> -----Original Message-----
> >
> > All of the hubs support Multi-TT. Based on this topology, I would
> > assume Hub #1 and Hub #2 perform the FS splitting, and the EHCI
> > controller on the USB host performs the FS un-splitting. Hub #3 would
> > only be passing high speed traffic between Hubs 1/2 and the PC.
> > Is this correct?
>
> Yes, pretty much. I'm not sure what you mean by "FS splitting" and "FS un-
> splitting", but it is true that all the split transactions would be sent to Hub #1
> and #2, and Hub #3 would see only high-speed traffic.
Poor terminology on my part. I was trying to confirm the middle hub (Hub #3) didn't need to perform any extra processing, and as you stated, it doesn't.
As a follow-up question, is the processing of Ssplit and Csplit handled by the EHCI hardware? Or does the kernel software need process the split transactions? If it matters, the our PC configuration has CONFIG_USB_EHCI_ROOT_HUB_TT=y.
> >
> > The PC has XHCI, EHCI and OHCI enabled. Test setup #1 is connected to
> > a USB 2.0 port and the lsusb output shows the FS devices on the USB
> > 1.1 bus. I think this means the OHCI driver is in use, but I'm not
> > certain.
>
> Yes, it does mean that.
>
> I don't see how you could have gotten more than 15 interrupt endpoints
> running at the same time unless the endpoints' bInterval value was larger
> than 1.
>
On all 7 devices, the IN and OUT interrupt endpoints have bInterval = 1 wMaxPacketSize = 64.
>
> The way it _is_ calculated is a mess. I can tell you the way it's _supposed_ to
> be calculated.
>
:)
>
> I suggest that instead of worrying about it, you divide your devices up among
> different buses. Most PCs nowadays have two EHCI controllers, and you can
> get more by adding PCI cards.
>
On the final hardware there will only be 1 EHCI controller available to service the full-speed USB devices in this topology. The other two controllers will be used for USB data from other parts of the system.
Thanks.
-Nate
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