Ignition Coil Overheat

Ah, we've got another guy who has never hooked an osciloscope between the ballast resistor and positive terminal of the coil on a breaker point ignition system. But he's gonna read a book and tell us all about it.

-jim

Reply to
jim
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Yea, don't bother reading any books or checking the links Dan gave you. It's all just a bunch of bullshit written by people who aren't nearly as smart as you.

And since I'm always willing to learn, what in the hell would you possibly expect to see when you hook an oscilloscope between the ballast resistor and positive terminal of the coil on a breaker point ignition system? I wouldn't expect to see much of anything if the circuit is good.

Reply to
Bob

Actually, if you hooked the positive lead of the labscope to the positive side of the ballast resistor and hooked the negative lead to the other side of the ballast resistor, you'd get a very nice rendition of the current waveform thru the ignition primary. Sort of a poor mans low-amps probe. Naturally, the voltage (amperage) values measured wouldn't necessarily be spot on accurate, the time values would be accurate though, and the waveform would be rendered accurately.

Ammeters were (maybe still are) just measuring a voltage drop across a calibrated resistance (shunt).

Reply to
aarcuda69062

I checked the links. I didn't spot anything that was incorrect. I didn't see anything that contradicted anything I have said either. Note: that if there is no resistance between the coil and battery all that a oscilloscope will show is a flat line at battery voltage.

Well you won't see much with a scope if the engine isn't running.

This whole thread has been a discussion about what happens to the voltage at that point in the ignition circuit. Now you jump in to reveal you have no clue - very helpful.

-jim

Reply to
jim

Hey Neil Like usual what you are saying is true. But what jim is doing is hooking up to both ends of the wire which connects the ballast resistor to the coil. That's why I said I wouldn't expect to see much if the circuit was good. Add a little resistance to the circuit and it would be much like hooking up on both ends of the resistor. Bob

Reply to
Bob

A good part of the thread has been directed at correcting your misunderstandings.

You came up with that all by your self and you say I have no clue? Lol... Ever heard of ohms law?

Reply to
Bob

Yes you could do that. I'm not sure what your reasoning is in suggesting that. The discussion was what happens to voltage at the positive terminal of the coil. To observe that you would hook the positive probe to the positive side of the coil and the negative to ground. But if you hook it up as you suggest you will see the same wave form - only now it will be inverted and shifted down by 12 volts (battery voltage).

-jim

Reply to
jim

No, that is not what I'm doing. If you had been following the discussion, it was about the voltage seen at the positive terminal of the coil when a resistor is in the circuit between the coil and the battery. I guess I assumed the readers would have enough knowledge of how to go about measuring voltage without a detailed explanation of where to attach the leads. But apparently I assumed too much.

To observe the voltage on a scope you need to connect the positive probe to the positive side of the coil and the negative lead to the negative post of the battery (or any good chassis ground).

When you do measure the voltage with the engine not running you will see steady battery voltage (about 12 volts) if the points are open and less than battery voltage (maybe 9 volts) if the points are closed.

When the engine is running you will see a dynamic waveform where some of the time the voltage is below battery level and some of the time it is above battery level.

-jim

Reply to
jim

I should add that the the shape of the dynamic waveform that is observed on a oscilloscope is influenced by a multitude of things. The condition of the coil, points, condenser, spark plug wires, spark plugs, cylinder compression, fuel mixture and timing all have some effect on the shape of the observed waveform (i.e. they all affect the voltage at various points in the ignition cycle). And of course the actual amount of resistance between the battery and coil will affect the scale of the observed waveform. Given all that, I can't imagine how anyone could read a book or a web page and thereby predict exactly what is going to be observed.

-jim

Reply to
jim

You won't see much waveform with the resistor out of the circuit because the battery has an extremely low internal resistance (on the order of about 0.2 ohms) and acts an excellent shunt to ground for any spike. That doesn't mean that the spike isn't being generated; it just means that the resistor holds back enough of it to measure. Again, the resistor does not cause the spike to form. The coil does that whether there's a resistor in the circuit or not.

Dan

Reply to
Dan_Thomas_nospam

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