12 volt conversion question

Ok, got a question on a 12 volt alternator conversion. Instead of using a light or a diode to prevent backfeed on the alternator, has anyone ever
hooked the excite wire to the solenoid term? Just wondering if this would work? I will be making a new wiring harness for the tractor. It is a
Farmall M that I am putting an electric solenoid to start instead of the foot switch.

Neither of the small terminals on a "Ford" solenoid would be usable places, the alternator puts 12v to the excite wire when it begins charging. This would cause the start relay (solenoid) to stay on running the starter if attached to the S terminal. The I terminal would probably turn off too soon to excite the alternator. The wire is best fed through a 10 ohm resistor, or a diode, or a side marker or dash light. (not LED type), from the ignition side of the ignition resistor if it has one, or directly on the ignition switch side of the coil if the coil does not use a resistor. Jim

No, it wouldn't work unless you were willing to hold the starter in until the engine got up to enough speed for the alternator to start charging.

Read mine. Worse!! Jim

ive seen a bunch of them wired thru a oil pressure switch wouldn't need any lights diodes or resistors.. you could jump from your coil to the pressure switch then to the alternator I myself just like the little indicator light in the dash

The little idiot light is a nice thing to have. Why don't you put one in?

Yeah, I didn't think that one through...

Been fighting a cold, taking mind altering drugs! LOL

The pressure switch must be wired to the load side of the amp meter getting voltage all the time. If hooked to the coil it would keep the engine running (still
has oil pressure till it stops, and it won't shut off at the coil. The load side just keeps 12v at the pressure switch, no issue with the ignition or the
alternator. Jim

I knew you knew. Jim

Once it "excites" a 10SI or 12SI Delco-style alternator will backfeed current OUT of the #1 terminal, which would be fed to the solenoid coil or the starter itself (depending upon where you connected the wire) and likely burn out the alternator's internal voltage regulator and/or diode trio while trying to operate the solenoid or starter.

If the solenoid has an unused "R" terminal (sometimes used for ignition resistor bypass) your plan MIGHT work, if it generates enough residual magnetism in the alternator during cranking to help it "excite" upon startup.

"Idiot lights" are handy to have, and a good diode for the job costs less than a buck (in quantity)!

Just use a double pole switch, one side to ignition, the side to alt, got a couple tractors like that, it works for me

I did my conversion before the Internet, no easy information available. I just used a push button switch to exite the alternator, and have since then bought another tractor that was done the same way.

Sure makes you feel stupid if you forget to push the button sometimes.

Why over complicate something so perfectly simple?

SMH.

We just put them on the ignition switch on when on and off when off. No special switches and no forgetting that way.

The ignition switch will turn off, but the regulator in the alternator will feed power to the coil side of the ""off"" switch keeping them running. It needs a 10 ohm resistor or a diode or a lamp filament in the circuit to keep the ignition from operating. I don't know what your solution is used on, but farmalls will keep running. Jim

(quoted from post at 17:44:55 02/04/19) The ignition switch will turn off, but the regulator in the alternator will feed power to the coil side of the ""off"" switch keeping them running. It needs a 10 ohm resistor or a diode or a lamp filament in the circuit to keep the ignition from operating. I don't know what your solution is used on, but farmalls will keep running. Jim

Click to expand...

aterpillar guy is probably running diesels

I fixed a one wire alternator by doing that

Ok, here is a question for you. I do not get how it would burn it out. It prevents back feeding with a light, because not enough power to light the lamp, why would a solenoid coil cause a larger draw? The resistance in the coil would accomplish the same thing as the filament in the bulb, or am I missing something?

(quoted from post at 20:50:16 02/04/19) Ok, here is a question for you. I do not get how it would burn it out. It prevents back feeding with a light, because not enough power to light the lamp, why would a solenoid coil cause a larger draw? The resistance in the coil would accomplish the same thing as the filament in the bulb, or am I missing something?

Click to expand...

With an indicator light, one side is powered when the ignition switch is "on" and the other goes to the alternator #1 terminal.

When the engine is running and the alternator is charging, the voltage at each terminal of the lamp is nearly equal, so the light does not.




After cranking ends, the solenoid terminal is at chassis ground potential.

If you were to connect the alternator's #1 (excite)terminal which is at battery voltage terminal to the starter solenoid coil, one of two things would happen, Either the starter solenoid would "pull in" or if the alternator's diode trio and circuit tracings in the voltage regulator can't handle the necessary current, $$$smoke will be let out.

Neither having the starer engage with the engine running, or burning out the diode trio and/or the voltage regulator sound to me like a good plan!

But try it, if you must!

Also, you wrote "The resistance in the coil would accomplish the same thing as the filament in the bulb, or am I missing something?"

The resistance of an indicator lamp is HUGELY greater than that of a solenoid coil, and they current they draw is just the opposite.

A properly connected indicator lamp draws a small current and lights when the engine is NOT running.

Connected as you propose, the solenoid coil would draw a relatively LARGE current once then engine starts and the alternator "excites".