The electric PTO clutch on my 1969 mini tractor is dead and discontinued.
Original winding is aluminum 18 gauge. Manufacturer specs were 2.88ohms, 237 turns. The manufacturer specs didn’t quite physically match what I found when I took apart the old clutch. If I understand this correctly, the 2.88ohms is the most important part and will pull 4.17 amps.
I just attempted a coil with 18 gauge copper magnet wire. I made it to the max dimensions I can get in the housing with a scramble wind. I’m getting 1.2 ohms, which would pull 10 amps or so. Not good.
Was able to get 187 feet given the resistance.
If I go with 20 gauge copper, assuming I can get 235 feet (1.26 * 187) and I should get 2.319 ohms. Probably get a little more than 235 feet and get the resistance up a little more.
What does this do to the strength of the magnetic field?
Would I be better off putting a power resistor in series with my 18 gauge coil?
Any advice greatly appreciated!
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Magnetic field strength is proportional to current * number of turns.
Using a resistor should work but so should using 20gauge copper if the number of turns is the same.
The wires need to be perfect so they align properly so the proper amount of turns can fit in the housing.
Coil was originally aluminum, so it had a higher resistance. Even with a ‘perfect’ wind, the resistance might still be low.
Scramble should be fine, it’s a DC application. So long as i get close enough: it should work.
Same number of turns as original than add a resistor to get current right. Extra turns shoud be okay but you will likely need a resistor
