Reply to: RE>Re: ballast resistor
>QUESTION :
>Does anybody know hoe the ballast resistor is wired in ignition
>circuit of a late Midget ?
I don't know what kind of electronic ignition system is involved here, but
with Capacitive Discharge ignitions, a ballast resistor should not be
necessary. See, the idea of the ballast resistor-lower-voltage-coil
combination is to get the coil to saturate faster. The ballast resistor
converts the 12V "Voltage Source", into a half-assed "current source". The
idea is to cram maximum operating current into the coil as fast as
possible when the points close.
A coil of wire ( an inductance ) has the property that it wants to keep
a constant current flowing through it. If the current flowing through it
previously was ZERO, the coil would like to KEEP it that way. It does this
by behaving like an open circuit at the instant that voltage is first applied.
Ballast-resistor systems make it possible to get the coil up to current
faster. When the points first close, the coil seems to have high resistance.
Therefore the current through the coil, and through the ballast
resistor, is small, and the voltage drop across the ballast resistor is also
small ( voltage-across-resistor = current THROUGH resistor * resistance OF
resistor ). So for that tiny fraction of a second, the FULL TWELVE VOLTS is
applied across the coil, bringing it up to saturation much faster than if you
had a higher-voltage-coil straight across the twelve volts. This system would
work even better if you had a higher voltage and a bigger resistor, but 12V is
what we've got, and it's adequate.... When the
coil saturates, its current stabilizes at the design point, which is determined
by the internal DC resistance of the coil, and the value of the
ballast resistor.
The ballast resistor is shorted out while starting because the battery
voltage typically goes down to 8-9 Volts anyway, due to the enormous
load imposed by the starter motor. Also, the motor is turning very slowly,
and there is plenty of time for the coil to come up to saturation anyway.
Now, the way CD ignitions work, is, they use a DC-DC converter to boost
the 12V to some 400V DC. They use this 400V to charge a capacitor. When the
points open, a transistor/SCR switch connects the capacitor to
your ignition coil, and dumps its charge into the coil. This typically occurs
an order of magnitude faster than the constant-current-ballast-resistor system
can charge the coil. Also, the current output is limited by the charge in the
cap, and the internal resistance of the ignition system. If you use a ballast
resistor with a CD
ignition, you will probably lose the main benefit of the ignition, which is
the extremely quick dump of energy into the coil. This fast energy dump
translates into a fast spark that tends not to "foul out". Of course, I don't
know if the ignition in question is a CD unit, or just a set of "electronic
points". In the latter case, the resistor might be necessary, and for the
exact same reason as with the mechanical points.
- Jerry Kaidor
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