Ballast resistor question

[Nowik35]

Little back story. When I first finished wiring the bronco, ballast resistor read 7 volts going to coil.
I have it wired traditionally. I have the i terminal from Start solenoid going to the coil directly and the run wire from the ignition through the ballast resistor also going to the coil. One day when the car was on run position, the engine randomly started cranking. I had to pull the battery. I disconnected the starter completely and tried this again couple days later. After a few minutes I start to see smoke coming from the ballast resistor and it gets super hot. When I check the voltage at the i terminal of the solenoid and at the coil it only re reads .5 volts. The battery reads regular 12.3 volts.
Question one.... Why is the ballast resistor drop almost 11.5 volts.
Question 2 ..... Is it possible the resistor failed and it supplied 12v to the i terminal of solenoid and that's why my engine cranked over magically for a few seconds??
Can I wire the starter directly to the coil and not use the i terminal to supply the 12v during start to the coil?

 

[Jdgephar]

The coil primary may have a short, pulling too much current and causing the resistor to get hot, and the voltage too low. That, or a shorted wire to the coil.

I terminal shoild only have 12v when the key is in start. Otherwise it will be the same voltage as the coil primary terminal in the run position.

You need the I terminal to provide power to the coil in the start position. If you remove it, you won't get spark while starting.

There could be an issue at the ignition switch that caused the starter to be stuck on.

You can disconnect the resistor and then start checking voltage in the different key positions. This will make sure the starter relay only has 12v in the start key position to rule that out. Then check that you have 12v on the I terminal out with the key in start. Then check that it is 0 in run. Next see if the wire to the resistor is 12v in run.

You can also measure the resistance of the coil primary to see if it is in spec.

Sent from my SM-S916U using Tapatalk

 

[Oldtimer]

When engine is running, current thru coil cycles on & off with firing of each spark plug.
If key is in RUN position AND engine is not running it is possible for the coil to be connected 100% of the time.
The constant power to coil may have;
1) caused coil to over heat
2) caused ballast resistor to burn out (hence .5 volt reading)
3) caused ignition switch contacts to get hot (causing starter circuit to engage)

There is a reason ignition switches have an ACCESSORY position.

 

[Broncobowsher]

There are 3 ways the starter can engage the and start cranking without turning the key to the start position.
The starter solenoid fails internally. Stray leakage to the S post internal to the solenoid,
Ignition swatch failed
Harness issue with a short from the S wire to a power source.

The ballast resistor is a heating element. Watts of heat are disposed of in the form of heat. The wattage is based on the voltage drop across the resistor and the amperage. In the case of an ignition system, as the points open and close, there is a duty cycle involved as well. If the points are closed (engine off) the load is part ignition coil and part ballast resistor. You have 7V at the coil, the coil is dropping 7V to ground and the ballast resistor is dropping the other 5V. (I'm ignoring exact voltages and other voltage drops from the wiring, connectors, the points, etc. as they are generally too small to matter unless there is a big problem). If the ignition system is pulling 1 amp (number picked just because it makes the math easy) the ballast resistor is a 5 watt load (5 volt drop x 1 amp current, volts x amps = watts). The coil is 7 watts (7 volts, 1 amp).
Watt is a watt is a watt. It happens that the unit of energy is also a unit of heat. If you have a 5 watt light bulb, that is the same amount of heat.

Now if the coil shorts out. You measured 0.5V, effectivly a dead short. The ballast resistor is now dropping all 12V. Sounds like you should be dealing with 12W of heat. But things don't actually work that way. The resistance was 2 items in series (coil and ballast), and it is now just one. I'm going to get a little more math lazy here, the resistance dropped in half. Now you are dealing with 12V and 2 amps. As the resistance drops the amperage goes up. That ballast resistor is now dealing with 24 watts. That is 5X the heat it would normally deal with. That is assuming it is only a 1A draw to start with. If it was 50% more, just 1.5A, the end result is 50% more as well.

Seeing a ballast resistor get nuclear hot and start burning up is a pretty good sign the coil (or coil power wire) is shorted to ground.

 

[Nowik35]

Seeing a ballast resistor get nuclear hot and start burning up is a pretty good sign the coil (or coil power wire) is shorted to ground.

Bingo! I found the root cause..( kind of a dumb one). So after I finished wiring the resistor to the coil and everything worked. Months later I took my valve cover off and when I reinstalled it I didn't know but I pinched the coil power wire by the corner of the cover and shorted it to the engine block...... So that means my coil wasn't getting any power and my resistor was dumping all 12 volts into heat. I just fixed the wire tonight and everything works as it should... But I found another issue ( it never ends)
On "run" my relay s terminal is 0v and I terminal is 7.... Great, when I had my son hold the key on start, relay clicked nice and load and I get 12v at the s and still only 7 at the I. So I removed the coil wire off of I terminal and still in start position my I terminal is 0v and s terminal 12v and it still clicks no problem with nothing hooked up to the i terminal. Is my relay shot? The relay is brand new never used on a running engine ...... Any ideas?

 

[DirtDonk]

Yes. It’s just faulty right out of the box. Essentially it’s shot. And also very common.
What brand and part number did you buy?
Did you go cheap and with a warranty? That’s typically the scenario where you might get two, three, or even four defective ones in a row.
Some may actually work for a couple of starts. But the failure rate is far higher than the free warranty replacements make it worth.
Spend big money on a USA built, or marine certified (Volvo/Penta comes to mind), or anything that seems of higher quality.

That’s the only reason I can think of for you to get a 0 V reading while the relay is energized by the “S” wire.

 

[Broncobowsher]

Might have fried the I-wire inside the solenoid when the ignition coil wire was shorted to ground.
It may have been a good part, damaged when the output was shorted. In which case it is a good design as it acted as a fuse before burning out the harness.

 

[Nowik35]

Ok on my 4th solenoid everything works fine as it should, now my ballast resistor is acting up again. Actually I have 3 , all 3 doing the same thing, meaning 12v goes in and 12v goes out on the other side. I'm so confused. It can't be the wiring because, I disconnected everything and have a 12v hot wire going to one side of resistor, and I just have the multimeter on the other side reading 12v . On all 3 resistors. Am I that unlucky with 3 defective resistors in my possession????? When I check the resistance across all 3 , they all read 1.8 ohms.....

 

[Nowik35]

I might of just answered my own question. All my spark plugs are currently out so I have an open circuit at the distributor. I assume that's what's causing no voltage drop across the resistor?

 

[Oldtimer]

What is voltage reading of battery?

 

[Nowik35]

12.3

 

[DirtDonk]

I thought we had talked about it earlier in this thread. But maybe it was another one somewhere.
The voltmeter does not pull enough current to drop the voltage. You can read voltage across the resistor like this, and it will read full voltage until a load is applied, such as with the ignition running.
So, don’t determine good or bad based on just a voltmeter reading. Get the thing running and then after a few seconds, take your reading and see if the voltage goes down.

 

[DirtDonk]

Another way to test would be to leave the ignition connected, and if it’s a points ignition, simply rotate the engine until the points are closed.
At this point, with the key on, the coil is pulling current and heating up. It should also be enough load to drop the voltage across the resistor.
The same can be done with basic electronic ignitions, but I’m not sure where the trigger has to be to accomplish this.

 

[Jdgephar]

I might of just answered my own question. All my spark plugs are currently out so I have an open circuit at the distributor. I assume that's what's causing no voltage drop across the resistor?

Both sides of the resistor will measure 12V when the circuit is open. Now, what is open? The next item in your system is the coil. Assuming the coil is good, then the next component is the points in the distributor. The points open and close all time as the motor runs. With the points closed, the circuit is complete, and you will be able to measure the voltage drop at the resistor. On Duraspark systems, the module has a circuit that opens and closes the coil.

There's your quick 1st semester of Electrical Engineering 101.

Bottom line, your resistors are fine.