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I am doing a school assignment that requires me to design a stable Integrator circuit. But after I completed the design, I found that the peak value of the output waveform (blue line) was unstable. Is there any way to make the output waveform become a straight line, just like the input? (Need to keep Crossover Frequency at 10KHz)

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  • \$\begingroup\$ Why are you using a resistor Rf ... if you want make an "integrator" circuit? What you get with it (Rf) is a first-order system. \$\endgroup\$
    – Antonio51
    Feb 22 at 11:56

2 Answers 2

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Is there any way to make the output waveform become a straight line, just like the input?

What kind of straight line ? ... No, you can't. The integrator do its job.

Do you want something like this ... Be aware that offset can occur.

enter image description here

It integrates between ~ 10 Hz and 10 MHz ...

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  • \$\begingroup\$ Thank you for your help. After seeing your design, I think I misunderstood the question. I will reconsider the question. I have benefited a lot from your design and data. \$\endgroup\$ Feb 22 at 14:06
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You've named \$C_i\$ rather oddly. In the convention of such circuits, the subscript \$f\$ means feedback, and \$i\$ is used for inputs. Why not call that capacitor \$C_f\$, as it's clearly feedback? From this point on, that's the name I shall give the capacitor, \$C_f\$.

You've used the wrong resistance to calculate the cross-over (break, or cut-off) frequency. The formula is

$$ f_c = \frac{1}{2\pi R_fC_f} $$

The actual cross-over frequency of your circuit is:

$$ f_c = \frac{1}{2\pi R_fC_f} = \frac{1}{2\pi \times 90\rm{\ k\Omega} \times 1.77\rm{\ nF}} = 1\rm{\ kHz} $$

If the frequency of square wave input approaches the cross-over frequency, then the circuit behaves less like an integrator, and more like a regular low-pass filter, which is why you are seeing unstable curved sections in the output waveform, instead of straight slopes. If you want those nice straight "integrator" slopes, then:

$$ f_{in} << f_c $$

Your input waveform does not satisfy this condition. Input frequency is 500Hz, which is way too close to 1kHz for you to see the expected straight slopes. 500Hz may be fine if the cross-over frequency were actually 10kHz, as required.

By the way, you also use the term "unstable" somewhat loosely. This circuit is not unstable, as you put it. The blue line is not "linear" or "straight", or is "curved" is probably what you meant.

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  • \$\begingroup\$ Thank you for your reply. It seems that I misunderstood the question. I will rethink the direction of the question. Thank you for your help. I have benefited a lot. \$\endgroup\$ Feb 22 at 14:02

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