I'm pretty sure there is only 12V on the small stud on the starter cable side when the key is in the "Start" position.
Steve
I looked at the shop manual and from what I can decipher from the schematics I think Steve is correct in his assessment (assuming the wire you guys are all talking about is the brown wire #262 on page 19-36 of the shop manual).
The world stands aside to let anyone pass who knows where he is going.
12 volts is correct. Ford designed its ignition system with a ballast resister wire as already spoken to. To drop the voltage to the points to 7 or 8 volts to reduce wear to the points in the "run" position
.
The wire from the starter solenoid identified, is a back feed to the ignition system. To aid in cold starting and is only energized when the starter solenoid is engaged and the starter is in operation. As the starter solenoid "throws" across to energized the cable to the starter it also energizes the small post to provide the back feed to the ignition system with 12 volts directly from the battery. Once the starter solenoid disengages the starter this post will be dead.
I still bet it is the ignition switch. Seen this before. The engine will run in the start positon but once you let the switch return to the run position it dies. I have also seen the converse. The engine will not start, but if the engine has enough momentum and in proper tune it will briefly try to start when the key is released to the run positon.
More Info: 1968 Gold Nugget Special, 302 J code 4 Barrel. Paint code Y5. Currently Wimbledon White with Black stripes and vinyl top. DSO 741111 as confirmed by Marti report. One of the earliest known Califor
12 volts is correct. Ford designed its ignition system with a ballast resister wire as already spoken to. To drop the voltage to the points to 7 or 8 volts to reduce wear to the points in the "run" position
.
The wire from the starter solenoid identified, is a back feed to the ignition system. To aid in cold starting and is only energized when the starter solenoid is engaged and the starter is in operation. As the starter solenoid "throws" across to energized the cable to the starter it also energizes the small post to provide the back feed to the ignition system with 12 volts directly from the battery. Once the starter solenoid disengages the starter this post will be dead.
I still bet it is the ignition switch. Seen this before. The engine will run in the start positon but once you let the switch return to the run position it dies. I have also seen the converse. The engine will not start, but if the engine has enough momentum and in proper tune it will briefly try to start when the key is released to the run positon.
Switch, switch, switch.
Rob
Rob
Just want to ensure we're on the same page here. It sounds like the present hookup is the 1.5ohm resistor wire (is it 1.5ohm for 6-cyl?) is hooked up to power both the *Pertronix* coil (with an internal 3ohm resistor for 6-cyl) and the ignitor1 module. This is not a stock coil (which, I believe, has no internal resistor). The Pertronix coil specs a full 12v to it with no resistor in series. Pertronix ignitor1 also wants 12v with no resistor. Now I am aware that on a V8 if you just replace points with an ignitor1, use the existing resistor wire, and retaining the *stock* coil that it "usually works". This is how my car was originally set up. But I'm not sure that this setup works with the Pertronix coil where you would get an effective resistance of 4.5ohms (1.5ohm resistor wire + 3ohm internal resistance) before you get to the internal primary windings of the coil. The key switch certainly may be at fault but from what I can tell from the discussion so far is that neither the Pertronix coil or the Pertronix ignitor1 is getting a non-resistor wire 12v source as they are required by their specs.
12 volts is correct. Ford designed its ignition system with a ballast resister wire as already spoken to. To drop the voltage to the points to 7 or 8 volts to reduce wear to the points in the "run" position
.
The wire from the starter solenoid identified, is a back feed to the ignition system. To aid in cold starting and is only energized when the starter solenoid is engaged and the starter is in operation. As the starter solenoid "throws" across to energized the cable to the starter it also energizes the small post to provide the back feed to the ignition system with 12 volts directly from the battery. Once the starter solenoid disengages the starter this post will be dead.
I still bet it is the ignition switch. Seen this before. The engine will run in the start positon but once you let the switch return to the run position it dies. I have also seen the converse. The engine will not start, but if the engine has enough momentum and in proper tune it will briefly try to start when the key is released to the run positon.
Switch, switch, switch.
Rob
I replaced the switch with a new one. They would need to both be faulty in the exact same way.
That said, there may be a problem with the female contact on the connector. I'll check that out tonight.
This getting older ain't for cowards. - John Mellencamp
I just installed the Pertronix Ignitor One on the 68 six shooter I just finished. Used the Pertonix 40511 coil as recommended. Pertronix systems DO NOT work well with stock coils, but I have seen some be ok. I always use a Pertronix coil in these installations.
My 68 six shooter is running on the stock ballasted wire and works great!
More Info: 1968 Gold Nugget Special, 302 J code 4 Barrel. Paint code Y5. Currently Wimbledon White with Black stripes and vinyl top. DSO 741111 as confirmed by Marti report. One of the earliest known Califor
See attached photos: Ign Sw in RUN position, 12V @ Solenoid Brown wire.
(I hate my sideways photos)
Neil
Neil,
I think that post is fed from the ignition circuit, and since there’s something wrong with my ignition circuit I can’t use the post as a coil feed.
This getting older ain't for cowards. - John Mellencamp
OK you electrical folks, how does the brown wire read 12 volts if the pink wire is 8.5 volts?
When the key is turned to start, the starter solenoid energizes, 12V is put through the contacts of the solenoid to the starter cable and the brown wire (#262). That 12V goes through plug D and then the flat 4 prong plug to the coil. It will also go through the resistance wire to the switch, but as the switch is in the start position, the run position contact isn't made and the voltage is just a potential there. When the key is released from the start position and goes to the run position, the starter solenoid disengages and 12V is no longer going to the starter motor or the brown wire terminal through the solenoid. Instead, 12V goes from the switch run position contacts, through the resistance wire, to plug D and the flat 4 prong plug to the coil. Assuming the engine is running, the points complete a circuit through the coil and voltage is dropped to 8-9V by the combined resistance of the coil primary winding and the resistance wire. Voltage will also go out along the brown wire to the starter solenoid terminal, but as the contacts in the solenoid are now open, is just a potential at that point. If the car isn't running, and the points are open, there is no circuit path, no current flow, and you would read 12V at the coil and the starter solenoid terminal. Once the car is running and the points complete the circuit path, you should see the 8-9V. Hopefully this helps. I can make a guess at how the pertronix is built and functions, that would involve a 12V transistor or a 5V transistor and a voltage divider circuit, with a .7 volt base voltage, which is why function is marginal if the resistor wire is used as a power source, but I would be guessing.
Yep, I get all that, so how does Neil get 12 volts at the solenoid brown wire terminal? That's what I can't figure out because the way I read the wiring diagram it shouldn't be possible.
This getting older ain't for cowards. - John Mellencamp
OK you electrical folks, how does the brown wire read 12 volts if the pink wire is 8.5 volts?
From the schematics I don't see how they can be at different voltages.
When starting, both "S" and "C" on the switch are at 12v (battery voltage). When running only "C" is at 12v and "S" is open.
The wires are both at 12v when starting because of the relay being closed by control signal "S" = 32 which forces 12v on wire 262 (= 16A = 16B = 16).
They are both at ~8.5v when running because "S" is open which leaves the relay open and the 12v is only being supplied from point "C" on the switch. This voltage is then divided down to ~8.5v by the resistor on 16A and the resistor in the coil.
The world stands aside to let anyone pass who knows where he is going.
Yep, I get all that, so how does Neil get 12 volts at the solenoid brown wire terminal? That's what I can't figure out because the way I read the wiring diagram it shouldn't be possible.
Well when I looked at the picture it (after rotating it) it sure looked like 10v to me. If the alternator were charging to say 14v then you would be seeing some sort of resistive drop. Caveat: I'm of the generation of digital readouts so I may have been misreading the picture...
Edit:
A little math (3ohm/4.5ohm)*14v = 9.3v. Now with the coil switching on and off the average resistance of the coil will *look* higher than 3ohm... that might make up the difference...
The world stands aside to let anyone pass who knows where he is going.
The check I did today was a static test: Ign Sw in the RUN position (Engine not running). My old reliable Simpson VOM read 12V on the Solenoid small stud.
Tomorrow I'll do the same test with the engine running, and see what I read.
BTW, this CS has a Pertronix system with a 40KV coil.
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