LT1 Based Engine Tech 1993-1997 LT1/LT4 Engine Related

Optispark theory of operation (and general ignition theory)

Old Apr 18, 2004 | 12:41 PM
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Optispark theory of operation (and general ignition theory)

So looking inside of an Optispark, it seems to serve two distinct purposes: a distributor with a traditional rotor and an optical crank trigger/sensor.

A real basic question: Why do engines need points or a computer triggering the ignition? It would seem to me that if you left the coil constantly energized the spark would fire when the rotor passed by the pickup for that plug wire.

Is that just too random and variable, and hence the points/trigger are introduced to add precision?
Old Apr 18, 2004 | 12:48 PM
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cam triggers and such are used so that you know when to fire the injectors.... this isnt a regular HEI!!!
Old Apr 18, 2004 | 12:52 PM
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energizing the coil won't create a spark at all... it's the colapse of the magnetic field in the coil that generates the high-voltages that travel to the spark plugs.

There's a pulse sent to the coil, this has to be generated somehow... it used to be by the points style, then it went to a magnetic for better reliability, and no necessity to re-gap and replace at regulat intervals...

Then once computers came around the magnet was integrated into the PCM controls to vary the spark timing. The Optical pickup was developed and integrated for even better resolution than the magnetic pickup...

need any more info???
Old Apr 18, 2004 | 01:19 PM
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Yes please

I'm still trying to figure out this whole "collapse of the magnetic field" and "an inductor resists change in current flow" bit....

So the transformer outside my house supplies a constant supply of stepped up voltage. But a coil needs saturation time and the produces a big surge when you break the supply to it?

What's the difference between the two, is it an AC vs DC thing? Or instead of this whole collapsing magnetic field thing why can't they just use the coil as a step-up transform and blast in a bunch of voltage when its the appropriate time?

Software guy asking hardware questions :-)
Old Apr 18, 2004 | 01:54 PM
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The transformer outside your house steps down line voltage to your typical 120/240 house voltage.. and it does this in a long period of time (in comparison) to your ignition coil...

The coil in your car is taking 12V and stepping it up to 30,000+ Volts... and doing it in a fraction of a fraction of a second...

I'm not a physics major so I can't explain exactly how it works..

Something about collapsing magnetic fields and the electron forces that it creates... It's beyond defentely beyond my scope of knowledge (And I'm not one to admit that I don't know something very often)
Old Apr 18, 2004 | 02:09 PM
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the inductor is a coil, the voltage across an inductor is l(di/dt) di/dt is the rate of change of the current. So when you switch the current through the coil on and off very quickly you generate a high voltage, low current signal.
Old Apr 18, 2004 | 02:52 PM
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Originally posted by davepl

Or instead of this whole collapsing magnetic field thing why can't they just use the coil as a step-up transform and blast in a bunch of voltage when its the appropriate time?
Your thinking is correct, there are actually primary and secondary windings in an ignition coil similar to a transformer. But the ratio of secondary to primary coil windings is something like 100:1 (stepping up the voltage 100 times), which is not enough. As mentioned earlier, you need around 30kV to fire a spark plug.
Old Apr 18, 2004 | 03:31 PM
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So when the magnetic field of a coil collapses, it generates a spike of voltage far in excess of what the normal stepup from its windings would be?
Old Apr 18, 2004 | 05:08 PM
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Originally posted by davepl
So when the magnetic field of a coil collapses, it generates a spike of voltage far in excess of what the normal stepup from its windings would be?
Yeah, basically. As someone mentioned, the voltage across an inductor (a coil) is

V = L * (dI / dt)

L is inductance measured in Henries.

The key is the (dI / dt); change in current with respect to change in time. So the faster you change the current through the coil, the larger the voltage across the coil. Now imagine flipping a switch and what that does to (dI / dt); it becomes huge, and so does the voltage.
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