this post was submitted on 24 Sep 2024
33 points (100.0% liked)

Ask Electronics

3325 readers
1 users here now

For questions about component-level electronic circuits, tools and equipment.

Rules

1: Be nice.

2: Be on-topic (eg: Electronic, not electrical).

3: No commercial stuff, buying, selling or valuations.

4: Be safe.


founded 1 year ago
MODERATORS
 

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!

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 12 points 1 month ago

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.