Electronic – Current flow priorities in PSU / PSU review

Let me redraw the central part of the circuit. Without the diode being present, current will flow clockwise. When D1 is inserted, it is reverse-biased, because V2 is higher than V1 (they are in parallel). Reverse-biased means no current flows.

From then on, you can treat this entire thing as its Thevenin equivalent - no matter what you do the path V1-D1 will always be inert, because V2 makes sure D1 stays reverse-biased.

^{simulate this circuit – Schematic created using CircuitLab}

Just to clarify this issue, let's make a thought experiment.

1. The circuit of V1, V2, D1 is clearly a sub-circuit of the circuit in question.
2. Voltage between NODE1 and NODE2 is guaranteed to be 20V, since V2 is connected directly between the two. (V2 being an ideal voltage source, voltage across it cannot change by definition.)
3. Since V1-D1 and V2 are in parallel, they share the same voltage across them.
4. When you subtract 5V from 20V (this is perfectly legal with voltage sources), you're left with 15V between D1's anode and cathode.
5. Diode will act as an open circuit, since the potential at its cathode is higher than its anode = it is reverse biased.

Why does current flow from PSU a to PSU b?

If they're ATX supplies then the output GND is connected to protection earth, so both GNDs are connected together to the power strip's earth (or the house wiring). So, no surprises here.

So after seeing the results of that, I had another idea. I wondered if the PSUs could sink current from a 9v battery. So I connected (-) of the 9v battery, and 12v from the PSU according to the following schematic

This sounds like your switching PSU is leaking a bit of high frequency common mode noise. This is normal and expected for a switching PSU, although the actual current should be very low to meet with EMC/EMI regulations.

To have current flow, you need a closed circuit. In your case, what closes the circuit would be the parasitic capacitance between the 9V battery and the ATX power supply metal case. If I'm right, the LED illumination will vary if you bring the 9V battery closer to the ATX supply. It would also work if you replace the 9V battery with any metallic object (like a kitchen spoon, whatever) since what matters isn't the fact it is a battery, but rather a conductive object large enough to form a capacitor with the PSU case.

Try it ;)

I find this question interesting, because the first part is really basic, while the second is rather high-level ;)