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Name: Lorenz
Status: Educator
Grade: Other
Location: ND
Country: United States
Date: February 2007


Question:
Engines consist of an iron core coil, a set of breaker points and a rotor inside the distributor cap. If the breaker points are closed and one accidentally leaves the ignition switch on when the engine is not running, the ignition coil will heat up (I would guess to about 300F.) and the engine ill not start until the switch is turned off and the coils cools for a couple of hours. Question: Why does the ignition coil fail to work when heated to only about 300F which is below the Curie temperature?



Replies:
Hi Mr. Lorenz

Interesting question. The usual thing that occurs when the ignition is left on, and the points in the ignition system are closed, (or their transistor equivalent is "on") is the resulting overheating condition causes a permanent open circuit in the coil, and the coil is useless after that. I recall permanently "frying" several coils as a result of this in my misspent youth!

I cannot think of a definitive answer to your question. You are right that 300°F is far below the curie point of the silicon-iron that is used to make the core, so that cannot be the reason (even if you account for the fact that the coil's internal temperature will be far higher than its external temperature of 300°F). One possibility is that the heat causes an open circuit where the ends of the coil wire are soldered to their respective terminals. As the coil cooled and things inside contracted slightly, a mechanical connection was made and the coil started working again. But frankly, I suspect this is grasping at straws!

A more likely reason relates to the fact that copper's resistance increases about 0.4% per degree Celsius. Without even taking into account that the coil's interior will be significantly hotter than the outer housing, the 300°F temperature you refer to, will increase the coil's resistance to rise 60%. If the higher internal temperature is factored in, one can see that the resistance could easily be nearly double, reducing the coil's primary current by nearly half, resulting in an identical reduction in stored magnetic field in the core.

An ignition system belongs to the "flyback" family of DC voltage conversion circuits. This coil does not act like a true transformer. Current flowing in the primary (when the points are closed) causes energy, in the form of a magnetic field, to build up in the core. When the points open and the primary current is suddenly cut off, the magnetic field starts to collapse, "dumping" its energy into the secondary. If the primary current were reduced nearly by half, the result would be a much lower amount of magnetic energy stored in the core, and a weakened spark; perhaps too weak to reliably start the engine.

Regards,

Bob Wilson.


Although I do not know the coil system you are referring to (sorry, I do not know cars too well), I could guess that the excess temperature is affecting the electrical conductivity of the ignition coil, thus allowing it to carry less current, and have less than the intended effect.

Ryan Belscamper



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