Subject: SMD (7053) Questions again.
To: None <port-sparc@NetBSD.ORG, port-sun3@NetBSD.ORG>
From: David Gilbert <email@example.com>
Date: 12/25/1996 01:08:02
OK... let's see if I can put these thoughts in order. I have
the backplane jumpers all configured properly, and I have all the
jumpers on the 7053 set properly.
Firstly, on bootup, I see xd0 but not xd1 (both are
connected). They are both D2363 drives, and I have them
daisychained. I'm wondering if someone can guide me with the order
that things get connected at the back --- is the 'out' control cable
the front or back one (looking down on the connectors). Likewise, on
the last drive, one cable comes in, and the other has a terminator in
it... does it matter which one.
That said, I'm observing a few other strange things. I can
boot just fine with a late 1.1-current kernel as long as I don't have
to fsck my SCSI disks. However, this config will eventually crash.
If I have to fsck my SCSI disks, the kernel crashes with an alignment
fault shortly after declaring sd0a fine.
Someone was also kind enough to send me a configuration guide
for the 7053, and it talks about only having one 7053 if you also had
SCSI (where you could have 4 otherwise?). Is this a limitation that
Sun just didn't want to work around, or is it and actual hardware
problem --- what I'm asking is if NetBSD can work around this...
Now... I have 3 D2363's and one DK815-10. I'm under the
impression that I can get better performance by putting them on two
controllers (I may be wrong there). I have cables to attach them to
three (I basically need more daisy chain connecters to attach them to
fewer than three).
Some help with this old hardware would be nice. I'm running a
4/260 (some of you have seen my postings before). It's very possible
that a more recent kernel would be in store... if someone could work
one up for my config file :).
|David Gilbert, PCI, Richmond Hill, Ontario. | Two things can only be |
|Mail: firstname.lastname@example.org | equal if and only if they |
|http://www.pci.on.ca/~dgilbert | are precisely opposite. |