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I'm designing a few boards that go in series with with each other.

schematic

simulate this circuit – Schematic created using CircuitLab

The PCB layout is the same for all, so PCBs with unpopulated PSUs will receive power from 5V rail, populated will provide its own power.

The ones that have a PSU(Vin to 5V) will supply 5V to the following boards.

The ones that do not have a PSU populated will use 5V from the former.

However the ones that have a PSU populated and also will receive 5V from former PSUs, and will have to cut the 5V input while allowing its PSU to supply the chain from this point on.

So we'll assume the CONSTANT 5V rail is permanently on at ideally 5V.

Will this power selection circuit correctly turn M2 on and M1 off whenever INTERMITTENT is on and vice-versa?

I have simulated this circuit on Circuit Lab and tweaked it a bit, have I missed something?

I have purposely put the FETs body diodes in this orientation (5V can be fed into the 5V rail but not into the INTERMITTENT PSU).

I have studied a few O-Rings but I'm not quite sure this will work.

p.s.: there is another higher voltage rail present for the PSU in all boards.

schematic

simulate this circuit

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  • \$\begingroup\$ I apologise, but I don't quite understand the question, especially "However the ones that have a PSU and also will receive 5V from former PSUs, and will have to cut the 5V input while allowing its PSU to supply the chain from this point on." Could you do a simple block/schematic diagram of the arrangement of a few boards to explain this? It seems on the face of it that a board with a local PSU could have a connector to the previous boards 5V out, but it would not use the previous boards 5V, it would be open circuit. Hence there is no need to switch anything. \$\endgroup\$ – gbulmer May 7 '16 at 3:42
  • \$\begingroup\$ I get your point, but I'd like to use the same connector for all boards since the ones connecting them might not know the workings of it. I'm drawing a block diagram now. \$\endgroup\$ – Wesley Lee May 7 '16 at 3:44
  • \$\begingroup\$ I was assuming you wanted to use the same connector for all the boards, that's why I phrased my comment "... a local PSU could have a connector to the previous boards 5V out ...". However, I am struggling with why a board which has its own PSU would need to have any electrical connection to the previous boards 5V out. Just because it has a connection to it, doesn't mean it uses it. It might be worth coming up with a more compact terminology than "a board which has its own PSU" :-) Maybe PSUd and Un-PSUd? \$\endgroup\$ – gbulmer May 7 '16 at 3:48
  • \$\begingroup\$ So a board with PSU might not always have its PSU turned on, right? If the PSU is turned on, it supplies power to itself and following boards only? And if the PSU is turned off, the board receives power from the previous board and passes it to the next board? \$\endgroup\$ – Bruce Abbott May 7 '16 at 3:52
  • \$\begingroup\$ @BruceAbbott thats it. \$\endgroup\$ – Wesley Lee May 7 '16 at 3:54
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The first question is: what for do you need to switch on and off power supply line dynamically? Who decides to bypass the 5 V line from the previous board or to use own power supply?

There is a simple and straightforward solution:

You reserve 2 pins on each connector for 5 V line: 1 pin for power "input" another one - for power "output".

Each board with PS does not use "input" 5 V; it makes "own" 5 V that it uses as a local power and sends out for next boards through "output" pin

Each board without PS has a jumper (or soldered zero Ohms resistor) that connects "input" +5 V straight to "output" 5 V.

Sure the 24 V to 5 V DC-DC converter should be protected against overload in your design, as we do not know how many boards it powers up.

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