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Old Feb 21, 2004 | 01:05 AM
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New Coil Setup

I just bought an ACCEL #8140C coil and I hooked it up but as I was sitting there admiring my work I realized that there is a positive lead coming from both the starter relay and from a ballast resistor to the positive terminal on the coil and I started to scratch my head. Shouldn't there be one or the other? And if so which one should I remove? Seeing ACCEL is so good about giving me an instruction sheet (yeah right) I am counting on you guys to pull me through this one. This is on a 76' 390 with a 2 bbl.
 
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Old Feb 21, 2004 | 02:26 AM
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Well, technically that's correct as far as the circuit is concerned but I would have thought they'd all be bundled in the wire harness. Is the ballast resistor really a resistor or a wire?
 
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Old Feb 21, 2004 | 08:21 AM
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I have an Accel supercoil on mine. I have a hot wire from the ignition and the hot wire the the distributor joined on one post of the balast resister. Then I have a hot wire from the resister going to the coil. Hope this helps.
 
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Old Feb 21, 2004 | 08:33 AM
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well, I assume it is what you call a ballast resistor, it is a three inch long white bar with two post on it
 
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Old Feb 21, 2004 | 08:37 AM
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yup
 
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Old Feb 21, 2004 | 12:10 PM
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4x4 your wiring scheme is fine but not stock, that's why I was asking about the resistor. The lead coming from the starter relay is the bypass to feed the coil when starting. The bypass goes directly to the coil or else it wouldn't be bypassing anything.
 
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Old Feb 21, 2004 | 05:44 PM
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what would happen if there was no resistor on the line to the coil

would it get really hot???

because that is how mine is hooked up but it works great but does realy get hot.

which wire is the resistor on. the pick up is a 63 so it had a generator and voltage regulator and worked differently. thanks for the help

P.S. sorry to ask a ? on your thread but it does pertain to the same thing
 
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Old Feb 21, 2004 | 07:09 PM
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You stock coil will burn up with a full 12V going to it. The wire from the ignition switch will have the resistor or should at least.
 
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Old Feb 21, 2004 | 07:13 PM
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corect me if im wrong but i think the balist takes 12V and turns it into 6V
 
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Old Feb 21, 2004 | 07:20 PM
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It depends. I think I measured 9V with the engine running. I don't think it takes it down all the way to 6 on any setups.
 
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Old Feb 21, 2004 | 07:27 PM
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Conventional ignition coils suffer the disadvantage of being designed to operate best at about 12 volts. Unfortunately, a 12 volt battery often produces as little as 7 volts when "run down" because of excess starter operation, especially in extreme cold. To produce optimum coil performance (and hence nice fat sparks at the spark plugs) under such adverse conditions, the "ballast resistor" or "ballasted coil" system was developed.

This system uses a coil which is designed to be most efficient at about 8 volts. For starting, full battery voltage is supplied! This makes this system as efficient at low battery voltage as a "conventional" coil is with the battery supplying a full 12 volts. (For any battery voltage above the coil's design voltage, it's even better - an "overboost" condition.) However, an eight volt coil cannot be run continuously at 12 volts without overheating and failing. As soon as the starter switch is released, the coil no longer receives full battery voltage. It is then powered through the ballast resistor which reduces the 12 volts (the generating system is now working) to the coil's design voltage.
 
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Old Feb 21, 2004 | 08:52 PM
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So when I hear people talking about running 10 guage to have a more steady supply of power to the coil does that mean they are running it to the ignition then to the ballast and then to the coil?

On a side note too, the VECI label said DWELL : NONE and after about 50 trys to adjust it the closest I came was 18 degrees, should I try and get zero or am I alright (please say no)lol
 
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Old Feb 21, 2004 | 09:15 PM
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Your coil is energized from both wires, but not simultaneously. The wire from the solenoid is just to start it, once you release the key, the solenoid is de-energized, and the circuit opens up. When the key twists back to the 'run' postion the wire from the resistor is feeding it the voltage.

I have never heard of using such a heavy guage wire, but it would be okay to use it, it's just that you will have to replace every inch of primary wire right back to the tap source or it will do no good. Run it this way. Voltage source to ballast, then ballast to coil. AND from the solenoid to the coil.

Here's the catch, the negative is ground and it runs to the distributor to the points. You need to get that one too.

The points have to have a dwell time or the engine can't run. I ahve no idea what a VECI label is. Dwell should be around 22 degrees. 18 is close enough.

Two reasons for accurate dwell: One is coil saturation, the other is it has to break at the right moment to collapse the field to induce the voltage into the inner windings to get that spark to the plug at the moment the rotor nears the respective terminal post for spark plug.

Hope I have helped you.
 
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Old Feb 21, 2004 | 09:37 PM
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Well I don't get it, Ratsmoker was saying in another post that with a Accel Super Stock coil you dont need the ballast? So should I just run the wire straight from the ignition? Ratsmoker, can we get some closure here.
 
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Old Feb 21, 2004 | 10:08 PM
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usually hotter coils designed for just what you need are designed to run without a ballast. also, some coils have internal ballasts.
 
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