16-bit sector addresses in CP/M 2.2

Hi, I'm attaching a hard drive to my 8080 computer and there seems to be something about the way the sector translation works that's caught my attention. Before I go wasting my time trying to figure it out myself I'll ask here because there's going to be someone who knows (it's 1:30am and I'm feeling a little lazy). Observe the following code: sectran: ;translate the sector given by BC using the ;translate table given by DE xchg ;HL=.trans dad b ;HL=.trans(sector) mov l,m ;L = trans(sector) mvi h,0 ;HL= trans(sector) ret ;with value in HL My question is does the BDOS use this as a 16-bit value, or does it cut it to an 8-bit value, like sectran does. H is loaded with 0 and L is loaded with the translated sector. I'm thinking that if this is the case, then a total of 256 tracks * 65536 sectors * 128 bytes = 2048MB. This could be done without a massive translation table by transferring the desired sector, BC, into HL. I've only just started writing the CBIOS for the hard drive interface and I don't really want to waste my time on pointless endeavours. It wouldn't make sense to use 16-bits for a sector register and 8-bits for a track register if you're only going to use 8-bits of the sector register, but perhaps the upper 8-bits are used for internal flags or error conditions? Another thing is that the total sectors per track in the Disk Parameter Block is a 16-bits, not 8-bits. Apologies if this has been discussed and I've missed it. (Perhaps someone can provide a link if it has been) Alexis.