XMOS uses port precedence; where there are multiple functions on a pin, the narrowest active port wins the pin.
Consider an XU216-512-TQ128 device. If Port 1M and Port 8D are both enabled as outputs, then XD36 is connected to output register of port 1M. You can still use the other 7 bits of 8D, however if you read from 8D, bit 0 will contain the output value of 1M.
So because the USB PHY uses 8A and 8B internally, you can still use 4A/4B/4C/4D to read/write from the pins. Unfortunately you do loose 1E/1F/1G/1H/1I/1J because the pin these appear on has no other ports connected to it, and besides, 1b is the narrowest port so would win anyway.
If you have a simple question and just want an answer.
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