Let's say I have a USB 3.0 card with four USB 3.0 ports on it. I just connected the card to my mother board.

When we say 640MB/s of bandwidth for USB 3.0, is that bandwidth per port (or 640*4 = 2560MB/s total), or is that the total bandwidth (or 640/4 = 160MB/s per port)?

  • \$\begingroup\$ USB3.1 is 10MBps \$\endgroup\$ Mar 10, 2017 at 19:39
  • \$\begingroup\$ Aight...but I have USB 3.0 \$\endgroup\$
    – Pototo
    Mar 10, 2017 at 20:52

1 Answer 1


The short answer from the USB spec, each port has a maximum throughput of 450 MB/s. But the USB 3.0 PCIe Host controller interface has a maximum throughput of 500 MB/S. USB 3.1 raise this to 1GB/s by changing the serial data from 5 Gbps to 10 Gbps and changing the data encoding.

The 450 MB/s is the maximum data transfer rate for the whole USB 3.0 interface (It is a single lane PCIe interface). Just like USB 2 each port supports transactions that while underway take the whole bus at the exclusion of all other devices on all other ports for this Host controller.
If you had a 4 port card and had disk drives connected to each port, then the aggregate throughput for all drives combined cannot exceed the maximum of 450 MB/S (ignoring other protocol overheads) for the Host controller (the one lane PCIe interface). For each individual drive the throughput rate would be 450 MB/S for each data transaction, but an aggregate of 450/4 if all the drives were transferring equal data loads.

  • \$\begingroup\$ :-(...I understand. That is not very good. Maybe I should try USB 3.1 some time. \$\endgroup\$
    – Pototo
    Mar 10, 2017 at 20:54
  • \$\begingroup\$ Same deal, just a faster Lane. 10Gbps PCIe instead of 5 Gbps. usb.org/developers/ssusb \$\endgroup\$ Mar 10, 2017 at 21:23
  • 1
    \$\begingroup\$ Actually, the max theoretical bandwidth of USB 3.0 is 500 MB/s. as well as PCIe Rev.2, not 640. Because of 8B/10B raw encoding, which takes 10 bit times to transmit 1 byte. So 5000 Mbits/s / 10 = 500 Mbytes/s. \$\endgroup\$ Mar 10, 2017 at 23:25
  • \$\begingroup\$ You are absolutely correct....altered answer \$\endgroup\$ Mar 11, 2017 at 1:13

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