Question 2.5 Sata SSD with NVME SSD

Zekke5577

Junior Member
Sep 3, 2020
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Hi all

Just wanted to ask my system an Aorus Z390 Master has an old 2.5 Sata SSD installed I wanted to get another drive a NVME 970 Pro most likely to use as a main drive and use the old Sata one as a backup for more space, can i use them both just was reading that certain M2 slots disable sata ports and havent had a M2 before can i use them both or if I get a M2 will it disable the Sata SSD.

Cheers
 

knght990

Member
Jun 3, 2006
178
9
81
On your board yes, you can use both with no loss of an Sata slot. But, depending on the slot you use, some bandwidth may be lost on specific SATA or PCIe slots. Page 34 of the manual explains you have 3 M.2 slots labeled M2M, M2A, and M2P.

M2M shares bandwidth for SATA 4 and 5.
M2P shares bandwidth PCIe x16 slot, labeled PCIEX4 (bottom slot). Both become 2x, instead of 4x, if cards are installed to both slots. (M2P also has a length limit on the M.2 drive)
M2A shares bandwidth for SATA 1.

Sharing bandwidth doesn't mean the slots won't work, or that you will have a noticeable loss in speed, it would depend on user demand of the items in these slots.

The manual doesn't indicate if any of the M.2 slots are associated with the CPU or if all are part of the z390 chipset so I assume all are part of the z390 chipset. The block diagram for z390 boards indicates if an M.2 is installed to the CPU it will reduce the PCIex16 to x8 and this seems undesirable to me.

In your case, I think I would use M2A and place a rubber dust plug in the SATA 1 slot to remind me to avoid using it.

Someone else may be able to provide better information, but that's how it looks to me.

Aorus Z390 Rev 1.0 Manual
 

Zekke5577

Junior Member
Sep 3, 2020
3
0
11
Cheers just didnt want to not be able to use it going to move backup data and less used programs to that drive and get a newer faster M2 for maindrive and most used programs/games, I dont have an other PCIE cards except for my 3090 so reduced bandwidth on the other slots wont be a problem.