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Old Aug 6, 2005 | 08:21 PM
  #1  
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Question about duraspark

The modules (boxes) have different grommets for identification purposes.
My question is which one do I want or need.
What are the differences??

429 w/9.5:1 comp.
c-6 auto
 
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Old Aug 6, 2005 | 09:44 PM
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From: Eustis FL
I think what you want is the Duraspark II. The first Duraspark ignition I believe came out around 1974 or so, but it had limited dwell capabilities and maybe other issues, but in the following years the D2 was released, and it works great. I never buy jobber modules. I went through 2 in one month. Bought a factory Ford unit and all was well. I even trust Motorcraft modules from the junkyard over the jobbers, and frequently grab them to keep as spares for my Fords.
What year vehicle is your engine in? There is a simple color code chart posted somewhere to help you wire it correctly to your ignition and power supply, coil, distributor, etc. 6 wires total, though I have seen 7. (4 wires in one harness, three in the other, instead of 4 and 2)
You will need all the components of the Duraspark II to make it work. Look for the larger diameter dist. cap if going used. You can convert the small cap to a large with a factory spacer, but change the rotor too.
Hope this is what you were after.
Mark
 
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Old Aug 7, 2005 | 09:43 AM
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I have the distributor. I am going with the blue grommet box. I have read that I can use the TFI coil with nothing else. Just hook it up the same as I would a regular duraspark. Is this true?
 
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Old Aug 7, 2005 | 11:14 AM
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I'm still not sure what you mean by 'blue grommet'. What does that refer to?
If you are installing a higher power coil, it is a straight replacement, wire for wire.
 
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Old Aug 7, 2005 | 06:04 PM
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Here's a blue grommet on a Duraspark module:

http://bellsouthpwp.net/1/b/1bad6t/duraspark-module.jpg

It's the blue connector that the wires are going into on the side of the unit.
 
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Old Aug 8, 2005 | 03:27 PM
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What I mean is do I need a ballast resistor to run a TFI coil?
 
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Old Aug 8, 2005 | 04:11 PM
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From: Houston/Hope BC
Yes, keep the resistor. My concern is the DSII box or coil overheating from the additional current you'll get without the resistor.
 

Last edited by Ecuri; Aug 8, 2005 at 04:35 PM.
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Old Aug 8, 2005 | 07:51 PM
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I thought that the TFI was made to run 12V?
By the schematics it does.
I don't see how the module would overheat, it just sends the signal.


I don't know, I am trying to understand.
 

Last edited by 8man; Aug 8, 2005 at 07:57 PM.
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Old Aug 8, 2005 | 08:25 PM
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From: Houston/Hope BC
All the spark energy that is stored in the magnetic field of the coil and dumped through the module. The module does the same thing as the points except with a transistor. No resistor = more current = more heat. The coil may or may not take it, you'd have to check with the MFG. I have serious doubts about the DSII module taking anymore internal heat than it allready does. (Wells seems to make the best replacement for longevity.)
MSD talks about it a little, that is one of the benefits of CDI, the spark energy is stored outside the coil in the capacitors instead of the windings of the coil.
 
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Old Aug 8, 2005 | 08:45 PM
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From: lino lakes, mn
dsII box has two wires coming in for power one gets 12 volts during start up the other get 7.5 for the run circuit there is a couple good threads in the electrical forum with schematics
 
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Old Aug 8, 2005 | 09:42 PM
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From: Houston/Hope BC
Hate to correct you, but that is not right. The module always gets +12V. (Red is run, and white is Start w/ spark retard.) The coil gets +12V on START and connects through the ballast resistor in the RUN position.
 
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Old Aug 8, 2005 | 11:30 PM
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Saw something in another thread that has me thinking.

Battery voltage with the engine off is lower, hence no need for resistor in the start position.

Battery voltage goes up with engine running requiring the resistor to lower it to acceptable levels. Prevents some of the problems.

I believe it read that this was originally done to reduce the load to the points on the old systems, but it makes some sense to keep it around for protection.

If the resistor wasn't bypassed during starting the voltage would be low. Once the engine is running the resistor keeps the voltage from going up to the same levels as the alternator/charging system puts out.

I think that it makes sense.
 
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Old Aug 9, 2005 | 11:13 AM
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From: Houston/Hope BC
The operation of an induction coil is a fairly complicated electrical process. The resistor is there, IMO, to limit CURRENT. Voltage is seconday to the power output. The coil is not a transformer! It is one part of a DC to High Voltage AC converter and depends on the collapse of the magnetic field, the current that collapses is much more important than the voltage. However, no resistor at start = higher voltage and current = more energy to start. I wish I understood the finer points. Points handle about 4 amps before failure, the DSII is prob around 8 amps, and without the arc, wearing parts, and dwell issues of points. Transformers exhibit a simple ratio of turns to change voltage in AC curcuits, obviously more is going on in an induction coil because even a 100:1 ratio at 12V would only make 1200 volts.
 
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