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How to limit the electromagnetic valve current? Answered

I know that the current is given by the voltage divided by the resistance.The resistance of the coil in the solenoid valve should be very small.In that case, if the voltage is applied directly, will the current be very large? For example, the voltage of 12V, the coil resistance of the solenoid valve is 1 ohm, is the current of this solenoid valve 12A? I wanted to build a simple model similar to a single-cylinder diesel engine, with solenoid valves to simulate piston motion and provide a compression resistance and push out power.

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Jack A Lopez
Jack A Lopez

12 months ago

Regarding this statement,

"I know that the current is given by the voltage divided by the resistance."

it seems to me you expect an electromagnet to be just like a resistor.

Certainly an electromagnet has resistance in its windings, and in the steady state, after it has been energized for some amount of time, the current that flows, is simply V/R.

However contemplating that steady state current, will not tell you anything about the other things you are wondering about, including the mystery of how a solenoid can convert electrical work, V*I*dt, into mechanical work, F*dx.

One sort of handwaving way to do it, is to imagine an electromagnet, after it has been turned on, it sort of becomes like a permanent magnet, and you could expect some pulling force F, on the steel plunger, that is a function of the displacement x, and the amount work the plunger can do is the integral of F*dx, a definite integral over the range, in x, the plunger moves.

A few weeks ago, someone else was asking this forum about electromagnet actuators,

https://www.instructables.com/community/NewStumpedNeed-Help-Solenoid-Simulated-Recoil/

and I think I pointed him towards a hundred year old, published in 1920, free textbook, authored by someone named C. L. Dawes.

I recall that text offered a graph of F (force) as a function of x (displacement), on page 23, figure 32, "Pull of solenoid on plunger."

However this text did not offer any explanation of the electrical physics of how this worked, and I am trying to think if I can come up with one, even just a hand waving kind of explanation.

It might be if you can calculate the energy in the magnetic field surrounding the solenoid and plunger both, that you will find there is more energy in the arrangement with a big air gap in the middle, and less energy in the arrangement where the steel plunger is centered home in the middle of the solenoid, and the difference between those quantities of energy is essentially work that can be done by the plunger.

Regarding your title question, about how to limit the current to an electromagnet, I think one way to do this, is simply to turn it on for only a brief amount of time. That is, a time shorter than the amount of time it takes for the current to climb to its steady state value. For an RL circuit, that amount of time is just a few multiples of (L/R).

https://en.wikipedia.org/wiki/RL_circuit#Time_domain_considerations

Also a solenoid is a kind of funny inductor, because the steel plunger can move, and L is different depending on plunger position. I am guessing L is largest when there is big air gap in the middle, corresponding to largest, I^2*L, stored energy, although I am not sure if I am guessing correctly about that.

By the way, RC circuits, a resistor and capacitor together, are maybe more familiar.

https://en.wikipedia.org/wiki/RC_circuit

The well known 555 timer IC, is a relaxation oscillator whose timing is based on the time it takes for a RC circuit to change, which is just a few multiples of (R*C), which like (L/R), has units of time, i.e.

ohm*farad = second
henry/ohm = second

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Downunder35m
Downunder35m

12 months ago

Might be a bit out of the regions you considered but anyway:
Have a search for electromagnetic valve.
These things use a solenoid like coil system that is optimised to move a select "piston" to work the actual valve.
Really strong and available for 12V - most that I have will draw less than 100mA and have about 1kg in pulling power - more and even push if proper neodymium magnets are used on the moving rod.

If price and size is of concern than try my old rubbish-tinker-trick:
Get 12 relais of the coil size you have in mind.
Discard all but the actual coil assembly.
They come with square or rectangular cores in most cases but should not be a problem if you can machine or file a suitable piece of soft iron.
The original core with the metal frame going around to the working end might need a drilling to be removed - try to only drill as deep as the frame is thick!
A bit of wiggling with force will seperate the two.

Alternative ideas and searches to try:
"12V electromagnet round" will give you quite a few results in the usual online market places.
Even better if you try "12V pull actuator" or instead of actuator magnet - these will all give you 12V coils with current requirements that are optimised for the task.
All you need is to select the lenght and pulling power you need - or the dimensions your project requires if you can deal with what a corresponding coil has to offer.
Last but not least: repurpose identical 12V relais:
If you can get 12V relais of suitable size in terms of diameter and lenght you can do a full teardown to directly rewind the wire onto a bobbin that fits your engine aAND as the right volume of free space to take the entire length of wire.
Bit time consuming if you can't make it fit onto a cordless drill but doable.
Benefit is that with similar dimensions you know the power it will need and how much pulling/pushing force you can expect, at least if you tried the relais first ;)

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rickharris
rickharris

12 months ago

I gather you want to build a solenoid engine. Yes Current is Voltage / resistance. Usually a solenoid will draw a fairly large current because this gives a big magnetic field and so more force.
I guess your really asking how may turns and what gauge wire to use to make your solenoids - easy way is to see what someone else has done.
https://www.youtube.com/watch?v=gncDJAV_Nhg
Otherwise you will have to look up the current your preferred/available wire gauge will carry and then work out how many turns ( and so resistance of the coil) for the current you want it to draw.
Starting with what someone else has done is a good place to be.