# Solenoid Charactaristics

Discussion in 'General Electronics Chat' started by abuhafss, Sep 12, 2014.

1. ### abuhafss Thread Starter Active Member

Aug 17, 2010
173
2
Hi

I am working on a solenoid powered from a 12V SLA battery and have a few questions:

1) Suppose the solenoid needs 5A to actuate, how can that current be reduced? By increasing the number of turns?

2) What will be effect of increasing the thickness of the coil wire?

3) If a solenoid is powered for longer duration, the internal resistance of the coil increases and thereby more and more current is drawn which causes self-heating of the coil. Can this be overcome using a fixed current supply?

2. ### ericgibbs AAC Fanatic!

Jan 29, 2010
3,222
554
hi,
Look here for Q1..

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html

Q2. Thicker wire has a lower resistance, so the same number of turns would draw more current.

Q3. If you check the temperature coefficient of copper wire you will that the resistance increases with temperature, so the current and magnetic field would be reduced.

E

Jul 18, 2013
12,872
3,516
Presumably this is DC, in any DC electromagnetic device that has an armature such as a solenoid or relay, the current can be reduced after the solenoid has become energized due to lower energy required to retain it than it does to shift the armature or operator in the first place.
Max.

4. ### Bernard AAC Fanatic!

Aug 7, 2008
4,531
476
Re, Max, add a resistor in series with solenoid of a low enough value to keep armature pulled in, connect a capacitor in parallel with R, large enough to pull in armature when at full extention.

5. ### abuhafss Thread Starter Active Member

Aug 17, 2010
173
2
So, if we increase the number of turns of the coil, the current drawn can be reduced......am I correct?

And the coil current can also be reduced (using a resistor in series) after it has been energized. Got it but, can you please show some practical schematic? Do we have to incorporate another relay to switch the supply thru resistor after the coil is energized?

6. ### crutschow Expert

Mar 14, 2008
16,166
4,317
Yes.

Bernard suggested how to do that with a resistor and capacitor. Note that the capacitor may have to be quite large to provide the initial 5A surge for the required time to pull in the solenoid. The value can be experimentally determined.

7. ### abuhafss Thread Starter Active Member

Aug 17, 2010
173
2
Actually, I meant to ask how do we switch from "direct" supply to "via resistor/capacitor"? Do we need to incorporate a timer + relay?

8. ### crutschow Expert

Mar 14, 2008
16,166
4,317
No, you just add the resistor and parallel capacitor in series with the switch that controls the solenoid.

9. ### abuhafss Thread Starter Active Member

Aug 17, 2010
173
2
Please bear with me. Could you please explain me how this combination would work?

Jul 18, 2013
12,872
3,516
There is an initial current to charge the capacitor that essentially shunts the resistor, once the cap has charged there is no more capacitor current, the resistor then comes into play to limit the current.
Max.

11. ### abuhafss Thread Starter Active Member

Aug 17, 2010
173
2
Hmm, got it. Thanks for the explanation.

For this case, the required wattage for the resistor would be about 70W.......Correct?

And what should be the starting value for the capacitor? 1000µF/16V, 2200µF/16V, 3300µF/16V.........?

12. ### crutschow Expert

Mar 14, 2008
16,166
4,317
The wattage and resistance of the resistor depends upon how low a voltage the solenoid will stay pulled in after it is energized. For example, for a voltage drop of 6V @ 2.5A the power would 15W and you should use a 30W resistor.