I don't think anyone has got the full explanation of the role of the
capacitor quite right.
First, consider the coil as a 'perfect' coil (no resistance).
The voltage in the secondary = voltage in primary times turns ratio
(turns ratio = # secondary turns /# primary turns).
Now before anyone says: 'but this means if you have a steady DC primary
voltage, then you will have a steady DC secondary voltage', I will say:
true, but, if you have a steady voltage in the primary, then the CURRENT
will be changing.
IMPORTANT POINT: in out 'perfect coil' the VOLTAGE is proportional to the
RATE of CHANGE of current.
So, what happens:
As the points open, the primary current suddenly begins to drop. Because
the current is dropping fast, there is a large voltage in the primary, and
a large voltage in the secondary. BUT WAIT: the points have only just begun
to open, so why don't we get an arc (spark) across the points? Well the
capacitor allows current to continue to flow after the points open and
hence limits the rate of change of current and hence, limits the voltage in
both the primary and secondary.
THis is why the capacitor is so important: it prevents the energy in the
coil being discharged in a spark across the points.
As the capacitor charges, the voltages rise in both primary and secondary.
As the voltage across the capacitor rises, the RATE of CHANGE of current
increases.
This continues until there is a spark across the plugs. Current can now
flow through the plug to ground. The supply path is ultimately via the
battery, which has almost zero resistance to ground, which combined with
the capacitances of various parts of the circuit, limits voltage spikes
elsewhere. The Points Capacitor may supply some of the current, but there
is no reason why it should supply all.
Oh, and why is there not enough voltage on the ignition switch to cause
problems: well, there are around 300 hundred volts (I think) on the points
and at the capacitor, but, remember that this is matched by the same
voltage drop across the primary winding of the coil. So the other side
(connected to the ignition switch) remains at the same almost voltage:
nominally 12v.
Simon
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Simon Matthews MailTo:simon_matthews@avanticorp.com
'57 MGA
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