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I have recently read of a dangerous USB drive containing a circuit instead of storage, that destroys the hardware of a computer by injecting a high voltage, high current pulse into the port it is plugged into. See this article

It should be easy to test an unknown USB drive for this sort of danger by plugging it into a separate test module before taking the risk of inserting it into a computer's USB port. Does such a test module exist? Is there a commercial version, a circuit or kit that can be obtained?

Having read that article I would routinely test every USB device before using it.

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    \$\begingroup\$ Remove the casing and check what is inside. In the article there are a couple of pretty big capacitors, in a USB drive should be nothing like that. \$\endgroup\$ – Bence Kaulics Dec 27 '15 at 18:04
  • \$\begingroup\$ I see this as paranoia. 1. Why would you plug in an unknown device? 2. The odds of actually coming across a "USB killer" are astronomical. 3. USB is probably one of the most hardened buses in existence. I find it very hard to believe that a bus designed for typical consumer use and abuse could actually be damaged by 110V pulses. \$\endgroup\$ – Matt Young Dec 27 '15 at 18:08
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    \$\begingroup\$ @MattYoung I am aware of that and just like the DO160/ABD100 ( 1500V:80A ) for lightning is for a SHORT PERIOD OF TIME (IEC6100: 100ns, DO160: 1us). The steering diodes/TVS can handle that for a short time. such USB killers are there for seconds if not longer. \$\endgroup\$ – JonRB Dec 27 '15 at 18:26
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    \$\begingroup\$ You are wasting your time. But according to the article, it applies a large DC voltage to the data lines. So you could easily design a test module which supplies 5V, just like a real USB port, and hook a volt meter to the data lines. If the death disk applies 100V to the data lines, don't plug it in to your computer. The other test method is to plug it into a disposable computer, then watch for smoke. \$\endgroup\$ – mkeith Dec 27 '15 at 20:06
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    \$\begingroup\$ A $5 USB hub is probably a good enough testing device. \$\endgroup\$ – rioraxe Dec 27 '15 at 22:47
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There is also the option of just not picking up any USB device you see and joyfully plugging it into whatever computer is nearest, as a general rule.

Back when Floppies (any size, really) were the thing and people owning a computer possessed some sensibility about it, this was actually general common sense. Go to an exhibition, get disks, think very, very hard about whether you want to put those in your main Atari/MSX/Whatever/PC.

Not because disks could generate high voltage and destroy all your hardware, but because they can also contain any kind of software that could do any kind of whatever.

Sure, modern operating systems are a bit more protective about "hey, this thing wants to install and it offers a driver itself, cool, here we go!" or "hey, this says auto run the software, and GO!". But funny thing, there are still plenty of holes and especially so in USB devices that combine multiple purposes.

Ever tried a Sandisk Cruzer on a fresh Windows install? Cool huh, that it offers to install software for you! If you have some security options disabled in up to Windows 7, it may even do that without telling you. Now think about someone like that monkey brain, but with more of a sense for software than hardware.

That said, here's your schematic for the destroyer detector:

schematic

simulate this circuit – Schematic created using CircuitLab

If D3 flashes there are spikes higher than 6.5V on the USB VCC, if D4 flashes there are spikes higher than 4.5V on the USB data lines.

The 1N4148 diodes are only to protect against high energy spikes forcing an insane base current through the relatively modest 20k resistor, now that flows through those diodes first. Still a lot can break, of course, but this could be mass produced at the one or two dollar range, excluding the power source. If you see it as a discardable tool, you can leave out the 1N4148 diodes and the 3.3k resistors.

The 1N4007 protects the power source against spikes on the power up to several hundreds of volts (though if you expect those, remove the capacitor or get one rated for hundreds of volts, or that one might pop).

When the diodes flash faintly the "violation" is small (10V spikes or such), when it flashes brightly the spikes are high voltage and probably high energy.

But, of course, mass producing this is an absolute and utter waste of your money and I see this as a simple course of "You don't always need complicated electronics to detect things" tutorialing, rather than "here's a real problem that needs a real solution".

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  • \$\begingroup\$ Thank you Asmyldof for a comprehensive and useful answer. Thanks to all who commented for an interesting discussion, especially to those who read my question carefully, and also followed the link. My statement "I would routinely test every USB device before using it" was a bit tongue-in-cheek -- for most USB sticks there would be good evidence that it is genuine: being new from a branded range; seeing it removed from a computer; or asking the donor to plug it in his own computer. A tester like the one described would be a last resort. \$\endgroup\$ – Harry Weston Dec 30 '15 at 17:27
  • \$\begingroup\$ This circuit does not help, if the USB Killer is a little bit more intelligent and activates the high voltage with a delay, or on an event (reading special block, or file). \$\endgroup\$ – Jonas Stein Nov 10 '18 at 13:47
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Get a cheap USB wall charger. Open and modify it so that the data lines are shorted together and add a current limiter to the Vbus pin. Now when ever you plug a drive in. It will be powered. If it's a USB killer, when it sends the pulse on the data line it should blow up quite nicely. Thus if the USB stick explodes, it was a USB killer.

But seriously, just use a cheap wall charger. If the charger stops working then you might want to question the USB drive. No computer anywhere to take damage.

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  • \$\begingroup\$ Oh goody, a down vote with no comment. \$\endgroup\$ – Dave Dec 28 '15 at 4:03
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    \$\begingroup\$ Presumably someone downvoted it because the linked article says it puts the pulse on the data lines, so it's not likely to damage a charger. \$\endgroup\$ – PeterJ Dec 28 '15 at 4:23
  • \$\begingroup\$ Presumably the killer USB device only has one pulse-charge in its super capacitor. Thus, it'd be spent on the first device it sees. \$\endgroup\$ – RoboKaren Dec 28 '15 at 4:33
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    \$\begingroup\$ @RoboKaren If you read the linked article about the device in question you can learn that it does not use supercapacitors, and will keep cycling high voltage + high current pulses until the target is fried. \$\endgroup\$ – pipe Dec 28 '15 at 4:36
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    \$\begingroup\$ @Dave, don't see that happening myself - the charger will just have a couple of resistors to set that voltage (it's up to the device to respect it) and if the USBKiller couldn't handle that voltage on the data lines it would self-destruct itself no matter what was connected. \$\endgroup\$ – PeterJ Dec 28 '15 at 5:21

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