Topic: How much voltage can my coils handle?

[Pinhead]

I was in the middle of building my home-brew rectifier/regulator for my bike and had an interesting idea. My RR is separate, regulator and rectifier in separate casings, so I've got a source of rectified DC around 50 or so volts. I know the coils couldn't handle that much voltage, but I could make an 18v regulator and power the coils with it... Would it be as simple as hooking the higher voltage to the coils, or do I have to wire it in somewhere else to get the hotter spark that I'm looking for? Before I do this I'll probably find an extra set of good coils, just in case they do fry. But could it work?

[TwoTired]

It's not the voltage but the power you need be concerned about.  Power is voltage times current.

The stock coils are 5 ohm coils.  If the points are closed with the ignition on, ohms law (E = I x R, R = E/I, I = E/R) states that 2.4 amps will be drawn through points and coil primary.  The power dissipated, assuming 12 v application, is 28.8 watts.  The coil will survive this application for a very long time in any environment you can survive in.

When running, the points are closed for 95 degrees of the 360 degrees crankshaft rotation, or a duty cycle of 26.3 percent.  So, the stock coil averages 7.6 watts consummed during run conditions.

Key points are how you trigger the coils or whether you use the coils to store the spark energy or use them simply as step up pulse transformers.  Capacitive Discharge Ignition systems  (CDI) apply 200-300 or even 400 volts on the primary, using the coil as a step up pulse transformer.  However, the time this voltage is placed on the coil is very small, (measured in microseconds) way smaller than the points closure duration.  And, the duty cycle or actual time the voltage is applied will equate to even less power consumed than a points system because the coils don't need time to store the energy, but merely convert the applied voltage pulse.

Another way to look at it, is how fast can you remove heat from the coil, and what it's heat transfer characteristics are.  If you bump that power dissipation from 28.9 watts to 43.2 (18V applied with stock points closed), can you remove enough heat from the outside of the coil to keep the inner temperature below the point where the internal insulation melts?  Anyone know what a stock coil's thermal transfer characteristics are?  Also, what are we to assume an ambient temperature environment is for the coil placement?  And, can we assume an airflow around the coil to help dissipate the heat?

If you don't know these design parameters, they can be empirically determined with a few destructive tests.  Then you can proclaim the same phrase that character in the "The Mask" uses.... "Smokin"!

Cheers,

[Pinhead]

Ok, since the 650 has CDI ignition, than the duty cycle of the coils is very low, as they're just acting as a step-up transformer. Therefore (correct me if I'm wrong), they should be able to take quite a bit of voltage without burning up. Step-up transformers are voltage multipliers, not voltage adders, so a slight increase in input voltage would equate to quite an increase in output. Since everything else stays the same, the power would also rise.

Now, where I'm not so sure is exactly how to wire it up. I know how to make the regulator. Just not so sure about the wiring of the CDI ignition. Can I just increase the voltage at the coils, or should I increase it somewhere else to equate to the hotter spark I'm looking for?

[TwoTired]

You're going to run your CDI on 18 V?  Hope you have spares!  I wouldn't worry about the coils.  It's your CDI I'd worry about!

And... if you've got a CDI, why are you trying to boost spark voltage?

Cheers,

[Boomologist]

>>>Key points are how you trigger the coils or whether you use the coils to store the spark energy or use them simply as step up pulse transformers.<<<
Please explain how coils can "store" spark energy. Aren't the coils a basic transformer?

[TwoTired]

>>>Key points are how you trigger the coils or whether you use the coils to store the spark energy or use them simply as step up pulse transformers.<<<
Please explain how coils can "store" spark energy. Aren't the coils a basic transformer?

Not in the Kettering system.
When the points close, current flows through the primary and a large electromagnetic field develops in and around the metal core of the coil.  At some point the core becomes saturated and the magnet can develop no larger with the given voltage. This is the minimum required dwell time and is stored energy.  When the points open, the magnetic field collapses and as the collapsing lines of flux cross the many turns of the secondary, a voltage is induced.  This voltage builds until it is large enough to ionize the atmosphere between the spark gaps of the plugs and turn it into conductive plasma.  This dumps the energy stored in the coils to cause a spark.

CDI sends a large voltage spike (or many) for immediate conversion or step up to the spark the plugs.  This is using the coil in pulse transformer mode and does not rely on stored energy for the spark voltage buildup.
Coils for CDI and other high energy ignition systems, do not need the high metal core mass required for the Kettering type.  In fact, high core mass dampens the pulse rise and fall times and this limits how many pulses that can be achieved in a single firing cycle.  It is one reason why high energy ignition coils can be physically smaller.

Cheers,