Measurement across the wires of a coil

[ECM1]

Did some searching of the forum, but didn't find an answer to my question.

If you used an ohm meter, and measured across a solenoid coil are you measuring the resistance of the coil or the reactance?

What I want to do is calculate the inrush and holding currents of a solenoid valve in a 24V AC 60 Hz application.

So if I take a measurement of the solenoid coil with the plunger in the open position and the closed position would I use this ohms value to calculate the inrush and holding current with ohms law? I'm currently reading 23? in the open position. Which should equal 1.04 Amp but the manufacturer rates the inrush at 0.41 Amp.

If not is there a way to calculate the inrush current and how do I do it? Also is there any possible way that the inrush current could be made larger. (Other than pulling the solenoid plunger further out of its open resting position) Say something like the plunger having a greater amount of pull, or a lower than 24 voltage applied to it. Or is the inrush current for a specific coil set by the resistance and just the amount of time the inrush current would be present before lowering to the holding current would be changed by these things such as voltage and pull.

Thanks for any help you can provide.

 

[IRstuff]

An ohmmeter measures a resistance. An impedance meter measures impedance. One would expect that inrush current would be affected by inductance as well as any other impedances you haven't mentioned yet.

TTFN

FAQ731-376

 

[DRWeig]

Hi ECM1,

Your ohmmeter measures only resistance.

Calculating inrush is tough -- you need to know all the geometry and an equivalent circuit of the driving power supply. Inrush is highly dependent on reactance.

Never calculate anything you can measure, I say! Hook it up and get a good storage o-scope and see what it does.

Good on ya,

Goober Dave

 

[itsmoked]

ECM1; Here is what I'd do. Measure the resistance and consider that the limit of the inrush. Design accordingly. The actual in-rush will indeed be less than this number.

Pulling the plunger out would likely increase the inrush current some. But again it won't exceed the pure resistance value.

Keep in mind that most solenoids meant for AC will absolutely fry in a second or two of operation when prevented from actuating, (sealing).

You could calculate till you were blue in the face and miss the actual value by 50%. If you have to know, instead of doing the conservative thing as I have described above, than you need to do as Dave suggests and measure it.

Just make sure you measure it in less than 2 seconds.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[BobM3]

Just curious, why do you want to increase the inrush current?

 

[ECM1]

Thanks for the suggestions, I could do that if I had such a scope. But I don't.

I can measure the current with a DDM and have. I don't get anywhere near the 1.04 Amps as I suggested above. That is what was confusing me. Even if I manually hold the solenoid from closing. What I read is 410mA and the manufacture rates the inrush current at 420mA. When the solenoid is allowed to close the reading is 230mA which is close to the holding current rating as well.

What I'm more worried about is what, would cause a inrush current if anything, to be greater than what the manufacture specifies the inrush current to be. I'm under the understanding that the inductance, reactance and impedance will determine the current and nothing will change that. I'm not changing the physical properties of the coil and other than decreasing the inductance by moving the plunger further out creating a larger air gap than normal would be the only way to change the inductance. But I'm not planning on doing that.

I'm I correct?

Here is a little example of what I'm getting at.

Lets say you have a solenoid is rated at 1 pound pulling force, but you only needed to pull in one ounce. The inrush current rating of the solenoid is 420mA. So the 420mA is what the inrush current would be pulling one ounce.

Then if you added twelve more ounces and had the solenoid pulling a total of thirteen ounces, which is still under it's rated pulling force. The inrush current would still be the same 420mA. But perhaps the amount of time the inrush current is present until it reaches its holding current would increase slightly due to the added weight.

I'm I right or wrong here?

I guess what I'm getting at is that a solenoid rated at 420mA is never going to see a greater inrush current no matter what you do to it. Other than physically altering the air gap by moving the plunger further out of the coil. Which I don't plan on doing.

Thanks again.

 

[ECM1]

I don't want to increase the inrush current. I just want to make sure it can't.

 

[itsmoked]

Yes you are correct. The current remains the same only the time varies. And not by much.

Yes if you remove the plunger you could see a small increase in the current over the stated inrush.

The reason the current is less than the current described by the resistance is that the coil still provides a bunch of inductance, however it's just wrapped around air until the plunger shows up.

The 10mA difference is totally within reason since temperature, number of winds, voltage, and frequency deviations, all play a part in the results.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

Inrush current is usually a transient effect. I'd be surprised that your DMM can accurately record the actual inrush current.

Again, you've failed to mention whether there are other factors, e.g., how is the solenoid driven? Are there no drivers involved?

TTFN

FAQ731-376

 

[ECM1]

Sorry about that.

I measured the inrush current by manually holding the solenoid from closing. Might not be exact but the only thing I could think of.

The solenoid is 24V AC. No drivers of anything like that.

It is a basic valve solenoid for opening a pilot actuated water valve.

 

[IRstuff]

At 23 ohms and 420 mA, that's almost 4 W of power dissipation.

What exactly is the concern about inrush current? Bear in mind that it's a transient phenomenon.

Also, maybe your supplier is measuring it the same way, or not. Are there any specification on the inductance, or the number of turns and size of the solenoid?

TTFN

FAQ731-376

 

[waross]

The current is limited by the impedance. In a coil the impedance is a combination of resistance and inductive reactance. The inductive reactance is varied by the iron core. In a solenoid, the movable plunger forme part or all of the core. As the plunger enters the solenoid cil, the inductive reactance increases, the impedance increases and the current decreases.
If you have manufacturers specs, use the specs.
If you want to see the effect first hand, Use a resistor to limit the current below the holding current. Energize the solenoid through the resistor with an ammeter in the circuit. Manually move the plunger in and out and watch the current change as the impedance changes.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ECM1]

There are no specs for the inductance, number of turns, ect. That I have found.


I have measured the current while moving the plunger to see what the effect would be. That is what actually started this whole thing. Because the ohms resistance added up to a higher Amps than what I was getting.

Thanks for all, and everyone's replies....

 

[roydm]

Yes, measuring the Ohms you will find no difference with the plunger in or out because your meter is DC but as several posters have pointed out on AC once the magnetic circuit is complete the inductive reactance goes up and current goes down. You can measure the reactance by using your meter on AC amp scale Reactance =Volts/Amps
Thats the beauty of AC solenoids they draw a large current to pull the plunger in then it drops off to hold. If you look at a large DC contactor you will usually find that they have a pair of N/C contacts that open to add resistance (economy resistor) in series with the coil to achieve the same result.
BTW that's the reason an AC coil will burn out on low voltage. The current may be too low to pull the plunger in but many times the normal holding current so it fries.
Personally I still prefer DC for small solenoids as the seem to last longer and are much quieter without the 60Hz buzz but you will find that the AC variety will handle more differential pressure (I'm referring to the direct acting, not pilot operated variety)
You know the resistance, have you calculated the inductive reactance with the plunger in/out?
Regards
Roy

 

[hacksaw]

the manufacturer usually specifies the inrush and holding currents

 

[itsmoked]

I think we've beaten this thread into submission... [flush]

Keith Cress
kcress -

http://www.flaminsystems.com

 

[PHovnanian]

ECM1 said:
--
Then if you added twelve more ounces and had the solenoid pulling a total of thirteen ounces, which is still under it's rated pulling force. The inrush current would still be the same 420mA. But perhaps the amount of time the inrush current is present until it reaches its holding current would increase slightly due to the added weight.

I'm I right or wrong here?
--
You are right. The transient inrush current will result in higher I^2R heating of the solenoid coil. This is ultimately what will limit its performance. Adding mass and/or friction to the mechanical system will increase the time the solenoid spends in this transient region and the resulting heating effect. This is a non-trivial problem in dynamics.

The solenoid manufacturer should provide two more pieces of data: The maximum duty cycle (% time energized) and number of operations per minute (or hour). These figures (particularly the latter) should be at rated mechanical load (1 lb). As long as you do not exceed these, you should be OK. It is possible to derate the solenoid (higher number of operations with a lighter load, for example). But this becomes trickier to calculate. Unless the economics of volume production dictates that you cut your design as close as possible, just buy a solenoid that exceeds your expected maximum demand or you'll spend $200 in engineering time to save $10 in parts cost.