Voltage to coil on 39 Ford

[mjdunn]

I have a 39 Ford with the original 6 volt set up. I have 6 volts going to the resister and 1.8 coming out of the resister and 1.5 at the coil with the ignition on and the motor not running. Shouldn't I have 3.5 at the coil?

[JSeery]

Have you tried measuring the resistance through the ballast resistor? And yes, the voltage should be much higher than 1.8 volts.

[JSeery]

From a recent Bubba post:

Ignition “ System” divided into 6 volts of electrical balance…….

In a perfect world using ohms law. The ignition coil winding would offer .6 ohms, the resistor would offer .6 ohms .
This would equal 1.2 ohms of resistance allowing a current flow of 5 amps.

The wattage ( volts x amps) of the circuit would be divided equally dividing the heat build up caused by the resistance.

This is fine until we start the engine , then the voltage becomes higher ( like 7-8 volts), lets use 7 volts.
7 volts divided by 1.2 ohms would be 5.8 amps .
This increase in amps must be taken up somewhere! The increase flow would increase the heat in each resistance and the small gauge wire to the distributor would take up the rest. This lead is a special resistance lead designed to take up the surges allowing good point life!!!

Testing with volt drop the drop across the coil would be .6 ohms x 5.8 amps= volt drop of 3.4 volts , the same would work for the resistor with a voltage drop of 3.4 volts…….the increase of resistor heat caused by flow would drop another 1 volts.
Using 7 as a spec , the distributor lead and the contacts could see approx. .2-.4 of a volt in this final voltage drop….

Moral of the story is use the factory designed set up and live happily ever after. A regular coil will supply 20-30,000 volts. A flathead with low compression and a richer than normal air fuel mixture would never need this much voltage from the coil.

[JSeery]

And from another older post:

What you are looking for is ~ 3.5A through the points. This is calculated by using Current (A) = Voltage (V) divided by the Resistance (R) in the circuit, A = V/R. Measure the coil resistance (I will use 1.0 ohm for an example) and add it to the ballast resister value (say 1.4 ohm), so total resistance would be 2.4 ohms. Battery voltage should be ~6.4V, it will be less when cranking the engine and higher when being charged by the generator, ~7V.

For this example, A = 6.4V/2.4 ohms or 2.67A, that is too low which indicates too much resistance in the circuit. At 7V the current would be 2.92A, so still on the low side. If you use the same coil the resistance would remain at 1.0 ohm (again, this is just an example you would need to measure your coil), then you need to change the ballast resister.

If you used a 1.0 ohm ballast resistor the current would be A = 6.4V/2.0 ohms or 3.2A and running with the generator charging A = 7.0/2.0 or 3.5A.

These are just example numbers the higher the system current the hotter the spark but the quicker the points burn and the capacitors go. Like everything it is a compromise, 3.5 to 4.0A is a good range to shoot for.

[G.M.]

Originally Posted by mjdunn

I have a 39 Ford with the original 6 volt set up. I have 6 volts going to the resister and 1.8 coming out of the resister and 1.5 at the coil with the ignition on and the motor not running. Shouldn't I have 3.5 at the coil?

Read Voltage Readings post. G.M.

[mjdunn]

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After reading all of the above, it sounds like I need to replace the resister because I am not getting enough voltage to the coil. Do y'all agree that is what I need to do?

Thanks so much for all of the help.

[BUBBAS IGNITION]

WHOA !
A couple questions here , is the engine running ??

What resistor are you using and the resistance is ??

Lets do a voltage drop across the resistor from one side to the other???

Are you trying to fix a no start???

[BUBBAS IGNITION]

Some numbers in detail from a GM post......

Battery 6.3
I.S. B.S. 6.3
I.S C toR 6.2 Less indicates dirty contacts in I.S
R.W. from I.S.Same as above
R.W. to coil Switch on points open 6.2 Points closed min. 3.7
Below 3.7 when starter is energized voltage starter current pulls
coil voltage down to where spark isn't enough to fire plugs. Weak
spark is blown out from compression. Some times will fire just as
starter button is released. Will also start when car is pushed and
jump started in gear. Bad or small battery cables or dirty connections
cause big voltage drops.
G.M.

[mjdunn]

The car was not running when I did the test and I checked it at the resister. 6 volts going in and 1.8 coming out. It takes too long to start and it misses out at low speeds after the engine is warmed up. It runs perfect when engine is cold.

[mjdunn]

See attached photo

[George/Maine]

Those resisters are usually good or bad. I would start with new points and condenser and if still doesn't start good replace or rebuild the coil.

[mjdunn]

George, I checked the voltage going into the resister and coming out of the resister and it went from 6 volts to 1.8 volts. Wouldn't that isolate the problem to the resister? If I am only getting 1.8 volts out of the resister, how would changing other parts help?

[flatheadmurre]

If you have a short in the coil you get a voltage drop like this to...more amps drawn makes the drop over the resistor higher.
Is the resistor getting real hot ?

[JSeery]

mjdunn, any reason you don't want to check the resistance as has been suggested?

[mjdunn]

I guess I don't understand what you mean by "as suggested". I tested it going in and coming out and there was too much drop in voltage. As you can probably tell, I don't know much about electrical current.

[flatjack9]

Originally Posted by George/Maine

Those resisters are usually good or bad. I would start with new points and condenser and if still doesn't start good replace or rebuild the coil.

Not always true. I had one where the resistance increased and the engine would only start when I released the starter switch. It was spinning fast enough to start as the voltage went up when the button was released.

[JSeery]

Originally Posted by mjdunn

I guess I don't understand what you mean by "as suggested". I tested it going in and coming out and there was too much drop in voltage. As you can probably tell, I don't know much about electrical current.

Ok, now I see the issue. You are checking voltage. I (and others) were suggesting to also check the resistance. You are checking a resistor (a ballast resistor) and it has a resistance value. It is measured in ohms (Ω). Most voltmeters (multi-meters) can check voltage (V), resistance (Ω) and amperage (A).

Use of the Ω function can be very handy in trouble shooting. Zero Ω across a wire or object would indicate there is no resistance to current flow. Any number above zero is an indication of the resistance in that item. In this case a resistor should have a value that you can check, something like the 1.3 ohm range.

Based on your response I'm assuming most, if not all, of what I posted on the previous post were of little to no value to you, sorry about that!

[fordwife]

Are you checking resistence with the points open? 0r with the small dist wire unhooked? that would give an accurate reading.

[flatjack9]

Originally Posted by JSeery

Ok, now I see the issue. You are checking voltage. I (and others) were suggesting to also check the resistance. You are checking a resistor (a ballast resistor) and it has a resistance value. It is measured in ohms (Ω). Most voltmeters (multi-meters) can check voltage (V), resistance (Ω) and amperage (A).

Use of the Ω function can be very handy in trouble shooting. Zero Ω across a wire or object would indicate there is no resistance to current flow. Any number above zero is an indication of the resistance in that item.

You must disconnect the leads to the resistor to measure the resistance.

[JSeery]

Originally Posted by flatjack9

You must disconnect the leads to the resistor to measure the resistance.

Yep, good point flatjack.