Say I have a device that accepts 10V @ 2A, and I have a wall-wart with two USB plugs that both output 5V @ 2A. Can I add a second male connector, connect both connectors to the wall-wart's two ports, and draw a cumulative 10V @ 2A?

This question's answer indicates that this would cause havoc for motherboards or other electronic devices, but there's a lot more going on with a motherboard than with a power block. Does this answer still apply?

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    \$\begingroup\$ No, the grounds will be shared and you'll short your system out. \$\endgroup\$
    – Majenko
    Oct 31, 2014 at 22:06

1 Answer 1


No you cannot! Both USB sockets are probably even the same power supply so the 5V of both USB ports is connected to eachother and the 0V/GND is also shared.

If you had two different adapters with internal galvanic isolation (transformer) it would be possible to make 10V 1A of two 5V 1A USB ports.

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    \$\begingroup\$ Or 5V 2A without isolation. \$\endgroup\$ Nov 1, 2014 at 0:01
  • \$\begingroup\$ My original question had a typo, the power outputs on the wall plug are 5V @ 2A, not 1A, but that doesn't change the answer in the original context. Thanks! \$\endgroup\$
    – Bryson
    Nov 5, 2014 at 19:14
  • \$\begingroup\$ Would a "power supply" that takes in both USB inputs (5V 2A) and outputs 12V at an amperage greater than 1A be difficult/impossible/complex to build? Or is it something I could do as a learning project? \$\endgroup\$
    – Bryson
    Nov 6, 2014 at 19:21
  • \$\begingroup\$ If I unstand you well, you want to make a device that converts 5V into 12V. This is normally done using a boost converter. However you need at least the same input power as output power. Output power is at least 12W (12V * 1A), and input is 10W (5V * 2A). So this is already impossible. When you have 5V * 4A it would be possible given an efficiency of 60% which is possible to make. To make a boost converter look for example at the LMR61428. If you don't have much experience with electronics I would however advice against building a boost converter because it is not very straightforward. \$\endgroup\$
    – Douwe66
    Nov 8, 2014 at 9:35

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