Hysteresis curve does not look like what is expected on the oscilloscope

Question

update: Based one the comment from @Andy aka, I tried to obtain the curve without using the integrator. It does give me a better looking graph.

Afterward, I switched the power supply to another larger one (which I can gradually change the input voltage by turning a knob. However, this time the curve looks so fat and unconcentrated. Why is this happening? (other than this issue, the curve looks good!)**

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I am using a 600turns 2A - 300turns 4A step down transformer with an iron core, AC voltage of 3V, and an RC integrator consisting of a 47 microfarad capacitor and an 8 ohm resistor (I got this from a variable resistor) This is my setup: This is my circuit diagram:

****This shape of the hysteresis curve looks like this. I observed that as I increase the voltage, the shape of the hysteresis curve does not actually change. From what I see, it is probably only the size of it that is changing. this is the curve I get when the input voltage is 1V. This looks the same as the one when the input voltage is 3V, but just a smaller in size... This might be why that I cannot get a saturation point on the curve...Why is this happening, how can I correct this?

I am unsure about why the curve looks round. I increased the voltage up to 15V. However, the curve still has not yet reached saturation.

I was expecting to see a curve like this when the curve has not yet reached saturation:

Also, the resistance of the resistor in the integrator is currently 8 ohm. If I increase it, for example, to 122 ohm, the shape turns really round like this:

Where is the mistake..?

Answer 1

This answer addresses the integrator, which accepts as input the sense winding, and provides an output voltage that's proportional to magnetic flux in the core.

A simple RC integrator can be used, but it does output a tiny voltage. Most oscilloscopes bottom-out at 2mV/div or 5mV/div. The RC integrator is most accurate when its output voltage across the capacitor is a tiny fraction of its input voltage. In some cases, where the sense winding has few turns, the simple RC passive integrator provides too-small output - as Hearth has pointed out in comments an active opamp integrator must be used in this case.

The OP's integrator uses a 47uf capacitor, which is likely polarized (either electrolytic or tantalum). Since output is AC having +ve and -ve polarity, a non-polarized capacitor should be used. Although 10uf non-polarized capacitors are available, 1uf is more common.
At 50 Hz., a 1uf capacitor has a reactance of 3183 ohms. The R of a simple passive integrator must be many times larger than this reactance, else the integrator has both phase and amplitude error: Ellipses appear on the oscilloscope display instead of straight lines. For this integrator running at 50 Hz, a resistor should be greater than 100k ohms. A larger resistor provides a better integrator, but smaller output voltage.

^{simulate this circuit – Schematic created using CircuitLab}
To saturate a magnetic core, sufficient ampere-turns must be applied. Voltage induced in the sense winding "spikes" to high amplitude when the core is saturated...when not saturated, this waveform is much more sinusoidal.
Below, the red waveform represents sense-winding voltage of a saturated core. In green, the voltage across C1 is very small, and has been "amplified" 50 times so that you can see its shape...it is integrating the 50 Hz waveform fairly well:

Using an opamp integrator releases you from the requirement \$R >> {1 \over {2\pi f C}}\$. A smaller C produces a larger output voltage. Keep R1 fairly large, so that little sense-winding current flows.
The opamp must be powered from dual DC supplies (one positive, the other negative) since input and output are AC above and below ground. A large-value resistor R2 is required across the integrating capacitor to maintain DC-stability, otherwise output voltage will slowly wander to one or the other DC supply rails:

^{simulate this circuit}