specific electrical Q's and diagnosis help.

LostDawg and I were trying to diagnose a problem with my CJ yesterday.
Found that the engine would run with the key in 'Start' but not in 'RUN' unless the coil was 'hot-wired'.So we ended up replacing the wire from the coil to the Key Switch..
End result was it ran GREAT for a while, but while we were out testing it (chronicled here: http://www.earlycj5.com/forums/showthread.php?t=46900)
It hiccuped once then died shortly after.
thought it was vapor locked, then out of fuel...blah blah, so we called in a fuel truck, but still couldn't get her started.
She now has NO spark from the coil. Has 12v power to the coil but no output.

Now the Question: The wire we replaced off the coil goes to the ALT, and the IGNITION SW. Red in color (at least in the diagrams). Is that wire a resistor wire of some kind? Seems like its too long for the run required, but all the diagrams we looked at said nothing about a resistor wire or ballast resistor in the system.
We ran it straight from the key switch to the coil. I'm thinking this burned up the coil by giving it full 12v (it was an aftermarket Jacobs coil) and possibly the Points eliminator kit in the Dizzy.

My solution for now is to reinstall the points set (if possible) and a new coil, wire in a new ballast resistor on the new wire, and see what happens.
Any other ideas?


SIDE NOTE: don't take off for a test run 10 miles from home without bringing any tools and being low on fuel....

 

Tried that. My shadethree mechanic skills said to replace the soleniod and the switch with new parts before troubleshooting it. R) The switch needed replaced anyways....
The problem is, we can't find a resistor wire (other than if the red wire is it) or a ballast resistor on the Jeep anywhere.
After the test run, now we have NO spark at the coil tower or the plugs...So I fried something in there, I'm sure...

 

w3srl has it right, but if somebody removed the ballast resistor it may very well have that special resistive ign wire. If that is it (res wire) it also came in pink and has a special firecoated rubber coating. By pulling on that wire along it's length you will get a rubberband effect at wire separation inside and it won"t show on outside. Hope this helps.

 

just a thought if You have a test light check to see if there is voltage at the coil with the key in run should be like 9v if nothing there Then check at the switch should be hot as well. That should give you an idea about the circuit If you do not get spark in the run it has to be that wire. The second wire is only hot to the coil when the solinoid is energized. It gives you the full 12v when the starter is engaged.

 

As I was reading your symptoms, first thing I thought was that you had a bad ballast resistor at first, and by jumpering around it you burned the coil; now you've got a bad resistor and coil.

I highly recommend you go down to sears and buy their ~$12 multimeter; works well and will pay for itself in replaced parts burned up during troubleshooting.

 

This is a classic case of ballast resistor.. The ballast resistor is needed to reduce the voltage to the points when the engine is running, otherwise the points burn up.. When starting the key shunts the full 12 volts to the coil. Hence it will run when starting and dies when the key is returned to the run position.. My ballast resistor broke on my 67 so I have experience. I got a new resistor at NAPA for like $5.

The ballast resistor was mounted onthe firewall, about in the middle. I am betting that if you look there you may see a single wire that has been spliced. More than likely that is the wire that used to go to the resistor.

If the wire is there then all you need to do is cut it and connect each end to the resisitor. Assuming all the rest of the wiring is correct.

 

Just a comment - it's a common misconception that the ballast resistor is there to reduce voltage to the points. The voltage would only be dropped if it were in series with another resistor, but it's not. The resistor acts as a current limiter, not a voltage dropper. The resistor and the capacitor (condensor) form a low-pass RC circuit that both limits current through the coil and reduces arcing across the points.

Note - online and book sources will tell you the ballast resistor drops voltage, but that's not correct.

 

Nort wanting to start an agument but, I am an eletrician, and the current reduction deal does not sound right to me. I would like to see a diagram of how the ballast resistor is connected in the circuit. If it is series then the resistor has a voltage drop across it that is subtracted from the total voltage. I believe the resistor is in series with the coil and points.The current is equal in all parts of a series circuit.

If it is in parallel, then the resistor and points see full volts and the current is dependant on the load in each branch.

I googled up ballast resisitor and came up with this: Ignition basics

Regardless, an open or missing ballast resistor will cause you some grief..

 

Hi Zila - I tried to phrase the reply in a way that wouldn't be argumentative.

The ballast resistor is in series with the coil and the points. The ballast resistor's purpose is to limit the current through the coil, not to drop the voltage. It is true that the coil has some resistance, and that the coil-resistor circuit acts like a voltage divider, but that is irrelevant to the ballast resistor's purpose.

This is not a DC circuit. You have to think about it in terms of all the passive components - resistance, inductance and capacitance - and their transient behavior.

When the points open, current is flowing, so the coil gives an inductive kick. This is the same inductive kick that fires the spark plug. There is a back EMF from the spark coil, acting on the coil primary. The resistor and capacitor in series dampens that back EMF. The high-frequency part of the back EMF is shorted to ground through the capacitor, and the resistor brings the coil back to 12V at a slower, current limited rate.

When the points close, there is no kick, because there was no current flowing. Here the resistor acts like a simple current limiter, preventing the coil from overheating.

The story of the voltage drop is only sort-of wrong, since the DC performance looks like a voltage drop. That's not why the ballast resistor is there though.

hth! Tim

 

Tim I aint gonna argue.. But you are wrong on several points.. The fact that the volts are DC is exactly why the points are needed in the first place.. Study up on tranformer theory.. Trust me I do know what I am talking about.

 

I used to know stuff. Been so long since I messed with 'leckity that I doubt if I could even shock myself anymore. Folks all talkin about their DC transformers n'such. What would ol'e Faraday think of that? He'd prolly just tell us to go see Otto Diesel and stop arguing about ignition system theory.

Chilly

 

Please don't leave me hanging. I'm really interested in your comments.

 

Where's the popcorn looks like there's going to be an altercationR)R)

 

Tim after reading your post I admit that you are not totally wrong in your thinking, but maybe you do not 100% understand what all the components do.. I will try to explain as best I can from an electrical standpoint. I should be fixin the leaky sewer for the lady but this is more interesting.

Firstoff. Your Jeep is using a 12 volt DC system.. The armatures of alternators and generators both put out AC and the difference is how the electricity is taken off the armature.. Slip rings and brushes = AC, a slotted commutator and brushes = DC. Now in the case of modern automotive alternator they generate AC but that is converted to DC by use of rectifiers and electronics in the alternator itself. This is a fact. Your battery is DC as well.

As for the points/coil/ignition deal, heres a fact: Any conductor moving through a magnetic field will generate a current and voltage. Key word: Moving. Vice versa if the conductor is stationary and the field is moving you will get current and voltage. Now in an AC circuit the current and voltage vary constantly in the form of a 60 hz sine wave, at least in the U.S., Euro's see 50 hz... Hence if you connect AC to the primary of a transformer the result will be current flow out the secondary. Due to the ever changing current and volts generating an ever expanding/collapsing magnetic field.

In the case of a DC system the transformer or coil in our case, will generate when it is first energized, rise and then stop and become flat. What we now have is basically an electromagnet and no volts or current flowing from the secondary. Now break the circuit and the collapsing magnetic field will generate a voltage and current.. The points serve the function of causing the magnetic field in the coil to collapse and expand at the proper moment to provide high volts and current for spark.. It does this by breaking the circuit and allowing the magnetic field to collapse and generate..


The capacitor is across the points is o help with controlling sparking at the points.. A capacitor is a lot like a battery in that it can store electricity. When the points break it cushions the arcing by charging up and diverting some of the current. It also discharges as the points close.

The ballast resistor is in series with the points and coil to limit the volts the points and coil see.. In series the voltage drop of the sum of the loads is equal to the source voltage. The reisistor and the coil are in series, so the resistance of the coil wire in the primary is the other load. The ballast resisitor limits the volts to around 8 in most circuits. With the ignition off you will see ~12 volts DC when measured with a meter. With The engine running you should see ~13-14 volts at the battery and other circuits. Still don't believe it? Then get out a DC volt meter and test for yourself.. Read across the ballast resistor and then across the primary of the coil. Also test at the tail lights etc.. Still don't believe it? Go get two ballast resistors and connect them in series,, Then to your battery.. Read DC volts across each one. Now across both.

Now would you like a real world demo of your DC coil and points in action? Yes? The find an old small step up transformer that is still good out of an old radio. Or you can make a small transformer using some iron for a core and wire. Next using a regular D cell connect the battery to the primary of the transformer. Now hold the wires of the secondary one in each hand. Have an assistant break the wires to the battery..

 

I guess I am wrong along with a guy named George Ohm and another named Kirchhoff and Faraday.

 

Well, I didn't think this was as much of a can of worms as it's turning out to be. Steve saw it coming...

As I see it, this is a classic RLC circuit (points open) or RL circuit (points close). Doesn't matter which side of the coil the resistor is on, it will have the same transient performance. Here R is the resistance, C is the capacitance, and L is the inductance of the coil. Howard, I wouldn't call it a tuned circuit because it's not resonant - an LC is a tuned circuit, an oscillator, like used in a radio - instead I'd call it a relaxer rather than an oscillator.

The transient is formed when the points open or close, and the circuit will relax to a 12V potential across the capacitor and no current (points open) or the voltage divided between the resistor and the winding resistance of the coil (points closed).

I really don't want to be a jerk about this, but RL, RC, LC and RLC circuits are sophomore level physics... not that complicated, but you have to use differential equations to predict their behavior accurately. I'm not sure we want to go there.

Zila, my friend, I appreciate your efforts and thank you for answering in such detail. My mistake for not making my reply clear, and maybe in posing this in such an analytical way. Transient circuits aren't AC or DC - instead they decay. Your example about the transformer is well taken - the connecting then disconnecting the battery from the transformer causes a transient. The transient in the primary causes an equal and opposite (except for resistive and hysteresis losses) in the secondary, multiplied by the turns ratio. The points circuit is a transient repeated over and over again.

Howard's right - you can express the problem in terms of current or potential. After all, they are related through Ohm's law, whether the circuit is AC, DC or transient. The resistor limits the power dissipation in the coil. I tend to think of resistors as current limiters, especially since the RLC problems are usually posed in terms of di/dt. Perfectly valid to pose the problem in terms of voltage or power, but the solution of these problems is easier in terms of current.

My, my, this dead horse is getting tender. I don't really want to, but if anyone is interested we can go deeper into the math side. I think most readers will find that deathly boring though. The Wikipedia article on RL circuits is pretty good, and will give you the flavor of the problem. The RLC article does not cover the step response, unfortunately, so it isn't very relevant to this discussion.