I was wondering if the following is possible:

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I was hoping to be able to do the above with my USB 2.0 10m active extension cable as I cannot find a Y version of the same cable, to simply cut the male USB head that only contains power to plug it directly into my external 5V power supply.

The reason I ask is because my device draws too much current exceeding my Raspberry Pi's USB current limit. This is only the case because I have other devices also connected to it which I cannot remove. I also cannot use an externally powered USB hub because USB suspension becomes a problem for me. I also understand there are cables with DC barrel jacks for external power however is not suitable for my use case as they are expensive and I will need to do this many times for different systems I make.

  • \$\begingroup\$ While this link answers a different question, the discussion there explains why VCC from host is important for enumeration. So, to answer your question, no, what you suggesting is not possible. But with some changes you can make it work. Usually, self-powered devices have all necessary schematics inside. An ad-hoc outside solution is possible. \$\endgroup\$
    – Maple
    Commented Dec 18, 2021 at 16:58
  • \$\begingroup\$ Alternatively, you can use off the shelf "power injector" cable that does it for you. Not USB compliant, but works in a pinch \$\endgroup\$
    – Maple
    Commented Dec 18, 2021 at 17:10
  • \$\begingroup\$ Yup, no problemo \$\endgroup\$
    – Codebeat
    Commented Nov 10, 2023 at 18:36

1 Answer 1


No, such a cable would be forbidden by USB standards and will break the rules how USB devices and cables are expected to work in a system.

The devices need 5V on input, but would expect the power to come from host. Devices pull one of the data wires high as a sign to be ready for enumeration, but that arrangement allows the Raspberry Pi to be unpowered.

So if the device pulls too much current, it means it is an incompliant device.

  • \$\begingroup\$ While this is true that such a cable is non-standard, it is possible to make an adapter that will provide power to the device from PSU, while still presenting pull-up/-down for the host using host's VCC. Self-powered USB devices are perfectly standard, OP just needs a little tweak to make existing device behave like one. \$\endgroup\$
    – Maple
    Commented Dec 18, 2021 at 16:50
  • \$\begingroup\$ @Maple sure it is possible to design a circuit that does it. All you need is a high side switch, controlling the power supply to downstream devices based on the host VCC - basically what an powered USB hub would already do. However, this question was about making a simple cable which is not recommended. Or simply do it at your own risk - if the device is already non-compliant device pulling too much current, or the active cable non-compliant if it can't power the devices. \$\endgroup\$
    – Justme
    Commented Dec 18, 2021 at 17:09
  • \$\begingroup\$ Yes, power switch is simple solution, although I think better solution is using re-driver with enable input, like TUSB217A-Q1 or a simple analog switch, like ADG787. At least these do not care how much current the device consumes. But you are right, OP did not ask for anything more complex than a hacked cable. \$\endgroup\$
    – Maple
    Commented Dec 18, 2021 at 17:28

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