Quoted from cottonm4:OK. The direction of current flow is from the column (ST) wire and current flows in this direction with my mock up, from the w/r wire, though the diode from the non-banded side, and the contact on the switch blade contact is sitting there hot and as soon as the hot contact makes contact with its opposite, the circuit is complete and power is fed to the w/g wire, correct?
Does this have anything to do with converting AC current into DC current?
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I will have one more question when this one is resolved.
No, it doesn't have anything to do with converting AC to DC. The strobes are pulsed one at a time. This is how the MPU is able to discern which switches are closed on any given "I" line.
For example, when strobe ST0 is pulsed and the MPU detects a closed switch on I4, it knows it's the spinner switch even though, say, the middle drop target is down and it's switch (also on I4) is closed. The MPU won't detect the closed drop target switch until ST3 is pulsed.
This is why diodes are important. Since all switches on any given "I" line are tied together, when the pulse on ST0 passes through the closed spinner switch, it will be present at EVERY switch along I4. So say the drop target didn't have a diode and the switch was connected directly to ST3, or the diode was shorted (same thing - a direct connection to ST3). If the drop target is down and it's switch is closed, the ST0 pulse is going to pass through the spinner switch, traverse I4, pass through the drop target switch and there will be nothing to prevent the signal from appearing on ST3. In this situation, if the drop target is down and you get a good spinner rip, then hit the Aquarius stand up at the intersection of ST3 and I7, the strobe signal from the closed spinner switch is going to pass through the DT switch, appear on ST3, go through the Aquarius switch and be detected by the MPU. But since the MPU is strobing ST0, the switch the MPU is going to think is closed is the slam tilt!