wiring meltdown

[fladeere]

my 4310 stopped suddenly...seemed like the fuel cut off. When I got into it...discovered that the fuel shutoff solenoid failed, and (since there is no fuse) the wires from that point nearly back to the dash completely melted. The ground was especially burned. Seems like a lousy, and potentially dangerous, setup given how the wires wind around in there. Wondering if anyone else has experienced a failure like this. When I called the dealer before I tore into it, they immediately suggested the solenoid, so I'm curious if the problem is widespread.

Thanks for any responses

 

[RRacing]

I had a simular problem with my JD 790. But I found I had a shorted battery And my solinoid was fine ( even though I had melted wires as well, and like yours I saw it more on the ground side) . Luckly the 790 has a 3 amp fuse that blew before any major damage. You might have a battery going bad as well. Being the current goes up as the voltage drops.

Jim

 

[TMcD_in_MI]

</font><font color="blue" class="small">( You might have a battery going bad as well. Being the current goes up as the voltage drops. )</font>

Jim,

I'm a little confused by this. (Normal for me) Don't you mean current goes down as voltage drops?

Tom

 

[RRacing]

Tom,
The current draw of the solinoid goes up as the voltage drops is what I was saying.not that of the battery. But what is using the current meaning the solinoid.Hope this clears it up some. /forums/images/graemlins/smile.gif
Jim

 

[TMcD_in_MI]

</font><font color="blue" class="small">( Tom,
The current draw of the solinoid goes up as the voltage drops is what I was saying.not that of the battery. But what is using the current meaning the solinoid.Hope this clears it up some. /forums/images/graemlins/smile.gif
Jim )</font>

Jim,

Thanks for trying to clear up my confusion. I don't mean to make a big deal out of this, and we can just drop it if you want, but I'm still confused. Ohm's Law says that when the voltage on a circuit drops, the current through that circuit drops proportionally. So when a battery starts going bad and its voltage drops, the current coming out of the battery and going through the solenoid must drop, right? Am I missing something?

Again, feel free to just drop this if you want.

Tom

 

[RRacing]

Yes Tom you are right about the battery current dropping with voltage. But the solinoid will draw more current as the voltage drops to keep it open. Once again I'm talking about the object using the current not the one feeding it. Just because the voltage and current are dropping at the battery doesn't have anything to do with the fact the the object useing the current acts differently not the same way . It actually increases current trying to stay open. If you have worked on any CNC equipment or even automobiles with DC systems You will have seen this type of problems were the power supply starts to go out and the other components will start to draw more current and fuses blow and wires will melt. I hope this helps.Kurchefs law( I know I butchered that up) /forums/images/graemlins/smile.gif

Thank you
Jim

 

[TMcD_in_MI]

</font><font color="blue" class="small">( Just because the voltage and current are dropping at the battery doesn't have anything to do with the fact the the object useing the current acts differently not the same way . It actually increases current trying to stay open. )</font>

Hey Jim, thanks a lot for answering. Just so I understand, are you saying that it's possible that the battery can be putting out, say, 300 amps, but the solenoid can have more than 300 amps passing through it? This sounds kind of impossible to me. Am I misunderstanding you?

Tom

 

[RRacing]

No Tom thats not what I'm saying. The solinoid is wired for 3 amps and would most likely only use 2. But if your battery shorts and voltage drops, a curcuit that is only designed for 3 amps is now drawing 6-8 amps.( Possible more) which would cause the wires to heat up and maybe melt.Also if the battery was shorted or plates buckled It would draw even more.I use to work at Beretta the gun company many moons ago and we would get brown outs ( not Black) were the lights would dim for up to a minute which was a drop in voltage and before anyone could hit the kill switches on the CNC machines the current used by the machines ( Motors , Encoders , Controls,Solinoids, ECT,ECT) would surpass the wiring and motor windings were designed for, and just burn many things up. We would spend days replacing, and repairing, and even rewiring machines /forums/images/graemlins/frown.gif
Does this help any?

Jim

 

[RRacing]

I guess in a nut shell what I'm tring to say is the wiring that is only designed to handle X# of amps becomes your fuse link and it heats up just like a fuse because your object( the solinoid ) is drawing more current than it was designed for. /forums/images/graemlins/smile.gif
By the way did you get your tractor fixed?

Jim

 

[TMcD_in_MI]

</font><font color="blue" class="small">( ... we would get brown outs ( not Black) were the lights would dim for up to a minute which was a drop in voltage and before anyone could hit the kill switches on the CNC machines the current used by the machines ( Motors , Encoders , Controls,Solinoids, ECT,ECT) would surpass the wiring and motor windings were designed for, and just burn many things up.)</font>

Hey, now this part is making sense to me. I do understand that AC motors will draw more current when the voltage on them drops, and that they can burn out because of that. But we were only talking about a simple DC circuit... a battery, a switch, and a coil of wire (the solenoid). I'm pretty sure that Ohm's Law applies here. Lower voltage = lower current.

Anyway, thanks a lot for taking the time to discuss this with me. And thanks for asking about my tractor, but I'm not the original poster with the problem. I guess I'm just the guy who has hijacked the thread. /forums/images/graemlins/blush.gif /forums/images/graemlins/blush.gif

Tom

 

[turbo36]

Actually the reason is as easy as PIE /forums/images/graemlins/smirk.gif as in P=IxE.
Power (watts or P) = amps (I) x volts (V). Since the solenoid requires a specific amount of power to operate, if the voltage drops the amperage must increase to maintain the same power. Now if we think of the flow of the electrons through the wire in terms of water through a hose then the voltage is like the pressure, and the amps are the amount of flow. The wire has resistance (measured in Ohms) to that flow that cause heat to be produced, so if the flow (amps) increases then the resistance to the flow increases causing the amount of heat to increase faster then the wire can dissipate the heat thus leading to melt down. :shocked
I left out a few things like the design of the coil voltage, temp rating of the wire etc. etc. so all you EE's can fill in the gaps if you must.But remember I'm just a dumb Mfg. Engineer /forums/images/graemlins/wink.gif so give me a break.

 

[TMcD_in_MI]

</font><font color="blue" class="small">( Actually the reason is as easy as PIE /forums/images/graemlins/smirk.gif as in P=IxE.
Power (watts or P) = amps (I) x volts (V). Since the solenoid requires a specific amount of power to operate, if the voltage drops the amperage must increase to maintain the same power. )</font>

Turbo - Thanks for jumping in here, and the power equation does sound pretty good at first. But ... /forums/images/graemlins/confused.gif there's something I don't understand (big surprise) about what you just said. You said the solenoid requires a specific amount of power to operate. Fair enough, but why would there be any guarantee that the solenoid will always get that amount of power? According to that reasoning, if the battery was fried and went down to 1 volt, the current to the solenoid would automatically ramp up and the tractor would still start. That would be cool, 'cuz we wouldn't ever have to worry about dead batteries. See my problem?

I think Jim is maybe taking a principle from AC motor theory, and applying it to a DC circuit. I'm sticking with Ohm's Law and the idea that a lower DC voltage can't result in higher current through the solenoid.

I appreciate your throwing in your thoughts. /forums/images/graemlins/smile.gif

Tom

 

[turbo36]

Hey teach, I think you are right on this, I overlooked the fact that the resistance is a constant and therefore the amperage is directly proportional to the voltage, now it the resistance changes then it is a different story. I hope you don't send a note home to my parents /forums/images/graemlins/laugh.gif
Here is a link to trouble shooting a 12 volt automative circuit.
Link

 

[turbo36]

Now you really have me thinking /forums/images/graemlins/confused.gif What happens if the voltage is so low that the solenoid does not close thus the iron core does not enter the magnetic field and satisfy the designed circuit loss, doesn't that in a sense raise the resistance??? Now we really do need the Double E -how about it any electrical engineers in thehouse?

 

[gates]

Turbo,

With out getting to technical I will try to help. There have been two equations talked about in this thread, p=iv & v=ir. The problem is that they must be applied together, not seperately. I believe you figured this out when you stated that the resistance was constant. If you apply the given numbers in the equations simultaneously you will find that some of the statements made above are incorrect.

RRacing,

The only way, with a drop in the voltage, that the current could increase is if the resistance of the solenoid changed. Once again both formulas need to be applied simultaneously.

Fladeere,

I believe what happened in your situation is the sol. shorted to ground, thus lowering the resistance, increasing the current and ultimately putting the wires in an over current situation.

Tom,

If you can come up with that battery where as voltage goes down the current increases we could immediately put MR. in front of your name. /forums/images/graemlins/grin.gif

Ron

 

[CAL123]

ok...just to follow up since I finally fixed the problem...here's the story on the wiring meltdown...

There is/was a defective "module" which failed, that did not cut off the higher voltage to the fuel solenoid once the solenoid opens (apparently the higher voltage opens the solenoid then converts to low voltage to keep it open). The module/fuse did not shut off the greater current which overheated the solenoid. There were no fuses in between the solenoid and the return wiring - so it heated up and melted the wiring all the way to the dash. Fortunately when I got in there the wiring finally burned through leaving enough ground to attach onto just behind the fusebox. In the repair I also wired in 2 in-line fuses after the solenoid so it doesn't repeat. What a mess. I ran a google search and apparently it's a known problem. I'm not sure all the models with the issue, but I would strongly consider putting in an in-line fuse or see which solenoid you have in your machine. BTW, I was quoted $800 for the repair, but was able to complete it with the help of a friend. /forums/images/graemlins/cool.gif

thanks to the posters for the input...frustrating to have a major repair at 125 hrs!

 

[hawk7]

OK Just some Info to freshen up every one from there school days. I've been an industrial Electrician for 10 years working on CNC equipment, 3Phase, single Phase, conduit, low and high DC circuits and so on. But we all tend to forget some of the basics from time to time. Remember one thing. If some one ever tells you their an expert, Get away from them as they will get killed or somthing. But this is the purpose of this forum, To learn from each other and this is why I joined. Anyway here is the theory. /forums/images/graemlins/smile.gif /forums/images/graemlins/smile.gif
<font color="red"> OHM'S LAW
Ohm's Law says: The current in a circuit is directly proportional to the applied voltage and inversely proportional to the amount of resistance. This means that if the voltage goes up, the current flow will go up, and vice versa. Also, as the resistance goes up, the current goes down, and vice versa. Ohm's Law can be put to good use in electrical troubleshooting. But calculating precise values for voltage, current, and resistance is not always practical ... nor, really needed. A more practical, less time-consuming use of Ohm's Law would be to simply apply the concepts involved:

SOURCE VOLTAGE is not affected by either current or resistance. It is either too low, normal, or too high. If it is too low, current will be low. If it is normal, current will be high if resistance is low, or current will be low if resistance is high. If voltage is too high, current will be high.

CURRENT is affected by either voltage or resistance. If the voltage is high or the resistance is low, current will be high. If the voltage is low or the resistance is high, current will be low.

RESISTANCE is not affected by either voltage or current. It is either too low, okay, or too high. If resistance is too low, current will be high at any voltage. If resistance is too high, current will be low if voltage is okay.

NOTE: When the voltage stays the same, such as in an Automotive Circuit... current goes up as resistance goes down, and current goes down as resistance goes up. Bypassed devices reduce resistance, causing high current. Loose connections increase resistance, causing low current.

</font>