I am finally getting around to building my new system. It will be an AMD Athlon XP1700 on an ECS K7S5A with 512mb DDR. I'll be installing Windows XP Pro & apps on the SCSI drive & data will be on a 40GB IDE drive jumpered as slave on the primary channel with a TDK CD-RW as master on the secondary channel.

I got a Western Digital WDE18300 18.3 GB - Ultra2 LVD 1700 RPM SCSI drive on Agora. I am using an Adaptec 2940 U\W Controller. When I first booted & went into the SCSI utility it showed the drive at ID 8 & wanted to configure it as drive "D". I went to Western Digital’s web site & found that on the jumper block no jumper on the SCSI ID Jumper Section is SCSI ID "O". It was already configured as SCSI ID "O" (no jumper) but on the "option pins" it had the only jumper over pins marked "D TRGT INIT".

My question is why was it coming up as SCSI ID8 rather than "0" if there was no jumper covering pins in the SCSI ID section & what set of pins in the Options section should the single supplied jumper be covering?

Thanks!