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Old 05-10-2007, 02:32 AM   #1 (permalink)
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What is the "S" terminal for and is it a constant 12 Volts when the car is running I just need to understand how the starter solenoid/relay works..
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Old 05-10-2007, 03:18 AM   #2 (permalink)
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On the older Fords, there are 2 large terminals and 2 small terminals on the selonoid.

The 1st large terminal is for the + side of the battery, and doubles as a connection point for the alternator, and power to the interior of the car.

The 2nd large terminal is the junction for the starter cable.

The starter selonoid is nothing but a heavy duty relay (it's often referred to as a starter relay). The red/blue wire on the "S" terminal goes to the ignition switch. When the key is turned to "start", the ignition switch applies 12v to this wire, and thus to the "S" post. This activates the relay, closing the circuit between the 2 large posts and the starter turns. When the key is released back to any other position besides "start", the power is then removed from the wire (and the "S" post) and the relay opens back up, breaking the circuit to the starter and turning it off again.

Now the 4th post (only on older Fords ... they did away with it once they went to electronic ignitions) is the "I" post. This wire (the brown one) ties directly to the red/green wire going to the + side of the coil. When the starter relay is activated and the circuit to the starter is closed, in addition to the power to the other large post, 12v is also applied to the "I" post, thus providing 12v to the coil. Like the large post, once the key is no longer in "start", power is removed from the "I" post and power to the + side of the coil only comes from the red/green wire, which goes through the firewall, and attaches to the resistor wire. This drops the voltage to the coil under normal operation, but allows a full 12v to the coil during starting, to make it easier to start.

So in summary ... there's always 12v on the battery side of the selonoid. The "S" post gets 12v applied to it only during starting, and has no voltage on it or it's wire at any other time. The Starter side gets 12v during starting only. The I post supplies 12v during starting only, so both it and it's wire have 12v during starting, and they have ~7v with the key in the on position, supplied by the resistor wire that's tied into the same wire.

There you go ... everything you ever wanted to know about starter selonoids!
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Old 05-10-2007, 07:41 AM   #3 (permalink)
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John, I don't think you have answered this question enough times. How about a drawing?

Todd
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Old 05-10-2007, 09:21 AM   #4 (permalink)
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Old 05-10-2007, 09:28 AM   #5 (permalink)
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Johnpro,

I thank you for taking the time to help me see the light. I really appreciate the explanation. I now understand how things work. I got several answers to this questions but no one was sure of the complete path of events. I will pass this knowledge on to members of my car group.
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Old 05-10-2007, 12:01 PM   #6 (permalink)
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I've also noticed in schematics that the I post applies 12V to the contact that bridges the two large posts that provide power to the starter. Since the starter is a large inductor, when the contact opens, an arc is likely to occur burning the contacts. Applying 12V to the contact provides a bias to reduce or eliminate burning of the contacts or a sticky contact (starter keeps going although the key was released).
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Old 05-10-2007, 08:02 PM   #7 (permalink)
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Quote:
Originally Posted by blkfrd
I've also noticed in schematics that the I post applies 12V to the contact that bridges the two large posts that provide power to the starter. Since the starter is a large inductor, when the contact opens, an arc is likely to occur burning the contacts. Applying 12V to the contact provides a bias to reduce or eliminate burning of the contacts or a sticky contact (starter keeps going although the key was released).

Not quite. Voltage comes OUT of the "I" post, during cranking. The voltage on the wire is the same voltage as the coil. You'd only have 12v on that wire when the 12v is coming OUT of the "I" post, of if you've elminated your resistor wire.

With the engine running, if you remove the brown wire on the "I" post, you will find no voltage on the "I" post, and ~7v on the brown wire (assumming the resistor wire is still intact ... 12v, if it's been by-passed).

You're not applying voltage to the "I" post, you're using the "I" post to apply more voltage to the coil during cranking.
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