The Doc
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| posted on 23/2/07 at 06:25 PM |
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Ignition, Solonoid, Coil Issues
Ford XFlow
Can someone tell me what goes on between the solonoid and coil. I can see that we have what is effectively a big relay powered up by the red/black
wire which becomes active when turning ignition.
What about the other terminal (yellow/black) that goes to the coil. At present there is no output from this. Should it become live once the engine is
running, remain live whenever the ignition is on - or what? Mine is dead. Does that mean the solonoid is US?
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David Jenkins
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| posted on 23/2/07 at 07:13 PM |
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If the solenoid gets the starter going, I wouldn't worry about it - simply take the coil's live feed to a convenient point in the
'switched' part of the 12v wiring (i.e. a circuit that's on when the ignition switch is in the run position).
My solenoid doesn't even have that extra terminal - just battery, starter and the start connector.
David
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flak monkey
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| posted on 23/2/07 at 07:16 PM |
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Its the 12v feed from the solenoid to the coil when the engine is being started. Should ony be live when you are cranking the engine.
You have a ballast resistor installed in the loom, which provides 9v to the coil when the ignition is on. This is over ridden by the 12v while
starting to give a better spark. Was designed to increase ease of starting.
David
Sera
http://www.motosera.com
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David Jenkins
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| posted on 23/2/07 at 07:24 PM |
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Good point - I'd forgotten the ballast resistor circuit! 
I suspect that you may have a mix-up in this area, thinking of your other post (fried coil).
I didn't have a ballast circuit in my car, so I can't help you any further - I'll leave that to others.
David
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The Doc
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| posted on 23/2/07 at 07:37 PM |
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As the coil is now buggered I think I'll get a 12V one and wire it to a convenient feed.
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