Bypassing or Removing Dual Ballast Resistor?

Started by Mosin, April 14, 2009, 04:04 AM

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Mosin

Hi Folks,
Because I am running a complete aftermarket (MSD) ignition system on Pandora's 440 I no longer need (or want) the voltage drop to the stock coil wire created by the dual ballast resistor. My question is, can I just remove the ballast resistor from the loop and jump the terminals together (top to top, bottom to bottom) with male-male spade connectors without harming something else in the ignition system? By something else, I mean the starter, alternator, voltage reg, or wiring, since the MSD system (coil, dist, and box) is totally removed from this wiring.

Right now I am runing the MSD coil from a different 12v ignition source (the stock coil wire from the harness temporarily capped) because I am not sure what the implications will be (if any) for non-coil items in the loop if I remove the resistor and jump the terminals together directly. I would like to use the stock coil wire (albeit at a full 12v) rather than the current external wire for simplification, which is why I am asking. I tried just unplugging both terminals and removing the resistor (leaving the terminals unconnected) assuming that nothing essential was still in this wiring loop, but the rig wouldn't start. This implies that the starter or alternator (? can't figure it out from the wiring diagram in the chassis manual) is in this loop, and that something else other than the coil uses the ballast resistor. If need be I can just leave it in place, but I would love to remove it entirely from the system if possible so it won't potentially fail down the road....
Any advise would be appreciated!
Thanks,
Dave/Mosin

Froggy1936

Do you mean the engine would not start ??  or do you mean the engine would not crank ??  with the wire disconnected   Totally different problems  Frank
"The Journey is the REWARD !"
Member of 15 years. We will always remember you, Frank.

DaveVA78Chieftain

If your using a complete MSD replacement ignition system, then the coil is supposed to be driven directly from the MSD control unit (see http://www.rowand.net/Shop/Tech/images/MSDWiring011.jpg
As such, the ballast resistor should not be in the circuit.
Below is a block diagram of the standard dodge electrical block diagram for the associated circuits.  Disconnecting the ballast resistor should not affect you unless your using one of the leads off of it in you msd wiring.
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Mosin

Quote from: Froggy1936 on April 14, 2009, 08:27 PM
Do you mean the engine would not start ??  or do you mean the engine would not crank ??  with the wire disconnected   Totally different problems  Frank

Sorry... the engine would turn over fine but not start up when I just removed the ballast resistor and left the connectors hanging. Hence my idea to just jump them together with spade terminals. I was hoping this would also give me a pure 12v via the stock coil wire.

Mosin

Quote from: DaveVA78Chieftain on April 14, 2009, 09:56 PM
If your using a complete MSD replacement ignition system, then the coil is supposed to be driven directly from the MSD control unit (see http://www.rowand.net/Shop/Tech/images/MSDWiring011.jpg
As such, the ballast resistor should not be in the circuit.
Below is a block diagram of the standard dodge electrical block diagram for the associated circuits.  Disconnecting the ballast resistor should not affect you unless your using one of the leads off of it in you msd wiring.


It is a complete MSD replacement system (Mopar 6AL box, Ready-to-Run dizzy, and Blaster 2 coil) and none of the 12v power leads or grounds are currently utilizing the existing stock wiring. Yet, when I unplug the resistor without jumping the conectors the engine will turn over but not start.  Since my goal is to eliminate the voltage drop to the stock coil wire, I figured jumping the connectors would not be a problem, but I'm not sure what other function/s the resistor serves.

I guess another way to put my question would be what besides a stock coil, if anything, needs the voltage drop (0.5) or increase (5.0) ohm provided by the ballast resistor. If the answer is nothing, then I think jumping the terminals together will be fine. :-\ (<== famous last words before harness meltdown?! LOL)

DaveVA78Chieftain

Regarding the wiring to your ignition switch from the MSD control module
You should be connecting to the red wire (I1) from the ignition switch.  It is the ignition switch engine run wire.

The pink (I2) wire only supplies voltage to the coil when the key is in the start position (bypassing the ballast resistor during start).

Sounds like your connected to the orange wire because:
1.  During start, the pink I2 wire is supplying 12VDC to the MSD.
2.  When you set the key to run, the reduced voltage of the ballast resistor is being supplied to the MSD module via the I2 junction on the ballast resistor output.
3.  When you disconnect the ballast, you loose the reduced voltage to the MSD ergo no start.

Make absolutely sure you are connected to the red I1 input side of the resistor.  To determine that, disconnect ballast resistor.  Turn key to ON (run not start).  Measure each lead to frame ground.  Only one ballast resistor lead will have 12VDC on it.  Thats the lead (red wire) that has to go to the ignition switch input of the MSD module.  Based on the MSD drawing I linked to above, and your stating that your using a MSD distributor also, your using the bottom drawing of that MSD link ergo that red wire goes to both the MSD module and the MSD distributor.
While the bulk of the power needed by the MSD is provided via the direct battery connection, the ignition switch I1 red lead controls the whole works from a ON or OFF perspective.
The I2 pink lead only has full 12VDC power to it during start (increases voltage to the coil during start).  However, it also see's the reduced output of the ballast resistor during run.
The ballast resistor dark green lead is always reduced voltage (2nd ballast resistor circuit) and is used for low rpm current regulation by the original Dodge ignition control module (not used by MSD module).

Hope that helps,
Dave
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Mosin

Many thanks for that detailed explanation! I will follow the procedure you describe to locate the I1 input, and then see if I can get the engine to run by using this as the sole 12v source for the dizzy and 6AL red leads (the Heavy Red and Heavy Black of the 6AL are straight to battery) with the ballast resistor disconnected.

On a related note, I am awaiting the return of my MSD RTR dizzy (year-old, 12K miles) from warranty repair, as it was misfiring heavily btween 3000 and 3500 RPM. It was the process of diagnosing the problem over the past several weeks that led me to the realization that the mechanic who did the installation after the rebuild last year broke every MSD ground, power, and RFI wiring rule there is: grounds were to aluminum intake manifold, 12v sources were all from stock coil lead via ballast resistor (!) and all wires were neatly bundled together and tightly zip-tied (with plug wires no less. plug wires in the same firing sequence even). What this means is that my fancy-schmancy MSD system never has had full 12v power or proper grounding, and has likely been misfiring due to RFI crossover/interference and with a weak spark for all 12K miles since the rebuild. IMHO this is the reason for the failure of the RTR dizzy module in the 3000-3500 band, which is where the vast majority of the miles have been driven. Whatever the cause of the dist failure, now that I have obtained a PhD in MSD I am religiously redoing the wiring according to their instructions... hence my interest in getting the BR out of the loop, yet potentially using the stock coil wire.

I'll post the results after I get the dist back, which will hopefully be any day now. Thanks again for the responses!

Dave


Mosin

Update: I'm not sure what, if any non-stock wiring in Pandora's system, but I tried it once again and even when the MSD is connected to a separate 12v igniton lead the engine will turn over but not run with the ballast resistor leads disconnected. Based on my layman's interpretation of the above schematic and explanations vs. my crusty stock wiring harness along with my color blindness I think either an alt, starter, or VR lead must be still passing through these leads (?). In any case, with no stock ignition system connected I am guessing that nothing in this loop actually needs the BR functions anymore so I tried just jumping the leads directly together with male spade connectors (thus removing the ballast resistor entirely from the loop) and she starts right up and runs great. The wires are not hotter to the touch than they were before, so for now I'm going to run with it, with a fresh balast resistor on hand just in case! Someday I hope to redo the alt wiring at which point maybe I will be able to determine what component/s are still in the BR wiring loop.

I say someday because for now I am still stuck trying to diagnose this ongoing misfire issue.  $@!#@! The MSD warranty repair yielded a brand new dizzy, but the exact same problem is still there.  :'( I bought yet another set of MSD plug wires to be abolutely sure it wasn't the wires and... no love. At this point I am thinking a valve spring or cam issue, but from what I know about spring or cam-induced misfires, they wouldn't happen randomly between 3000-3500 with these characteristics (i.e. perfect start up, idle, and acceleration until the misfire, which is intermittent). I am considering starting a new thread after exhausting all other leads here at CW and on the Internets (Moparts board, MSD tech boards etc) but I'm going to give the archives another one-over to make sure I'm not missing anything previously discussed.

BrianW21

Well I undertook putting a MSD 6A in last night and frankly did get very far.

First does someone have the colors that correspond to the terminals of the ballast resistor? There are four terminals, there are two red wires at one end, and orange, pink, and green at the terminals at the other end.

Based on what I read here, I connected the 6A's 12v RUN wire to the red wire of the ballast. Didn't work. I tried both.

Symtom: I checked both red wires and found +12V at both when the key is in RUN. When the key is turned to start, both go from 12V to 0V.

In fact, everything that I could find that was a +12V Run lead (even in the interior fuse block) goes to 0V during start.

Now I've read that the pink wire is +12V during start? If I connect the MSD to the Red and Pink wire, would that work? I'm at the point where I'm so frustrated that I might just put a switch in the cab connected to the battery with a toggle switch.


I have a 76 Dodgle chassis with the 440-3.

BrianW21

And I just tested my assumption and, it works!!!

So here is what you need to do to install an aftermarket sparkbox and delete the ballast resistor.

1. Remove the ballast resistor.
2. Connect the Red I1 wire to I2 pink wire (same side of resistor)
3. Connect the SparkBox RUN wire to step #2 (I1+I2)

-Brian

DaveVA78Chieftain

Both red wires are fed from the I1 (RUN) terminal of the ignition switch.  Should have 12VDC applied for both RUN and START positions.  Provides power to the ballast resistor and the ECM.  If 12VDC is missing in START position, there is a problem with the ignition switch.  I1 is the main fed for the ECM and Alternator/Regulator circuits.

Pink wire - Fed from the I2 (START) terminal of the ignition switch.  Should have 12VDC applied for START position only.  Bypasses the ballast resistor.

Green wire (aux resistor output) - used by the ECM (pin 3) as a voltage reference.

Orange wire (Ballast resistor supply to Coil plus
  • terminal;  Black wire on a Class A chassis) - Supplies voltage to coil.

    MSD can be used either with or without the ballast resistor.  Thats why the MSD documentation tells you that for Chrysler Electronic Ignition using Magnetic Pickup, to connect the MSD red (RUN) wire to the original coil plus
  • wire.  You should have only had to connect to one of the 2 red (I1) wires. Jumpering the red and pink wires together is Ok for a MSD setup.

    Dave
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