets and requirements to figure out what I need to get so that I can drive an LED strip with the correct voltages and current. I've been looking at the data sheet for the WS2812 NeoPixels here: https://cdn-shop.adafruit.com/datasheets/WS2812.pdf

Specifically at enter image description here

I'm quite confused about the Input current listed there and what exactly it means. Firstly it says in the Condition VI=VDD/VSS but VSS is 0 so is that not actually a division going on there? And I'm a little confused about it being a max of 1 micro Amp for the input, is that really the max current it can take?

I'm trying to figure out what sort of pull-up resistor rating I need so that I can drive the data line with an external 5v power supply and control the switch by turning on and off sending the 5v to ground with the GPIO pins of my raspberry pi, something similar to #4 in this answer on another post: https://electronics.stackexchange.com/a/82149/166893

Perhaps I'm going about this wrong and I don't need this? If there is any guidance I can get in determining the pull up resistor rating that I need that would be great. I can see that I need at least 3.5V but that doesn't seem to be all the information I need.

csteifel
  • 123
  • 2

1 Answers1

0

That is an "or". \$V_I\$ is supposed to be a digital input, so the valid states are either 0 (\$V_{SS}\$) or 1 (\$V_{DD}\$), and the input current value is valid for both in steady state (the input is supposed to have an impedance of at least \$5\ \mathrm{M\Omega}\$. It's most likely connected to the gate of a FET, so there is your high impedance.

Pulling up the GPIO on the RasPi is a problem. The GPIOs are not designed to see higher voltages than the processor supply voltage, which is \$3.3\ \mathrm{V}\$.

Instead, I'd just connect the WS2812 directly, through a small (\$33\ \mathrm{\Omega}\$) series resistor if you have one. The voltage drop across the resistor is negligible at \$1\ \mathrm{\mu A}\$, but it protects the Pi if the first WS2812 fails and shorts D to ground.

This is also out of spec (\$3.3\ \mathrm{V} < 5\ \mathrm{V}\centerdot0.7\$), but usually works.

Simon Richter
  • 12,031
  • 1
  • 23
  • 49
  • My throught was that if I do something like #4 on that linked answer that the GPIO pin wouldn't actually ever see that voltage it would just be "actuating a switch" essentially to allow the 5v to go to the data line, is that still a problem? – csteifel Oct 25 '17 at 20:23
  • @csteifel, yes, that would work and be within spec, at the cost of an extra component. – Simon Richter Oct 26 '17 at 14:36