Changed from points to electronic ignition

No. The goal of both ignition systems is to develop a voltage considerably higher than what is needed to make a spark jump the gap for the spark plug designed to be used in your engine. The spark plug does not know or care which type of ignition system is firing it.

What you will have to do is eliminate the ballast resistor. If you were to turn on the ignition switch to listen to the radio, as an example, and the points were closed, (turned on), you would have a pile of current flowing through the ignition coil and the points continuously. That is very hard on both. To reduce that current to a safe level, the ballast resistor is added in series with the coil and points. With the lower current flow when the points turn on, a weaker magnetic field is built up in the ignition coil, but, the coil is designed to produce the needed secondary voltage with that ballast resistor in there.

The problem comes during cranking the engine. The ignition coil operates on an average of roughly nine volts thanks to the ballast resistor. That is with a charging system keeping system voltage at around 14.0 volts. During cranking, you are starting with 12.6 volts from just the battery, and that is drawn down to as low as 9.6 volts thanks to the heavy load from the starter motor. So now you are starting with, lets say ten volts, and the ballast resistor drops the average ignition coil voltage another, ... Oh, ... Five volts, you are down to five volts or less to run the ignition coil. Spark voltage would be very weak. Probably too low to fire the spark plugs. For that reason, the ballast resistor is always bypassed during cranking. That puts the full ten volts from the battery to the ignition coil to provide a stronger spark. GM did that with a tap on the starter solenoid. Chrysler did that with a separate tap on the ignition switch, then later with an extra terminal on the starter relay. Ford did that with a tap on the starter solenoid that sat on the inner fender near the battery. That was the second smaller terminal on top of that solenoid.

Electronic ignition systems do not use a ballast resistor so you do not need to bypass anything during cranking, but you cannot have that resistor in there when you do this conversion. While this might sound complicated, what you are actually doing is making your wiring slightly less complicated by eliminating the resistor.

There is always a resistor, but it may not be a solid block unit. GM and Ford used resistor blocks and resistor wires and some also use a coil with a built in dropping resistor. The easiest way to tell is to connect a voltmeter to the coil power feed. Turn on the key. If you see 6 to 7 volts then you turn the key to start and see battery voltage you have a resistor inline somewhere. If you see full battery voltage both times you have an internal resistor.

There is a clinker to this however. To see 6 to 7 volts with the ignition switch on, the points would have to be closed so current would be flowing through the circuit. Then, the resistor would cause some voltage to be dropped across it and you would measure the remaining 6 to 7 volts at the ignition coil's positive terminal. If the points are open, there would be no current flow, no voltage dropped across the ballast resistor, and you'd end up with the full 12 volts at the ignition coil's positive terminal. That test by itself would incorrectly indicate there is no resistor in the circuit.

The other problem is during cranking, battery voltage is drawn down a lot by the high current the starter motor draws. That could result in finding 10 to 11 volts throughout the entire electrical system. If you do not understand what Steve W's test is doing, you could again incorrectly conclude there is no resistor if you just looked at the voltage.

I propose doing this test a different way. Instead of a digital voltmeter, use a test light. Hook it right across the battery terminals to get an idea of what "normal" looks like. Now disconnect the wire from the ignition coil's positive terminal. That eliminates the open or closed breaker points variable. Connect the test light to that wire, then turn on the ignition switch to the "run" position. If you see the test light is full brightness, there is no resistor in the circuit, at least not in the wiring harness. If there is a resistor wire or a ballast resistor you have not seen, the test light will have completed the circuit, meaning there is a path for current to flow, and that current will cause some voltage to be dropped across the resistor. There will be only part of the original 12 volts left to run the test light, so it will be less than full brightness.

If you are not sure about the brightness of the test light, you can use the light and the voltmeter at the same time. The voltmeter by itself, for all practical purposes, does not draw any current, so it will not cause a voltage drop across a resistor. The test light will draw current, then the voltmeter will show the lower voltage if there is a resistor in the circuit.

Basically the test light is taking the place of the ignition coil and points. Now there are two possibilities. There is a resistor or there is no resistor. If there is no resistor, you will measure full battery voltage of 12.6 volts. If you remove just the test light, you will still have 12.6 volts.

If there is a resistor in the circuit, current will be flowing through the test light, some of the 12.6 volts will be dropped across the resistor, and the voltmeter will see the remaining voltage. For this sad example, lets say that's 10 volts. Now remove just the test light and you will see voltage pop back up to 12.6 volts.

Basically, if you leave the voltmeter connected, then connect, remove, connect, remove, ... The test light, you'll see the voltage change significantly if there is a resistor in the circuit.

I am sorry that I made that sound so confusing. The tests are actually quite simple. Explaining it is what is so time consuming.

There is one more comment of great value I should add. With the engine running, if you were to look at a waveform pattern of the voltage at the ignition coil's positive terminal, you would see full system voltage of around 14 volts, (charging system running), when the points were open, then typically 2 or 3 volts when the points were closed. (This equates to high current flow through the coil when the magnetic field is building up). As soon as the points opened again, voltage would jump instantly back to 14 volts.

Digital voltmeters work by taking an instantaneous voltage reading, analyzing it, then displaying it while it takes the next reading. It could take one reading when your ignition coil's voltage is 2 volts, then it might take the next three readings when it is at 14 volts. The display would be jumping around and you would be unable to follow it or make sense of it. If you have an older or a cheapy analog voltmeter with a pointer, that pointer cannot move instantly, so it ends up averaging the readings it is taking. This is where you can legitimately take a voltage reading at the coil while the engine is running. The problem is this still is not going to tell you if there is a resistor in the circuit. There's no magic number to say one way or the other. The average voltage displayed will be a factor of point gap/ well angle. The longer percentage of time the points remain open, the higher will be the average voltage. RPM is not a variable.