Ignition Coil Overheat

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)

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

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

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. 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.

Dan

So the long and short of that is the OP can just put the key in run with the engine off and measure the voltage between the positive pole on the coil and the negative of the battery and tell from the drop if it has a resistor in the circuit or not.

No resistor would mean he would have battery voltage. With the jeep

1.35 ohm resistor in the line, he would have somewhere around 10 volts.

It is a simple way to use a multimeter to tell if he is smoking out the coil from too much power or if the Jeep resistor wire is still in there to match the Ford application coil.

Mike

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Yup. But he would have to make sure that the points are closed. If it's an electronic system, there *might* not be any flow at all until he starts cranking.

Depending on the coil primary's resistance. A lower resistance would increase current flow, and in that case (according to Mr. Ohm) the ballast resistor would drop the voltage more than if the coil resistance were higher. A good manual will give the coil's resistance range. Or the Accel people should have it.

Exactly.

Dan

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

-jim

Thank you.

Thank you.

With low amp probes being relatively inexpensive now days, we don't need to break the line anymore. The part about easier to measure voltage reinforces the fact that teaching that the ballast resistor is there to reduce voltage and 'oh look a voltage drop' was merely a path of less resistance in making the circuit understood.

Yes we can, but this isn't about calculating resistances and voltages, it's about what purpose a component serves.

No matter at all, the method just measures the voltage drop across the coil power wire from the battery. The coil negative can be disconnected for this static test.

Real life says the meter reads around 10 volts depending on the condition of the battery. Jeep uses the Ford ignition system.

Not so fast! If the points are open any measurements you take at the coil or ballast resistor will show battery voltage. The points do need to be closed...... Bob

Without current flow, there's no voltage drop. A better way to check for a resistance in the circuit to the coil + would be to measure voltage from battery + to coil +.. hook a ground wire up on coil - and see what happens to your voltage reading. This would give you actual voltage drop and eliminate variables like changes in battery voltage.

Jim

You are right, that should just be a resistance reading from the battery with the key to run and the engine off to see if the line has continuity and/or a resistor in it.

Mike

I was thinking a plain resistance reading, I spit that out wrong.

A resistance reading from the battery to the coil with the key in run will tell if he has a ballast resistor in the line or not.

Mike

86/00 CJ7 Laredo, 33x9.5 BFG Muds, 'glass nose to tail in '00 88 Cherokee 235 BFG AT's Canadian Off Road Trips Photos: Non members can still view! Jan/06 (More Off Road album links at bottom of the view page)

I know that, and the presence of any condenser adds a reactive factor to it. But we were talking here of key on, engine not running, where the coil, once flow is established, is nothing more than a resistive load, and we can use Ohm's Law. We use that characteristic to see what the ballast resistor, if any, is doing. Some manuals with good troubleshooting sections will call for such measurements. Inductive calculations are useful only in make/break or AC activity, not in established and steady DC flow. When the engine is running, the readings make much less sense because of the coil's inductive qualities, as I said here before in this thread, and I quote myself:

"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."

Dan

Well yes, if you never plan on starting the engine that's great. But if you do ever run the engine you will find that due to the inductance of the coil that the voltage at that point where you are measuring voltage is higher than battery voltage much of the time (you will need an oscilloscope to make this observation accurately). From that fact should we conclude that the purpose of the resister in front of the coil is to increase the voltage to the coil while the engine is running?

-jim

Where did ANYONE suggest that the resistor was there to increase the voltage while running? And maybe you could explain why the voltage is higher some of the time. The inductive spike is negative, not positive, and will tend to reduce the positive voltage at the coil's positive as the points open, not increase it, though a proper measurement across the coil would indeed find a higher voltage, just backwards to normal polarity. The presence of the resistor has nothing to do with that; that spike is there in non-resistor systems, too.

Dan

I asked a question - why aren't they saying that since increased voltage is what happens (at least part of the time).

EMF, oscillating magnetic fields the usual suspects

You are measuring voltage from the battery to ground over time. You really need an oscilloscope to observe this. If you connect the coil directly to the battery with no resistance between then all you will ever see at the coil is battery voltage (around 12 volts). If you put resistance in there, when the engine is running, you will not see battery voltage but instead see a dynamic wave form where the voltage is higher than battery for part of each cycle. If you don't believe me lick your finger and put it on the positive terminal of the coil while the engine is running. If the voltage was lower than the battery you would feel nothing.

-jim

jim wrote:

The spike is negative, not positive as the battery's feed is. You sure will feel a jolt, but it's a negative jolt, not a positive, and it works against the battery's voltage and will reduce it at that point right through zero and into a negative value, not increase its positive value. If there's any increase in voltage at that point it will be negative, and the resistor has nothing to do with it. The resistor did not generate or increase that voltage spike; the coil did. The resistor's only function is to reduce overall current flow though the coil during the dwell, which it will do in any AC or DC circuit. The coil is an inductor and by definition will resist any attempt to increase or decrease current flow through it. When the points open and the flow is arrested, the field in the coil collapses and tries to keep the flow going in the same direction. Since the coil no longer has the battery pulling electrons through it (electron flow is from negative to positive), and the coil is now pushing electrons out of its positive terminal, it's a negative spike and will read as such on any scope. It would read negative on any meter that could resond fast enough; instead of the positive pulling electons OUT of the meter's positive lead, the negative jolt would push electrons INTO the lead and show a negative. I have scoped 600 volts from the master solenoid coils in our aircraft when they're shut off. That's the reason your ignition switch shuts the radio off during start: so that the starter solenoid (and the starter's windings themselves) doesn't fry the radio when you release the starter.

Dan

Not on mine, not on yours either if you had actually tried it. A ignition coil is not a solenoid. You're boring me with your lack of knowledge. bye,

-jim

A coil is a coil and will behave as a coil whether it's in a solenoid, an ignition coil, an RF choke, or a transformer. Faraday's and Lenz's Law applies to them all. See

and

and study up on it.

Dan

It looks like he'd rather remain ignorant.