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Old Apr 15, 2008 | 11:21 PM
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Default Starting-- Coil---spark Question

When you crank the motor over do you get one spark from the coil when you hit starter or you you get multiple sparks from the coil as you crank over the engine? How can you tell if the coil spark is the correct strength or just a weak spark. If you get a spark from the coil why not spark from the ignition wire?
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Old Apr 16, 2008 | 07:44 AM
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The voltage to the coil is the same in Start as it is in Run, the only difference is the starter solonoid is getting 12V to engage and turn the engine. Once in Run, and into Start, the IGN system is energized.
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Old Apr 16, 2008 | 12:24 PM
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You'll get mulitiple sparks from a conventional ignition.
On a electronic ignition you'll get just one spark if the module is bad.
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Old Apr 16, 2008 | 10:58 PM
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Originally Posted by Durango_boy
The voltage to the coil is the same in Start as it is in Run
Not true for a stock points type distributor system.

On a points type system in START, full battery voltage is applied to the coil. In RUN, voltage goes through a resistor (usually mounted on the firewall) before it gets to the coil. The resistor knocks the voltage applied to the coil down while it is in RUN.

On a points type system you should get a spark out of the coil each time the points close. When the points are closed, a current is induced through the coil causing it to produce a spark. You will get a stronger spark out of the coil while in START compared to while in RUN. The rotor directs the spark from the coil to each spark plug wire when its their turn.

On an HEI type system, full battery voltage is applied to the coil in START and RUN. You should get a spark each time the poles of the rotating star line up with the poles on the stationary magnet inside the distributer.

When the star poles and the magnet poles are aligned, the resulting magnetic field induces current in the small pick-up coil located under the rotor. That current travels to the module and then to the big coil. (I'm not sure what function the module actually performs.) However, the big coil and the rotor in an HEI system provide the same function as their counterparts do in the point type system.
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Old Apr 16, 2008 | 11:11 PM
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I didn't really mean the intensity of the voltage, just that there was a constant 12V to the coil in both run and start.
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Old Apr 17, 2008 | 03:55 AM
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Originally Posted by Durango_boy
I didn't really mean the intensity of the voltage, just that there was a constant 12V to the coil in both run and start.
Not on a 74 and down points style ignition - More like 9V at the coil during "run" and 12V at the coil during "start".


HEI is constant 12V.

-Mark.
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Old Apr 17, 2008 | 08:04 AM
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Originally Posted by stingr69
Not on a 74 and down points style ignition - More like 9V at the coil during "run" and 12V at the coil during "start".


HEI is constant 12V.

-Mark.

Very interesting, now that...I didn't know for sure.
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Old Apr 17, 2008 | 10:41 AM
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On a points type system you should get a spark out of the coil each time the points close. When the points are closed, a current is induced through the coil causing it to produce a spark.(Quote)

Technically when the point OPENS is when the field collapses and a spark is induced. When the points are closed is the period when the magnetic field strength increases.
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Old Apr 17, 2008 | 11:03 AM
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Originally Posted by stingr69
Not on a 74 and down points style ignition - More like 9V at the coil during "run" and 12V at the coil during "start". No. Actually it's closer to the opposite of that. (12 volts during run, and 9 volts during cranking.)
HEI is constant 12V.

-Mark.
No. There is always, I repeat, always 12 volts powering the ignition system (Except during cold cranking when the starter sucks so much current out the battery that the voltage drops). Points or HEI, it doesn't matter. What is happening in a points system is that there is a parallel circuit connected to the starter solenoid that bypasses the current limiting ballast resistor so that even at lower battery voltages (during cranking) the coil will see somewhat similar primary current levels (voltage divided by resistance). The ignition system (points or HEI) always runs off of whatever voltage the battery happens to be. Always.
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Old Apr 17, 2008 | 11:13 AM
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Originally Posted by 67-73-76
...........................

When the star poles and the magnet poles are aligned, the resulting magnetic field induces current in the small pick-up coil located under the rotor. That current travels to the module and then to the big coil. (I'm not sure what function the module actually performs.) However, the big coil and the rotor in an HEI system provide the same function as their counterparts do in the point type system.
The module is where the intelligence and the hard work happens. The module reads the AC signal from the reluctor coil to determine engine position and RPM. It then uses that signal to determine not only when to fire the coil (end of dwell), but also when to start the dwell. Dwell time is determined by measuring the primary current passing through the heavy duty transistor in the module, and then varying the start of dwell time/position so that the primary current has time to reach the desired level (5.8 amps IIRC). The internal transistor performs the function of the points and ballast resistor, but in a much more efficient manner.
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Old Apr 17, 2008 | 12:43 PM
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Originally Posted by Donald #31176
On a points type system you should get a spark out of the coil each time the points close. When the points are closed, a current is induced through the coil causing it to produce a spark.(Quote)

Technically when the point OPENS is when the field collapses and a spark is induced. When the points are closed is the period when the magnetic field strength increases.
i'm thinking this is right. you can put a scope on the secondary system and see one spike per cylinder, less than 10kV normally, followed by at least 5 coil oscillations and dwell.
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Old Apr 17, 2008 | 08:06 PM
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Good Discussion Guys

Technically when the point OPENS is when the field collapses and a spark is induced. When the points are closed is the period when the magnetic field strength increases.

I stand corrected. I had the concept right, but the details wrong.



There is always, I repeat, always 12 volts powering the ignition system (Except during cold cranking when the starter sucks so much current out the battery that the voltage drops). The ignition system (points or HEI) always runs off of whatever voltage the battery happens to be. Always.

I'm not buying this at all. On a points system the ballast resisitor knocks the voltage down to More like 9V at the coil during "run". When the switch is in RUN, the parallel START circuit to the coil is disabled.



The module is where the intelligence and the hard work happens. The module reads the AC signal from the reluctor coil to determine engine position and RPM. It then uses that signal to determine not only when to fire the coil (end of dwell), but also when to start the dwell. Dwell time is determined by measuring the primary current passing through the heavy duty transistor in the module, and then varying the start of dwell time/position so that the primary current has time to reach the desired level (5.8 amps IIRC).

So the function of the module is to control the dwell. That makes sense. I did not know that.


The internal transistor performs the function of the points and ballast resistor.

Thanks to your expaination, I can see how the tranistor performs the same function as the points with respect to controlling the dwell.

However, the only function the ballast resistor performs is to knock the voltage down to the coil from 12V to 9V or so, thus prolonging the life of the points. If the points saw 12V all the time, they would tend to arc as they continually open and close which corrodes the faces of the points.

I'm not seeing the comparison for the transistor and the ballast resistor. Since the HEI system has no points to corrode, there's no reason to knock the voltage down. Thus the ballast resisitor is eliminated and full battery voltage is applied to the coil.
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Old Apr 17, 2008 | 08:27 PM
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Originally Posted by 67-73-76

However, the only function the ballast resistor performs is to knock the voltage down to the coil from 12V to 9V or so, thus prolonging the life of the points. If the points saw 12V all the time, they would tend to arc as they continually open and close which corrodes the faces of the points.
Close, but no cigar.
The average 8 cyl. coil is 1.5 ohms. Average ballast resistor is 1.5 ohms. 3 ohms total. 12volts/3 ohms is 4 amperes.
The ballast resistors job is to limit the CURRENT flow to 4 amperes during "run." The bypassed ballast resistor at "start" enables 8 amperes of current. Votage does not pit the points. CURRENT does.
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Old Apr 17, 2008 | 08:54 PM
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Originally Posted by 67-73-76
Good Discussion Guys

Technically when the point OPENS is when the field collapses and a spark is induced. When the points are closed is the period when the magnetic field strength increases.

I stand corrected. I had the concept right, but the details wrong.

Your statements above are correct.

There is always, I repeat, always 12 volts powering the ignition system (Except during cold cranking when the starter sucks so much current out the battery that the voltage drops). The ignition system (points or HEI) always runs off of whatever voltage the battery happens to be. Always.

I'm not buying this at all. Well, it doesn't really matter if you are buying it or not. It's a fact. Put an oscilloscope on the system and you'll see 12 volts pushing the current through the system. On a points system the ballast resisitor knocks the voltage down to More like 9V at the coil during "run". When the switch is in RUN, the parallel START circuit to the coil is disabled. No. Please take a first semester electronics course. A resistor is a passive, current-limiting device. It is not an active, voltage-regulating device.


The module is where the intelligence and the hard work happens. The module reads the AC signal from the reluctor coil to determine engine position and RPM. It then uses that signal to determine not only when to fire the coil (end of dwell), but also when to start the dwell. Dwell time is determined by measuring the primary current passing through the heavy duty transistor in the module, and then varying the start of dwell time/position so that the primary current has time to reach the desired level (5.8 amps IIRC).

So the function of the module is to control the dwell. That makes sense. I did not know that.


The internal transistor performs the function of the points and ballast resistor.

Thanks to your expaination, I can see how the tranistor performs the same function as the points with respect to controlling the dwell.

However, the only function the ballast resistor performs is to knock the voltage down to the coil from 12V to 9V or so, Again, no, it isn't, and no, it doesn't thus prolonging the life of the points. You are correct that the ballast resistor prolongs the life of the points (and coil). It does that by limiting the current passing through the coil and the points contacts, not by somehow changing the system voltage, which it is unable to do. If the points saw 12V all the time, they would tend to arc as they continually open and close which corrodes the faces of the points. Twelve volts is nothing for the points. The point contacts have to regularly switch anywhere from 100 to 300 volts when the points open (due to the inductive flyback voltage of the coil primary winding). That's what tries to cause the arc. The condenser is then used to quickly absorb or divert this voltage rise so that the points are protected while they're opening.

I'm not seeing the comparison for the transistor and the ballast resistor. Since the HEI system has no points to corrode, there's no reason to knock the voltage down. Thus the ballast resisitor is eliminated and full battery voltage is applied to the coil.
Transistor is a acronym for "transfer resistor". Essentially, it is a variable resistor. At the start of dwell, the transistor turns on fully, resulting in a low resistance pathway for the primary current. This is equivalent to a set of points and no ballast resistor. The primary current will rise quickly (a good thing at higher RPMs) because the only system resistance is the coil wire resistance (we're temporarily setting aside the whole inductance factor for these basic descriptions). Once the module transistor detects that the primary current has hit the desired level (5.8 amps), it turns off slightly, which increases the transistor's resistance, keeping the current from increasing any further. This is the same function as a ballast resistor. Limiting the primary current. Additionally, quality electronic systems will not use a condenser either. A condenser slows down the current shutoff, resulting in potentially lower available voltage at the coil output. A quality transistor is able to deal with higher primary currents (a good thing), along with being able to quickly shut off the primary current (a good thing also). High currents and fast shutoffs are conducive to higher energy and voltage levels available from the coil.


Last edited by 69427; Apr 17, 2008 at 09:20 PM. Reason: Refreshment break.
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Old Apr 17, 2008 | 09:32 PM
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Originally Posted by FB007
Close, but no cigar.
The average 8 cyl. coil is 1.5 ohms. Average ballast resistor is 1.5 ohms. 3 ohms total. 12volts/3 ohms is 4 amperes.
The ballast resistors job is to limit the CURRENT flow to 4 amperes during "run." The bypassed ballast resistor at "start" enables 8 amperes of current. Votage does not pit the points. CURRENT does.
FB's info is on the money.

An additional bit of info is regarding cold (winter) starting. Due to the extremely high starter currents resulting from cranking over an engine with thick oil, the battery voltage will drop substantially. At worst case, the battery may drop to six volts. This situation is the reason for the ballast resistor bypass wiring. Using Ohm's law again, we have 6v/1.5 ohms equals 4 amps. This yields the same coil energy as when the system is at the normal 12 volt level. Any cranking voltage above six volts just yields a bit of bonus energy to the plugs until the key is released and the starter bypass kicks out.

Last edited by 69427; Apr 17, 2008 at 11:20 PM. Reason: Spelling correction.
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Old Apr 17, 2008 | 10:10 PM
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Originally Posted by FB007
Close, but no cigar.
The average 8 cyl. coil is 1.5 ohms. Average ballast resistor is 1.5 ohms. 3 ohms total. 12volts/3 ohms is 4 amperes.
The ballast resistors job is to limit the CURRENT flow to 4 amperes during "run." The bypassed ballast resistor at "start" enables 8 amperes of current. Votage does not pit the points. CURRENT does.
Got me again by the details.

I'll rephrase.

The function of the ballast resistor is to knock down the current from 8A to 4A thereby helping the points live longer. By knocking down the current, the ballast resistor also knocks down the voltage which is easier to measure.

Tying this into the Wallifishmn's question about weak spark, voltage available to the coil is more important than current through the coil.

My overall argument remains valid. Full battery voltage is not available to the coil during RUN as DB and 69427 claimed.
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Old Apr 17, 2008 | 11:19 PM
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Originally Posted by 67-73-76
Got me again by the details.

I'll rephrase.

The function of the ballast resistor is to knock down the current from 8A to 4A thereby helping the points live longer. Correct. By knocking down the current, the ballast resistor also knocks down the voltage which is easier to measure. Uhhhh, what???

Tying this into the Wallifishmn's question about weak spark, voltage available to the coil is more important than current through the coil.

Wrong, wrong, wrong. Coil energy (in milliJoules) is determined strictly by the inductance of, and the current through, the primary windings. Voltage is nowhere in the energy formula.

My overall argument remains valid. Full battery voltage is not available to the coil during RUN as DB and 69427 claimed.
A RUN mode quiz: What is the voltage at the ballast resistor and the coil + terminal right before the dwell starts (points open)?
Answer:12 volts.
What is the voltage at these same points the moment the points first close?
Answer: 12 volts. The ballast did not magically change any voltages.
You are erroneously using a voltmeter reading (on a DC scale) to measure a fast, continuously varying signal. Using the wrong tool will only give you the wrong answer. Fast moving electrical signals require an oscilloscope. The 'scope reading will dispel the 9 volt theory.
I will give you credit for standing your ground. I can't give you credit for understanding electronics, though. I understand you don't believe me. I can certainly live with that. A suggestion: Please ask an electronics savvy buddy of yours to explain how a (simple series) circuit with a voltage supply, resistor, inductor, and a switch works. He will be unable to locate a (9 volt) voltage regulator in that circuit, and all current passing through the series circuit is driven by the 12 volt source.
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