# AC Solenoid current calculation

#### brianmk

Joined Dec 23, 2016
59
This should be simple, but...

I have a 24V 50Hz AC solenoid.
The DC resistance of the coil measures 70 Ohms at ambient temperature.
The solenoid has a U shaped iron core with a moveable armature.
When energised, the core and armature form a closed magnetic circuit.
The core includes a copper shading ring.

I measured the inductance with the armature closed using a Marconi bridge.
If I measure using the internal 1kHz oscillator I get 180mH.
If I use an external low frequency oscillator in the region of 50Hz, I get 440mH.
I assume the difference is due to the permeability of the iron core changing with frequency?

I want to calculate the expected rms current when the solenoid is connected to a 24V 50Hz AC supply.

Using the 440mH figure, the reactance of the coil at 50Hz is 138 Ohms.
That gives a total impedance of 155 Ohms, phase angle 63deg.
Hence I would expect the current to be 24 / 155 = 155mA rms.

When I hook up a test circuit I actually measure around 220mA rms (using an AVO which approximates rms).
The current drops to about 200mA as the coil warms up.

I don't understand why the measured current differs from the calculated result.
What have I overlooked?
Is it because the shading ring distorts the current waveform?
If it's no longer a true sine wave then the current as measured on an analogue moving coil ammeter won't be accurate.

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#### ronsimpson

Joined Oct 7, 2019
3,034
The solenoids I have change inductance & current depending on if they are open or closed.

Joined Jul 18, 2013
28,667
A wise practice is to use DC solenoids where ever possible.
The only time the AC version has the edge, is at turn on time, after that the DC wins.

#### crutschow

Joined Mar 14, 2008
34,386
Could be that the core is saturating some at the peak current level of the sine which would increase the average current above the theoretical.
You could determine that if you use an oscilloscope to observe the current through an added small resistor in series with the coil.

#### brianmk

Joined Dec 23, 2016
59
The solenoids I have change inductance & current depending on if they are open or closed.
Yes - realise that.
Inductance increases when the core is a closed magnetic loop.
That's why I stated...
'I measured the inductance with the armature closed'.
When the armature is open, the inductance is unimportant since the solenoid is un-energised.

#### brianmk

Joined Dec 23, 2016
59
A wise practice is to use DC solenoids where ever possible.
The only time the AC version has the edge, is at turn on time, after that the DC wins.
In my case the solenoid is from a gas valve. It runs from a 24V AC transformer. Using DC is not an option.
Using DC would result in too high a current and increase the temperature of the coil.
As far as I am aware, almost all gas valve solenoids use AC.

#### brianmk

Joined Dec 23, 2016
59
Could be that the core is saturating some at the peak current level of the sine which would increase the average current above the theoretical.
You could determine that if you use an oscilloscope to observe the current through an added small resistor in series with the coil.
That may be a possibility - although a current of only 200mA seems unlikely to saturate a solid un-laminated soft iron core.
I was planning to try and view the current waveform on a 'scope. Maybe that will show something.
When I tried searching for the expected current waveform in an AC solenoid that includes a shading ring, I came up blank.

#### crutschow

Joined Mar 14, 2008
34,386
although a current of only 200mA seems unlikely to saturate a solid un-laminated soft iron core.
It is not determined by just the current but by the ampere-turns of the coil which generates the flux level in the core.
With enough turns you can saturate any magnetic core at low currents.

#### brianmk

Joined Dec 23, 2016
59
It is not determined by just the current but by the ampere-turns of the coil which generates the flux level in the core.
With enough turns you can saturate any magnetic core at low currents.
OK understood. I will check it out using a scope as you suggest.

The solenoid in question is a home-wound replacement for one that failed open circuit after many years of service.
I was unable to count the number of turns in the failed coil because it was encapsulated in potting compound and fell apart during the extraction process.
However I managed to find an identical replacement gas valve & solenoid so was able to measure the dc resistance (= 70 Ohms).

The bobbin that the coil was wound on also fell apart when extracted from the potting compound.
I managed to make a replacement bobbin the same size as the original.

I found enamelled wire of the same diameter (for both bare copper and when enamel coated) and wound as many turns as would fit on the bobbin.
I then removed some of the turns until the resistance measured 70 Ohms to match the replacement. The final turns count was 1435.

I now knew that the total length of wire used was the same as the replacement (even though I could not be certain that the number of turns matched exactly).

I measured the inductance of the replacement solenoid at 1kHz : 160mH unenergised.
My home-wound replacement measured 120mH @ 1kHz unenergised.
I'm not sure why the measured inductance is lower - perhaps a difference in permeability of the core? Perhaps the cores have different permeability vs frequency characteristics? The original coils could be more closely wound allowing more turns for the same wire length?

The point is that I am trying to match the characteristics of the original failed solenoid.
If the original did not saturate the core then (assuming I get it right) it is unlikely that my replacement will do so.

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Joined Jul 18, 2013
28,667
Using DC would result in too high a current and increase the temperature of the coil.
Not sure how you came up with that?
AC devices of this nature tend to run hotter due to lower efficiency and introduction of shading ring etc.

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#### brianmk

Joined Dec 23, 2016
59
Not sure how you came up with that?
AC devices of this nature tend to run hotter due to lower efficiency and introduction of shading ring etc.
If you run a solenoid using DC then the current is determined purely by the resistance.
When using AC, the inductance is a factor so the current is lower.
Remember Phasor diagrams?

Z = R + jwL

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Joined Jul 18, 2013
28,667
I thought you were referring to a comparison to DC solenoid, Not an AC ran on DC!

#### crutschow

Joined Mar 14, 2008
34,386
If you run a solenoid using DC then the current is determined purely by the resistance.
Yes.
If you use an AC designed coil with DC then you need to either reduce the applied voltage or add a resistor in series to reduce the current to about the same as when AC is applied.

#### MrAl

Joined Jun 17, 2014
11,448
It is not determined by just the current but by the ampere-turns of the coil which generates the flux level in the core.
With enough turns you can saturate any magnetic core at low currents.
Hi there,

I think he is talking about an AC coil which would be the opposite of that as I am sure you know. With AC the fewer turns you wind on the bobbin the more the core saturates due to less inductance. With DC it saturates easier with more turns.

#### MisterBill2

Joined Jan 23, 2018
18,388
This should be simple, but...

I have a 24V 50Hz AC solenoid.
The DC resistance of the coil measures 70 Ohms at ambient temperature.
The solenoid has a U shaped iron core with a moveable armature.
When energised, the core and armature form a closed magnetic circuit.
The core includes a copper shading ring.

I measured the inductance with the armature closed using a Marconi bridge.
If I measure using the internal 1kHz oscillator I get 180mH.
If I use an external low frequency oscillator in the region of 50Hz, I get 440mH.
I assume the difference is due to the permeability of the iron core changing with frequency?

I want to calculate the expected rms current when the solenoid is connected to a 24V 50Hz AC supply.

Using the 440mH figure, the reactance of the coil at 50Hz is 138 Ohms.
That gives a total impedance of 155 Ohms, phase angle 63deg.
Hence I would expect the current to be 24 / 155 = 155mA rms.

When I hook up a test circuit I actually measure around 220mA rms (using an AVO which approximates rms).
The current drops to about 200mA as the coil warms up.

I don't understand why the measured current differs from the calculated result.
What have I overlooked?
Is it because the shading ring distorts the current waveform?
If it's no longer a true sine wave then the current as measured on an analogue moving coil ammeter won't be accurate.
The greater current is at least partly due to the shading coil dissipating energy as current flows in it, that energy comes from coil current. In addition, the current reading is from the maximum current, not the RMS current, while the voltage reading is RMS-equivalant volts.

#### MrAl

Joined Jun 17, 2014
11,448
Also, the current is not always sinusoidal, it can be very different. That would easily mess up the reading.

Joined Jul 18, 2013
28,667
The AC solenoid has the edge at turn on due to the low resistance - high current, until the armature moves over and Inductive reactance takes over.
The shading ring makes for low efficiency due to it's shorted turn nature & effect. & consequent heating.

#### brianmk

Joined Dec 23, 2016
59
By connecting a scope across a small series resistor I was able to monitor the current waveform.
It looks more like a triangular wave than a sine wave. See attached photo.

That may explain why the current measured using an AVO is not as expected.
An AVO full-wave rectifies an AC signal and the moving coil meter responds to the average.
A calibration factor is applied to the scale to display the rms value assuming a sine wave.
The calibration factor for a sine wave is 0.707/0.633 = 1.11
For a triangular wave it would need to be 0.577/0.5 = 1.15

That means that the true rms value is actually higher than indicated on the AVO.
i.e. 228mA rather than 220mA.

I verified this by measuring the current with a true rms DMM.
The DMM measures 250mA - presumably because it's not a perfect triangular wave.

A triangular waveform means that calculating the current becomes a lot more complicated.
I imagine it would involve Fourier analysis and summing the harmonics?

I have attached a photo of the rewound solenoid that shows the shading loop.

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#### ericgibbs

Joined Jan 29, 2010
18,826
Hi Brian,
Reading through your posts, may I ask why it is so important that you know the exact solenoid current?

E

#### Alec_t

Joined Sep 17, 2013
14,310
Attached is a simple LTspice sim of your solenoid to play with. I modelled the shading ring merely as a secondary winding on a transformer, but I suspect there's more to it than that in the real world. That secondary's inductance, although much smaller than the primary's inductance, has a big effect on the current through the primary.

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