You are blowing a fusible link, not a diode!
Something is drawing too much current.
I'd suggest looking for a chaffed wire that is shorting to ground first.

It is most definitely a diode that runs between the small wire from the positive battery terminal and the switch. the part description on the John Deere package I was sold to replace the diode says 'rectifier kit". Before I remove the wiring harness I would like to know if there are any other possible solutions or known things to look for. Thanks for your help.

"Small wire from positive battery terminal to switch"? I can't think of any reason, or need for a DIODE in that part of the circuit. A diode is an electrical "check valve" in that it will only allow current to flow in ONE direction, or polarity. The heavy duty (relatively high AMP rated) type of diodes used in automotive charging systems rectify the AC output of the alternator into DC to charge the battery (batteries are unable to use raw AC). When a diode of this type is used elsewhere in an automotive electrical system it is put there to prevent interconnected circuits from becoming energized (being "back fed") completely whenever various switches are closed. One other reason that a diode might be put into an automotive circuit is to prevent a "back flow of self-induced current" that occurs whenever a switch feeding a large "electrical coil device" is opened. Can someone come up with the OEM wiring diagram for this specific application?

What part # is your kit and exactly where is it installed? IE is it where the red/ orange come together. The only diode I am familiar with is the rectifier itself. As stated before there is a fusible link. Has this ever had the low voltage boost kit installed in the starting circuit? I don't recall if there is a diode there or not.

I have a schematic for the LX172-LX186. I don't know where I got it, so I can't link to it.

Battery to ignition switch "B" terminal shows a fusible link.
Alternator to ignition switch "B" terminal shows a Voltage Regulator (diodes?) in series.

I think if that rectifier is not in the circuit you will be applying a.c. to the battery. Did you make sure you installed the replacement correctly. If you put it in backwards it probably caused it to blow again. I think we would all agree that the manufacturer wouldn't have put it there if it were not needed.

Does your machine have an AMPS meter?

Has it recently been reading really high?

You could have a bad (partially shorted) battery.

Got a photo? I also don't remember a diode there. I also don't remember any kits being involved there. JD did a recall for the whole "keyswitch module" for the early LX172's, but that included the whole module. No "add on" parts. ???

Part number and or a photo would really help.

"I think if that rectifier is not in the circuit you will be applying a.c. to the battery.:

The voltage regulator on this tractor takes the place of the diode.
In any case, it would be located between the alternator and switch, not the switch and the battery.
Else, the PTO, fuel solenoid etc. would be getting AC.

The part number is AM101006. The mower will only run with the diode installed in one direction. The current has to run away from the battery for the mower to start. When you put it in backwards the mower will not start. When instaled correctly the mower only ran about 5 minutes before it blew again. I don't have the wiring diagram but have seen it at the dealer and it shows a diode in the small red wire coming off the positive battery terminal. This wire then goes into the harness at the bottom of the battery which then is fed up toward the switch. I don't know anything about a low voltage kit being instaled. The yellow battery indicator on the dash has flashed on a lot in the past but always went back off. This is an early mower and had the original keyswitch module still instaled when the diode first blew. I had the replacement keyswitch module JD had sent years ago and instaled it thinking this might correct my problem, so it had the new one in when it started and blew the new diode. The dealer could be misinformed and this may not have been a diode but I was told it was by the service manager who looked up the diagram. As I said before, it will run fine when you jump the wires and leave the diode out. I just don't want to take a chance of starting a fire in the garage by doing this.
Thanks for your help and suggestions.

I would suggest putting the battery on a charger (set the charger to the 10 amp range) and observe the current meter on the charger. Now wait 24 hours. Is the charger still putting out a high amount of charge? In that case your battery is shot and *may* be the cause of the diode burnouts.

Are you saying to let the battery charge for 24 hours at 10 amps? I can do this but my charger only shows red, green, and yellow areas that are small. Battery was only about 2 months old when the diode burned out.

Yes, my point is that if should go significantly further into the green (relatively speaking) after charging.

If it doesn't then you know it is bad (unless it starts way into the green at first).

It is not unusual to get new batteries that are defective, so it is worth a shot.

Best of luck.

I have a electrical background. You need to check your battery if it is good!!!! You need to charge it up and do a load test on it; auto places that sell batteries should be able to do this for free; possible customer. Charging diodes operate at about 3 amps bad battery will increase amps drawn possibly over the diode rating causing failure.

"Charging diodes operate at about 3 amps bad battery will increase amps drawn possibly over the diode rating causing failure." Huh???
Which specific "charging diodes" operate at about 3 amp? AND....If the "charging diode" is matched to the application......where are the additional amps (over and above max output of alternator) coming from?

Hello mownie
What I am saying is an example of what is normal; If battery is down in voltage amp will be a bit higher than 3amps if battery is fully charged amps will be a bit lower. Additional amps for starting for example come from the battery; a battery stores electrical energy. It is a storage device . Diodes come with many different voltage and current(amp) ratings and have to fit the application. Diode charging circuit like this are very simple charging circuit. Charging amp output is mainly determined by engine rpm.

Hey Guys!

Semiconductors are destroyed by heat. A diode rated at X amps can easily handle 150% of it's rating for a short time.

My "theory" was that if the battery is bad it may be drawing more than x amps all the time and eventually the diode dies.

Delbe was thinking about this same theory.

Now, if the charging system is capable of outputting >x then the diode should be rate for X (or X + a safety factor). This may or may not be the case.

I will admit, it is not a perfect theory, but other than a short in the harness it is the only one I can think of that would explain the multiple dead diodes.

Tennessee - Worst case you can use a larger diode from an electronic supplier, but I would test the battery first

We see it the same way wheelhorse. I Know I could install a diode in there that would never burn up; but it could cook the stator or the wiring; the diode burning up is a indication of a problem and it is acting as a fuse preventing more serious damage! Check the battery!

You guys are missing the point-
This machine ALREADY has a Voltage Regulator on the engine, or at least it did originally.
The "diode" that he is referring to is in the wrong physical location to be a charge diode.
A charge diode would be between the alternator and the key switch, not between the battery and the switch.
IF it were in the latter location, the fuel solenoid, electric PTO etc. would be getting ACV instead of DCV! How well would an electric PTO work when it was switching polarity 100's? of times/second?

Has this "diode" been actually verified to be a diode with an Ohm meter?

Hi Bill

OP did mention it worked only with the diode pointing in one way, so that does imply it really is a diode.

I agree there are pieces that don't add up in this equation.

tennesseepowerstroke
Are you the original owner?

IF NOT, I'm thinking someone may have "modified" things.
IF SO, does the engine still have the VR on it.
The wires "of interest" would be a brown/white & brown/yellow coming from the alternator and a red going to the key switch B terminal and on to the battery-

"The mower will only run with the diode installed in one direction. The current has to run away from the battery for the mower to start.

Then how could the battery possibly charge, since current wob't go TO the battery?"

I purchased the mower new in 1991. Don't know what a VR is. I would think the battery charge runs through the larger positive cable.

The diode might be in the circuit to prevent the battery from discharging when the mower is not operating. This is only a wild guess, without a circuit diagram it isn't possible to know for certain.

VR = Voltage Regulator. "Charging diodes" DO NOT "operate at about three amps", nor are they RATED at about three amps. They "operate" by design, within a range of current from zero amp.....to the maximum current that can be produced by the alternator of a specific application. If the maximum output of a given alternator was three amp, then the design would require that the rectifying diodes be rated for a minimum of "3 AMP". If the maximum current output of a given alternator was 25 AMP, the diodes would need to be rated for a minimum of "25 AMP" on so on ad infinatum. Now to the part about "bad battery will increase amps drawn possibly over the diode rating causing failure". Let us use an example scenario where the alternator can produce a maximum output of 25 AMP . If the battery was "bad" (let's say it has a "dead cell"), or just completely discharged (really makes no difference here), the voltage regulator would be operating to send the maximum design current to the battery in an attempt to raise the voltage level of the battery to the optimum design voltage for the type of battery used. At this point, the alternator is "maxed out" and the diodes are carrying 25 AMP (as they were chosen to do in this application). The fact that the battery is bad, or dead can only cause the alternator to produce the 25 AMP max output it was built to produce. High AMP Draw cannot cause an alternator to create more output than it was designed to.

Why not pull the battery and take it to your auto parts retailer or Walmart - have them test the battery condition - if it's good it's your time and milage. If its bad, add the cost of a new battery $25 to $40 depending.
Either way, you have eliminated one of the possible problems. And just maybe the only problem.