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by ggm 1062 days ago
Is there a royalty on the chipset to do it better than resistor pullups? The article implies its a relatively high BoM cost to have the negotiation. I can see its pinouts, board design, test but the actual chip.. surely is still down in the 0.0001c range? or is this one "pay the cartel" expensive?
2 comments

The chip is very expensive, sadly. If it were just a matter of the cartel, that would be one thing (and would have Shenzhen-made workarounds). No, the USB lunatics made USB-PD one of the most insane specifications you'll ever find, so that no one can implement it correctly period. (Seriously, it's actually the description of the actual behavior of an old TI part, bugs and all, turned into a spec. It's horrid. It's the genuine worst specification ever.)

So you end up needing a 48MHz Cortex-M0 microcontroller just to do the god damned power delivery. At least, I've never seen it done by a less capable part. And processors of that class are, alas, just not that cheap.

As someone who has read the PD spec and agrees with your assessment- I’m curious what that chip is?
I don't, I'm sorry. I also don't think I ever tracked down that part myself (but who knows, it's been years), so it may not actually be true. But it's sure plausible!
The chip lets you draw more than the minimum USB-A standard power. And that’s what, 500mA?
Yes, thats it's role, but it does it "intelligently" where the resistor pullups do it the "dumb" way. His complaint is they cheaped out on even the resistor path.

I just wondered if some hypothesized USB-C consortium decided to "patent" how the signals work, and charge $ for compliance/conformance to the chipset for a logo badging and green tick.

It's more usual higher up the complexity food chain like MP4 decoding. Frauenhoffer wants its IPR respected. Philips made coin on conformance to the audio cassette form factor. Somebody made book with CD-ROM size, encoding. It's normal.

Without such a chip you can use resistors to configure up to 5V, 3A.