Hi.
After reading the very detailed solution below, my question
is: does the standard care about the inexistent 8 bits (for
a 16 bits transfer on the 68 pin disk from the 50 pins
controller, would have to do instead two 8 bit transfers),
or that doesn't work at all?
I have never tried, but neither seen that this arrangement
indeed works, although with the performance penalty.
************************************
Javier Vizcaino. Ability Electronics. [log in to unmask]
http://www.automodelismo.com http://ability53.hypermart.net
  Starting point:        (-1)^(-1) = -1
  Applying logarithms: (-1)*ln(-1) = ln(-1)
  Since ln(-1) <> 0, dividing:  -1 = 1     (ln(-1) is
complex, but exists)
-----Mensaje original-----


On 30 Nov 98 at 18:50, Dennis Thiel wrote:

> I have an Ultra SCSI PCI adapter with 50 pin internal
connector.
> If I buy a 68 pin SCSI hard drive can I use a 68 to 50
adapter to
> connect it to my card? Will I lose any functionality on
the hard
> drive? Any problems I should be aware of?

  You *may* lose a little bit of throughput.  A 68-pin
("wide")
interface is able to move 16 bits of data at a time, but a
50-pin
interface only moves 8 bits of data.  Ultra SCSI (50-pin)
provides up
to 20 Mbps, whereas Ultra Wide can go to 40.
  The thing is, these bandwidth numbers are higher than any
single
drive made needs.  Odds are that you'll never use up your
SCSI
bandwidth until you add multiple fast devices.
  So go for it.

  Oh, one other thing:  The 68-pin drive will have four bits
for
"device ID".  Only three of these are connected through the
50-pin
interface.

David G

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