Michael,
A coil that is intended to be used without a ballast usually has around
three ohms primary resistance. A coil that is intended for ballast
applications usually has around 1.5 ohms resistance with a ballast
resistor (either external or as a part of the wiring loom) to give about
three ohms total resistance. In a way, you could say that an extra 1.5
ohms of resistance is built into the non ballasted coil.
For it's original purpose of the ballast was in the coil primary circuit
under normal running conditions but bypassed to give temporarily higher
coil voltage via a separate contact in the starting circuit. The
thinking is that during starter operation the battery voltage is
somewhat lower than when running & the temporary extra coil voltage
helps starting.
On cars without the ballast & starting bypass contact, it isn't believed
that the extra voltage while starting is needed so they just used a
three ohm coil. Original Healeys for example.
In your Mallory example, the main electronics portion of the ignition
electronics is designed to run on full battery voltage & the coil
primary/trigger electronics is designed to run on the reduced voltage
provided by the ballast.
Note: the resistance values given are approximate & will vary with the
exact ignition application.
Dave Russell
BN2
Awgertoo@aol.com wrote:
> In a message dated 4/2/2004 7:23:08 PM Eastern Standard Time,
> Rick@genomictechnologies.com writes:
> What does the term ballasted mean for a coil?
>
> If I purchase a non-ballasted coil for my 100-4, which is what is specified,
> does that mean I have to add a resistor? Does the term ballasted coil mean the
> coil has the resistor built in
> As long as we're talking about this, I have a Mallory Unilite Ignition on my
> 100 with the ballast resistor installed before the coil as per the first
> diagram on the page found at:
>
> http://www.taperformance.com/ignition1.htm
>
> The statement is made that if the ballast resistor is not used between the 12
> volt source and the coil positive terminal the igniton module will eventually
> be destroyed. However, since there is also a 12 volt lead from the hot side
> of the ballast resistor to the module would not the same amount of power be
> reaching the module? Can someone please explain this?
>
> TIA--Michael Oritt, 100 Le Mans
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