Re: Ignition Coil Overheat





Dan_Thomas_nospam@xxxxxxxxx wrote:

aarcuda69062 wrote:
In article
<1158161749.021298.286760@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Dan_Thomas_nospam@xxxxxxxxx wrote:

With the points closed, the voltage measured at the coil
positive will be lowered. It's measurable because the resistor's
decreasing of the current though it will cause a voltage drop.

So, the resistor drops the current.

Which do you suppose is the intent and which do you suppose is an
observed reaction?

Consider;
One part number ignition coil fits two vehicles of the same make,
one vehicle has a firewall mounted ballast resistor, the other
doesn't. Why does one need the voltage reduced to the coil and
the other one doesn't? (both vehicles have electronic ignition)

The electronic module may have an internal means of reducing the
amperage to the coil during run, while permitting full flow at start.

I went through all this recently while restoring my '51
International. I had to install a 12-volt system but stayed with points
and condenser ignition, and used a nichrome heater motor resistor to
get the 5-volt drop. I don't like resistor wires (or fusible links,
either) because they fail due to vibration and can be hard to find.
And I teach it in college aircraft systems courses. You might
want to review Ohm's law.

Will Ohms law teach what is an intended action and what is a an
observed inconsequential action?

Ohm's Law says E=I*R (Voltage = current times resistance), or
E/R=I (voltage divided by resistance = current), or E/I=R (voltage
divided by current = resistance), or I*R = E (current times resistance
= voltage).

If we have a nominally 12 volt system and a coil with a 3-ohm
primary, the current flow through it will be 4 amps.

The problem is Ohm's law only applies to a purely resistive load. A coil
is not just pure resistance.

-jim


We will be able to
read 12 volts across the coil when the points are closed. If we install
a 2 ohm ballast resistor in this circuit, the circuit's total
resistance will be 5 ohms, and the amperage will drop to 2.4 amps (12/5
= 2.4). That 3 ohm coil with the 2.4 amps running through it will have
a 7.2 voltage indication across it (3 x 2.4 = 7.2), which is what we
wanted.
None of this is inconsequential. We wanted a reduced current
flow through the coil, and do do that we had to reduce voltage as well.
Leaving 12 volts at the coil would give us the initial 4 amps. It's
easier to measure voltage than amps (don't have to break the line to
put an ammeter in series), so we can calculate the current flow like we
did above using the coil's resistance, the resistor's resistance, and
the supply voltage.

When the engine is running, the points (or EI module) are
constantly breaking the flow. The meter won't read accurately because
of the lag and the constantly changing flow, not to mention the
inductive, negative spike emitted by the coil's primary at every break.
The condenser will absorb most of that spike, but it would still mess
up readings. We need to use Ohm's law, or a good oscilloscope, to find
out what's really going on. Of the two, Ohm's Law is cheaper.

http://www.the12volt.com/ohm/ohmslaw.asp
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html
http://www.grc.nasa.gov/WWW/K-12/Sample_Projects/Ohms_Law/ohmslaw.html

Dan

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