In a message dated 4/12/98 4:20:30 PM Eastern Daylight Time,
prubrew@ix.netcom.com writes:
> I figure while she is down for a nap, I'll swap out my questionable
> original solenoid for the new one I just got. Sounds like a fairly easy
> swap...no? Justunplug wires and plug in on new solenoid, fasten down
cables
> and solenoid to firewall.
>
> Turn the key...cranks just fine, but only wants to catch when I release
> the key. Pull off coil wire and test for spark. Get initial spark as key
> turns to starter motor position, but no spark while cranking. Get another
> spark as I turn key off.
>
> How did I do this? What wires did I inadvertantly misfit? A little
> background re: ignition:
>
> Mallory Dual point dizzy
> Mallory coil
> Crane EI Sytem
> Mallory ballast resistor
>
> All fully functional this morning before I swapped out the solenoid. Help
> me Dan Masters...wherever you are
Hi, Chris,
Well, I'm back, and just in time, it seems!
Your solenoid has four terminals on it -- two large ones, and two small ones.
When you apply 12 volts to one of the small terminals (via the white/red
wire), the solenoid is energized, and all three of the remaining terminals are
connected together inside the solenoid. Starter current is applied to the
starter from the battery via the large terminals, and 12 volts is applied to
the positive side of the ignition coil via the other small terminal, using the
white/yellow wire. This 12 volts to the coil bypasses your ballast resister,
giving you a hotter spark for starting. As soon as you turn the key from the
start position to the run position, the solenoid is de-energized, and the 12
volts is no longer applied to the coil. At this time, the coil gets its
current through the ballast resistor.
If you reverse the leads to the small terminals, one of two things will
happen:
1) The solenoid will energize as soon as you turn the key to the on position,
and the starter will crank. The solenoid will not get the full 12 volts in
this situation, as it is being fed from the ballast resister, so the voltage
may or may not be enough to actually energize the solenoid, but it will try
to. What is happening here is the solenoid is being "backfed" from the
ignition lead to the coil.
2) If the voltage with the ballast is not enough to energize the solenoid,
and you turn the key to the start position, nothing will happen. As long as
the solenoid is not energized, the terminal that is used for ballast bypass is
not connected to anything inside the solenoid. Since the leads are reversed,
you are now applying starting voltage to a dead ended contact.
I'm sorta surprised that the car would not start without the full 12 volts on
the coil, but then again, that's why the factory installed the extra terminal
on the solenoid to bypass the ballast resister. Evidently, you have a bad
solenoid, and the ballast bypass contact is not connecting inside the
solenoid, since you said in a later post that the car started when you
replaced the old one.
Two things to try:
1) Put the new solenoid back in and measure the voltage at the bypass terminal
when you crank the engine. I'm willing to bet that it is zero.
2) Or, with the new solenoid in place, just before you try starting the car,
jumper from the positive post of the battery to the positive post of the coil,
and then crank the engine. I'm betting that it will start. This does
manually what the solenoid normally does automatically. Don't leave the
jumper in place very long after the engine starts -- the coil is not designed
to take the full 12 volts for very long. No need to remove it in a panic mode
-- anything less than 5 minutes should not be a problem.
Dan Masters, with a very happy wife, who is enjoying her new kitchen cabinets,
appliances, and acres of freshly laid ceramic tile, and me very happy that
most of it is behind me. I do not like to lay tile!
Glad to be back on the list!
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