# Determine inductance of large inductor

I asked a similar question before but i am still struggling with this problem.

I am trying to measure the inductance of a large (34cm diameter) electromagnetic brake.

Important findings to keep in mind:

• The inductance changes when the armature plate gets attracted to the electromagnet due to changing reluctance of the electromagnetic circuit.
• 100Hz measurements are used since the coil gets powered using a 50Hz rectified mains voltage.

I tried using 3 methods to determine the inductance. The ESR is 869Ohm and the DCR is 23.9Ohm.

## 1.LCR meter

I used an LCR meter at 100Hz measurement mode. The measured series inductance is: 1.586H

## 2.Function generator

To calculate the complex impedance. Which resulted in an inductance of 1.6H

## 3.Current rise time

I also measured the current waveform to calculate the inductance using the rise-time of the inductor according to the formulas shown in the figure below:

My current waveform is shown below:

The maximum current is 9A as can be seen on the oscilloscope screenshot (measured using 0.2Ohm shunt) The 1 timeconstant current is:

IL(L/R) = Imax(1-e^-1) = 5.6A

Using the current waveform the measured time to reach 5.6A is 0.70s This results in an inductance of:

L = t * R = 0.73 * 23.9 = 17H

Can someone explain the large difference between method 1&2 and 3?

• What happens if you measure via (1) or (2) with the brake held in the operated position? I have a feeling that (3) is operating the brake halfway through and decreasing the reluctance (increasing inductance)
– user16324
Commented Nov 12, 2021 at 15:37
• Its true the brake gets activated visualized by the 'bump' halfway in the current waveform. It's also true the inductance changes. But it does not explane the large difference between 1&2 and 3
– Rens
Commented Nov 12, 2021 at 15:51
• I potentially disagree with "But it does not explain" ... if the airgap reduces from 1mm to 0 (just wild guesses) that could easily vary inductance by an order of magnitude. This is easily testable per prev comment. However, also note that (3) is not a 100Hz measurement, per Andy's answer.
– user16324
Commented Nov 12, 2021 at 15:56

Can someone explain the large difference between method 1&2 and 3?

The most obvious difference is possibly due to eddy currents. In your DC test (3. Current rise time shown using a DC source), eddy current induction into static metal parts will be quite small and this won't affect the inductance value very much.

However, when operating the device at 50 Hz (or 100 Hz), induction of currents into static metal parts will be significantly higher and possibly account for the much lower inductance value. The eddy currents will represent shorted turns of a transformer and thus make inductance much less.

• In test 3 i am operating the coil at 50Hz rectified sine wave. So there is an 100Hz voltage applied. Not DC.
– Rens
Commented Nov 12, 2021 at 15:14
• Not true. You may be operating at this frequency but the rectifier reduces the AC content significantly as you can see in your own current waveform graph. I'm just pointing out information that is in your question. Commented Nov 12, 2021 at 15:19
• Thanks for your clarification. But do you think that the eddy currents can lower the inductance by this much seems a bit high to me?
– Rens
Commented Nov 12, 2021 at 15:54
• Why do you think regular 50 Hz / 60 Hz AC power transformers have laminations? They are not electrically bonded to each other - each lamination is insulated from its neighbour to prevent massive eddy currents. So, yes, I really do think that eddy currents in your measurements are responsible for the seemingly low values of inductance @Rens Commented Nov 12, 2021 at 16:04