Wiring question

[turbinedoctor]

I am trying to help out a friend who has a GM 120MR engine in his boat. We are converting from point ignition system to electronic with a Pertronix's ignitor kit. While looking over the wiring diagram for his engine I have a question or two about the Resistor wire and the Purple/Yellow wire to the starter. I think the Purple /Yellow wire is called a Ballast bypass wire but am unsure as to it's job. The resistance wire is to drop the voltage to the coil ineffect lowering the current in the circuit to about 3 or 4 amps, is this correct? The kit said to connect the red wire to the 12 volt before the resistor wire for full voltage. He connected it after the resistance wire and the engine only back fires when trying to start. Please forgive the second and third hand information here as I am trying to help and have only seen the engine once.

 

[turbinedoctor]

Re: Wiring question

I think I figured out the resistance wire. It is to limit the current to the coil while the engine is running. Correct?

What I want help with is the Ballast bypass wire. I think it is to deliver a full 12 volts to the coil only while starting and is fed from a contact on the starter soleniod. Correct?

Durwood

 

[achris]

Re: Wiring question

I think I figured out the resistance wire. It is to limit the current to the coil while the engine is running. Correct?

What I want help with is the Ballast bypass wire. I think it is to deliver a full 12 volts to the coil only while starting and is fed from a contact on the starter soleniod. Correct?

Durwood

Give the man a carrot.... Correct on all but the smallest detail...

The resistance wire is to lower the VOLTAGE to the coil, not the current. Current is a factor of voltage and resistance (I=E/R).

To get full 12 volts to the ignition system while the engine is running pick up any purple wire. From the diagram you posted, you have one on the back of the alternator or the electric choke. Once you've done that, remove the resistance wire bypass from the starter and the coil and the resistance wire from the coil... no longer required.

The reason you're getting backfiring is because the ignition module is only being fed with 7 volts.

Chris........

 

[turbinedoctor]

Re: Wiring question

Chris while I greatly respect your knowledge and thank you for your reply, I do have to disagree with you on the resistor wire verse current.

The resistance wire is to lower the VOLTAGE to the coil, not the current. Current is a factor of voltage and resistance (I=E/R).

When you lower the voltage with a resistor dont you also lower the current. Example Resistor wire is 2 ohms and the coil is 1.2 ohms which equals 3.2 ohms. 12 / 3.2 equals 3.75 amps. Now if you take out the resistor wire you now have 12 volts divided by the 1.2 ohms of the coil which equals 12 / 1.2 = 10 amps.

Although it has been a while since I have had to use my Ohms Law I think I am correct on this but am open to corrective comments.

We will be leaving the resistance wire in place per manufacturers instructions but moving the red to the 12 volt side as you mentioned.

I hope to be posting updates as we progress.

Thanks again,

Durwood

 

[achris]

Re: Wiring question

Durwood...

You are absolutely correct on all aspects ^^^. By putting the resistance wire in and lowering the voltage you automatically lower the current, yes, but that wasn't the original design concept. The original design concept was to cater for the lower voltage available to the coil while cranking. Hence the bypass wire from the starter solenoid.

I only bother to remember 2 electrical formula. Ohm's law (E=I.R) and the power law (P=E.I). EVERYTHING else I need can be derived from those 2. I also remember that resistances in series get summed and resistances in parallel are the inverse of the sum of the inverses. Capacitors are opposite.... That's IT.

Cheers,

Let us know how you go....

 

[turbinedoctor]

Re: Wiring question

Thanks Chris, I enjoy reading your posts and will keep you informed of the progress.

Durwood