Hello, I'm using a continuous solenoid: Current @ maximum stroke is 6.50 amps. It has 14.9 ohms coil resistance. Coil voltage rating @ 60 hz 120 VAC, seated current 0.48 ampere, seated power 19 watts And the problem I'm having with it is its overall life. It's a strong solenoid and it basically beats its self to death. So I want to reduce the supply voltage so it's not so strong. I thought about just putting a resistor in series in the hot line side but I'm not sure how big it should be.
Does this sound like a good idea? Feasible? How do I calculate the resistor if I want to achieve maybe 75% load? Or how do I just reduce the AC voltage by 15%?
I would find the necessary pull force voltage for your opperation. If it does not seat, then you have a major problem, current will stay high and may burn something out. Calculating or even using a resistor may be impossible. A transformer can be used to reduce voltage.
Like Greg says, you'll have to do this experimentally.
If you use a transformer to reduce the voltage, it will cost less to use a 12.6VAC, 8A secondary transformer to buck down from 120 to 108 than using a variable autotransformer.
Check output voltage before you attach the solenoid. If it's 132VAC (you're boosting), just attach L1 to the other secondary lead instead.
If you do use a resistor, divide down based on the coil resistance (15 ohms). Almost certainly a 1 ohm 10 watt resistor will be OK. That's what I would try. A 2 ohm 20 watt resistor will reduce the voltage, but it might not leave enough pull force. But if you divide down based on rated current, you'll never get enough force to pull it in.
Make sure to check your work by applying a countering force to the solenoid, and _briefly_ applying power. Make sure it will still pull in hard with the lower voltage and your countering pull. You need a safety factor here for possible increases in friction and variations in line voltage. If you can't test, you shouldn't do it.
Solenoids are mechanical wear parts, and are meant to wear out. You may extend its life a bit, but you should accept that you're periodically going to have to replace the part. You might also want to contact the manufacturer, and see if they have another version of the solenoid available which might better suit your needs.
You could try putting a suitably rated incandescent lamp in series with the solenoid. This will allow a high startup surge to pull in the solenoid, then the current will ramp down as the lamp heats up. It really depends on how often the solenoid is operated, and if the lamp is given a chance to cool down sufficiently
Update... It turns out that a 100 Ohm resistor (25 W) does the trick. It resuces the voltage from 120 to 100 VAC. Still giving me enough force to pull in but not as hard. Although the resistor does get too hot to touch.
Also, apparently as the voltage changes so does the resistance of the solenoid.
Not exactly. What happens is, when voltage is initially applied, there is no current, so the voltage drop across the resistor is zero, and the solenoid sees the full applied voltage. Then, the current increases by whatever that equation is - something about the time constant, and the voltage drop across the resistor increases, reducing the drive on the solenoid. I seriously doubt if the resistance of anything changes, except for the normal tempco.
And you might want to get a higher-wattage resistor. :-)
If this is the AC case, the impedance of the solenoid is not resistive so the math is not the usual adding and subtracting. There can also be things that change the effective resistance and inductance of the solenoid.
The current in the solenoid lags behind the voltage across it by almost 90 degrees. If it was exactly 90 degrees sqrt(R^2+Z^2) would be the way to find the total magnatude not just (R+Z).
Some of the impedance of the solenoid is resistance of the copper wire. There is another couple of resistive things going on. The magnetic field of the coil causes AC currents to flow in the metal of the plunger. It looks a like a poorly coupled transformer with a low resistance on its secondary. This looks like a resistor was placed in parallel with the solenoid.
The iron of the whole assembly is not a perfect magnetic material. Every time the field changes, some energy is lost. This also kind of looks like a resistor in parallel. This resistance will vary with frequency and voltage however.
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