BX2200 intermittant ignition failure

   / BX2200 intermittant ignition failure #12  
Hi,

I just received my John Deere AM107421 Starter Relay Kit AM107421.

First thing I did was do a quick check of relay operation using my battery charger. Unfortunately I do not have a variable voltage DC supply, but I do have a true RMS reading Fluke meter. I found that using the 6 volt setting on the charger, that the relay has a solid pickup at 8 VDC and will stay energized at 5 VDC. These are the voltages supplied by my battery charger on its two 6 volt settings. Operation may be a little different with ripple free DC as supplied by a battery, but I think my results are representative.

The important thing is that whatever relay is used, it should be rated for 12 VDC but be capable of energizing at voltage lower than 12 VDC, to compensate for voltage drop across the safety circuit switch contacts.

Those relays KennyV pointed to look like good 12 VDC relays. But if they need close to 12 VDC to activate, then they may not work for the purpose of compensating for voltage drop across the various contacts in the safety circuit feeding the relay coil.

Bill
 
   / BX2200 intermittant ignition failure #13  
I installed the John Deere relay. Plugged right in and I have not had a problem since.
Good luck with it.
 
   / BX2200 intermittant ignition failure #14  
I got curious and found a data sheet for the JD kit relay.

http://www.onlinecomponents.com/datasheet/vf4a55f11s01.aspx?p=12120483

Tyco electronics VF4-55F11-S01 is the part number of the relay that came in my kit.

The second "F" in the part number designates a relay coil with a "Must operate" voltage of 7.2 VDC, and a "Must release" voltage of 1.2 VDC. So this relay will work even is there would be about 5 VDC lost across the safety circuit contacts...

Just posting this for future reference if anyone should be interested.

Bill
 
   / BX2200 intermittant ignition failure #15  
Hi,

I just received my John Deere AM107421 Starter Relay Kit AM107421.

First thing I did was do a quick check of relay operation using my battery charger. Unfortunately I do not have a variable voltage DC supply, but I do have a true RMS reading Fluke meter. I found that using the 6 volt setting on the charger, that the relay has a solid pickup at 8 VDC and will stay energized at 5 VDC. These are the voltages supplied by my battery charger on its two 6 volt settings. Operation may be a little different with ripple free DC as supplied by a battery, but I think my results are representative.

The important thing is that whatever relay is used, it should be rated for 12 VDC but be capable of energizing at voltage lower than 12 VDC, to compensate for voltage drop across the safety circuit switch contacts.

Those relays KennyV pointed to look like good 12 VDC relays. But if they need close to 12 VDC to activate, then they may not work for the purpose of compensating for voltage drop across the various contacts in the safety circuit feeding the relay coil.

Bill
The circuit loss is due to the relatively large current [Amps] drawn by the solenoid coil. You will be feeding this coil thru the closed relay. Your IG switch circuit will only see the small current the relay coil draws [fractional Amps]. Thus the relay coil will see full OCV [>12] until the starter kicks and pulls the whole system down a V or so. So no problem with activating or holding Voltage.
 
   / BX2200 intermittant ignition failure #16  
The circuit loss is due to the relatively large current [Amps] drawn by the solenoid coil. You will be feeding this coil thru the closed relay. Your IG switch circuit will only see the small current the relay coil draws [fractional Amps]. Thus the relay coil will see full OCV [>12] until the starter kicks and pulls the whole system down a V or so. So no problem with activating or holding Voltage.

It is not exactly clear to me what you are trying to express...but when you say there is "no problem with activating or holding Voltage." I can tell you that this portion of your statement is denied both by electrical theory and practical experience of members here at TBN.

Specifically, a relay coil that is activated through a series string of contacts, whatever those contacts may be, will have a voltage impressed across its operating coil that is equal to the value of the voltage applied to the circuit, MINUS the voltage drop across each contact that is in the series string. This is fact not fiction.

Historically, in the archives here over the past ten years or so, there have been multiple posts regarding the starter motor solenoid just clicking and NOT pulling in enough physically to engage the starter motor. The reported final cure has been to install a relay that will pick up at lower voltage than the factory equipment, to be activated by the circuit that originally was hooked to the starter solenoid terminal. This is the one in the JD kit. (but as I think you may be saying, different relays that pull lower current may also work) Then contacts from this new relay were inserted directly between the battery terminal on the starter, and the starter solenoid terminal. This way full battery voltage was applied to the starter solenoid terminal when the ignition switch was turned to the start position( provided that all permissive contacts in the safety circuit were closed).

The relatively large current that you refer to I assume is that required by the starter motor solenoid. This high current is what causes relatively large voltage drop to develop across contacts in the safety circuit that have deteriorated somewhat over time. The new relay requires less current that the starter solenoid for proper operation. Therefore there is relatively less voltage drop lost across contacts in the safety circuit, and a greater percentage of applied voltage is felt across the relay coil. Any 12 VDC relay that draws less current than the starter motor solenoid may work. Having a coil rated at 12 VDC that will operate at 7.2 VDC gives an even greater margin or error to the benefit of proper operation. Of course the power circuit, between the battery terminal of the starter motor and the starter motor solenoid, in both cases will have only ONE contact in use.

It is that simple. Any relay that operates at 12 may work; one that works between 7 and 12 volts can only be better in the long run.

Spyderlk, I am not sure if you are saying that any 12 volt relay would work for this modification; there certainly is a chance this could be the case, at least in some instances. A new small relay would likely draw less current than the starter motor solenoid does. Less current through the safety circuit would result in less voltage drop across the problem contacts.

However, a relay coil that both draws less current AND activates at a voltage significantly less than the 12 VDC battery voltage will always be superior in this application, especially since contacts that have increased there resistance over time will likely continue to worsen.

This post is an attempt to ensure that the reasoning behind this proven fix is reviewed (if anyone cares to). I installed the JD relay in my BX2200 yesterday and it works perfectly. No more "Click, click, click....start!" I am expecting this to be the case even as the safety circuit contacts worsen.

If I had a suitable standard 12 VDC relay on hand to use in this application I probably would have tried that first. Why not, the price would be right and it might be enough to do the job. But since I did not I decided to maximize my odds over the long run, and buy a relay that offered greater potential long term benefit... :)
 
   / BX2200 intermittant ignition failure #17  
Hi,

I just received my John Deere AM107421 Starter Relay Kit AM107421.

First thing I did was do a quick check of relay operation using my battery charger. Unfortunately I do not have a variable voltage DC supply, but I do have a true RMS reading Fluke meter. I found that using the 6 volt setting on the charger, that the relay has a solid pickup at 8 VDC and will stay energized at 5 VDC. These are the voltages supplied by my battery charger on its two 6 volt settings. Operation may be a little different with ripple free DC as supplied by a battery, but I think my results are representative.

The important thing is that whatever relay is used, it should be rated for 12 VDC but be capable of energizing at voltage lower than 12 VDC, [to compensate for voltage drop across the safety circuit switch contacts.]

Those relays KennyV pointed to look like good 12 VDC relays. But if they need close to 12 VDC to activate, then they may not work for the purpose of compensating for voltage drop across the various contacts in the safety circuit feeding the relay coil.

Bill

The circuit loss is due to the relatively large current [Amps] drawn by the solenoid coil. You will be feeding this coil thru the closed relay. Your IG switch circuit will only see the small current the relay coil draws [fractional Amps]. Thus the relay coil will see full OCV [>12] until the starter kicks and pulls the whole system down a V or so. So no problem with activating or holding Voltage.

It is not exactly clear to me what you are trying to express...but when you say there is "no problem with activating or holding Voltage." I can tell you that this portion of your statement is denied both by electrical theory and practical experience of members here at TBN.

Specifically, a relay coil that is activated through a series string of contacts, whatever those contacts may be, will have a voltage impressed across its operating coil that is equal to the value of the voltage applied to the circuit, MINUS the voltage drop across each contact that is in the series string. This is fact not fiction.

Historically, in the archives here over the past ten years or so, there have been multiple posts regarding the starter motor solenoid just clicking and NOT pulling in enough physically to engage the starter motor. The reported final cure has been to install a relay that will pick up at lower voltage than the factory equipment, to be activated by the circuit that originally was hooked to the starter solenoid terminal. This is the one in the JD kit. (but as I think you may be saying, different relays that pull lower current may also work) Then contacts from this new relay were inserted directly between the battery terminal on the starter, and the starter solenoid terminal. This way full battery voltage was applied to the starter solenoid terminal when the ignition switch was turned to the start position( provided that all permissive contacts in the safety circuit were closed).

The relatively large current that you refer to I assume is that required by the starter motor solenoid. This high current is what causes relatively large voltage drop to develop across contacts in the safety circuit that have deteriorated somewhat over time. The new relay requires less current that the starter solenoid for proper operation. Therefore there is relatively less voltage drop lost across contacts in the safety circuit, and a greater percentage of applied voltage is felt across the relay coil. Any 12 VDC relay that draws less current than the starter motor solenoid may work. Having a coil rated at 12 VDC that will operate at 7.2 VDC gives an even greater margin or error to the benefit of proper operation. Of course the power circuit, between the battery terminal of the starter motor and the starter motor solenoid, in both cases will have only ONE contact in use.

It is that simple. Any relay that operates at 12 may work; one that works between 7 and 12 volts can only be better in the long run.

Spyderlk, I am not sure if you are saying that any 12 volt relay would work for this modification; there certainly is a chance this could be the case, at least in some instances. A new small relay would likely draw less current than the starter motor solenoid does. Less current through the safety circuit would result in less voltage drop across the problem contacts.

However, a relay coil that both draws less current AND activates at a voltage significantly less than the 12 VDC battery voltage will always be superior in this application, especially since contacts that have increased there resistance over time will likely continue to worsen.

This post is an attempt to ensure that the reasoning behind this proven fix is reviewed (if anyone cares to). I installed the JD relay in my BX2200 yesterday and it works perfectly. No more "Click, click, click....start!" I am expecting this to be the case even as the safety circuit contacts worsen.

If I had a suitable standard 12 VDC relay on hand to use in this application I probably would have tried that first. Why not, the price would be right and it might be enough to do the job. But since I did not I decided to maximize my odds over the long run, and buy a relay that offered greater potential long term benefit... :)
Youve introduced lots of confusing factors without saying anything wrong. ... The succinct point is that you dont lose voltage thru contacts and wiring that is not carrying current. And the current to the relay coil is very low.
E = IR. 1 Amp thru one Ohm uses a Volt, 2A and 1 Ohm loses 2V, etc. So the more Amps the solenoid coil pulls the less voltage it has to produce the magnetism to close the big contacts to the starter -- closing is the hard part - holding always requires much less. These starter solenoids draw several amps and therefore need a supply having low resistance. Thats the relay youre putting in. It has a tiny pull in coil that draws low current. 1/4A at 1 Ohm only loses a quarter V .

,,,,,, The relay closes easily with the IG switch on Batt+ reduced by only 0.25V . ... And then its contacts switch a good solid [LOW OHM] B+ line to the starter solenoid and it snaps the starter contacts closed. Problem solved. No need of the relay being able to operate at well below 12V. Having it able to pull in with a near flat batt wouldnt start anyway.
 
   / BX2200 intermittant ignition failure #18  
"Youve introduced lots of confusing factors without saying anything wrong. ... The succinct point is that you dont lose voltage thru contacts and wiring that is not carrying current."

Of course you do not lose voltage in a circuit that passes no current. BUT we are not discussing a circuit that passes no current and therefore can do no work. Discussing such a circuit is meaningless...

IF you reread my comments above with an open mind you will see the logic is sound and the theory is correct.

By the way, I forgot to mention that the JD kit I received was not plug n play. The connectors that were on the ends of two of the wires were not compatible with the BX2200. I had to cut them off and reuse one of the original BX connectors, and use a butt connector for the other wire. No big deal. There may be a number of different JD kits; The one I bought did not specify Kubota and that could be the reason for the different connectors.

Happy camper here. No more clicking sounds, just immediate starts!
 
   / BX2200 intermittant ignition failure #19  
Hi,

I just received my John Deere AM107421 Starter Relay Kit AM107421.

First thing I did was do a quick check of relay operation using my battery charger. Unfortunately I do not have a variable voltage DC supply, but I do have a true RMS reading Fluke meter. I found that using the 6 volt setting on the charger, that the relay has a solid pickup at 8 VDC and will stay energized at 5 VDC. These are the voltages supplied by my battery charger on its two 6 volt settings. Operation may be a little different with ripple free DC as supplied by a battery, but I think my results are representative.

The important thing is that whatever relay is used, it should be rated for 12 VDC but be capable of energizing at voltage lower than 12 VDC, [to compensate for voltage drop across the safety circuit switch contacts.]

Those relays KennyV pointed to look like good 12 VDC relays. But if they need close to 12 VDC to activate, then they may not work for the purpose of compensating for voltage drop across the various contacts in the safety circuit feeding the relay coil.

Bill

The circuit loss is due to the relatively large current [Amps] drawn by the solenoid coil. You will be feeding this coil thru the closed relay. Your IG switch circuit will only see the small current the relay coil draws [fractional Amps]. Thus the relay coil will see full OCV [>12] until the starter kicks and pulls the whole system down a V or so. So no problem with activating or holding Voltage.

It is not exactly clear to me what you are trying to express...but when you say there is "no problem with activating or holding Voltage." I can tell you that this portion of your statement is denied both by electrical theory and practical experience of members here at TBN.

Specifically, a relay coil that is activated through a series string of contacts, whatever those contacts may be, will have a voltage impressed across its operating coil that is equal to the value of the voltage applied to the circuit, MINUS the voltage drop across each contact that is in the series string. This is fact not fiction.

Historically, in the archives here over the past ten years or so, there have been multiple posts regarding the starter motor solenoid just clicking and NOT pulling in enough physically to engage the starter motor. The reported final cure has been to install a relay that will pick up at lower voltage than the factory equipment, to be activated by the circuit that originally was hooked to the starter solenoid terminal. This is the one in the JD kit. (but as I think you may be saying, different relays that pull lower current may also work) Then contacts from this new relay were inserted directly between the battery terminal on the starter, and the starter solenoid terminal. This way full battery voltage was applied to the starter solenoid terminal when the ignition switch was turned to the start position( provided that all permissive contacts in the safety circuit were closed).

The relatively large current that you refer to I assume is that required by the starter motor solenoid. This high current is what causes relatively large voltage drop to develop across contacts in the safety circuit that have deteriorated somewhat over time. The new relay requires less current that the starter solenoid for proper operation. Therefore there is relatively less voltage drop lost across contacts in the safety circuit, and a greater percentage of applied voltage is felt across the relay coil. Any 12 VDC relay that draws less current than the starter motor solenoid may work. Having a coil rated at 12 VDC that will operate at 7.2 VDC gives an even greater margin or error to the benefit of proper operation. Of course the power circuit, between the battery terminal of the starter motor and the starter motor solenoid, in both cases will have only ONE contact in use.

It is that simple. Any relay that operates at 12 may work; one that works between 7 and 12 volts can only be better in the long run.

Spyderlk, I am not sure if you are saying that any 12 volt relay would work for this modification; there certainly is a chance this could be the case, at least in some instances. A new small relay would likely draw less current than the starter motor solenoid does. Less current through the safety circuit would result in less voltage drop across the problem contacts.

However, a relay coil that both draws less current AND activates at a voltage significantly less than the 12 VDC battery voltage will always be superior in this application, especially since contacts that have increased there resistance over time will likely continue to worsen.

This post is an attempt to ensure that the reasoning behind this proven fix is reviewed (if anyone cares to). I installed the JD relay in my BX2200 yesterday and it works perfectly. No more "Click, click, click....start!" I am expecting this to be the case even as the safety circuit contacts worsen.

If I had a suitable standard 12 VDC relay on hand to use in this application I probably would have tried that first. Why not, the price would be right and it might be enough to do the job. But since I did not I decided to maximize my odds over the long run, and buy a relay that offered greater potential long term benefit... :)
SPYDERLK said:
Youve introduced lots of confusing factors without saying anything wrong. ... The succinct point is that you dont lose voltage thru contacts and wiring that is not carrying current. And the current to the relay coil is very low.

... E = IR. To illustrate by example -- 1 Amp thru one Ohm uses a Volt, 2A and 1 Ohm loses 2V, etc. So the more Amps the solenoid coil pulls the less voltage it has to produce the magnetism to close the big contacts to the starter -- closing is the hard part - holding always requires much less. These starter solenoids draw several amps and therefore need a supply having low resistance. Thats the relay youre putting in. It has a tiny pull in coil that draws low current. ~1/4A at 1 Ohm only loses a quarter V .

,,,,,, The relay closes easily with the IG switch, on Batt+ reduced by only 0.25V . ... And then its contacts switch a good solid [LOW OHM] B+ line to the starter solenoid and it snaps the starter contacts closed. Problem solved. No need of the relay being able to operate at well below 12V. Having it able to pull in with a near flat batt wouldnt start anyway.
"Youve introduced lots of confusing factors without saying anything wrong. ... The succinct point is that you dont lose voltage thru contacts and wiring that is not carrying current."

Of course you do not lose voltage in a circuit that passes no current. BUT we are not discussing a circuit that passes no current and therefore can do no work. Discussing such a circuit is meaningless...

IF you reread my comments above with an open mind you will see the logic is sound and the theory is correct.

By the way, I forgot to mention that the JD kit I received was not plug n play. The connectors that were on the ends of two of the wires were not compatible with the BX2200. I had to cut them off and reuse one of the original BX connectors, and use a butt connector for the other wire. No big deal. There may be a number of different JD kits; The one I bought did not specify Kubota and that could be the reason for the different connectors.

Happy camper here. No more clicking sounds, just immediate starts!
I have highlighted and emphasized the part of my true quote that likens the relay activation circuit to a near "zero" current case. That established it follows thru orderly thought that such [high activation impedance] relay capable of pulling in at well below 12V has no added value in this case. Its OK that it can ... and could be critical in a different circuit situation - just not of any real benefit here.

,,,,This is what I was bringing out in my response to your initial post about the relay characteristics. From future posts I see that you know the pertinent principles but did not convey them in the 1st post. This could lead some, less sophisticated in electrical things, to simplistic conclusions.

larry
 
   / BX2200 intermittant ignition failure #20  
You seem to want to ignore that there is some kind of voltage drop across contacts in the safety circuit. This is a fact since the starter motor solenoid does not properly activate, but does work as designed when other means are used to apply battery voltage to the solenoid.

You appear to want to claim that any 12 VDC relay will work to fix the problem. But you have no specific measurements or data to back up this claim.

Why would one recommend use of a relay that MIGHT work, when a relay designed for use in 12 VDC systems, and which is guaranteed to activate at voltage as low as 7 VDC is readily available and proven to work well in this application?

Your whole argument, which offers great opportunity to confuse, is based on the likelihood that the chosen interposing relay will draw less current than the starter solenoid. This is almost certainly true. But without taking measurements of circuit characteristics and having a specification sheet of the relay being applied, one can only guess, but not be sure, that a given relay will work.

So one can do one of two things:

1) Use a relay that is proven to work in this application, and will likely still continue to work if the safety circuit contacts continue to degrade, OR,

2) Try any 12 VDC relay available, and see if it works. If it works, hope that it will continue to work if the safety circuit contacts continue to degrade.

Confusing? Not really.

Saving a couple bucks by following choice 2) was not worth the risk to me personally. Others may feel different, no problem with that. Hopefully the decision will be made with full understanding of costs and risks of each choice...

I think we beat this horse to death. You are welcome to have the last word... :)

Bill

Edit: I guess there is a choice 3) Do nothing like I have done for the last three years and enjoy performing the "click, click, click...Click" Ritual.... :D
 
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