What the hell did I blow up this time? The answer and now a question

[SJ]

Yeppers. I forgot about that simple ohm meter check.

The coil is a transformer . . .basically two coils in close proximity to
each other. It works on the principle that a moving electric field cutting
across a conductor produces a current in that conductor.

That being the case, you put two conductors near each other . . .the
"primary", one end connected to ground thru a switch, the other end
connected to 12v battery . . . . the "secondary", having about 100 times
more wire than the primary, one end connected to ground, the other end
connected to spark plug.

Turn the switch "on" and the primary coil saturates . .reaching max current
. . maximum electric field expansion. "Open" the switch and the current
stops and the electric field collapses . . the collapsing field cuts across
the secondary coil wires producing current and voltage.
Because there are so many more wires in the secondary coil, the resulting
voltage is proportional to the length(turns) of wire in the secondary.
12volts at primary turns into 25K volts at the secondary.

The resistance of wire is proportional to its length. Assume for arguments
sake that the primary has 10 ft of 16 awg wire and that the secondary has
100 feet of 28 awg wire. If 16 awg wire measures 0.1 ohms/foot, and if 28
awg wire measures .20 ohm/foot . . .then the Bentley will state that the
primary  should measure 1.0 ohm, and the secondary should measure 20 ohms.

If you measure more than 1.0 ohm at the primary . . .it could mean that you
have poor internal connections, partially melted wire(reduced cross
section).
If you measure less than 1.0 ohm . . . it could mean that there is an
internal short inside the primary . . .some of the wire is shorted out and
out of the circuit.
The result of higher resistance is less current in the primary . . .hence
reduced magnetic field . . .hence reduced secondary voltage.
The result of lower resistance is more current and most likely ending up in
melted primary wires and an open circuit. If the wires did not melt . . .the
higher current should result in greater magnetic field and greater resulting
secondary voltage. But, the coil is a a bit more complicated than two coils
positioned near each other . . .the primary and secondary wiring is
intertwined . . I think.

Somebody else can elaborate on that. Something tells me that a lower
resistance would give you higher secondary voltage. But, this is just a
precursor to coil failure . . . .open at the primary or failed insulation at
secondary and shorts. If you are measuring low resistance at the primary . .
.thats a red flag waving.

In the secondary winding, a high resistance would indicate bad connections,
etc. That would produce a higher than normal voltage drop across the
internal coil resistance and less voltage at the spark plug.
A lower resistance would indicate an internal short . . effectively reducing
the amount of wire being cut by magnetic field . . .lower output voltage.

I think the most common coil failures are higher than normal primary
resistance(aged internal connections), and shorted coils(insulation
breakdown) in the secondary.

I had nothing else to do this afternoon . . . I wanted to practice my
penmanship. Too damned cold to work on the rear door locks.

Over to you Huw.

SJ
88 5kq
90 100Q

> > How does one check the coil anyway.
>
> You can measure the resistance of the primary and secondary coils, per
> Bentley.  I don't know how well those two numbers determine if a coil is
> *good*, but if they are wrong, it surely must be bad.  Also, examining
> for arcing or cracks at the output tower I suppose.
>
> Or you can swap in a known good unit.
>
> --
> Huw Powell