alt Hacker NewsAssuming this is just standard 2.0 speeds, it should be nothing. It adds a slightly more expensive connector and 2 resistors… and that’s it.
Replies
chipweinberger • last Sunday at 8:25 PM
If you consume power over USB-C, 2 resistor are enough.
But for providing power over USB-C, you generally need a dedicated IC to handle VBUS switching. It wasn't worth it.
Also, once you ship a consumer product with a USB port on it, you'll realize a lot of people really don't understand how USB works, and USB-C doubly so. Square holes for square pegs goes a long way.
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bloudermilk • last Sunday at 4:39 PM
These are the hardest cuts to make in design, being so incremental, but they add up!
It is likely to be 1.1 speeds, and considering that USB-B connectors are chunky and dying out it is likely going to be cheaper to go for USB-C.
USB-C connectors from well-known brands start at less than $0.20, USB-B ones are $0.30. Same story for the low-end: C can be found for as little as $0.011, B starts at $0.05. Resistors are basically free - about a cent a piece if you splurge, more like $0.0007 if you really push it.
With the EU practically mandating USB-C on all portable electronics, the market is rapidly shifting towards it being the default for all 5V-powered gear: either you ship a cable/brick with it, or you give it USB-C. Even with a premium connector it is quite obvious which option is cheaper, so even bargain-bin gear is incentivized to adopt USB-C, which in turn pushes the connector manufacturers to develop dirt-cheap USB-C options.