Reduce the temperature of solenoid valve for continuous driving

For most application driving  the solenoid valve we just apply the proper voltage to the coil. But we found the coil became very hot for continuous driving for a while.Here I had a solution to reduce the power consumption when it driving It worked like this,when the power DC24v apply it and the capacitor charging for large current and the coil same get the current and the valve activate, capacitor charging for a while the charging  current reduced and the voltage across the coil devide by the resistor low down the power of the coil and the coil get only a little heat.

Step 1:

First step you should test the solenoid valve "holding"voltage and "inrush" voltage. Connect the solenoid valve and adjust the voltage from 0v to 24v and you should hear the activated noise "ka". For this example it about 20v, it means the voltage must be large than 20v the valve activate.(This we call holding voltage). And then turn the voltage from 24v to 0v, same as before you can hear another "ka" noise, it means de-activate (or released).  For this example it about 3v.(This we call inrush voltage)

Comments

[tendo]

Well, one more question: my task is the same - to reduce temperature. I've the same solenoid (visually from your picture), 3/4'', but it is DC12V and only 4.8 Ohms by my measurement. "Inrush" voltage is about 7.5V. Now I'm trying to find correct resistor and capacitor with no success. So, my question is - is it ok 4.8 Ohms resistance for the solenoid coil (vs Yours 66 Ohms)?

[paiwayne]

Wou! Good luck.There must be something wrong,maybe the coil resister 4.8ohm was too small,because it will consumption 30W. Oh! it to large for normal solenoid valve power consumption.

[paiwayne]

If you change the power supply from 12v to 15v are inconvenient. I re-calculate the resistor it must less be than 3.2 ohm(@12v) so I choice 3 ohm 5W, but the capacitance of the capacitor you should make more test. 2200uF/16v, 3300uF/16v maybe 4700uF/16v.

[paiwayne]

I think the 7.2v that a little higher, so I suggest you change the input voltage from 12v to 15v, and the resistor less than 4.8 Ohm, 4.3 Ohm 5w maybe a good choice.Because of coil is low resistance, it means need more current to drive it. So the capacitor must be larger than my job. I think may be 2000uF/16v or more.(and the change input voltage from 12v to 15v it also make the driving current larger).
Here you NO success means the solenoid deactivate, it mean coil current not enough(very low resistance of coil)to activate , so the capacitance of capacitor should be larger than my job.

[tendo]

Finally I’ve find a working solution in my case. I use the regulated DC 12 2A power supply, 4 ceramic resistors 10 Ohm x 5W (in series) and 1 electrolytic capacitor 10000uF/35V (6600uF is minimum in my case). In this case current I=0.26A (by multimeter) is enough to hold my coil in open position, voltage drop on a coil was 0.95V, total voltage drop on 4 resistors was about 10.5V.
Testing: after 14 hours of continuous operation the coil was absolutely cold (and still in open position), resistors was warm (so I can freely hold my fingers on all 4 resistors).
So, thank you once more for the help!

[tendo]

And, after shaking by hand coil still remains in open position.

[tendo]

Morning! Good instructable! And what about capacitor on the first picture?

[paiwayne]

Where I use 470uf/35v. This capacitance is about the switching time and coil driving capacity. If you use larger capacitance capacitor, it means switching time will slower because it need discharge time to the resistor.And you use smaller capacitance capacitor the charge current must be enough to activate the valve.So thank you for your question.

[tendo]

Thank you for so clear answer!