Measure the resistance of the primary circuit starting at the battery post to the coil wire. If its like any other american car there are about 12 connections and between the battery and the coil wire, and 30 years of oxidization on each of these rack up the total resistance. I suspect the issue is here, and bypassing the ballast is a band-aid. The resistor is in place to limit the current in the system. You need to limit the current to 6 or 7 amps. Without the resistor you don't know the current, and it could be 20amps or more. If it is that high the best you could hope for is a burned up coil. Transformers and ignition coils work on different principles and the transformer ratio doesn't apply to this system.
Thius is from kurt02gt: The Unilite really doesn’t care too much about supply voltage, as long as it is in the 6-18v range. What it DOES care about are coil current, and coil energy. The coil driver in the Unilite is a transistor that has certain maximum ratings, some which are mutually incompatible. For example – the maximum continuous current rating for the transistor is 14 amps, but you can’t have that much current in the ignition circuit with standard coils, because the energy level would be too high. The purpose of ballast resistance is NOT to control voltage, but to limit the current in the coil primary (which is what the transistor switches on & off). Ideally, for “normal” coils (with primary inductances in the 5-8 milli-Henry range) the Unilite wants about 2 to 2.5 ohms total primary resistance. That means that the combination of coil primary resistance and ballast should be 2-3 ohms. This will limit the current to 5-6.25 amps with a nominal 14v electrical system. If you look at the latest instruction sheet, they show just a single ballast – whether it is in the vehicle loom wire or a separate piece. Older instructions showed two ballasts for those applications that didn’t have a loom resistance wire. At any rate, our (and most other) ballast actually increases in resistance when it gets warm. This allows a bit more current for cold starts, then lowers the current to reduce stress on the coil driver transistor. In addition, virtually all coils’ primary resistance goes up as they heat up, so that drops the current even more. Unilites normally fail because of other things. A number of years ago, we did a fairly comprehensive analysis of returned Unilites. What we found was that half of the returns worked without any problems! Of the half that didn’t work, about half of them had internal damage consistent with miswiring – two major problems were getting the red and green reversed, and getting the red and brown reversed. In both of these cases, you got significant excess current that either blew out the coil driver transistor, or melted the brown wire. Of the final ¼ of the returns that weren’t obvious wiring errors, they had a variety of problems, many of which have been addressed in the latest redesigns. I’ve been with Mallory almost 25 years, and as far as I know, we have never had a Unilite fail of the installs we’ve done here. We have also built (I’m approximating, as I don’t have hard numbers) close to 1 million Unilite modules over the 30 years they’ve been in production, most of which worked (or are working!) just fine. Unfortunately, with that many, you are bound to have problems related to installations, random part failures, electrical problems in the vehicle, etc. and those failures are held up as examples of how “bad” the product is, when in fact, it has been very reliable. So, to wrap this up – the things that kill Unilites are bad installations, using a low-resistance coil with no ballast resistor, heat, and vehicle electrical gremlins. One of the worst is to have the battery connection come loose or have a high resistance. This causes the alternator output to shoot very high for a short period of time, and often, this will be enough to pop the control circuit in the Unilite. There are some other charging system problems that can cause similar glitches. The in-line power filter we sell helps clean up the power to the module, but it also has limited capacity for severe charging system failures. The active power filter option protects the red wire (12 volt) against voltage spikes and protect the green wire from feed backs and load dumps. You should not have to use the active filter but if you have a lot stereo equipment or a voltage regulator going bad it can help protect the module from voltage spikes or surges. Will it prevent it, no. So to answer your question, if your electricial system is working correctly you do not need the power filter.
Here's my suggestion on electronic ignitions an ballast resistors: I hate ballast resistors. They create heat, and they waste energy. What to do then, beside putting a fan on the resistor and using it for a heater... The answer is a CD ignition. Since the late 20th century, all "hot" rod ignitions are dependant on "hot" spark. There is just no way to get a hot spark at the right time without a CD. Second, the output transistor of the electronic distributor is perfectly matched to the input of a CD box (MSD, whatever). Since the output of the distributor module can feed the CD, there is no need for a ballast resistor in-line with the coil. On a CD, the CD unit itself drives the coil. In review: points and ballast - works, has normal spark. Electronic module and ballast - works, has normal spark. Electronic module, no ballast, and a CD - works, has a very large spark at perfect timing. Hot spark at perfect timing = hot rod.
Yep, the CD would be the way to go, but I've always baulked at the cost (remember, I'm in Australia so you can double whatever you pay over there). FWIW, I have one of the Circuit Guards in place just for insurance. I'll check everything again and maybe run that wire from the solenoid to give the coil 12V and then see how it runs.
Recardo is right, it is a current limiting resistor. So make sure you put it in the circuit. FWIW: You can put a smaller value resistor in series with the I (Furd starter relay) or R (GM solenoid) to the coil side of the ballast resistor. This will give you a hotter spark at startup. The CD is the way to go if you want to solve all your issues. But the best thing to do with a car that old is to make sure all connections are clean and tight.
Cool. Was an interesting discussion and I definitely learnt some stuff. Looks like the Bosch Australia website is now back up. Check out the footnote at the bottom of Page 1. http://apps.bosch.com.au/products/saa/ignition_coils.pdf N.B. Coils which are for use with Ballast resistor have the letter R included in the part number, e.g: SU120RT, GT40RT Looks like those crazy Germans like doing things backwards! LOL
LOL! HA HA! Looks like we're back to square one. I love it! I learned a lot too. When you got it going and know what fixed it, please let us know.........
I would put the newer Rambler scrambler aside and work on this beauty: Just kidding, good luck with the ignition thing.
Most starter solenoids have two posts, when wired correctly one post gives the coil 12 volts when starting and the other runs through the ballst resistor and give the car 6 while running.
You mean inductance doesn't apply to one or the other? Collapsing magnetic fields inducing voltages only works on one of them? Perhaps you're making your point with poor terminology. Welcome to the post.
The ideal Vs/Vp=Ns/Np is valid for transformers having infinite self inductance on both the primary and secondary windings, a coefficient of coupling of unity, of negligible parasitic resistance and under sinusoidal steady state conditions. A keen observer might say "Thats impossible! you can't have a coupling coefficient of 1, and there are always parasitic losses, and infinite inductance isn't possible!". This is true. But lots of real world transformers come pretty close. This is an approximation, and before using it you need to know its limitations. The most important difference between a car ignition coil and the transformer you plug in to power your model trains is that the model train transformer operates under sinusoidal steady state conditions. The car ignition coil does not.
Yes, but the principle of magnetic inductance is identical between the two... Your train transformer runs at 50-60Hz while a V8 at 6000 RPM will be running the coil at 400Hz (the AC standard of the aviation industry) - extraneous trivia which means squat.
Now were getting close very close not exactly very close .... kinda the opposite way..... now factor in Dwell and Saturation... and that throws it all to crap..... The above statement puts it in the best laymans terms I could think of. I could not explain it any better and be understood.. Now run it without a ballast resistor and the unused DC current willFRY your coil
Also the ONLY way your going to run AC current in an Ignition system. is a flying magnet, induced Reluctor, capacitive discharge or magneto
Sorry guys you don't understand electricty. He had three devices in series r1 plus r2 plus r3 ballast plus "protector" plus module. Only current is equal everywere he want divide voltage by three to divide voltage by two....
The Mallory wiring diagram does show "all other wires originally connected to the + coil terminal," one of which (I will bet) is a bypass-ballast-during-start hot wire. If you are using an original era coil, bypass the ballast resistor during start-up as others have said.
Where is Nikola Tesla when he's needed...? Somebody's got to switch the big breaker before the overbearing redundant magnetic flux density of this thread zaps us all into oblivion.....
They both use inductance, yes. The automotive ignition system does not operate under sinusoidal steady state conditions. An automotive ignition system is more like a boost converter than an ideal transformer. http://en.wikipedia.org/wiki/Boost_converter
Ah-hah! You have conceded my point - I WIN! Yay! ...now what do I win? Interesting stuff here...for EEs!
I never said they didn't use inductance. They use inductance in different ways and the design and analysis of the two circuits is completely different.
The collapsing field of the primary puts about 3-500 volts on the coil winding and this is multiplied by the coil turns ratio. When the points are closed and the ignition is on the current in the primary will heat the coil to failure. coil will draw about up to 10 amps while closed and that is somewhere around 100 watts. The ballast resistor is simply used to limit the amperage through the coil when the engine is not running. I believe that the bulb in the Mallory takes the place of the ballast resistor and has the advantage of being a variable resistance. http://www.imperialclub.com/Repair/Electrical/coil.htm " [FONT=Times New Roman, Times, serif]When the engine is running, you can get by fine with a lot less current at the points. That's the purpose of the ballast resistor. If the points were to receive full battery current/voltage at all times, they would only last a few hundred miles before needing replacement.[/FONT] [FONT=Times New Roman, Times, serif]At the time of cranking (when the starter is operating), a separate wire gives the points the full 12 volts, bypassing the ballast resistor. When you release the key from "start" to "on", all the power to the points now has to flow thru the ballast resistor, preserving the points.[/FONT] [FONT=Times New Roman, Times, serif]So that's its purpose....for those curious. And I know I'm kinda mixing volts/amps in the above....which isn't fair....but you get the idea.[/FONT] [FONT=Times New Roman, Times, serif]Addition from Bob:[/FONT][FONT=Times New Roman, Times, serif]Let's be clear that there is no separate wire on the early models. The points get the current directly through the ballast. In order to provide the full 12 volts during a cold start, the ballast is also cold and has nil resistance. As the ballast heats up (some of them get VERY hot!) the resistance gradually climbs up to 1.5 th 1.9 ohms thus reducing current to the points.[/FONT] OK, that should solve everyone's problems
Note: my advice is don't do it. The electronic module needs the ballast for the load resistor. If you have points, this is how you would do it: A GM starter solenoid is all set-up for this. I think it is the R terminal and you run it to the + coil. If you have something else, then wire the ignition switch start position to the + side of the relay coil, and wire the - side to ground. Then run the contactor wire to a source of 12 volts (solenoid) from one side of the relay, and to the + coil position on the other side of the relay. You need a normally open (NO) type relay.
I wish people would stop talking about POINTS! The MBI = Magnetic BREAKERLESS Ignition, so I don't have any points to fry. Comments such as: "When the engine is running, you can get by fine with a lot less current at the points..." etc, are probably valid for stock engines, but I am telling you, my car has picked up at least another 500 RPM, so there's definitely a stronger spark. I'm tempted to keep the bypass until the coil burns out (I've got a spare) but I'm a bit wary as I don't want to kill the Mallory module... You Electrical Engineers crack me up though. I've got my EE exam on Monday night, so I'll consider all this stuffing around on the HAMB as 'studying'. LOL
You got a Mallory, I got a Mallory, but I also have an MSD box, so the ballast resistor isn't used in my application. If you put an ohmeter lead on the - terminal of the coil, and hook a ballast to the + terminal of the coil, and then the other ohmeter lead on the opposite terminal of the ballast, you should read about 2 to 3 ohms. So if you measure a lot more than that, then you probably have a resistor internal to the coil, and don't need the ballast. Just in case you have a resistor wire coming from the ignition switch, you can measure the wire with the ohmeter, and replace it with a straight wire if needed. Definately don't use the ballast bypass from the starter, or with a relay. P.S. Being able to work on ignitions, only gets you to polishing the shoes of an EE, so your studies are deep shit compared to this crap :=)
I've got my ballast mounted inside the car (don't worry, not near anything flammable) so it's a bit hard to measure the total resistance between it and the coil. Measuring them separately though showed the BR as 1.9 Ohm and coil as 1.8 Ohm, for a total of 3.7 (and that was with the BR stone cold). My wife took the car out today, so I hooked the BR back up and ran a wire from the starter solenoid. Started OK and when running had between 9-10V at the + side of the coil. Didn't want to take the chance that the coil or module might crap out with the wife driving.
Not to beat this thing to death, but I played with my adding machine and voltage divider theory... 1.8 Ohms + 1.9 Ohms + [1.9 Ohms] = 5.6 Ohms Total In order to measure 9.5 Volts at the + Coil, there would have to be more resistance than just the BR. I calculate about 1.9 Ohms more. Thus, 14 Volts divided by 5.6 Ohms Total = 2.5 Amps. 3.8 Ohms (BR + Mystery) times 2.5 Amps is 9.5 Volts. Since we know the BR measures 1.9 Ohms, that means there is a mysterious [1.9 Ohms] somewhere else. Anyway, what I'm getting at, is maybe there is a 1.9 Ohm resistance wire in the path from the ignition switch, or a high resistance connection from a poor connector. Or maybe my adding machine is totally gimped...
Definitely no resistor wire, the whole car's been pretty much rewired. I will have to check all of the connections as I did notice the wires to the BR were a little loose, so I gave them a gentle squeeze with the pliers. Wire going to the + on the coil also looks a little hinky, but not corroded or anything. Wife had no problems with the car today, so it might all be OK with the current setup (BR and wire from starter solenoid), but just to satisfy my curiosity I'll check the resistance between the coil back to the 12V side of the BR. That should tell me if there is another resistance (of significance) in the circuit somewhere. In a kind of weird geeky kind of way, this has been fun.
The Jalopy Journal
