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I was planning on making a usb switch and I figured the most convenient way to power it and any connected devices would be to draw power from any and all connected hosts. From the searching I've done already I determined that I'd want to connect the grounds together, but I'm not sure if my planned way of automatically handling the 5V power is correct. I thought using a diode for each voltage source to prevent backflow would be correct, but I'm not sure if this is the case. Attached is a circuit diagram for how it would work.

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  • \$\begingroup\$ This is not wrong and will work. Keep in mind you'll lose some voltage on each diode (0.7V for silicon diodes, 0.3V for a Shottky), so make sure your load is tolerant of something less than 5V getting to it. The load would only source current from the higher-voltage source (if two are on at the same time). The purpose of the diodes is to prevent a higher-voltage source from smoking a lower-voltage source. If you want to actually SHARE the load between sources, you would use some low-value resistors instead. (Like a couple ohms). But that will NOT protect one source from another. \$\endgroup\$
    – Kyle B
    Aug 26, 2020 at 0:52
  • \$\begingroup\$ The goal would automatic power switchover as long as at least one device is on, so that should be fine. \$\endgroup\$ Aug 26, 2020 at 1:39
  • \$\begingroup\$ Yep, should work fine for that. You might drop a capacitor from the diode common node to ground, something like 100-470uF depending on your load. "Just because" It won't hurt anything and may stabilize the power supply at the moment one kicks off and the other kicks on. Caps are really cheap. \$\endgroup\$
    – Kyle B
    Aug 26, 2020 at 1:42
  • \$\begingroup\$ Out of curiousity, what would be needed if I wanted to be able to draw power from both sources simultaneously for an increased amount of current? \$\endgroup\$ Aug 30, 2020 at 9:07
  • \$\begingroup\$ The easiest method is to replace the two diodes with resistors. However, for this to be successful the amount of load current needs to be considered. Is this a continuous load, or does it vary alot??? If its is widely varied, you would probably need something more sophisticated than just 2 resistors. \$\endgroup\$
    – Kyle B
    Aug 31, 2020 at 1:40

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Yes that would work but you will dissipate power due to the diode voltage drop from the hosts. You will also loose some voltage due to the diodes. Best to use schotky diodes to minimize voltage drop and reduce dissipation. Otherwise if you cannot tolerate much voltage drop you should consider the Power Multiplexers that will do very similar voltage multiplexing but with very, very small voltage drops, and also offer current limiting etc.

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