A pic is providing the endpoint, so it's also PIC related.  I read about
interrupt xfers in the USB spec and would like to confirm my
understanding.

For an int endpoint you specify a packet size, and the frame interval.
So for a packet size of 64 bytes (FS max) and an interval of 1 frame (1
frame=1ms) you are guaranteed (64*1000)=64,000 bytes per second of
throughput.

Depending on bus traffic, you may get more than one packet per frame, up
to 90% of the available bytes in a frame.  A frame is 1500 bytes but
some of these are for packet overhead and spacing.  That plus the 90%
limit means you can get as many as 19 64 bit packets in a frame,
resulting in a max possible bandwidth of (64*1000*19)=1,216,000 bytes
per second

Finally, for a protocol converter (protocol X to USB), if protocol X is
up to 40 kbits/sec, then I should choose a packet size of 8 bytes and a
frame interval of 1, resulting in (8*1000*8)=64,000 bits per second on
USB.  The small packet plus low frame interval reduces buffer
requirements in the PIC.

Does the above look right?

thanks,
--steve

-------------------
Steve Ravet
ARM
steve.ravet@arm.com

-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist