Balast resitor.

[BouncinBetty]

I was supprised to find ballast resistor on ignition of my 42, thought it was a 70's thing.

Does anyone know the reason for fitting a resistor in the circuit .

[Tinker]

Coil, stock resistor knocks it down from the 6v to around 3-4v at the coil.

The ceramic ones typically are used to run 12v on a stock ignition. Then it's 12v down to 6v (or so).

[V8COOPMAN]

Originally Posted by BouncinBetty

I was supprised to find ballast resistor on ignition of my 42, thought it was a 70's thing.

Does anyone know the reason for fitting a resistor in the circuit .

A good explanation from Bluebell in another thread:

"The Ford 6 volt system runs a 4 1/2 volt ignition coil.
What happens is that when you first turn the key on, the coil gets 6 volts. This means that when you hit the starter button and the starter drags the battery voltage down, you are still getting a good 4 1/2 volts to the coil, maybe as much as 6 volts. This means the ignition system is not robbed and probably gets a voltage boost while the engine is starting.
What happens though is that very quickly the resistor heats up and as it does so, it drops the voltage to the coil to the voltage that the coil is designed to run on.
It is a very simple system, and credit to Henry Ford, that he was a fan of simple .
The fact that his coils have not stood the test of time is a different thing."

Then, a little "icing on the cake" in a follow-on from Kevin Koates:

"Bluebell, 10 out of 10 for your explanation of the original Ford V8 ignition resistor and coil. There are a lot of explanations put forward about this very simple system ,most of which are incorrect. The .5 ohm Ford 18-12250 wirewound resistor is made from a special resistance wire and is designed and matched to go only with the original Ford V8 Bakelite ignition coils and NO OTHERS necessarily. Apart from overcoming the problem of the starter motor dropping the battery voltage whilst cranking the engine, and robbing the ignition coil of voltage the resistor also controlled the coil supply voltage and current at low and or higher engine RPMs. The distributer contacts opened and closed slower or faster depending on the RPMS of the engine and thus had the effect of the coil current draw going down and up as well. The resistors resistance increased and decreased with the varying current flow through it and there by evening out the performance of the ignition coil and protecting the primary windings from overheating at slower RPMs. Its only a simple resistor but understanding its several functions is sometimes difficult. Best to always use the original FORD part and not the reproduced ones which can vary quite a lot in their build quality and performance. Regards, Kevin."

[rotorwrench]

Marion Mallory of Mallory Electric was another close friend of Henry Ford. He and Henry worked together to design the initial coils for the 1932 V8 and subsequent coils through 1948. If Henry like someone and he could do business with them he would stay with them. In this case FoMoCo and Mallory Electric had a close relationship right up to Henry's death and on into the 50s there after. The Mallory plastic case coils had a unique design that required a ballast to control the current. This design may have been due to patent rights for other designs that Henry didn't want to deal with. In any case the system was a good design and as reliable as any on the market at the time. It's biggest draw back was that it was sensitive to static current flow when a person forgot to turn the ignition switch off and the points just happened to be closed. Many a coil bought the farm from incidents like this.

[big job]

Ok, then whats up with this. I have 6vts at the ign switch, I have 6vts at the coil.
Start it up and now have about 4vts at the coil the way it suppose to be. So I look
under the dash at the resistor and somebody put the out wire on the in wire
terminal (if you know what i mean) terminating the resistor. So whats dropping
the coil down to 4vts? The coil is a standard Napa non internal resisted??
Then checked my tractor (no resistor) 6vts stopped and 6vts running. checked
50 F6 no resistor same thing 6vt stopped and 6 running. I always questioned why
Ford used a resistor nobody else did Until 12vts came in the picture..? sam

[flatheadmurre]

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Originally Posted by big job

Ok, then whats up with this. I have 6vts at the ign switch, I have 6vts at the coil.
Start it up and now have about 4vts at the coil the way it suppose to be. So I look
under the dash at the resistor and somebody put the out wire on the in wire
terminal (if you know what i mean) terminating the resistor. So whats dropping
the coil down to 4vts? The coil is a standard Napa non internal resisted??
Then checked my tractor (no resistor) 6vts stopped and 6vts running. checked
50 F6 no resistor same thing 6vt stopped and 6 running. I always questioned why
Ford used a resistor nobody else did Until 12vts came in the picture..? sam

Either you have a coil with to low resistance that draws way to much current for the wiring or you have resistans in ignitionswitch that drops the voltage.

[rotorwrench]

Big Job
You didn't identify the year of the vehicle in your example of the bypassed resistor. Cars that have been converted to the can type coils don't need the resistor so it can be bypassed in that situation. Not every can type coil is the same. There are different primary winding resistance levels for these coils depending on the type of system it is designed for.

[koates]

big job, Well I would say when you did the voltage test at your coil with the engine not running and a reading of 6.0 volts, then the timing contacts on the distributer were open. Contacts have to be closed to complete the coil and resistor circuit to ground so that the resistor can heat up to reduce the voltage. You will then get 3 to 4 volts at the coil. The longer you leave the ignition switch on then the hotter the resistor will become and the lower the voltage. After a couple of minutes you might end up with about 2.8 volts (engine not running). Don't leave the ignition ON longer than 2 minutes without starting the engine. Regards. Kevin.

[flatheadmurre]

Originally Posted by koates

big job, Well I would say when you did the voltage test at your coil with the engine not running and a reading of 6.0 volts, then the timing contacts on the distributer were open. Contacts have to be closed to complete the coil and resistor circuit to ground so that the resistor can heat up to reduce the voltage. You will then get 3 to 4 volts at the coil. The longer you leave the ignition switch on then the hotter the resistor will become and the lower the voltage. After a couple of minutes you might end up with about 2.8 volts (engine not running). Don't leave the ignition ON longer than 2 minutes without starting the engine. Regards. Kevin.

He said resistor bypassed and Napa std coil.
So either a coil with to low resistance drawing more current then the wiring can take or a ignitionlock thats not good.
If it is a non internal resisted coil it is built to use an external resistor, check the resistance in the coil might be very low.

[big job]

Originally Posted by rotorwrench

Big Job
You didn't identify the year of the vehicle in your example of the bypassed resistor. Cars that have been converted to the can type coils don't need the resistor so it can be bypassed in that situation. Not every can type coil is the same. There are different primary winding resistance levels for these coils depending on the type of system it is designed for.

1946 V8 that had original square coil that had a melt down & I put
a regular coil. I was just curious so I took a Volt test but still can't figure
why volts drop while running. so be it I guess..?

[flatheadmurre]

Have you measured the resistance of the coil ?
Is it still 6v with engine not running and points closed ?

[JSeery]

Is the wiring original or does it have a resistive wire added ahead of the coil?