Should you fear your USB cable?
cnet.com
cnet.com
>When you connect your USB-C-equipped Nexus 6P phone to a MacBook Air port designed to only provide 1.1A ... the phone can pull too much power ... and potentially destroy internal components.
This is just nonsense, as far as I know. No device can "pull" more current than you can provide. Even if you short the pins, the port will just be shut down, as noted later in the article itself.
>But Saunders, chairman of the USB-IF standards body, says he's not aware of any protection against a completely miswired cable like the one that struck down Leung's Chromebook Pixel. "That was something that the circuits of his notebook can't protect against," he says.
I highly doubt that. What's the worst that can happen? You deliver 20V to the pin expecting GND. Trivial to protect against. And this is also only for things like chargers. If it is any peripheral device, the voltage drop will never exceed ~5V I'm guessing. All this just comes down to bad engineering on the part of laptop manufacturers. Bad cables, bad devices, reverse polarity chargers have always been present. And our computers have always had protections against them. And that is how it should be. A completely miswired cable should simply not work.
Now if you're talking about differentials of >100V, then I would understand. But we're nowhere close to those levels. Google and Apple: Your ports are badly designed and you should feel bad!
Apparently polyfuses can take days to properly un-trip.
When I took the offending cable out, and tried with a different device, it worked straight away, no reboot required (not sure about earlier versions/alternative scenarios though, it's possible they do require a reboot).
Apparently the resistance falls quickly at first but can take a long time to fully return to pre-trip level.
This has caused trouble in the Raspberry Pi. Users were getting odd effects when voltages didn't quite return to spec after polyfuse trip. [2]
[1] http://www.te.com/content/dam/te-com/documents/circuit-prote...
This explains why my computer has a bunch of USB3 ports that work... sometimes. (It also very sporadically has a massive wibble and hard-crashes with an error relating to USB.)
In USB 1 and 2 (probably also in 3, but I don't know about this), the device has two power modes. Initially it is only allowed to draw a few mA. This allows to power up enough of the device (the USB controller) so that the device descriptor can be read. The device descriptor tells the computer that the device requires some additional current. If in a chain of several non-externally powered hubs there's not enough capacity, the COMPUTER will NOT TELL THE DEVICE TO POWER UP. The computer can not turn off the voltage on the port. Typically USB interface chips will have an output pin that's only providing additional ("high power" circuitry) with power, by switching a mosfet, when the computer gave the OK to proceed.
sudo lsusb -v
(...)
Device Descriptor:
(...)
Configuration Descriptor:
bmAttributes 0xa0
(Bus Powered)
Remote Wakeup
MaxPower 500mA
...which means that you deliver -20V to the +5V rail.
Right here I have a roll (1000pcs) of chips that do exactly this (usb power management ICs) that cut power off and report this incident to controller. Maybe notebook manufacturers do not use them due to cost or space constarints...
I'd really be surprised if there would be anything but polyfuses on the "old WinXP PC" my original comments's parent comment referred to.
In theory, original USB1.1 expected the trip-point to be programmable, but most USB hosts are not implemented that way, because it requires some kind of low-drop high-side power switch that tends to be expensive.
Not very expensive ($.5 a piece), but at scale may become costly.
USB started with four pins - GND, Vcc, Data+, and Data-. USB-C has 24. Yes, 12 of them are just the other side of the connector flipped over. And there are two ground and two power lines on each side, for heavier current. But there are 10 different pins, with 3 differential pairs for data (the original bidirectional pair, plus two unidirectional pairs), and the SBU line, which is used to tell the ends that some lines are not being used for USB, but for some other protocol, such as DisplayPort or PCIe. (Coming soon to an airport near you - power outlets with enough smarts to load exploits through these features.)
A true USB-C cable is simple - it's just wires. It's the ones that are adapters to other connectors that give trouble. Here's a USB-C to DisplayPort connector.[1] There's far more in those devices than a programming resistor.
Controlling USB-C ports is so complicated that Cypress's IC has a 32-bit ARM processor just to operate the port.[2] Either end can be the master, and either end can be the power source. (These need not be the same.) If all the complicated setup fails, the interface drops back to USB 2.0 and pretends to be a "Billboard device", which gives diagnostic info.
The scary part is that USB-C can potentially deliver 100 watts, as 20V at 5A. USB-A was rated for 5V at 0.5A. Botch the power handshake and Bad Stuff is likely to happen.
[1] http://www.ecnmag.com/product-release/2015/05/adapter-soluti... [2] http://www.cypress.com/products/ez-pd-ccg2-type-c-port-contr...
https://thunderbolttechnology.net/blog/thunderbolt-3-usb-c-d...
...it's a Cortex-M0, 4k RAM, 32k Flash. By today's standards, this is a very small controller. For the Arduino-Crowd, it's roughly equivalent to the ATmega32u4 on the Arduino Leonardo, but with a Cortex-M0 instead of the 8-bit AVR. Both, by the way, run 16bit opcodes. -- But I get your point, it's nevertheless a lot of "brains" for what people wish would be a dumb cable.
> Either end can be the master, and either end can be the power source.
Unfortunately, this is what people expect from a USB device these days. The poor little "ID" pin on micro-USB was overloaded with functions already: There are interface chips that can distinguish 40 or so different resistor combinations on various lines (see the ancient TSU6721, which was used in the Galaxy-SII). So, I think requiring some intelligence at least for the more complicated combinations really makes sense.
Regarding the number of pins, I don't think it's that bad: USB2+/- has stayed the same, just bridged to two pins each. USB3 Tx/Rx is still two differential pairs, with two extra differential pairs for "future use", Display-Port, ... The ID-pin with resistors to Gnd/Vcc just has been massively upgraded to a serial protocol spoken on the CC1/CC2 pins.
So, for me, after reading a little bit about it, I now see USB-C as Micro-USB3 plus upgraded ID-Pins plus two spare differential pairs... But I was very skeptical at first, too.
[1] http://www.cypress.com/documentation/application-notes/an965...
>When you connect your USB-C-equipped Nexus 6P phone to a MacBook Air port designed to only provide 1.1A ... the phone can pull too much power ... and potentially destroy internal components.
Does anyone know what the hell they're talking about? There is no electrical force to 'pull' power. The source 'pushes' power at a certain voltage, and can supply a certain manufacturer-set amount of current. If the charging device can use more current, nothing happens. There is no way to pull "unsupported" amounts of current out of something.
With purely passive loads, the power "pulled" from a fixed-voltage power supply is related to the impedance of the load, and can be calculated with a fairly simple formula.
Phones, however, are active, and can regulate how much they draw to charge the battery.
Perhaps a simpler explanation: if you plug a night-lite and a vacuum into an identical outlet, they will clearly pull different amounts of power.
A DC power supply can still largely be modeled as a DC voltage plus some small amount of resistance, capacitance and inductance. This will give it a characteristic impedance for a short-circuit which is the only limit on how much power can be drawn from it.
Aha! I found where the confusion is in this sentence:
> If the charging device can use more current, nothing happens. There is no way to pull "unsupported" amounts of current out of something.
This is wrong. If you draw too much current, then something bad will happen. The UL just makes sure that "something bad" isn't "will start a fire or electricute someone" it could very reasonably be a component of the DC power supply that fails with an open-circuit, which would then need replacing for the power supply to work again. In some cases it may be a part specifically designed to be a fuse, in other cases they may just take a diode, MOSFET or resistor that they needed anyway and choose a part that will die if too much current is drawn.
Dell is claiming that their USB ports will not die permanently in this case, but will shut-down temporarily, more like a self-resetting circuit breaker than a fuse.
That's not true -- if you put a power supply capable of supplying, say, 10 amps at the end of a wire rated for 0.5A, and then you connect a 10 amp device to that wire, the wire will get very hot and then melt.
If this wire is on a circuit board inside a device, the resulting heat may destroy the components nearby and/or start a fire. If the wire is a USB cable, the cable and/or connector may melt.
Most of the wires / traces in an computer or smartphone aren't capable of carrying the full amount of current that a battery or charger can supply. That's why protection circuits are required, and why USB 3 has a complicated negotiation process around how much power the bus-powered device can draw.
Basically it had the data and power pins switched, thus feeding high amps where it should not be.
For myself, i managed to crosswire a case port (the supplied wire required me to attach individual pins rather than socket a fixed block).
The only major difference would be the amount of As bouncing around (the Chromebook, hooked up to a dumb wall wart, would get some 2-3As rather than the .5A my thumbdrive got).
Anyways, there also exist a proof of concept "badthumb" out there that is basically a capacitor that will reverse the polarity of the USB connection after having built up a sizable charge. This will likely take out at least the controller, if not more.
(I once bought a noname car charger, "it's for 12V so how bad can it be?". Turns out "very" is the answer as that phone never took charge again.)
Edit: that was a 12V powered hub, not "a cable", so maybe that's why.
Seriously though, why would you assume this? The article doesn't state it - it gives examples of both devices being damaged in different ways, but cites only Dell as actually designing their hardware to tolerate bad cabling.
Perhaps Apple has built in some mechanism to protect the rest of the circuitry from a USB failure, but without further information it's more reasonable to assume that the MacBook owner just got (comparatively) lucky here.
Years ago, I shorted out a USB port on a desktop PC, with the same results - system booted and worked just fine, except the USB port was dead. Upon opening the box, I found a popped cap and some scorching on the motherboard; I'd be... reluctant to call this a well-designed failure mode.
The cable attached to the Chromebook was totally miswired and flipped Vbus and ground, supplying -5V where 5V was expected. As the article said, nothing is really designed to handle this kind of failure because it should be blatantly obvious to the cable manufacturer if there were any sort of quality control.
Reverse polarity protection is very common, and often only requires a single diode in series with the +V line, so that current can only flow in one direction.