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In this review (that quickly received significant public attention), a Google engineer pointed out a flaw in the cable, that, according to him can be "potentially damaging the USB hub or charger".

From the same review, "a value of 56 kΩ ± 5% shall be used, in order to provide tolerance to IR drop on V BUS and GND in the cable assembly". Could someone shed some light on this and explain how a bad cable could actually damage a hub or charger and why the 10 kΩ resistor used in this particular cable does not suffice to prevent this?

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Basically, that resistor is used to tell the device what kind of interface it's connected to.

If the resistor is 56K, it indicates to the connected device that it's actually connected to a USB-2.0 port, and it has to negotiate with the remote host to learn the amount of power it can safely draw.

If the resistor is 10K, it should indicate that it's connected to a host that is able to provide 3 amps of power without any negotiation being required.

The concern here is that even the highest-power USB-2.0 interfaces were only specified up to 1.5A (higher for dedicated chargers without data).

Basically, the concern is that there is no way to safely predict what will happen to every piece of hardware out there when the connected device draws 2X the current it was originally designed for. Depending on the hardware design, it could fail in a multitude of ways, some of which are capable of potentially damaging any connected devices.

If you're lucky and have a properly designed host, the failure would just be the open-circuit disconnection of a PTC which would prevent charging, but reset once the overload is disconnected. .

However, the above implies you have a host-device which has a PTC, which is often not true in many ultra-cheap hubs. If your device does not have any protection facilities, you instead get to discover the failure mode of it's power supply for yourself.

Again, ideally, the power-supply would just shut down on overload. However, again, this implies proper safety considerations have been taken into account in the power supply, which generally raise it's price. On the other hand, it's possible (for example) the host uses A DC-DC converter to convert a local 12V rail to 5V for the USB interface, and the overload causes the switching MOSFET to fail short-circuit, resulting in 12V on the USB 5V line, which would almost certainly damage both the host's internal circuitry, and the electronics of the device.

The absolute, pathological worst case would be the above, only the connected device/charger then catches on fire, and burns down your house, killing your cat/cat-substitute (significant-other, kids, dog, ferret, small angry squirrel etc...).

This is a unlikely scenario, and the failing system described above is certainly not well designed, but it would work normally otherwise, and is standards-compliant. The adapter manufacturers cannot know that there are not devices like that out there, and as such selling a device which intentionally violates the USB specifications just to make a connected device charge faster is highly irresponsible.


In effect, the 10KΩ resistor doesn't "prevent" it, it's actually used as part of a system for informing the connected device about the host it's connected to. It's effectively lying to the connected device, and claiming the host can do things it was likely never designed to do.

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    \$\begingroup\$ But you're assuming that the power source will have PTC. If user connects their high-current device to one of the ultra-cheap $1 shipping included Chinese chargers, that use sellotape for insulation, the consequences may be much more drastic than just loss of the device. \$\endgroup\$ – AndrejaKo Nov 9 '15 at 1:49
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    \$\begingroup\$ @AndrejaKo - Did you miss the If you're lucky portion before the description of a PTC saving things? \$\endgroup\$ – Connor Wolf Nov 9 '15 at 1:53
  • \$\begingroup\$ Of course! Maybe it should be pointed out better? \$\endgroup\$ – AndrejaKo Nov 9 '15 at 7:15
  • \$\begingroup\$ @AndrejaKo - I've done a fair bit of editing. What do you think now? \$\endgroup\$ – Connor Wolf Nov 9 '15 at 19:17
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    \$\begingroup\$ In a laptop it could of course initiate the SDI command \$\endgroup\$ – Aron Nov 10 '15 at 1:39

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