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!)**

enter image description here


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: integrator This is my circuit diagram: 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... curve when Vin is 1V curve when Vin is 3V 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: expected curve 2

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: hysteresis curve when the resistance is 122 ohm

Where is the mistake..?

  • 1
    \$\begingroup\$ This looks like a reasonable saturation curve to me. What's the transformer's rating? You may just be nowhere near saturating it. \$\endgroup\$
    – Hearth
    Jun 29, 2021 at 14:58
  • 2
    \$\begingroup\$ Why are you using an output filter at all? You should be able to get an XY plot without the filter. Try it. I'm interested in seeing what you get. Please keep me informed by commenting to @andyaka \$\endgroup\$
    – Andy aka
    Jun 29, 2021 at 15:18
  • \$\begingroup\$ Your transformer has a big core? To get a saturated BH display, you'll likely need a LOT more current from your signal generator. The RC integrator you're using seems to have too-small R with too-large C. It is likely loading the winding it is connected to. I've found that this kind of RC integrator CAN WORK with larger R and smaller C...in my setup it provided a tiny 2mV signal for Y-axis. In my setup, a small toroid core was used with equal # turns PRI:SEC. Have done tiny memory cores too, with very square BH - in these only 1 turn PRI:SEC will fit. \$\endgroup\$
    – glen_geek
    Jun 29, 2021 at 16:18
  • \$\begingroup\$ @Andyaka I did what you suggested and updated the post. Why did you suggest that I should be able to get a curve without the filter? In almost every paper or video that I have watched, people all used integrators. I think it is necessary, isn't it...? Also, do you have any idea of why the graph looks so ... diffused? \$\endgroup\$
    – Juye C
    Jun 29, 2021 at 16:21
  • \$\begingroup\$ @glen_geek Yes the core is pretty large, it is about 5cm x 5cm x 10cm. (I will try out a larger R and a smaller C tomorrow!) Do you mean that you used equal turns of primary and secondary coils? Mine is a step down one, though. Will it still apply to my case? \$\endgroup\$
    – Juye C
    Jun 29, 2021 at 16:23

1 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: integrator input voltage, output voltage

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

  • \$\begingroup\$ Thank you. What kind of op amp would you recommend to use with a 1uF capacitor and a 100k ohms resistor? Does an AC op amp integrator work? \$\endgroup\$
    – Juye C
    Jun 30, 2021 at 6:10
  • \$\begingroup\$ Have added an opamp integrator. Many opamps are appropriate. The default opamp shown has FET input (very low bias current), and can take DC supplies up to +/- 18V. Low bias current, and large DC supplies are two of the more important specs. \$\endgroup\$
    – glen_geek
    Jun 30, 2021 at 13:00
  • \$\begingroup\$ mV/cm? Do you mean mV/div? I haven't used a scope with divisions anywhere close to a cm in years. \$\endgroup\$
    – Hearth
    Jun 30, 2021 at 13:56
  • \$\begingroup\$ @Hearth Thank you. I've got another quick question: The smallest capacitor that I have is 10uF and it is polarized. Though it is polarized and C is larger, is there any ways that I can make it work? \$\endgroup\$
    – Juye C
    Jun 30, 2021 at 14:35
  • 1
    \$\begingroup\$ @Fisher Not really; this circuit kind of requires a non-polarized capacitor. You could do it with a larger capacitor, but you need it to be non-polarized. I'm surprised you don't have anything smaller than ten μF, that's fairly large for anything other than a decoupling cap! I recommend you get an assortment of ceramic and film capacitors between 1 nF and 10 μF at least, if you plan on doing much more electronics tinkering. \$\endgroup\$
    – Hearth
    Jun 30, 2021 at 15:27

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