| rdevarga | -- 02-25-2013 @ 4:33 PM |
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In the process of doing a complete rewiring of a 1940 Pickup and modifying to a 12V system. While under the dash I came across the "Headlight Circuit Breaker & Ignition Resistor Part no. 01A-12250". I am wanting to replace this item. Q1: Does this Circuit Breaker need to be 12V as well? Q2: If so, what should I replace it with? Note: I am not adding any electronic this or that. Just a utility/play truck. Thanks alot for the continual help gentlemen. Richard deVarga Austin Texas
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| ford38v8 | -- 02-25-2013 @ 6:28 PM |
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Richard, the circuit breaker serves all the 6v lighting systems in the car. The ignition resistor drops voltage from 6v to 4v for your ignition coil. Whether you use them or not depends on whatever your plan calls for. Don't chuck them out, though, as someone else will know exactly what to do with them. Alan
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| supereal | -- 02-25-2013 @ 7:14 PM |
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In addition to Alan's info, you will need to obtain a 12 volt coil, which will not need a resistor. The "circuit breaker"is a sort of a buzzer which was designed to keep the headlights only illuminated in order to get home. It is not a true circuit breaker, and can be eliminated, as well. If you wish, you can disconnect them and leave them in place to preserve appearance. Converting to 12 volts is a lengthy project involving any part of the vehicle to which wires are attached.
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| TomO | -- 02-26-2013 @ 6:50 AM |
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The circuit breaker will still open when you have a direct short in the headlight circuit, so I would leave it in. Most 12 volt coil;s have a built in resistor, so you can bypass the original one. Tom
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| juergen | -- 02-26-2013 @ 9:35 AM |
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A circuit breaker trips on current, not voltage. And the appropriate current limit is based on the wiring size and load. With 12 volts replacing 6 volts, your current needs should drop so the present breaker should work OK. As for the resistor, it all depends what you are going to do for the ignition circuit including coil. If you are transitioning to a 12 volt system, my first choice would be to bypass this resistor and go to a ballast type to drop the voltage to a 12 volt coil. Select a coil and ballast for a 12 volt car that you prefer. Also change to a later 12 volt starter solenoid which has the terminal to bypass the ballast resistor while starting.
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| supereal | -- 02-26-2013 @ 6:26 PM |
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When we do a 12volt conversion, we have a Ford coil rewound for the voltage by Skip Haney. That eliminates the need for a ballast resistor, and preserves the appearance.. If you use a modern coil, determine whether it has a built in resistor, as many round coils do. As for the breaker, you can compute the headlight current load when the wattage of the headlamps is known. Only the headights are connected to the "breaker", and as the voltage doubles, the current drops to half the former value. As noted above, the unit is current based to briefly open and close in the event of a short in the headlight circuit. Unless the headlight wattage doubles, the unit will likely be useless. You can leave it connected, but it is probable the wiring will be smoked before it protects. We prefer to protect wiring with modern resetable breakers, mounted on a panel where the breaker/resistor was located on the inside of the firewall..
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| juergen | -- 02-27-2013 @ 5:06 PM |
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We all have our preferences based one our experience but I would keep the circuit breaker in. About 50 years ago I bought a 41 Ford in the winter. Whenever I shifted at night, the lights would go out, then go on again. It was just a nuisance as long as there was snow on the ground and no garage. An 18 year old has more important things on his mind. When it warmed up, I crawled underneath and found that depressing the clutch shorted the headlight wire which was frayed by the clutch lever. The breaker protected the wiring and it never burned up. A little friction tape and tieing the wire out of harms way and the repair was done. Also I always like the coils with a ballast because it allows higher voltage to the coil when the starter is drawing so many amps and the internal resistance of the battery drops the voltage. The coils without an external ballast get reduced voltage just when you need it most, when the engine is cold and the gas is poorly atomized. Just my story and I'm stickin to it. This message was edited by juergen on 2-27-13 @ 5:08 PM
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| supereal | -- 02-27-2013 @ 7:05 PM |
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The extra pole on the solenoid requires access to the coil side of the ballast to be effective. We connect a diode between the starter motor side of the solenoid and the coil input to compensate for the voltage drop caused by the starter. The diode serves as a "check valve" to prevent the ignition circuit from trying to run the starter. If anyone wants me to repost the diagram, I will. It is an easy, effective, and cheap way to produce faster starts.
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