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Elite Veteran
Posts: 1102
 
Location: Hayward, Calif | Hey Guys, I put a 76 360 in wife's 59 Dodge. I finally got it running after replacing all the electronic components. Seems really strange since the engine was running before I pulled it out of the motorhome. I took the complete ign. harness out with the eng to keep the circuits correct. The test procedure in motor manual, I carefully adhered to step by step. I have 3 different references that show the dual ballast resister having a 5.0 ohms resistence on one side and a .5 resistance on the other side. The ballast resister that came with this ign sys when I removed it has 5.6 on one side and 5.4 on the other side. Since I have already replaced all electronic components, I have Napa trying to find the 5. / .5 resister. He does have a 5. / 1.2 ohm resister. The 1.2 side goes to the run side of the coil. If I use the 5. / 1.2 resister, will this make a difference ???? Sure would hate to have to buy all those components again. Thanks, Jerry |
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Elite Veteran
Posts: 1102
Location: Hayward, Calif | Hey Guys, I did a little more checking around and testing. With a voltmeter I found a voltage drop of .2 of a volt across the 5.0 ohm side of the resister. Which is practically nothing. I found that ohms will restrict amps. But I can't find anything to tell me how many ohms will affect how many amps. Must be someone out there who knows a little more about electrical sys's than I do. |
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Extreme Veteran
Posts: 444
Location: North of the 49th |
Hi
Chances are your NAPA boys won't find the "early" 5/.5 dual ballast resistor as of 1979 Chrysler only made the 5/1.2 resistor available, as announced in Chrysler service bulletin 8-12-79 , the bulletin announced that the "later" resistor can be used on all previous model years so its no big deal. I have four Chrysler products with this system and all are as dependable as the sun, I've replaced the Dual Ballast resistor on one of them using the later type with no ill effects . I've installed quite a number of these systems including my own 1959 Dodge Custom Sierra Wagon and when properly installed they just plain work.
Here is a link to a good wiring diagram for the system: http://www.chargerr.com/Ignition/mopar_5pin_ign.jpg I'm sure if you recheck your wiring all will be well. |
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Elite Veteran
Posts: 1102
Location: Hayward, Calif | I finally contacted a local shop who is familiar with this electronic sys. He said because of the higher alt. output ( 60 amps ) I should use the 5.6 / 5.4 ohm resister. He also confirmed that ohms restricts amps. So I was on the right page there. That wiring diagram you suggested is the same one I have other than the ohms value. I have a 5. / 1.2 ohm resister but am a bit uneasy about using it as it may allow too many amps to the coil and the sensor in the dist and overload them. Sure would be nice if I could find out how many amps are coming out of that resister. Thanks, appreciate your help. Jerry |
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Expert
Posts: 3575
 
Location: Netherlands | I don't quite see what the alternator output has to do with the coil in the ignition system.
The ballast resistor keeps the voltage to the coil in a certain range.
This is all to prevent the coil from overheating.
You can try the lower rated resistor but I would keep a sharp eye (or infra-red temp gauge) on the coil and see if it doesn't overheat.
If you don't care much for originality AND want to do away with the resistors and extra connections, you could consider replacing the ECU and resistor with a GM HEI module. You might even replace the coil for a modern, better designed e-core coil. These GM modules and e-core coils run on full 12-14v and don't run hot.
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Expert 5K+
Posts: 8443
Location: Perth Australia | The bloke that told you to use the bigger resistor because of an alternator is a fool and has no idea what he is talking about.
The main purpose of a ballast resistor ignition system was to make a way of boosting the high tension spark voltage on crank, thats it, no magic, just that
There is a reason for the twin resistors (which, with a little research you should find) that I cant remember right now, but look up ready to run distributors on ebay, yes, they are mostly Chinese and yes, they are not oem, but for the price, they are cheap and easy (just like an old w^^^^) dont think I can say that 
If you must stick with the old mopar ignition, look up what the resistors do on google and see what the reason for the differing values are for
The twin resistor system came out in the mid to late 60s I think, before then it was single or none
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Expert 5K+
Posts: 9653
Location: So. Cal | Your ignition likely won't run with 5.5 ohm resistance in the line. If it does run, it will run terribly. The most it needs is 1 ohm extra, or even less. The idea is that you would like your ignition system to get as much power as it can handle. The stock chrysler system needs around 1.5 ohms to survive (1 ohm extra from the ballast + .5 ohm from the coil). GM HEI systems (HEI module + E-core coil) don't need any resistance. If you would like to mount an HEI module to your current distributor, you can get one from me, here: http://www.designed2drive.com . This would allow you to eliminate all the original Chrysler wiring, ballast, & module. Externally, it only needs a power line, that's it. But an original Chrysler system can be made to work OK too, just not as hot of a spark. Just use the 1.2 ohm resistor version and you should be fine. |
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Elite Veteran
Posts: 1102
Location: Hayward, Calif | Hey Guys, I'm no electronic guru, and maybe I'm over thinking this thing. But this is what I'm finding. The ballast resister has practically no effect on Voltage. But since it deals in Ohms it does have an affect on Amps. I read somewhere in all this there was a 4 amp draw wothout eng running, 3amps at idle, and 1 amp at higher RPM's. Not sure if this is specific to this application or generic figures. Theroretically, the alt will only put out the amps that are called for in the sys. I just got the T/F back from the trans shop yesterday, so am putting that back in. I think once I get it running again I'l wire a test amp gauge into the sys at the resister, on both sides and see what it reads. I'l let ya'll know what I come up with. Jerry |
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Expert 5K+
Posts: 8443
Location: Perth Australia | Jerry, if you take your voltage readings with the system open (no current going through the coil) it will have the same reading on either side of the resistor, with it closed, you will have a very different reading.
Generally, resisted ignition systems used an 8V coil (some are marked 12V use with resistor) and the idea was that when cranking, the battery voltage would drop to 9 or 10V, so with a bypass system fitted to bypass the resistor when cranking, the coil will actually see more voltage than normal and then provide a hotter spark for starting.
Bypassing the resistor will show more amps than with the resistor in circuit purely because your feeding more voltage to the coil than its designed to get.
The lower the coils designed voltage, the heavier the wire used inside the coil and it has less turns as well.
Without typing a wikipedia description on electrical winding basics, that should give you an idea whats going on.
If you go HEI or Bosch electronic or even Luminition (pertronics) whatever, you will need to get a 12V coil, bypass the resistor and live happily ever after.
Thinking about the twin resistor, I have ideas (= vague memory) that the second resistor has to do with the starter bypass side as a way of the ignition not feeding enough voltage back to keep the starter engaged, but this is possibly wrong
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Expert 5K+
Posts: 9653
Location: So. Cal | Jerry, The relationship is the voltage drop across a resistor is V=IR. This means that the amount of votage drop across a given resistor is proportional to the current going through it. So measuring the voltage drop doesn't tell you much since you don't know what the current is. But just think of the extremes. If you have no resistor, R=0, and a device that draws 10 amps, then there will be 10 amps in the line, with 0 voltage drop - or it will see the full 12 volts in this case. As the resistance increases to infinity, which is the same as an open wire, the voltage drop across that resistor is the full 12 volts so the device sees 0 volts, and the current is 0. So both the current and voltage change at the device in proportion to the amount of resistance that you have there in the fixed ratio V/I = R. An increased resistance cuts down both the voltage and the current that the device will see. |
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Elite Veteran
Posts: 1102
Location: Hayward, Calif | Guys, The thing that got all this started was that when the eng was in the motor home it used a 5.6 / 5.4 ohm resister. The folks drove it around for a few years until Dad decided he was getting too old to handle it anymore. Then it sat for a few more years. Dad decided to get rid of it, that when I got into the picture. After I put the eng in the 59 and couldn't get it to start, I tested all the components and had to replace them all. That didn't make any sense. Rechecked my wiring circuits and found I had put the starter bypass on an ign terminal instead of the starter terminal at the ign switch. OK, I'l accept that as a reason for replacing all the components. I have done all my testing up to this point without the eng running. The point of confusion comes when all sources including Motors Truck Repair Manual, of test procedures show a 5. / .5 resister. When the engine has been running for years with a 5.6 / 5.4 resister. Dad would've mentioned that he had ign problems. I finished installing the rebuilt T/F today, so will see what happens when I get it started tomorrow. Thanks Guys, appreciate your help !! |
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Expert 5K+
Posts: 9653
Location: So. Cal | Ballast resistors do go bad. It is pretty common. The Chrysler module certainly doesn't need this much resistance. If you properly measured 5 ohms there with a multimeter, you can rest assured that the ballast resistor is bad. Replace it. |
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Dodge electronic ignition
