I recently did an experiment aiming to verify the magnetization curve of the cores of a single-phase electrical transformer. I applied a sine within one core and I observed this one and the other one responses over the time. The circuit builded for the experiment and the output of this, that is a snapshot from the oscilloscope that I've used to make the measures, we can see at the images below.

enter image description here

enter image description here

I then exported the graphics raw data into CSV files to manipulate it with some computer software, like Matlab or Excel. I mixed the data from both graphs in a manner that I could plot one as being the X-axis (1st CSV column) and the other being the Y-axis (2nd CSV column), and indeed it worked with Matlab. With the data organized I executed the following within Matlab:

r1 = Array(:, 1);
r2 = Array(:, 2);

The result is the transformer hysteresis graph as expected and showed bellow, and we can see a lot of noise within the signal.

enter image description here

Finally, I could calculate the area from the graph through:


The obtained result is 4.3768e+09, analyzing the data I would say that, if the X and Y axis was expressed in terms of Volts units, the resulting area would be 0.43768.

My doubt is:

I know that the encountered area represents somehow the energy loss within the transformer core, but... How do I express it in terms of energy? Is the encountered result (0,43768) expressed in [N/m^2]?

  • \$\begingroup\$ Is the yellow trace actually the output voltage or the input current? \$\endgroup\$ – Andy aka Oct 31 '18 at 12:57
  • \$\begingroup\$ @Andyaka The yellow is the output one. As we can see this delay and wrong shape are the energy loss. And by what I know, this is exactly what the hysteresis graph shows us. \$\endgroup\$ – Gabriel Magri Oct 31 '18 at 15:39
  • \$\begingroup\$ The thing is, with a transformer, I wouldn't expect to see signs of saturation on the output voltage given that the primary is connected to a voltage sine wave. If the input signal were a current sine wave then yes, this would look more like what you see. I am confused by this. \$\endgroup\$ – Andy aka Oct 31 '18 at 16:08
  • \$\begingroup\$ @Andyaka Sorry, I made a mistake. The yellow signal is the input indeed. I updated the question with an image of the circuit used to make the experiment, on it we can see the points where I made the measures. Thanks. \$\endgroup\$ – Gabriel Magri Oct 31 '18 at 21:30

No, it is not.

First of all, you must measure the input current, not voltage; The R1 resistor lets you to do that, the CH1 measurement divided by the R1 shunt resistor value is the input current of the circuit.

After dividing the CH1 measurement by the R1 value and plotting the CH1/R1 and CH2, now you can find the area inside the hysteresis curve and get the energy losses per period - in Joules, if you've plotted the CH1/R1 in amps and CH2 in volts. I have no idea where do you derive the N/m^2 units from, the energy units, joules, are N*m by definition.

To get the energy losses per second (that is - the power losses in watts) you need to multiply the energy losses per period by the signal frequency.

  • \$\begingroup\$ Yes, thanks a lot for your explanation! It make a lot of sense, indeed. \$\endgroup\$ – Gabriel Magri Nov 2 '18 at 2:21

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