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I have a high power (700ma+) USB device I wish to use on a single front-panel USB port.

The USB specification in Wikipedia states:

The USB 1.x and 2.0 specifications provide a 5 V supply on a single wire to power connected USB devices.

A unit load is defined as 100 mA in USB 2.0, and 150 mA in USB 3.0. A device may draw a maximum of 5 unit loads (500 mA) from a port in USB 2.0; 6 (900 mA) in USB 3.0.

When using the device in a normal USB port the voltage drops to 4.7~ Volts, which is dangerously low for my device.

My front-panel dual USB ports are connected to a USB header on the motherboard which supplies the power and the data transmission lines.

My PSU is a 750w that can supply up to 25A on the 5V rail.


Is it possible to remove the power lines (Vcc and GND) from the front panel and then replace them with the 5V and GND from the PSU?

Will this procedure be safe for my PC, power supply and device being powered?

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    At those power levels you should implement some kind of short circuit protection. As long as you don't back feed the 5v@25A to the motherboard your PC and power supply don't care. Now the device being powered is whole different story. It might be fine, but if there were a short circuit or a number of other things it could easily damage the device being charged. – cybernard Jan 25 '14 at 23:03
  • by the way 25A requires you use 12 or less gauge wire size. The smaller the gauge the larger the wire. – cybernard Jan 25 '14 at 23:43
  • Thanks for the info. So, there is really nothing stopping me if I just make sure I use the correct wire gauge? – suit Jan 26 '14 at 0:56
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    Nothing is stopping you, but you need to take some safety precautions. I would get a 5A fuse as a fail safe. – cybernard Jan 26 '14 at 1:31

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