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A PCB circuit must have capability to be powered from a bench power supply, or USB supply. The bench power supply can be replaced with an external power supply of fixed voltage as well.

The external supply range is 5V to 12V. The USB supply is 5V. What is required is that, both can be connected at same time without malfunction. Only one of them can be attached for normal system operation.

The way to do this, as my knowledge, is to use diode OR operation with the two DC sources. I am not exactly sure why this works in theory or practice. I should choose shottkey diodes since they can have lower Vf. Also, the reverse breakdown of these should be larger than 12V. The current rating of both should be what the circuit power demands if one of them is active i.e if the circuit demands 1A, then both of these must be rated at 1A.

Is this approach correct? Also, does it matter if the USB power is coming from a "floating" device e.g laptop running off batter and not connected to the mains.

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  • \$\begingroup\$ I actually just answered a question earlier with a solution that would easily work for your use case as well, if you want to take a look at this: electronics.stackexchange.com/a/670041/57556 \$\endgroup\$
    – Hearth
    Jun 12 at 3:30
  • \$\begingroup\$ The problem is there is not enough info to answer. Schottky diodes have lower Vf but higher reverse leakage current. And device that blindly consumes 1A from USB without negotiating is a bad thing. There may not be 1A available any more if other devices are already consuming almost all available current. And depending on what the device is or what it does and where it connects to, it can't be said if floating or grounded supply is better. \$\endgroup\$
    – Justme
    Jun 12 at 5:58
  • \$\begingroup\$ OK the circuit I am designing right now will only take a few hundred mA at most. However, I will extend this circuit in the next revision and this is where the current draw will become larger. \$\endgroup\$
    – gyuunyuu
    Jun 12 at 13:20

2 Answers 2

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Here is a commercial USB hub that has an external power adapter and can run off of host USB power:

enter image description here

Ignore that someone has pulled out the diodes (D3-D6) and soldered on a regulator. As you can see from the silkscreen, there is a diode to the USB socket and 3 parallel diodes (to increase current handling) to the barrel jack. Whenever the barrel jack is plugged in it will provide a higher voltage, forward biasing D4-D6 while reverse biasing D3. Thus the barrel jack provides power with USB as fall back.

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  • \$\begingroup\$ Since the main's supply coming into the barrel connector can be in range 5V to 12V, is it possible that, this shall cause a reverse current to flow into the USB port when both are connected at the same time and the mains power is higher voltage than the USB 5V? \$\endgroup\$
    – gyuunyuu
    Jun 12 at 13:24
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    \$\begingroup\$ @gyuunyuu Diodes only conduct forward so no reverse current. \$\endgroup\$
    – user1850479
    Jun 12 at 14:17
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You are close, the area I would be concerned about is the 1A on a 1A diode. Use something that will do a bit more current such as a 3A or whatever is convenient. Just be sure the grounds are connected. You can also use MOSFETs to isolate but the diode is the simplest and straightforward to understand and design.

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  • \$\begingroup\$ The biggest concern I have is with the USB port, the USB port could be a floating port of a laptop running off battery, or it could be a laptop connected to mains or a mains power supply that has USB output. What will happen if I connect the USB power from floating source when the other power source is already connected to the PCB. \$\endgroup\$
    – gyuunyuu
    Jun 12 at 13:22

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