02-11-2021, 09:51 AM
(02-09-2021, 11:40 PM)dsimic Wrote: My apologies for the delayed response.
I'd say that it should be all fine and that you're on the right track. Otherwise, how is the battery in the recently announced PinePhone keyboard case going to be (dis)charged? The case connects to the phone only through the pogo pins.
No worries, I'm not in a rush :-) Haven't gotten the extra testing in yet that I wanted to do, beyond what I already shared.
Yes exactly, the keyboard case with battery is yet another example of a device that maybe draws power from the Pinephone, but that also _supplies_ power.
Our findings here are basically that one pogo pin is a power input, and the other is a power output. (And it's maybe also acceptable to draw power from the power input pin when it's not at 0 V. For example, you could draw power from the USB C connector through there.)
(02-09-2021, 11:40 PM)dsimic Wrote:(01-18-2021, 08:40 PM)bokomaru Wrote: Yeah. I'm still curious why they are named "DCIN" and "DCIN ", but we'll probably never know. Maybe the hardware designer had to work around something. Maybe it means something, I don't know.
That's probably due to the current-carrying capacity of a single pogo pin. As the single pogo pin probably cannot carry the required current, the only good solution was to use two pogo pins in parallel.
Nope, the two "DCIN" and "DCIN " nets are both connected to only the one pogo pin. The other pogo pin is "USB-5V" and connects somewhere else and has a different purpose. There aren't any two pogo pins that are connected to the same place to increase current-carrying capacity.
When we were talking about "DCIN" vs "DCIN ", we were only just trying to guess that maybe the USB C connector wasn't actually on the same net as the one pogo pin. But we were wrong; "DCIN" and "DCIN " are in fact connected together. That's indicated both by the schematics and an experimental connectivity test.
Otherwise, it's also still not clear _what_ the current-carrying capacity is. The wiki says 500 mA in one case, but why is that? Did the author of that just assume that it's 500 mA because that's what one of the USB standards is supposed to provide? Or can somebody show that some component in the schematics actually provides 500 mA (or more, or less)?