Hooking up a Coil on Positive Ground

Ok, I realy want to say that I understand completely, BUT. I understand DC wiring and some AC wirering, but this is so far over my head that I even got the ladder out and still could not get it LOL. I am picking up my coil tonight and will post back with the specs on it.

I greatly appreciate your patience and help with this matter. THANKS BUDD

As I explained below, a limiting factor in the old coil/points ignitions is the current you can switch with mechanical switching contact points before they run too hot and arc too much and burn up. HOWEVER, with a solid state switch (uses a transistor like device to conduct and switch coil current), you dont have to worry about burning points and can run higher current and energy through a coil primary thereby discharging more energy (current x voltage) across the plugs gap (fatter, hotter, wider, more surface area current arc) to better ignite the fuel.

Actually, your new coils primary may be all wire and not even have any resistor ballast if its intended for a solid state switch only (just no way for me to know) in which case as I stated below, I would then wire it + to the distributor. Otherwise not knowing and to be safe, I had to advise it - to distributor. If the specs or any tech support people from the coil company can tell you about any internal coil ballast resistor (if it has one and where its wired internally), then we could answer your question.

John T

1) My best guess, since you say its a 45,000 volt coil, is that its for a high performance type auto ignition which is negative ground, and to be safe, I would wire it neg to the distributor. HOWEVER if I knew for sure it had no internal ballast (like if its near 3 ohms total with primary winding resistance only and no ballast), then I would wire it + to distributor which is how you wire an unballasted coil (like a 6 volt non internal ballasted coil) at pos ground.

I CAN ANSWER YOUR SECOND QUESTION BETTER.

2) You asked how many "volts" on the distributor, but thats not the right and valid question, cuz the distributor is a current (not a voltage) device that is a mechanical switch (points) which closes the conduction path so the coil can conduct current to ground when points are closed, then open the path when the points are opened. The correct question is how many "amps" can I allow to pass through the distributors ignition points (mechanical switch).

The basic answer (how much current through distributor) so the points do not prematurely burn out is around 4 amps give or take. That means on a 12 volt system, the total resistance of the coil plus any ballast (internal or external) is around 3 ohms. Some 12 volt JD's used a 6 volt coil (around 1.5 ohms) and a series ballast resistor which drops 6 volts leavign 6 across the coil as it was designed for. The voltage at the distributors input (lil side hook up terminal where coils primary output wire attaches) would be 6 or 12 volts (dependign if 6 or 12 v battery) when the points are open, then drop to zero when they are closed, since the points closing makes that point a dead ground. So, the disributor is basically the same, regardless if a 6 or 12 v battery, and its just a open/close current switch with the input voltage dependant upon what the battery voltage is.

I hope this un cornfuses you some Budd, post back or email any more questions.

Good Luck 2 ya and God Bless

John T Nordhoff in Indiana