3 ms·
Absolutely. On top of that, I feel like USB-C is deliberately engineered to sell more ICs. Which wouldn't be that huge of an issue except as a result, all rele
by kanetw 4y ago
Absolutely.
On top of that, I feel like USB-C is deliberately engineered to sell more ICs. Which wouldn't be that huge of an issue except as a result, all relevant ICs are sold out (or unsuitable).
CC/PD use a custom transmission standard so you can't implement it without a PHY, which basically means at least one dedicated IC.
- Evidlo 4y agoWell you can still do USB2.0 over USB-C for all your low-speed, low-power gadgets.
- kanetw 4y agoYeah, that's about the only thing you can do without ICs. 500mA max, high-speed lanes unused.
- gbraad 4y agoMarker chips are 'optional' for USB-C, but essential for Lightning You can also directly feed the needed voltage without the negotiation. So this is only partially true.
- kanetw 4y agoA compliant USB-C source will not provide more than 5V/3A without PD negotiation period, and is only guaranteed to provide default USB power. So if you can potentially draw more, you need to check whether it's allowed.
- Dylan16807 4y agoYou need a chip in the power supply for that, but you already have a chip in the power supply because switching supplies are massively cheaper. You don't need a marker chip in the cable unless you intend to go over 20V/3A.
- kanetw 4y agoI'm just talking about end devices here (source/sink).
- deleted 4y ago[deleted]