# Reduce the Temperature of Solenoid Valve for Continuous Driving

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## Introduction: 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)

## Step 2:

Second step we measure the coil resistance use the multimeter. For this example it about  66 ohms at coil resistance.

## Step 3:

Decide the ceramic resistor, There is a voltage devider of coil and ceramic resistor. So we choice the 220 ohm resister and make the voltage across the coil about 5.5v large than 3v , It means the valve always  activate at 5.5v(because it large than 3v), but the coil consumption less power (less heat).
Attention: this method only for DC driving valve, (for AC driving valve it maybe use TRIAC or SCR structure)

## 14 Comments

Novice, aka layman here. Can someone please give me a elementary schematic or drawing of what tendo stated worked?

tendo stated:

"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!"

Thank you for anything that helps.
J

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)?

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.

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!

Interesting article. But the resistor calculation can be much simpler:

1. measure current with coil in hot state (usually 30 to 50% lower than "cold" current)

2. reduce voltage of the power supply until the coil drops out (usually 10% of nominal voltage).

3. Multiply this voltage by at least factor 3 (safety margin); I would take this value, or at least half the nominal supply voltage: the "actual voltage"

4. Measure the current at that "actual voltage"

5. R = (difference between nominal and actual voltage) / actual current

6. Power dissipation in resistor: P = R x (square of current); choose a resistor at least twice that value.

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.