Any good electrical mechanics out there? There seems to be plenty of info in the service manuals regarding the TI system... voltages and trouble shooting procedures..etc... but very little on the breaker system...

First question, what voltage should be present at the + terminal of the coil when the ignition is in the run position?

Second, what is the source of the voltage when in the run position? (open ended question... I know)

Thanks

 

The voltage comes from the ignition switch in the run position. The current passes through a resistor to the coil. That resistance is in the wire on a 69. The voltage at the coil varies, but is usually 8+- Volts. The current passes through the coil to the points, to ground when the points are closed. During the starting proceedure the solinoid bypasses the resistor and applies 12+ volts to the coil. As soon as the starter is released, that path is switched back off.

 

Thats exactly how I thought it works. However, the 69 wiring diagram in the service manual shows a 20 W-R/B running to the bulkhead conn... then I loose... If I were looking for the wire comming from the ignition switch, what color would it be? Pink... changing over to a white resistor wire? I also found a pink wire comming directly from the fuse pannel... pluged into the IGN connector. But, it doesn't go anywhere... its taped up??? Any suggestions appreciated!

 

I dont have access to a wiring diagram here at home, but yes, they are usually pink coming from the ignition switch, and the resistor wire is a different color, probably white.

 

From the bulkhead connector the wire changes to a 12 gauge Pink wire that runs up the steering column to the ignition switch. There is another wire running to the + terminal of the coil, it is a 20 gauge yellow wire that runs down to the "R" terminal of the starter solenoid. On points type ignitions it was necessary to reduce voltage to only 7 or 8 volts, in order to not fry the points. This reduced voltage was obtained by running the routine power to the points through a ballast resistor or a resistor wire. This reduced voltage is fine under normal operation, but inadequate for a cranking start up. So the yellow wire was attached to a terminal on the solenoid, when the solenoid is activated, a set of contacts close in the solenoid, routing 12V directly to the coil. Once cranking has stopped the contacts in the solenoid open and the 12v is cut off from the coil. The coil then operates on the reduced voltage from the resistor wire.
I don't know if this answers your questions, but it's about all I know about the 69 ignition system.

 

Thanks!

 

I'm going through the same thing with my '68. I decided to rebuild my engine harness rather than pay $175 for a new one. The wire leading from the firewall connector is indeed a resistance wire that is white cloth covered and drops the voltage to about 8v at the coil. The insulation on this wire was shot so I started looking around for wire that I could replace it with. After checking with dealers, auto-electric shops, auto-parts stores and some of the harness reproduction places, I could not find the wire I needed. The harness places wouldn't sell it separately (after all, they'd much rather make $175 and sell me the whole thing). One of my favorite auto parts stores recommended I either fit a ballast resistor like the old Mopars used or use one of the later type coils that are designed to work with the full 12 V. I chose the latter since it seems a waste to bleed off the extra voltage to heat (from the resistor). Might as well use the extra voltage for it's intended use--TO SPARK THE PLUGS! The coil looks just like the stock one so nobody but an NCRS judge will be able to tell the difference. Besides it will be all covered by the ignition shielding when I'm done anyway.

Good Luck on getting your ignition sorted out!