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RE: New Mystery

To: tigers@autox.team.net
Subject: RE: New Mystery
From: "Theo Smit" <theo.smit@dynastream.com>
Date: Fri, 4 Jun 2004 09:16:28 -0600
Steve, and Steve,

For the problem you're trying to resolve, the DC behaviour is what you're
trying to determine. I'm going to assume that you have the coil and
Pertronix wired according to the instructions: The coil (+) and the
Pertronix red wire are connected to the ignition switch, the coil (-) is
wired to the Pertronix black wire, and the Pertronix unit itself is properly
grounded inside the distributor (i.e, it is snugly bolted down onto a clean
breaker plate). The Pertronix acts as a replacement for the stock points,
which is to say that it functions like a switch that connects the coil (-)
to ground periodically. When that ground connection is broken, a high
voltage is generated at the secondary coil terminal, and the high voltage
pulse is conducted via the coil wire, distributor, rotor, and spark wires,
to the appropriate spark plug. The distributor rotor has to be properly
timed relative to the magnetic sensor in the Pertronix unit, otherwise the
high voltage pulse has nowhere to go, and it will discharge through the
insides of the coil, or through the side of the distributor cap, and that's
bad. Steve, note that this is NOT the same as your description of the coil
operation. When the primary current is interrupted is when the voltage pulse
is generated on the secondary - not whenever the rotor happens to come by.
The coil acts as an energy converter, not an energy storage device.

Anyway... Without the engine running, your ignition circuit consists of the
battery, the wiring to the ignition switch, the switch itself, the wiring
between the switch and the coil/Pertronix, the coil and Pertronix units
themselves, the engine block, the block ground strap, the chassis, and the
battery ground strap. Any of the wires, switches, and wire junctions can
easily act as resistances, and that is most likely the problem. A resistance
causes a voltage drop proportional to the current and the resistance: V = I
x R, where I is the current, and R the resistance. If you have just the 0.6
ohm coil connected across the battery, then from 12 volts you would get 20
amps of DC current: 12 = 20 x 0.6.
When you turn on the ignition, the Pertronix quickly figures out that the
engine isn't turning over, so it doesn't turn on its coil charging switch
(transistor), and no coil current flows. Because there is no current flow,
there is no actual "circuit", and you won't measure any voltage drop. You
can measure the voltage at either coil terminal, and you'll just see the
battery voltage: 12 volts.

Now when you ground the (-) terminal on the coil, you complete the circuit,
and current flows from the battery, through all the wire and ignition switch
(remember, these could be like resistances if the contacts aren't clean
anywhere), to the coil (+), through the coil, then from the coil (-) to the
chassis or engine block, and back to the battery through its ground strap.
Steve, you're getting 7 volts when you do this, so that tells me that if we
assume 0.6 ohms coil resistance, we have about 11 amps coil current, and
somewhere in the remainder of the circuit we're getting 5 volts drop. 5
volts divided by 11 amps is 0.45 ohms - the resistance in the remainder of
your coil circuit (i.e. all the wiring, the chassis, the ignition switch,
and the battery).

To find out where the problem is you need to establish an absolute voltage
reference and measure the voltages relative to it. With the coil grounded,
measure the voltage at the battery terminals. Write it down. Then, get a 20
foot piece of 14 gauge wire, and firmly clamp it onto the battery negative
terminal, and use this as the (-) meter connection for all the other
voltages you measure. Measure the voltage at the chassis ground connection,
the engine block, the distributor, the coil (-), the coil (+), both
terminals of the ignition switch, and the voltage at the fuse box. Bits and
pieces of your missing 5 volts will show up at each of these connections,
and you need to deal with each of them in turn.

By the way: Your multimeter measured the coil as 0.9 ohms - that's not
surprising. What does the meter measure when you short the two test leads
together directly? Most meters can't correctly measure impedances below
about 5 ohms. To do this you need to connect the coil and some precision
resistors in a bridge type arrangement and then measure the voltage drops
across the branches - not hard to do, but not that easy for me to explain.

Best regards,
Theo





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