I have an older Gigabyte Technology GA-MA770T-UD3P motherboard with an on-board network card described like this:

Model: "Gigabyte Onboard Ethernet"
Vendor: pci 0x10ec "Realtek Semiconductor Co., Ltd."
Device: pci 0x8168 "RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller"
SubVendor: pci 0x1458 "Gigabyte Technology Co., Ltd"
SubDevice: pci 0xe000 "Onboard Ethernet"

The BIOS of that board is described like this:

Vendor: "Award Software International, Inc."
Version: "F12"
Date: "03/22/2011"

However that BIOS has an interesting feature: It can determine the length of the wires of a network cable being plugged into the network card (if the other end is "open" (i.e..: not plugged into a switch), and I wonder how that works.

enter image description here

I've seen no other motherboard/networks card/BIOS combination that offers a similar feature, while I wonder: If one network card can do it, aren't all the others able, too?

  • 2
    \$\begingroup\$ Many are able, but not all; and the feature might be only available through Windows drivers (especially if it's not an onboard NIC -- and even then, it is rare for BIOSes of that age to have Ethernet support at all). The same feature is also relatively common in managed switches. \$\endgroup\$
    – user1686
    May 6, 2021 at 19:01
  • \$\begingroup\$ youtube.com/watch?v=Il_eju4D_TM \$\endgroup\$
    – AndreKR
    May 6, 2021 at 19:59
  • 1
    \$\begingroup\$ TBH, the surprising/unique/interesting part here is not so much that the NIC can do that (TDR for cable fault detection is not an uncommon feature in Ethernet PHY chips) but the fact that BIOS has support for querying that information from the PHY chip. \$\endgroup\$ May 8, 2021 at 3:59
  • \$\begingroup\$ @alex.forencich It's probably the combination of the NIC being onboard and ethernet being often used for booting. \$\endgroup\$
    – AndreKR
    May 8, 2021 at 10:23

1 Answer 1


Many network PHY chips can measure the flight time of a pulse. When a sent pulse is reflected back at the open end and the echo arrives back, the time it took can be used as an approximate estimate of the cable length.

It is called TDR for Time Domain Reflectometry.

  • 2
    \$\begingroup\$ One place where this is probably explained in detail are Microchip's docs for the KSZ line of phys and switches, where the feature is called LinkMD. I'm only aware it exists, but have not read up on it yet. \$\endgroup\$
    – jaskij
    May 6, 2021 at 8:29
  • 17
    \$\begingroup\$ One place where it's useful (and the reason why it gives you information per-pair, not just for the whole cable) is that it can help you find if there's a break in one of the wires, and if so, approximately where. \$\endgroup\$
    – hobbs
    May 6, 2021 at 16:49
  • 11
    \$\begingroup\$ @Eptin the impedance mismatch between the intact cable and the open end causes the reflection. \$\endgroup\$ May 7, 2021 at 0:15
  • 3
    \$\begingroup\$ @Eptin At low-frequencies you are correct in thinking that electricity doesn't flow into an open end. But when you remember there's a speed of light, then you start thinking about the electricity knows not to flow. The answer is it starts flowing anyway, bounces off the end, bounces back "upside-down" off the start, bounces off the end, a bunch of times, and after a few tens of nanoseconds, it settles down into an equilibrium where it's not flowing. but you can't see this unless you can see nanoseconds or you have a really really long cable! Humans have the tech to see nanoseconds now \$\endgroup\$
    – user253751
    May 7, 2021 at 10:20
  • 2
    \$\begingroup\$ @Eptin the signal reflects from open wires similarly like how waves behave in a diving pool. Waves propagate nicely in the water until they hit the pool wall so they can't continue and get reflected back from pool wall. \$\endgroup\$
    – Justme
    May 7, 2021 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.