- Joined
- October 12, 1999
- Messages
- 6,595
- Reaction score
- 71
- City, State
- Pennsylvania
- Year, Model & Trim Level
- 2024 F-150 Tremor
Bill,
Not using the PIAA switch. The OEM switch has a line in and a ground out to break the circuit. The switch itself pulls no power, and does not have a "power out" prong per se (though I assume it is, simply, like a U where line in and ground out don't matter as much as generally the breaking of one leg of the circuit). But, I suppose I could run power in and tie the ground out on the switch (here, it would function as power out) to the coil power in (essentially having the OEM switch break the power prong of the line, instead of the ground), but I also assumed since the OEM switch was built to break the ground side of the circuit (I think this is how Ford wires it), I would run and break the ground wire instead.
I think--breaking the ground side of the circuit has the same effect as the power, right? Essentially, unless the circuit is completed, that is, power flows to item and to ground, the circuit will not fire. In this case, the on/off of the OEM switch makes or breaks the ground leg of the circuit, either allowing or prohibiting it from grounding, and thus allowing or prohibiting the coil to fire.
I wish I paid more attention in Hon. Physics. So... long... ago...
Not using the PIAA switch. The OEM switch has a line in and a ground out to break the circuit. The switch itself pulls no power, and does not have a "power out" prong per se (though I assume it is, simply, like a U where line in and ground out don't matter as much as generally the breaking of one leg of the circuit). But, I suppose I could run power in and tie the ground out on the switch (here, it would function as power out) to the coil power in (essentially having the OEM switch break the power prong of the line, instead of the ground), but I also assumed since the OEM switch was built to break the ground side of the circuit (I think this is how Ford wires it), I would run and break the ground wire instead.
I think--breaking the ground side of the circuit has the same effect as the power, right? Essentially, unless the circuit is completed, that is, power flows to item and to ground, the circuit will not fire. In this case, the on/off of the OEM switch makes or breaks the ground leg of the circuit, either allowing or prohibiting it from grounding, and thus allowing or prohibiting the coil to fire.
I wish I paid more attention in Hon. Physics. So... long... ago...