Can someone please verify that my understanding is correct? I'm making a dumb device that will charge by plugging into a USB-C port. I only need a max of 500mA at 5V. I believe that the only thing I need to do is put 5.1k pull-downs to both CC pins on my device. The host port will detect the pull-down and turn on default USB 2.0 power levels, which is 500mA @ 5V. Is this correct? And do I need to do anything special to be sure the host port doesn't go to sleep and turn off the power?

 

hi,
This PDF gives more information.
Page #8/17
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Table #2 & #3

Link for PDF
https://www.google.com/url?sa=t&rct...00496853.pdf&usg=AOvVaw160A2g9xqMyxXHDbN7ps4W

 

@ericgibbs I don't see the pdf file.

 

hi Dick.
It shows it uploading 0% thru 100% then I get this shot.???

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Update:
Try this web link
Reloaded;

https://www.google.com/url?sa=t&rct...00496853.pdf&usg=AOvVaw160A2g9xqMyxXHDbN7ps4W

 

It is all in the timing! I was looking at your post before the link became visible. Reloading the page made it visible. Thank you Eric.

 

Thanks for the doc. Table 4 shows 500mA @5V as the default USB 2.0 current, but it doesn't say explicitly that you get that just by pulling down the CC lines. I believe this is the case, but would love to hear confirmation from someone who knows for sure. The consumer can monitor voltage on the CC line (see section The CC1 and CC2 Pins of this doc) to determine how much current the source can provide, and it looks like 500mA is the lowest option so logically I'm thinking I'm good here, but having confirmation always feels nice.

 

You might find a definitive answer on the USB Implementer's Forum website.
https://www.usb.org/usbc