# Solenoid Help

#### Erik_Swiger

Feb 25, 2023
6
Hi, this is my first post, hope I'm in the right place. I need help with a home project. I have 2 solenoids, but only one operates at a time, with about a 16% duty cycle each. The solenoids use 12.6 volts, and they're about 3.5 ohms each, 18-gauge copper enamel wire. The solenoids are switched with rotary switches, with (2) 10-amp diodes each for flyback protection. It seems if I could get the resistance exactly right, the solenoids would pull with a lot of force (enough to make things work). The problem is that there is some fine line of enough current, versus too much current, and I don't know what to do. I tried to calculate it out and thought I needed a 1-ohm resistor for each solenoid, but I can't make it work. Any help would be appreciated, TIA.

#### Minder

Apr 24, 2015
3,479
The solenoids use 12.6 volts, and they're about 3.5 ohms each, 18-gauge copper enamel wire. TIA.
I assume that is 12.6vdc?
Or?

#### Erik_Swiger

Feb 25, 2023
6
Yes, 12.6 vdc, a car battery. I'd really hate to re-wind the solenoids, but maybe I should just change the wire gauge. I don't have a good way to wind them, so I can't really count the turns as I'm doing it. So, I thought it would be easier just to added a measured resistance to the circuit. I'm winging it here.

#### kellys_eye

Jun 25, 2010
6,514
It seems if I could get the resistance exactly right, the solenoids would pull with a lot of force (enough to make things work).
You mean they aren't working? In what way? Not pulling hard enough? Pulling too hard?

#### Erik_Swiger

Feb 25, 2023
6
Pulling too hard. By my calculation, 12.6 v / 0.35 ohms = 36 amps. When I touch power to the solenoid, the armature flies across the room, and there's a huge spark. So, I'm hoping if I add a 1-ohm resistor of the right wattage, that it will give me 12.6 v / 1.35 ohms = 9.33 amps, which is more reasonable. I just need to be sure I'm doing the math and formulas correctly. This is my third attempt to get the solenoids right. Also, is there a formula or rule-of-thumb to calculate the pull of the solenoid?

#### Externet

Aug 24, 2009
893
Hello. What is the use if the solenoid, what will it pull, how strong is the return spring ?
...is there a formula or rule-of-thumb to calculate the pull of the solenoid?
Ampere turns tells the magnetic strength.
If you can show pictures, great.

#### Erik_Swiger

Feb 25, 2023
6
The solenoid pulls a caliper against a brake rotor. The intent is to slow the rotor momentarily, not stop it completely. This is applied at up to 20 times per second. It is a third-class lever system; pivot on the left side, solenoid armature pulling in the middle, and load/brake surface on the right. The return spring is weak, just a tiny half-inch length. First picture is a solenoid 1 inch in length and 1 inch in diameter, with the core outside dimension at 7/16 inch. The other pic is a diagram of the arrangement. The full solenoid is not shown, and the bottom half contains other views of the upper part. It's all I can do, I can't get a good picture of it now.

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#### kellys_eye

Jun 25, 2010
6,514
If you can stand the loss of 'pull' then adding a resistor is fine and will reduce the current accordingly.

#### Alec_t

Jul 7, 2015
3,598
Another way of reducing/limiting the solenoid current, albeit more complicated than just a resistor, is PWM (pulse width modulation). This has the advantage of dissipating less power than simply a resistor.

#### Minder

Apr 24, 2015
3,479
That might work on an AC version, bit the efficient way is to wind on a lot more turns for a DC solenoid.

#### Externet

Aug 24, 2009
893
I would go with Minder suggestion. More turns of smaller diameter wire. Visually, it seems the solenoid is not original, or was hand-wound; perhaps with the wrong coil in a previous attempt to fix something.
I think I have seen similar buzzer coils used in clothesdryer machines...

--> https://www.ecosia.org/images?q=coil for electromagnet

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#### Bluejets

Oct 5, 2014
6,992
If you want to wind your own coils and make your own dc solenoids, some study on how to design the complewte unit would be your first stop over.
Just winding a coil will not cut it, there is a lot more to it than that.

Usually best and cheapest all around is simply buy off the shelf what you need.

As a quick example, the solenoid from an older type vehicle would suit I'm sure.
They are sometimes used to engage the starter gears as well as switch the starter and would be around in their millions at auto wrecker and the like.
Example of this type shown below.
Possibly even still available new on some of the common online places.

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#### Erik_Swiger

Feb 25, 2023
6
That might work on an AC version, bit the efficient way is to wind on a lot more turns for a DC solenoid.
You're saying, create the needed resistance with more turns, it's more efficient, right?

#### Erik_Swiger

Feb 25, 2023
6
Thanks to everybody, this has been a great help.

#### Bluejets

Oct 5, 2014
6,992
it's more efficient, right?
Efficiency doesn't come into it (well, not so far as power out versus power in)....it's the required turns of the correct size wire to create the correct magnetic flux to pull in a solenoid of a given size and load, all the design being for DC operation.
This includes the armature design and the coil housing and the right current for the given job.

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