intermittent starter or ignition issue?, where to look next? Options

[DRPHIL914]

So this issue has happened 3x now in the past few months, and now today after driving 300 miles up to Brevard, NC on the way , stopped to get gas, going to start it, no crank. all other electronics work, fuel pump priming just no starter engagement. it’s happened cold too, i can get it to start by push start.. so about an hour ago i pulled the steering wheel and replaced the plastic switch still no starter , i also checked the yellow wires under passenger seat (seat buzzer by-pass) , i don’t think that’s the issue.

I bought and installed a new Bosch starter about 6 months ago and installed it on this newer motor a few months back, so now i am wondering if it’s a bad solenoid? wond not crank . the last time it happened was a months ago, push start got me home from the store, and when i got home and pulled into the garage after turning it off it started right back up and has for the next 10 times —- until tonight.

what are the other possibilities? does the power go from the ignition switch to the relay board and then to the starter? i have done a few searches , but looking for someone that’s maybe had this issue or understands the whole schematic better . i don’t have a multimeter with me but i can get one tomorrow at O’Reilly’s and i might see if they can get a starter., i’m just guessing and grasping at straws ..

- oh, for those that don’t know it’s a ‘75 2056 d-jet.

Phil

[r_towle]

yellow wire under the passenger seat...
Is it still going through the "logic circuit" box under the seat?
If yes, I would strongly suggest you remove the two yellow wires and directly connect them.

Grounds, go to the parts store and get yourself a 36-40" ground cable, with bolt tabs (female) on both ends.
Bolt on end to the large upper starter bolts that physically bolt the starter to the transmission.

Bolt the other end to the negative terminal at the battery at the bolt that tightens the battery terminal.
Leave the battery terminal connected to the normal body mount so everything else works.

Remove any doubt with this method.
If it solves it, you need a new ground strap.

I have built a starter button for testing things...and hotwiring cars from the engine bay.
Its simply a 30amp push button switch with two long leads.
One goes on the battery positive terminal, the other goes on the starter at the little lead (yellow wire) terminal.

Side of the road, scary yet effective.
Use a quarter... (ed showed me this) and in neutral..get under the car on the drivers side.
Use a quarter to touch the large positive terminal on the starter to the yellow wire terminal at the starter.
Car will start.
Neutral....ebrake on (or block the wheels)

This shows you if the starter itself may be at fault when hot.
If it works fine with the hand held switch, you have wiring issues.
If it does not work..you have a starter issue.

If you add the ground strap as I have described (do it how I am describing) and your starter works when hot...you have a ground strap issue.

Eliminate the issue of heat creating resistance by going all the way around the stock wiring in both cases to determine where the issue may be.

I am a fan of the aftermarket relay to help eliminate the issue, but if you fix the root cause, the relay is more of a way to protect the NLA ignition switch from melting.

the OEM design is running something like 20 amps (now 30 amps) from the battery, up through the ignition switch (failure point) back down under the passenger seat to that "logic circuit" (failure point) and out to the starter.

the logic circuit is all designed so you cannot start the car if the seat belts are not engaged, and it fails.
Thus, wire around it.

the old wiring we have gets corroded inside, thus can no longer move 20 amps...and the starter may pull 30+ amps...which can Melt the wire somewhere you cannot see, or melt the ignition switch.

Method 1) add relay at starter location that now moves the 20 amps through new wires, directly from the battery to the relay. Use the stock yellow wire to trigger it (now 1-3 amps)
Method 2) replace the chassis ground solution (again..30 amps now) with a direct connect ground strap from the battery to the starter body.

These two methods are cheaper/easier (arguable) than fixing it properly.
Fixing it properly may be a new wiring harness, eliminate the logic circuit , and a brand new ground strap between the transmission and the chassis (and clean the ground from battery to chassis)


Rich

[Superhawk996]

the old wiring we have gets corroded inside, thus can no longer move 20 amps...and the starter may pull 30+ amps...which can Melt the wire somewhere you cannot see, or melt the ignition switch.

Rich

I’m sorry to have to point this out, but a circuit that was previously designed to draw 20A (as in example) will not suddenly draw 30A due to an increase in wiring resistance increasing from corrosion. This is physically impossible as dictated by Ohms Law.


However, I’m in rabid agreement with Rich’s suggestion to 1/2 split the system by basically hot wiring the starter B+ to the yellow solenoid control. Whether you do that with a screwdriver or a remote starter switch doesn’t make a bit of difference in the end. You then know if the problem is in the starter solenoid assembly or in the wiring and can focus on the appropriate solution.

[r_towle]

the old wiring we have gets corroded inside, thus can no longer move 20 amps...and the starter may pull 30+ amps...which can Melt the wire somewhere you cannot see, or melt the ignition switch.

Rich

I’m sorry to have to point this out, but a circuit that was previously designed to draw 20A (as in example) will not suddenly draw 30A due to an increase in wiring resistance increasing from corrosion. This is physically impossible as dictated by Ohms Law.


However, I’m in rabid agreement with Rich’s suggestion to 1/2 split the system by basically hot wiring the starter B+ to the yellow solenoid control. Whether you do that with a screwdriver or a remote starter switch doesn’t make a bit of difference in the end. You then know if the problem is in the starter solenoid assembly or in the wiring and can focus on the appropriate solution.

This is why contribute our suggestions in the forum to help others correct us.
Its also why (DM me or give me a call) is not the way to help others.

a 20 amp circuit design will indeed melt wires that are designed to support 20 amps.
I do understand ohms law...but I was trying to explain things in a public forum , of non electrical engineers, to help.

Happy to hear criticism, but I am also quite comfortable in my own skin to help others in a public forum, so others can refine the answer and create a positive outcome.

Rich

[Superhawk996]

the old wiring we have gets corroded inside, thus can no longer move 20 amps...and the starter may pull 30+ amps...which can Melt the wire somewhere you cannot see, or melt the ignition switch.

Rich

I’m sorry to have to point this out, but a circuit that was previously designed to draw 20A (as in example) will not suddenly draw 30A due to an increase in wiring resistance increasing from corrosion. This is physically impossible as dictated by Ohms Law.


However, I’m in rabid agreement with Rich’s suggestion to 1/2 split the system by basically hot wiring the starter B+ to the yellow solenoid control. Whether you do that with a screwdriver or a remote starter switch doesn’t make a bit of difference in the end. You then know if the problem is in the starter solenoid assembly or in the wiring and can focus on the appropriate solution.

This is why contribute our suggestions in the forum to help others correct us.
Its also why (DM me or give me a call) is not the way to help others.

a 20 amp circuit design will indeed melt wires that are designed to support 20 amps.
I do understand ohms law...but I was trying to explain things in a public forum , of non electrical engineers, to help.

Happy to hear criticism, but I am also quite comfortable in my own skin to help others in a public forum, so others can refine the answer and create a positive outcome.

Rich

As am I. So let’s help others by understanding what is going on.

So let’s be clear that a circuit designed to pull 20A will not melt when pulling 20A through good wire. It can actually melt when pulling LESS than 20A through degraded wiring.

How?

Let’s start with ohms law and a 12v system

To draw 20A, how much resistance was in the circuit as designed?
12v / 20A = 0.6 ohms of resistance
This 0.6 ohms of resistance would represent normal solenoid resistance + the wiring resistance which should be almost, but not quite equal to 0.

So now let’s say the wiring has corroded and increased to 0.5 ohms of resistance by itself. We now have a circuit with 0.5 ohms of wiring resistance + 0.6 ohms of solenoid resistance. This is 1.1 ohms of total resistance.

So how much current will we pull?

12v / 1.1 ohms = 10.9A

Circuit is now unable to pull full current. In fact, it is only pulling slightly more than 1/2 of the design current. So if in this example, the solenoid needs 20A of current to operate the solenoid, it can’t pull that current and the starter solenoid won’t energise properly. This will be true regardless of how many grounds are attached to the solenoid.

So why does the wiring melt when pulling less than 20A it was designed to pull?

The power (I.e heat) the circuit has to to dissipate is Power = current ^2 (squared) * Resistance

When wiring is near 0 ohms, there is no heat created . This is the design condition of healthy wire.

But we have 0.5 ohms of resistance, internal to the wire, created by corrosion.

So how much power (heat) will be created by pullling that 10.9A though the corroded wire?

Power =10.9A * 10.9A * 0.5 ohms
Power = 59 watts

So now we are asking a wire to dissipate 59 watts of heat. Remembering that it was not designed to do this. 59 watts may not sound like much but it is quite a bit. Trying to pick up a 10 watt ceramic resistor that is operating normally will be quite hot to the touch and uncomfortable to hold. We are talking about roughly 6 times hotter.

So this is how wiring melts from corrosion while actually pulling LESS current than what the circuit was originally designed for.