Did you take all ohmic loads of circuit your circuit into account? E.g. the coil and your measurement device have a specific impedance. This could change your cut off frequency. Maybe you can provide a schematic of your circuit.

When you are asking a question about the misbehaviour of a circuit it is impossible to give any comment without seeing the circuits diagram.

Hello dear Jiayi Geng ,

As has been mentioned by other respectable researchers, without the circuit it is difficult to provide help. Also, your plot is not clear. What is it? Is it the Bode plot of the integrator? If so, it does not even look like an integrator's Bode plot (magnitude response).

Nevertheless, by making some "guesses", I will try to give you some advices that you may consider to avoid the problems you mentioned.

I guess, you are having DC bias problems (DC stability of the opamp's output voltage), so you will not be able to obtain a proper Bode plot for any frequency range (Since the opamp is already saturated, it cannot function as an amplifier as desired).

Of course, I am assuming that you are using the classical opamp-based integrator, for which you should keep in mind that the DC gain (gain at f=0) is infinite for ideal integration. This means, the capacitor will never stop being charged or discharged due to a DC input voltage and the charging/discharging can stop in practice only when the output is saturated eventually. However, I guess you are using a resistor (let's say RF) in parallel with the capacitor, which limits the low-frequency gain (hence, the DC gain), i.e. avoids integration below the corner frequency 1/(RFC). If this corner frequency is low then the low-frequency gain (hence the DC gain) will be large - as you know. This means, a small non-zero DC voltage at the input can still cause the DC output voltage to be very large - which again can saturate the output voltage. If the signal at the input of the integrator has a DC component, then the opamp typically saturates. In practice, Even with a zero DC input component, in practice, the offset voltage and the non-zero DC input current of the opamp will still provoke this problem jointly. Reducing value of RF can help to solve this problem but the integration frequency range will be narrower.

If the integrator is used within a larger loop (e.g. 2-integrator loop), the global feedback helps to maintain a stable DC level at the output of an integrator (Even for the ideal integrator). However, I guess for your case there is no such global feedback, so your integrator will suffer from the saturation problem.

There are various techniques to avoid this; but some of them may be complex and yet may not be fit your application. I can suggest the following links, which provide some basic warnings and advices:

https://en.wikipedia.org/wiki/Op_amp_integrator

https://berdisoft.com/op-amp-integrator-design/

(Keep in mind that, each solution typically can give rise to some other performance degradations)

If you are exploring the circuit using simulations, you can apply an offset voltage to the opamp to set the DC output voltage to zero Volt (or approx. zero Volt) to guarantee that it will function properly. However, for a practical circuit, you may need to take more precautions.

Best regards...

Dear Jiayi Geng ,

It's very difficult to analyse without the actual circuit.

But the difference in the cut-off frequency might be due to effect of intrinsic/junction capacitance and internal resistances which might be affecting the actual cut-off frequency of the circuit.

Is it magnetic core kind ogf Rogowski coil? Then the saturation might be happening in the ferrite core e.g. Article Core Saturation Detection and Calibration of a Current Probe...

Alternatively, low freq integrators might be saturated by their own offset voltage.