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To play a joke on this question, we came up with the following image on the Super User Chat.

Of course this will not work, but now my question is:

What happens if you charge a laptop using it's USB power?

Assume there is a fully charged battery in the laptop to begin with. ;)

I see three possibilities:

  1. A short circuit happens, the cable might map - on + and vice versa.

  2. Nothing happens, like you were to connect the + and - of a battery to itself.

  3. Power goes around in a circle and you lose faster than usual due to heat.

But I can't see which one of those would really happen...

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    \$\begingroup\$ you're mental. \$\endgroup\$
    – Majenko
    Commented Jun 11, 2011 at 22:04
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    \$\begingroup\$ "What happens if you charge a laptop using it's USB power?" You can't. \$\endgroup\$
    – endolith
    Commented Jun 12, 2011 at 2:28
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    \$\begingroup\$ @studiohack - I like folloing better. It's shorter and is easy to understand by everybody. \$\endgroup\$
    – stevenvh
    Commented Jun 12, 2011 at 7:03
  • \$\begingroup\$ What about foloing? :) \$\endgroup\$ Commented Jun 12, 2011 at 10:40

3 Answers 3

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Humourous, but once you look a little bit below the surface, the absurdity of the idea is hard to ignore.

(1) is possible, but USB ports to my knowledge are current-limited, so no permanent damage should occur.

(2) is the most likely scenario, for a different reason than you describe. The 5V coming out of the USB port is well below the 12-18VDC that a typical laptop wall-wart provides to the computer. The negatives are most likely already common (connected together) inside the unit. Many laptops require communication with the power supply before power is allowed to flow as well (locking out knock-offs and other, incompatible equipment).

(3) could only be possible if the laptop was intended to take +5V input, and if the USB port could source whatever current the laptop wanted. Even if these two things were made 'possible' by the electronics gods, the major issue still remains. Energy transfer is lossy. Any benefit gained from power coming in is more than negated by the energy wasted getting it there. Not to mention, the energy that arrives at the inlet would have to be converted / transferred before going back to the batteries / USB ports, wasting more energy.

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  • \$\begingroup\$ Due to the current-limiting and the low possibility that it swaps - and + around I think we can cancel out option 1. As for option 2 and 3, I think it lies somewhere in the middle. It will result in a leak that's either noticeable or not, but at least we know that nothing bad should happen... :) \$\endgroup\$ Commented Jun 11, 2011 at 20:15
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Your battery will discharge. How much faster depends on how much loss is in your "magic cable"'s circuitry (as noted in other answers, USB voltage is lower than the typical charging voltage, so the cable would need some circuitry).

The simplest way to look at it is that even if you dealt with the issues in lower-than-needed voltage and current, every watt of power that comes from the USB port is coming out of the battery, so even if you had zero loss on the path back to the battery, you'd be at scenario #2, "nothing happens" (i.e., putting in the same numbers of watts that you took out). Of course, your laptop is running during this time (most, if not all, do not power the USB port when not running), so the battery is discharging.

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If you were to build a tiny step-up converter into the cable (taking the USB's 5 V and converting them to to whatever higher voltage the laptop accepts (12...19 V), and if your laptop is of the type that doesn't negotiate weird special secret protocols with its power supply (or if you were to spoof it with just the right protocol), it would maybe work while wasting energy, given the notebook accepts a low-current charger, because typical currents are 3 A at 19 V (i.e. a power in the region of 50 W), while max. USB power is some mA at 5V (i.e. around 1 W, don't know the exact value). I assume the last "if" makes it impossible.

The original question that triggered your idea will therefore, too, have to be answered with: Does very likely not work.

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  • \$\begingroup\$ It's 5*100 mA for USB up to 2 and 6*150 mA for USB 3. \$\endgroup\$
    – AndrejaKo
    Commented Jun 12, 2011 at 10:41
  • \$\begingroup\$ I'll bet you could sell such a step-up converter. I think it would be better to do DC->AC conversion and power the original power supply, though. (Note: A brief search turned up a lot of AC->DC converters, but no USB->AC converters that I could find...) \$\endgroup\$ Commented Jun 13, 2011 at 1:56

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