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A lot of aftermarket pick ups will go bad with 12 volts constant.

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Not sure which aftermarket pickup coils you mean. The factory style pickup is a pulse generator(I believe MSD is also)it sees NO voltage..it creates an AC signal used as a trigger by the ECU

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The ballast protects the coil and the ignition module. If you want to bypass you should replace both of these with units that can withstand the current passing through them at 14V.

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The ballast protects the coil and the ignition module. If you want to bypass you should replace both of these with units that can withstand the current passing through them at 14V.

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Its protecting only the coil. If you check the wiring diagram, the ECU gets full battery voltage from the vehicle harness.

As mentioned earlier, the stock distributor pickup doesn't care as it's a Variable reluctance sensor and is not powered. The magnetic field creates AC voltage as the reluctor passes by the pickup.

I ran an Accel 8140 coil for years with a stock replacement ECU and never had an issue.

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True the ECU is powered from the harness but it also carries the same current as the coil primary winding because it controls the coil negative on/off. When the coil is "on" current flows through the coil primary to ground provided by the ECU

Here's a pretty good description of system operation copy/pasted from the tech archives..just remember, any current flow through the coil primary also goes through the ECU:

The + wire supplies the power to the coil and the ECU grounds the negative part closing the circuit. This creates a magnetic field in the coil. The coil is what we call an autotransformer. When the ECU receives a signal from the magnetic pickup to fire a spark, the ECU OPENS the circuit, the magnetic field collapses and a spark is generated. A collapsing magnetic field creates a high reverse polarity voltage (energy).

The ballast resistor knocks the voltage down by a few volts to the coil (+) from 12 volts. The ballast resistor is bypassed during starting to produce a hotter spark. Run without the resistor and you will burn the coil up in a matter of minutes.

Simple actually.

Inside the distributor is the pickup and reluctor. The pickup has a magnetic field and a coil of wire. As the edges of the reluctor pass thru the pickup's magnetic field it generates a voltage in the coil, a small voltage and not enough to drive the coil.

This signal is sent to the ECU which uses it to control the coil. The ECU transforms the small pulse from the pick up into a signal of proper duration (this is where the dwell is created) to fire off the coil. It will generate a ground to the coil to allow it to charge up then remove the ground causing the spark.

The coil generates it's high voltage spark by collasping a magnetic field as described above.

The ballast resistor is not there to knock down voltage but to limit current thru the system. It's the excessive current that will burn out a coil. However, being that voltage = resistance times current any drop in current will affect voltage. But it is still the current that does the damage.

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Any of you ever burn your finger on the ballast?

Without it there to limit current that heat would be added to the coil.

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Exactly! AND the ECU!!

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The ballast protects the coil and the ignition module. If you want to bypass you should replace both of these with units that can withstand the current passing through them at 14V.

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Its protecting only the coil. If you check the wiring diagram, the ECU gets full battery voltage from the vehicle harness.

As mentioned earlier, the stock distributor pickup doesn't care as it's a Variable reluctance sensor and is not powered. The magnetic field creates AC voltage as the reluctor passes by the pickup.

I ran an Accel 8140 coil for years with a stock replacement ECU and never had an issue.

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True the ECU is powered from the harness but it also carries the same current as the coil primary winding because it controls the coil negative on/off. When the coil is "on" current flows through the coil primary to ground provided by the ECU

Here's a pretty good description of system operation copy/pasted from the tech archives..just remember, any current flow through the coil primary also goes through the ECU:

The + wire supplies the power to the coil and the ECU grounds the negative part closing the circuit. This creates a magnetic field in the coil. The coil is what we call an autotransformer. When the ECU receives a signal from the magnetic pickup to fire a spark, the ECU OPENS the circuit, the magnetic field collapses and a spark is generated. A collapsing magnetic field creates a high reverse polarity voltage (energy).

The ballast resistor knocks the voltage down by a few volts to the coil (+) from 12 volts. The ballast resistor is bypassed during starting to produce a hotter spark. Run without the resistor and you will burn the coil up in a matter of minutes.

Simple actually.

Inside the distributor is the pickup and reluctor. The pickup has a magnetic field and a coil of wire. As the edges of the reluctor pass thru the pickup's magnetic field it generates a voltage in the coil, a small voltage and not enough to drive the coil.

This signal is sent to the ECU which uses it to control the coil. The ECU transforms the small pulse from the pick up into a signal of proper duration (this is where the dwell is created) to fire off the coil. It will generate a ground to the coil to allow it to charge up then remove the ground causing the spark.

The coil generates it's high voltage spark by collasping a magnetic field as described above.

The ballast resistor is not there to knock down voltage but to limit current thru the system. It's the excessive current that will burn out a coil. However, being that voltage = resistance times current any drop in current will affect voltage. But it is still the current that does the damage.

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