Just bought a 12v Conversion...coil?

Hello. I just bought a 12v conversion kit for my Farmall M, It was 100.00, I probably could've pieced it together cheaper, but oh well. Anyways, I haven't got the kit yet, but one thing I'm a bit confused on is the coil. It comes with a Delco 10si, wiring, resistor, and brackets. The seller told me the only things I would need are a 12v battery (obviously), 12v bulbs, new cables, and a 12v coil. It comes with a resistor, so do I use this with my stock 6v coil? Or do I just switch to a 12v coil without using the resistor like he says? I was always under the impression you could keep the stock coil with the resistor, but I'm pretty new to electrical systems. Thanks!

The secret is to limit primary ignition current to the range of 3.5 to a little over 4 Amps, (points CLOSED engine stopped, ignition switch "on") to prevent overworking the points so they quickly burn and/or overheating the coil.

The CORRECT combination of the original 6-Volt coil AND external ballast resistor with the proper value will achieve this and even allow for a "starting bypass circuit" to give full voltage to the coil for cranking, if desired.

OR, a "true" 12 Volt coil that needs NO ballast resistor will work, instead. (NAPA IC14SB, for example.)

Im basically with the Mayor on this, heres my take on it. The picture looks like the kit has a garden variety external series voltage dropping (12 to 6) Ignition Ballast Resistor, if so it should measure around 1.25 to under 2 or so ohms, does it?????

So to your questions:

"It comes with a resistor, so do I use this with my stock 6v coil?"

YES you sure can

"Or do I just switch to a 12v coil without using the resistor like he says?"

You can (toss the old coil and buy a new coil rated for use at 12 volts nominal, no ballast required), if you wanna spend a bit more money

BUTTTTTTTT Id use the ballast and the old coil as it 1) Saves money 2) Gives you the option (with proper switch and wiring) to by pass the ballast during start up for a better starting spark

Pay your money and take your options

John T

Hi John T...Can you help me understand why the stock 6 volt coil would benefit from having a resistor in the circuit? My limited electrical sense makes me think that the coil would get less electrical power with that resistor in place then without it. I didn't think a stock 6 volt coil (like one from a tractor designed for a 6 volt battery) used a resistor in the circuit.
Pete

Pete,

You're going from 6V to 12V. The coil is designed to see 6V. At 12V it will get very hot very quickly, and will soon fail.

Put simply, the resistor "reduces" the voltage back down to 6V so the coil doesn't burn up in the first 10 minutes of run time.

Thank you guys so much! This is the only board I've ever been on where I get a quick answer, you guys are a huge help.

Andy, I will be sure to do that, thank you.

(quoted from post at 14:24:17 12/20/10) Pete,

You're going from 6V to 12V. The coil is designed to see 6V. At 12V it will get very hot very quickly, and will soon fail.

Put simply, the resistor "reduces" the voltage back down to 6V so the coil doesn't burn up in the first 10 minutes of run time.

Click to expand...

I've always wondered if that was the reasoning for the resistor. Thank you for the explination. I know next to nothing about electrical components, but at least I'm picking up on stuff here and there thanks to you guys.

You put the resistor in the circuit that FEEDS the coil...cutting the 12V input down to the 6V that the coil can handle. You are NOT using the resistor to cut the voltage below 6V.

If you keep your electrical system 6V, there is NO NEED for a resistor in the ignition circuit. AND if you switch to a 12V coil, there is no need to have a resistor in the circuit.

YOU ONLY USE THE RESISTOR WHEN USING A 6V COIL WITH A 12V ELECTRICAL SYSTEM...AND THEN ONLY TO REDUCE THE INPUT VOLTAGE TO THE 6V COIL DOWN TO THE 6V INPUT IT WAS DESIGNED TO HANDLE. YOU DO NOT USE THE RESISTOR ON THE OUTPUT [DISTRIBUTOR/POINTS] SIDE OF THE COIL.

Does that make it easier to understand?

Can do sir,,,,,,Okay iffffffff you have that external series ballast resistor YOU CAN BY PASS JUMP AROUND IT WHEN STARTING/CRANKING, WHICH GIVES YOU MORE SPARK ENERGY at the crucial starting time because the high current draw of the starter motor reduces battery voltage and spark energy. When the temp is cold the battery efficiency is greatly reduced PLUS the starter draws much higher current thereby reducing battery voltage causing a very weak spark when its needed the most in the straight unballasted 12 volt coil situations.....

If the 6 volt coil with external ballast is used, you can run a wire off a starter switch or solenoid etc to supply unballasted voltage to the coil ONLY while cranking, but then when running it goes back to the ballasted or 6 volts for a 6 volt coil as its designed. You cant use 12 volts on a 6 volt coil, she would overheat so the external ballast drops 6 volts leaving only 6 on the coil, buttttttt you can apply higher voltage for a short time like while shes cranking to make up for the reduced battery voltage

Some say they start on 12 volts and run on 6 although thats not exactly right since the battery voltage drops some while youre cranking.

BUT IF THERES NO EXTERNAL BALLAST RESISTOR (you use a 12 volt coil) its not laying there in the open avaialble to be by passed like a 6 volt coil with an external ballast can be.

OKAY if you have a 12 volt tractor but use that voltage dropping resistor ahead of and before the coil IT DROPS 6 VOLTS LEAVING 6 ON THE 6 VOLT COIL SAME AS IF IT WERE ON A 6 VOLT TRACTOR. It sees the same voltage and current as a 6 volt coil on a 6 volt tractor AND YOU GET THe SAME POWER AND SPARK ENERGY

Got it??????

Good question

John T

A resistor adds a load to the circuit...which
creates a voltage drop, and usually generates
heat. A resistor of the proper value will drop
the voltage coming into the coil from 12V to 6V.

A proper understanding of Ohm's Law and the
relationship between resistance [R] in ohms,
current in amperes, and electromotive force
[EMF, or E] in volts makes the concept simple.
E=IxR...R=E/I....I=E/R... It's really not rocket
science, but a lot of people tend to think of it
that way.

That cleared things up for me and now I got it! Thanks for the explanation from you and all the others.
Pete