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This is the second article on these cables that I've read saying the charging device "pulls too much power" >When you connect your USB-C-equipped Nexus 6P phon

This is the second article on these cables that I've read saying the charging device "pulls too much power" >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.

aidenn0 11y ago
The limit to how much power you can pull is just the ESR of the power source. Don't take a car battery and connect the two terminals together with large-guage wire, unless you want to see some impressive fireworks. 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.
- awakeasleep 11y ago
  Does that explain things when you're talking about a DC power supply? Isn't that what we're talking about?
  - aidenn0 11y ago
    DC and AC don't matter for generalities (though there are significant differences in how a given device will draw power from an AC or DC source, the underlying principle is the same). 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.
rayiner 11y ago
Remember that the flow of electricity is just electric potentials evening out.
wzdd 11y ago
> If the charging device can use more current, nothing happens. 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.

aidenn0 11y ago

The limit to how much power you can pull is just the ESR of the power source. Don't take a car battery and connect the two terminals together with large-guage wire, unless you want to see some impressive fireworks. 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.

awakeasleep 11y ago
Does that explain things when you're talking about a DC power supply? Isn't that what we're talking about?
- aidenn0 11y ago
  DC and AC don't matter for generalities (though there are significant differences in how a given device will draw power from an AC or DC source, the underlying principle is the same). 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.

rayiner 11y ago

Remember that the flow of electricity is just electric potentials evening out.

wzdd 11y ago

> If the charging device can use more current, nothing happens. 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.