Starting circuit question “coil energization”

[Moparfiend]

After the starting circuit has ended (key no longer cranking) does the + side of the coil float or does that stay at 12V?

It seems to me it needs to have 12V but I also read that when you come out of cranking to Ign1 state that the 12V is cut off the coil/ balast resistor.

I am asking because I may want to hot-wite the car to test out all the non-electrical work I have done. So not sure if I leave that balast resistor connected to 12V while running or if I should pull it after it starts....

Thanks for yiur help on this!

 

[Photon440]

After the key is in the run position, the coil resistor is no longer bypassed and the voltage drops., typically to about 9 volts.

 

[JimKueneman]

Depends on if you have sacrificial set of points ...... You can run it for a bit at 12V but it will eat up the points fairly fast.

 

[Moparfiend]

So after it starts if I drip it to 9V I should be OK?

Your comments is with the resistor in series right?

I have a high current DC power supply I canbuse to drive this if needed.

I am going to run points.....cant afford an MSD yet..

 

[Dragon Slayer]

Why? The Ballast resistor protects the coil from overheating and burning out. The ignition switch in run position supplies 12V to the ballast resistor which is about .5ohms. When you place the ignition in start the 12V is still at the front of the ballast resistor, but a wire on the coil end gets 12V just to ensure good spark for the start. Why not leave it alone?

 

[JimKueneman]

Dropping the voltage is a bit different. The resistor will vary the voltage on the coil as a function of current as the coil charges. If you look a the negative you will see battery voltage on the coil until the points close. It then starts to drop with a linear slope as the current ramps up and more voltage is dropped across the resistor. I would just put the resistor in and dial the power supply up to 14V and back down to 12 V once it starts. The bypass is because the voltage drops to way below normal when cranking. If you have a voltage that won’t drop when cranking you don’t need the bypass anyway

 

[JimKueneman]

And if your doing what I think you are make sure you remove the voltage when not cranking it. If you have a supply on the coil and the engine stops in a place where the points are closed you will cook the coil, resistor and points if it sits there to long

 

[Moparfiend]

Why? The Ballast resistor product the coil from overheating and burning out. The ignition switch in run position supplies 12V to the ballast resistor which is about .5ohms. When you place the ignition in start the 12V is still at the front of the ballast resistor, but a wire on the coil end gets 12V just to ensure good spark for the start. Why not leave it alone?

 

[Moparfiend]

Hum ok Jim so is the voltage to the coil should not be present long if the engine is not rotating (either cranking or running) because the current will overdissipate through the points/ignition circuit path correct?

 

[JimKueneman]

Hum ok Jim so is the voltage to the coil should not be present long if the engine is not rotating (either cranking or running) because the current will overdissipate through the points/ignition circuit path correct?

If it stops with the points in a closed position. Same can happen if you have the key on to long in that state.

 

[Dragon Slayer]

Yes kind of. For electrical power with basic DC resistive circuits you have 2 formulas. V=IR and Power =VxI in watts. V =volts, I =amps and R = resistance in ohms.
The coil primary is about 1 ohms, but it is more like an inductor, it also is in a oil bath for cooling.

So if we add the .5ohm ballast resistor with the 1 ohm coil we have a 1.5ohm resistive circuit. The points are really a switch. Open or closed. But if they are dirty partially closed they can add another resistive measurement to this circuit.

If points are open you have no path to complete circuit and the coil has 12V on it because there is no current flow hence no voltage drop over the resistors.

When the points close you have 12 divided by 1.5 ohms or 8 amps flowing. When 8 amps flow through the ballast resistor you drop 4V. 8amps x .5 ohms. SO the coil sees 8V. Calculating power the ballast resistor drops 4V at 8 amps or 32 watts. The coil drops the remaining 8V at 8 amps or 64 watts. In a DC series circuit current flow of the total circuit flows through all components. Total energy dissipated is 96watts (8amps x 12V). Think light bulb. The points are a switch at near 0 ohms. 8 x 0 is 0 volts so the power dissipated is 0 watts. But lets say your points have about .2ohms resistance. You will see 8amp x.2 ohms or about 1.6Watts heat at the points.

So none of this will be fast, you will just have components getting warm, as long as you are not pushing high volts directly to the coil. If you bypass the ballast and put 14v directly on the coil you would wind up with 14V and 14Amps (14V divided by 1 ohm coil) or 196Watts all dissipated by the coil. So instead of 64 watt bulb you have a 200watt bulb.