Having problems with the starter solenoid output terminal going to case ground as pointed to by my lovely assistant

This is measured as a 1.4 ohm short as measured by a meter ..

This when I have the pigtail wire to the motor brushes lifted . Otherwise when connected the motor will engage and turn the engine /

The problem arises when the grounding path brings the voltage line down to 9.5 volts , not enough for the engine ECU to run for ignition .

From my understanding 100 % of the power from that output post should go down the pigtail wire .

Battery load tested on propper large load tester .

What I'm I not understanding ?





This is a print from the wiring guide .

The drawing in red square is the solenoid .

The Orange line is the unexpected ground path causing the problem .



And the big picture of the starter circuit .



I/m stuck on the one point of the oramge ground path should not exist .

Can someone walk me through this to get her started again . I worked electronics for the airlines so I have the skillsets

 

I'll need to check fusable link under the rear seat to cover that item

But it shows 9.5 volts down the line toward the ECU first power

 

I can't help at all with electrical issues but good to see you are recovering

 

Good to have you back Lady P! Where exactly are you measuring the 9.5 V? Right there at the 250A fuse (bad) or downstream nearer the ECU itself?
If downstream the path to ECU via RH front fuse box is a long and winding road, not sure why Jag put so many connections in the path to critical component like ECU. If you see battery voltage drop significantly along this path may be resistive losses and need to clean connections at Left and Right hand front bulkhead terminals for example

 

9.5 volts on the right terminal post with a meter on the firewall besides the instrument cluster showing the large sag

Ironically the 250 amp fuselink at the battery downturn does not blow ( with a 1.0 ohm short ) but that may be from the limited current avail from the battery although it was tested with a large load tester along with terminal shims to address stretched terminal clamps ..

Cleaned up the firewall terminals once before recently but now showing signs of rust after previously applying dielectric , Must be engine bay washing .

Will go through the whole right engine fuse box path again ( and the points you circled ) cleaning the terminal post along the way .

I can understand the mode of failure of my original starter solenoid , a new from stock solenoid with same results after full car installation . a different sources solenoid that failed bench shorting fault test , and a donor starter starter assembly failing a bench test before installing .

I can understand quality control missing in the manor in which the solenoids fail but passing in porting power down the pigtail , and the donor failing and the owner punting and sending the car to the salvage yard .

But I can't get past the point in how the starting system should work and % 100 of the power should go down that brush pigtail .

'Thanks for helping me on this as it stumps me .

 

Hi Parker, good to see that you are recovering.
I think that the starter solenoid has 2 windings -
1. Pull in winding using a higher current to initially energise the solenoid - resistance typically 0.4 ohms
2. Hold in winding to hold the solenoid in the energised position - resistance typically 1.2 ohms.
They are both connected at one end to the Control Connection ST1.
The other end of the pull in winding is connected to the starter brushes (the pigtail)
The other end of the hold in winding is connected to ground - hence your 1.4 ohm reading.
The wiring diagram for the solenoid actually shows there are 2 windings.
Measuring the resistance at the solenoid output with the starter brushes (pigtail) disconnected you are measuring both windings in series to ground.
When the solenoid is activated 1/3 of the current flows through the hold in winding and 2/3 through the pull in winding and this provides the initial impulse that is required to move the pinion rapidly out towards the flywheel ring gear. The pull in winding is also connected to the starter motor so 2/3 of the current is supplied to the starter motor. The pinion gear therefore begins to turn slowly as it moves forward. This aids engagement with the flywheel ring gear. When the pinion is engaged the main solenoid contacts are made. This connects 12V directly to the motor to spin it at full power and also bypasses the pull in winding. The pull in winding now has (effectively) the same voltage on both ends and therefore draws very little current so that the solenoid is now drawing 1/3 of the current when cranking and the solenoid is protected from overheating.
A = common end of both windings (Control Connection ST1)
B = hold in winding (welded to solenoid body)
C = pull in winding (connects to the solenoid output pigtail)

 

Thanks , I'll have to review and wrap my head around this solenoid information later

 

Here is a schematic that explains why you get the reading to ground.

 

Thanks for the reply on the business of how the specific solenoid works with the current paths and measurements . Now we're talking .

I'm assuming off the top of my head without measuring in this example the coils should measure around 30 ohms .

I break the problem of getting the 9,5 volts back to the 11.4 for ECU operation to get ignition / current supply as a whole car from the battery . a short somewhere on the car other then the specific solenoid , and the specific solenoid . ,The first one being separably worked on and the second short being doubtful based on whole car meter evidence of it not occurring until solenoid engagement .

I do have engine rotation so I have the solenoid physically moving and engaging the engine flywheel . and enough current going down the pigtail to the brushes for motor rotation . And that is a good thing /

It is only during this solenoid engagement when the voltage drops to 9.5 so it points too that .

Now a talk on the solenoid drawing and it's markings .

I see the current paths working but only if the voltage on the control line is sequenced off . This can happen from the Body Processor Module CM even though you have the key in the ignition switch held on the start detent .

Only if you remove the voltage force on the control wire would you have the current though the " pull in " coil to function .

Once you have that sequence you would have successful starter assembly operation .

I respectfully think the pull and hold markings are backwards I think as my brain works .

This poses the question on how the starter engagement is uncommanded to release by removing the control voltage .

But this still poses the problem of the short to case ground as the output post exits the green square with the pigtail lifted in the drawing that I can only explain as solenoid failure in my original and and donor starter assemblies and a manufacturing defect in acquired 2 separate solenoids .

Still thinking how to word this

Editing
I

 

I think I hot it figured out with your drawing in the current paths .

With B + voltage applied to the control wire by way of the starter relay commanded by the body processor module the top coil pulls the contact closed and causes power coming around gtom the large battery terminal to the bottom coil it sits their until the voltage force is relaxed by the BPM . This allows current to flow holding the latch coil . this accures while you still have the key in the start detent .

This makes the top coil the pulling coil and the bottom coil the latch .

During the moment of starter engagement and rotation along with engne block rotation there would be no voltage on the small control wire .


To stop the starter would require a voltage " shot " commanded by the BPM as you relax off the key start detent . This shot only has to last for a second ending in no voltage .By providing the shot there would be a blocking voltage force stopping the current though the bottom latch coil and the moment the control wire relaxes the current to the pulling coil goes to zero opening the solenoid contact .

Now I see how I would get a resistance reading ( thinking it is a faulting short ) at the tip of the large output post to case ( by lifting the pigtail off ) by going through both coils , but the value of my measured 1.6 ohms seams too low to be reasonable as a engineering comprimize vs. having % 100 of the current going down the pigtail .

Anyone welcome to correct this if wrong and sounds like a pile of poohoy or at least better wording .

Editing complete .

 

Hi Parker,

The solenoid does not latch. It is only activated as long as there is +B on the ST1 (Control) terminal from the starter relay when the ignition key is in the crank position and deactivates as soon as +B is removed. It does not require a pulse to reset.


A solenoid requires a high current to initially get it move but a much smaller current to hold it in the activated position. If the higher initial current is maintained the solenoid will overheat and burn out. By using two windings the high current can be supplied initially to pull in the solenoid and then one of the windings which has a higher resistance and therefore draws much less current holds the solenoid on. This reduces the current through the solenoid while it is activated and reduces the heat generated.

The top winding is the hold in winding because it is still active when the solenoid is fully engaged.
The lower winding is the pull in winding because it is only active until the solenoid is fully engaged.


On first turning the ignition key to the crank position ("pulling in" the solenoid)



When the solenoid is first activated with +B on the ST1 (Control) connection with the ignition key in the crank position both windings are energised. The current flows through the top winding to ground and through the lower winding and the starter motor to ground.

The solenoid starts to pull in and because there is a current through the lower winding and starter motor the motor starts to turn the pinion gear as it moves out to engage with the flywheel ring gear.


Starter motor cranking (solenoid fully "pulled in")



When the solenoid is fully engaged the main solenoid contacts are closed providing full cranking current directly to the starter motor. At this point the solenoid has fully "pulled in". The lower winding connection to the starter (pigtail) is now directly at +B as the starter motor is connected directly to the battery. The other end of the lower winding is also connected to +B on the ST1 connection (Control) because the ignition key is in the crank position. Because both ends of this winding are now at +B there is no current flow through this winding. It is now effectively out of circuit now that the solenoid has fully "pulled in".

The upper winding is still active as it has +B on the ST1 connection (Control) with the key in the crank position and the other end of the winding is permanently connected to ground. This is the hold in winding that holds the solenoid in the fully engaged position.


When the ignition key is no longer in the crank position +B is removed from ST1 (Control) connection. There is now no power to the upper winding and the solenoid is deactivated.

 

I can see how you could think that the lower winding was a "latch" but when fully engaged both ends of this winding are at +B so there is no current flow and it is not providing any power to the upper winding which is holding the solenoid engaged as long as there is +B on the ST1 (Control) connection while the key is in the crank position.

Although the pigtail end of the lower winding is still connected to +B when the key is released the current through the lower winding at this instant is effectively zero and it cannot increase fast enough to maintain +B to the upper winding to allow it to "latch on" before the solenoid releases.

The upper winding is the only load due to the solenoid itself on the battery and is only present during cranking. With a typical resistance of 1.2 ohms the current is around 10 Amps. This load is insignificant compared to the load imposed by the starter motor itself where the current would be at least 100 Amps.

Hope this explains how it works.

 

Thanks , canalizing your advice and mush appreciated