After replacing the coil, points & condensor I got our 880 back running. Something concerns me tho...before I replaced the burned up points & condensor I wanted to see if I had current to the points so I bridged them. It didn't take long to see some smoke & the spring on the points arm started to glow red!! If there were a inline fuse would I still get 12v with less amps? Lets say I put in a 20 amp fuse. Would this be ok? Right now there's nothing & I'm guessing that's why this tractor is so hard on points. How many amps does a typical 12v battery put out? Thank you for your inputs here.
- Thread starter elkcagg
- Start date
You need to check the coil resistance, if it is too low the points will burn. Then to fix it you will need another proper coil or a resistor in line before the coil. If you put 12 volts to the points without the coil that is the same as direct short to ground. If it has always burned the points the voltage to the points is too high.(same as too low resistance) Also the condenser can cause points to burn.
oliver90owner
Member
- Location
- UK
DickW is on the money.
Fuses are irrelevant as is current from a 12 volt battery (it could deliver over 500A for the starter motor at initial switching or thousands of amps if short circuited across the terminals).
Seems to me that you short circuited the contact breaker to ground directly from the battery! You perhaps need to understand Ohm's Law before doing thigs like that, if that was what you did.
The coil is an inductive load, but while final current could be 5A, it would have a rising current, the rate of which depends on the time constant of the coil. Average running current would be far less as the dwell angle will also be taken into account.
Fuses are irrelevant as is current from a 12 volt battery (it could deliver over 500A for the starter motor at initial switching or thousands of amps if short circuited across the terminals).
Seems to me that you short circuited the contact breaker to ground directly from the battery! You perhaps need to understand Ohm's Law before doing thigs like that, if that was what you did.
The coil is an inductive load, but while final current could be 5A, it would have a rising current, the rate of which depends on the time constant of the coil. Average running current would be far less as the dwell angle will also be taken into account.
The more help I ask for the more I'm getting confused. So what should a coil resistance measure? How many ohms...under current or at rest? This mess all started when I left the key switch on after running out of gas & all my ignition parts & wires got ruined. I bridged the points...as in closing them by moving the adjusting plate over & closing the old points just like they do in normal operation with key switch on. I was just replicating what happens when I leave a key on with the points closed. That's when I seen things heat up in just a few seconds. This was all done before I replaced the coil, condensor & the new points. But what I need to know is why did all this get wrecked in the first place? I just want to prevent this mess from happening again. Where should I put a fuse if I need one & how big?
oliver90owner
Member
- Location
- UK
Most certainly not.
Hello Bison,
(assuming negative ground here)
If you are seeing 12 volts at the distributor,
You need either 1. a coil with internal resistance built in, or 2. an inline ceramic resistor between ignition and the coil positive
The resistors are pretty common, you can order one for a 77 Chrysler LeBaron, 6 or 8 cylinder, doesn't matter. I have one for a Ford tractor on my 880, replacing the original which crumbled away. I believe the internal resistance in the coil is about 3.5 ohm.
The 6-12 Volt coil you found has the internal resistance, it can be used in a 6V or 12V system to create spark based on the signal from the points.
For negative ground:
Battery>Ignition>Resistor>Coil +>Coil ->Distributor points terminal
This should drop your voltage at the distributor terminal to the 8V range.
Hope this helps. By the way, I'd toss your old coil after allowing that much amperage through it. Probably burned it up. Suggest also check any upstream wiring and also the ignition switch. Is off REALLY off?
- Landon
(assuming negative ground here)
If you are seeing 12 volts at the distributor,
You need either 1. a coil with internal resistance built in, or 2. an inline ceramic resistor between ignition and the coil positive
The resistors are pretty common, you can order one for a 77 Chrysler LeBaron, 6 or 8 cylinder, doesn't matter. I have one for a Ford tractor on my 880, replacing the original which crumbled away. I believe the internal resistance in the coil is about 3.5 ohm.
The 6-12 Volt coil you found has the internal resistance, it can be used in a 6V or 12V system to create spark based on the signal from the points.
For negative ground:
Battery>Ignition>Resistor>Coil +>Coil ->Distributor points terminal
This should drop your voltage at the distributor terminal to the 8V range.
Hope this helps. By the way, I'd toss your old coil after allowing that much amperage through it. Probably burned it up. Suggest also check any upstream wiring and also the ignition switch. Is off REALLY off?
- Landon
J.Wondergem
Well-known Member
- Location
- Rockford, Mi.
I think you have Bison mixed up with elkcagg.
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