At 12:43 PM 7/26/00 -0700, Dave Wood wrote:
>.... there is a formula to calculate the output based on the input that
depends on the number of windings on the input side versus the number of
windings on the output side. So if 12 volts puts out 40,000 volts, 6 volts
should put out 20,000 volts and any proportion in between for other changes
in input voltage. Anyone care to correct this?
Yes. Your formula would be correct for a circuit with a fixed load and
continuous current on the output side, but, ....
In a very simplistic explanation (much detail omitted), in a high energy
ignition coil, when the contact points are opened the magnetic field
"collapses", which induces a high potential in the output windings of the
coil. As the field is collapsing the output voltage rises. When the
output voltage is sufficient to overcome the resistance of the spark plug
gap the spark jumps the gap dissipating the energy from the secondary
windings of the coil such that the output voltage of the coil falls off and
does not exceed that voltage reguired for it to junp the gap.
As such, the output voltage at the time of the spark will be (aproximately)
the same regardless of input voltage, assuming that there is in fact enough
voltage to make the spark. The output voltage at the time of spark is
determined by the pressure in the cylinder and the resistance of the spark
plug gap at that time. The number of turns of wire in the coil can
determine the maximum attainable output voltage, but it does not determine
the actual voltage at the time of the spark. You either get a spark at the
required voltage, or you get no spark.
Barney Gaylord
1958 MGA with an attitude
http://www.ntsource.com/~barneymg
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