6 ms·

TL;DR; USB-C cable had GND tied to Vbus and Vbus tied to GND (opposite as it should be) Honestly, I'm surprised that this isn't something that thelaptop and ot

by earlz 11y ago

TL;DR; USB-C cable had GND tied to Vbus and Vbus tied to GND (opposite as it should be) Honestly, I'm surprised that this isn't something that thelaptop and other such USB adapters aren't equipped to deal with. This is only slightly worse in my book than a USB adapter being fried from shorting the Vbus and GND pins together

krapht 11y ago
Why would you design in reverse polarity protection? It adds cost, and the mechanical connector already makes it impossible for VDD to short to GND in USB type C because of its rotational symmetry.
- andrey-g 11y ago
  I constantly have Vdd shorts due to ground loops when measuring stuff. But a more frequent case is worn out cables (every lightning cable I've owned has eventually broke). Its not rare enough of a failure mode to ignore.
  - krapht 11y ago
    Yes, and that's why the USB spec puts the burden of protecting against electrical shorts on the controller. The device side shouldn't have to worry about trying to shunt 100 Watts of power somewhere. I mean this kind of protection is way beyond typical ESD protection due to the power involved. It would add a lot of cost to do this, maybe even a dollar or two in raw material.
    - mikeash 11y ago
      If you stop the current from flowing, you don't have to shunt any power anywhere. You'd have to open the circuit in the face of 5V of potential hooked up backwards, not really a tall order.
      - krapht 11y ago
        I don't want to be a broken record, but you're right, not a tall order, but no it's unreasonable on account of cost. It's impossible for USB C to be hooked up backwards by its mechanical design, so why guard against it? The high-side transistor to prevent reverse polarity faults could be 3 to 4 cents in manufacturing cost! It's also not trivial to make sure that transistor trips fast enough to prevent all your low-voltage ICs from being destroyed.
        mikeash 11y ago
        Sure, but the magnitude of that cost matters a lot. You were saying a dollar or two in raw materials to protect against 100W, now you're saying three or four cents. A few cents to protect hundreds of dollars in equipment from electrical faults seems worthwhile to me.
        teraflop 11y ago
        That makes sense from the consumer's point of view. But imagine you're a manufacturer, producing USB device controller chips by the million. Would you really be willing to spend tens of thousands of dollars to protect against a failure mode that can only happen due to some other company's extreme negligence?
        mikeash 11y ago
        Yes, but of course it doesn't matter what I think. It's not abnormal to add protection that isn't needed when everything works properly. As noted elsewhere, the USB spec already requires resettable overcurrent protection, for example.
        knughit 11y ago
        Tens of thousands of dollars total is far less than it costs a company just to moot the question.
        deleted 11y ago
  - klodolph 11y ago
    Shorts and polarity reversal aren't the same thing, and the protection circuitry is different.
- earlz 11y ago
  unless the connector were cut, or some other catastrophic failure. I tend to think ruining a devices USB controller should be near impossible short of something like the USB killer (http://arstechnica.com/security/2015/10/usb-killer-flash-drive-can-fry-your-computers-innards-in-seconds/ http://arstechnica.com/security/2015/10/usb-killer-flash-dri...)
  - lucaspiller 11y ago
    I was playing with a FTDI (a clone :D) and an Arduino and accidentally mixed up VCC and GND from an external power source. I only realised because Minicom was failing to connect to the device. Then I realised what I had done, I tried removing it from the USB port and putting it back in, but it wouldn't work. I then tried a mouse too, but that wasn't working either. Thinking I'd killed a USB port on my Mac (well there are two, so...) I then tried rebooting and that fortunately fixed it.
    - earlz 11y ago
      Yep, did almost exactly this, shorted a converted 3.3V from a different USB port (same machine) to ground on a different USB port and end result is my motherboard lost all USB connectivity until I rebooted
    - nerfhammer 11y ago
      the usb ports on macs are supposed to have overcurrent protection. You actually get a dialog popup in the OS when it happens. I know because I've triggered it doing something with an arduino.
      - klodolph 11y ago
        USB overcurrent protection is pretty sophisticated. Devices get 100 mA for free, then have to ask the host for more current. Well, it's a bit more complex nowadays, but that's the basics.
        rcxdude 11y ago
- marcosdumay 11y ago
  Well, it protects against some metal object getting inserted into the USB port.
- digi_owl 11y ago
  This was a C-to-A adapter cable. what was wrong was that the Vbus on one end was hooked to the GND of the other end.
- deleted 11y ago
  [deleted]
- vacri 11y ago
  Look for a long response near the end of the article comments, from Mike Crawford. It goes into detail as to why that protection isn't designed in - it's not cost (diodes are cheap), but power drop, which has side effects.
paulmd 11y ago
Electrical faults are just a fact of life. Wires fray, cables get tugged, or are miswired from the factory. And you just know that sooner or later some genius is gonna jam a butter knife in there for the hell of it. The basic USB spec requires that all ports be protected with a small self-resetting polyfuse for exactly this reason. I don't know if the Type-C or USB 3.1 specs includes something similar - I would assume so but perhaps the additional current capabilities made that infeasible. http://elinux.org/Polyfuses_explained http://elinux.org/Polyfuses_explained A standard USB polyfuse usually resets pretty fast (as soon as it cools down) but they could take up to a few days. It's also possible the manufacturer cheaped out and didn't include a polyfuse, of course.
- qb45 11y ago
  This was an A-C cables, which means it was used to charge the laptop, not to power something from the laptop. Current limiting doesn't magically protect powered devices from voltage reversal (-5V instead of +5V), so both the laptop and power delivery analyzer went poof. They may have taken enough current to trip charger's current limiter, but then it was a bit too late.
- ackfoo 11y ago
  I found a USB Cellular data adapter in the trash and plugged it in to my MacBook Pro, which instantly shut down. Rebooted, and the USB port did not work. Shut it down and left it for half an hour, and, yay, all working again. Disassembled the adapter. Turned out to be a dummy used for display. No guts, just a steel bar across all four pins of the USB connector. Looked real from the outside. It was a very, very stupid thing to do. Thank-you Apple for putting a self-resetting fuse in my MacBook Pro. Saved my idiot butt.
  - culturestate 11y ago
    Plugging random USB devices into your computer is a bad idea, full stop -- miswired or not. "I found this in the trash and plugged it in" is the bane of CISOs everywhere.
cesarb 11y ago
That's worse than a short. A short pulls the Vbus to 0V, leading to high currents on the source, so whatever overcurrent protection it has kicks in. What he had here, instead, was -5V on the Vbus pin, way outside the acceptable voltage range (which is probably something like from -0.5V to 5.5V). It's no wonder everything fried. The protection against reversed voltages like that would be a crowbar diode between Vbus and GND, to convert the negative voltage into a short circuit. But even if it had that reverse protection, if the current was high enough (since the negative voltage was from a charger, it's not unlikely) the diode could burn open, and then you have negative voltages again. And I don't know how common these crowbar diodes are; they probably are rare, since the USB connector is keyed so it can't be plugged "backwards" (the type C connector is mirrored instead, either orientation connects the pins the right way).
- Zardoz84 11y ago
  A diode crowbar or a single would imply a drop of ~0.6v on the line so you always would get 4.4 volts . However there is other kind of protections like this : http://www.edn.com/design/analog/4368527/Simple-reverse-polarity-protection-circuit-has-no-voltage-drop http://www.edn.com/design/analog/4368527/Simple-reverse-pola...
  - cesarb 11y ago
    From what I understand, a crowbar diode is a reverse-biased diode put in parallel with the bus, not in series. In normal operation (correct polarity) it doesn't conduct, since it's the "wrong way", and there's no voltage drop since it's not in series with the circuit. But when the polarity is reversed beyond the diode's voltage drop (that is, beyond -0.6V in your example), the diode is now the "right way" and conducts, limiting the reverse polarity voltage to the diode's voltage drop (in your example, it won't get below -0.6V). However, that means that all the reverse-polarity current goes through the diode, in effect a short (like putting a physical crowbar across the battery terminals). If the diode holds long enough, the voltage source's short-circuit protection (be it a fuse, a polyfuse, or something else) should cut the current and all is (mostly) fine.

by earlz 11y ago

krapht 11y ago

Why would you design in reverse polarity protection? It adds cost, and the mechanical connector already makes it impossible for VDD to short to GND in USB type C because of its rotational symmetry.

andrey-g 11y ago
I constantly have Vdd shorts due to ground loops when measuring stuff. But a more frequent case is worn out cables (every lightning cable I've owned has eventually broke). Its not rare enough of a failure mode to ignore.
- krapht 11y ago
  Yes, and that's why the USB spec puts the burden of protecting against electrical shorts on the controller. The device side shouldn't have to worry about trying to shunt 100 Watts of power somewhere. I mean this kind of protection is way beyond typical ESD protection due to the power involved. It would add a lot of cost to do this, maybe even a dollar or two in raw material.
  - mikeash 11y ago
    If you stop the current from flowing, you don't have to shunt any power anywhere. You'd have to open the circuit in the face of 5V of potential hooked up backwards, not really a tall order.
    - krapht 11y ago
      I don't want to be a broken record, but you're right, not a tall order, but no it's unreasonable on account of cost. It's impossible for USB C to be hooked up backwards by its mechanical design, so why guard against it? The high-side transistor to prevent reverse polarity faults could be 3 to 4 cents in manufacturing cost! It's also not trivial to make sure that transistor trips fast enough to prevent all your low-voltage ICs from being destroyed.
      - mikeash 11y ago
        Sure, but the magnitude of that cost matters a lot. You were saying a dollar or two in raw materials to protect against 100W, now you're saying three or four cents. A few cents to protect hundreds of dollars in equipment from electrical faults seems worthwhile to me.
        teraflop 11y ago
        That makes sense from the consumer's point of view. But imagine you're a manufacturer, producing USB device controller chips by the million. Would you really be willing to spend tens of thousands of dollars to protect against a failure mode that can only happen due to some other company's extreme negligence?
        mikeash 11y ago
        Yes, but of course it doesn't matter what I think. It's not abnormal to add protection that isn't needed when everything works properly. As noted elsewhere, the USB spec already requires resettable overcurrent protection, for example.
        knughit 11y ago
        Tens of thousands of dollars total is far less than it costs a company just to moot the question.
        deleted 11y ago
- klodolph 11y ago
  Shorts and polarity reversal aren't the same thing, and the protection circuitry is different.
earlz 11y ago
unless the connector were cut, or some other catastrophic failure. I tend to think ruining a devices USB controller should be near impossible short of something like the USB killer (http://arstechnica.com/security/2015/10/usb-killer-flash-drive-can-fry-your-computers-innards-in-seconds/ http://arstechnica.com/security/2015/10/usb-killer-flash-dri...)
- lucaspiller 11y ago
  I was playing with a FTDI (a clone :D) and an Arduino and accidentally mixed up VCC and GND from an external power source. I only realised because Minicom was failing to connect to the device. Then I realised what I had done, I tried removing it from the USB port and putting it back in, but it wouldn't work. I then tried a mouse too, but that wasn't working either. Thinking I'd killed a USB port on my Mac (well there are two, so...) I then tried rebooting and that fortunately fixed it.
  - earlz 11y ago
    Yep, did almost exactly this, shorted a converted 3.3V from a different USB port (same machine) to ground on a different USB port and end result is my motherboard lost all USB connectivity until I rebooted
  - nerfhammer 11y ago
    the usb ports on macs are supposed to have overcurrent protection. You actually get a dialog popup in the OS when it happens. I know because I've triggered it doing something with an arduino.
    - klodolph 11y ago
      USB overcurrent protection is pretty sophisticated. Devices get 100 mA for free, then have to ask the host for more current. Well, it's a bit more complex nowadays, but that's the basics.
      - rcxdude 11y ago
marcosdumay 11y ago
Well, it protects against some metal object getting inserted into the USB port.
digi_owl 11y ago
This was a C-to-A adapter cable. what was wrong was that the Vbus on one end was hooked to the GND of the other end.
deleted 11y ago
[deleted]
vacri 11y ago
Look for a long response near the end of the article comments, from Mike Crawford. It goes into detail as to why that protection isn't designed in - it's not cost (diodes are cheap), but power drop, which has side effects.

paulmd 11y ago

Electrical faults are just a fact of life. Wires fray, cables get tugged, or are miswired from the factory. And you just know that sooner or later some genius is gonna jam a butter knife in there for the hell of it. The basic USB spec requires that all ports be protected with a small self-resetting polyfuse for exactly this reason. I don't know if the Type-C or USB 3.1 specs includes something similar - I would assume so but perhaps the additional current capabilities made that infeasible. http://elinux.org/Polyfuses_explained http://elinux.org/Polyfuses_explained A standard USB polyfuse usually resets pretty fast (as soon as it cools down) but they could take up to a few days. It's also possible the manufacturer cheaped out and didn't include a polyfuse, of course.

qb45 11y ago
This was an A-C cables, which means it was used to charge the laptop, not to power something from the laptop. Current limiting doesn't magically protect powered devices from voltage reversal (-5V instead of +5V), so both the laptop and power delivery analyzer went poof. They may have taken enough current to trip charger's current limiter, but then it was a bit too late.
ackfoo 11y ago
I found a USB Cellular data adapter in the trash and plugged it in to my MacBook Pro, which instantly shut down. Rebooted, and the USB port did not work. Shut it down and left it for half an hour, and, yay, all working again. Disassembled the adapter. Turned out to be a dummy used for display. No guts, just a steel bar across all four pins of the USB connector. Looked real from the outside. It was a very, very stupid thing to do. Thank-you Apple for putting a self-resetting fuse in my MacBook Pro. Saved my idiot butt.
- culturestate 11y ago
  Plugging random USB devices into your computer is a bad idea, full stop -- miswired or not. "I found this in the trash and plugged it in" is the bane of CISOs everywhere.

cesarb 11y ago

That's worse than a short. A short pulls the Vbus to 0V, leading to high currents on the source, so whatever overcurrent protection it has kicks in. What he had here, instead, was -5V on the Vbus pin, way outside the acceptable voltage range (which is probably something like from -0.5V to 5.5V). It's no wonder everything fried. The protection against reversed voltages like that would be a crowbar diode between Vbus and GND, to convert the negative voltage into a short circuit. But even if it had that reverse protection, if the current was high enough (since the negative voltage was from a charger, it's not unlikely) the diode could burn open, and then you have negative voltages again. And I don't know how common these crowbar diodes are; they probably are rare, since the USB connector is keyed so it can't be plugged "backwards" (the type C connector is mirrored instead, either orientation connects the pins the right way).

Zardoz84 11y ago
A diode crowbar or a single would imply a drop of ~0.6v on the line so you always would get 4.4 volts . However there is other kind of protections like this : http://www.edn.com/design/analog/4368527/Simple-reverse-polarity-protection-circuit-has-no-voltage-drop http://www.edn.com/design/analog/4368527/Simple-reverse-pola...
- cesarb 11y ago
  From what I understand, a crowbar diode is a reverse-biased diode put in parallel with the bus, not in series. In normal operation (correct polarity) it doesn't conduct, since it's the "wrong way", and there's no voltage drop since it's not in series with the circuit. But when the polarity is reversed beyond the diode's voltage drop (that is, beyond -0.6V in your example), the diode is now the "right way" and conducts, limiting the reverse polarity voltage to the diode's voltage drop (in your example, it won't get below -0.6V). However, that means that all the reverse-polarity current goes through the diode, in effect a short (like putting a physical crowbar across the battery terminals). If the diode holds long enough, the voltage source's short-circuit protection (be it a fuse, a polyfuse, or something else) should cut the current and all is (mostly) fine.