I found a link http://www.electronics-tutorials.ws/opamp/opamp_6.html that discusses about AC Op-amp Integrator with DC Gain Control

I am wondering how I derive the frequencies (1/CR2 and 1/CR1) ???

AC Op-amp Integrator with DC Gain Control


The lower frequency is \$F_{lower} = \dfrac{1}{2\pi C R_2}\$

The upper frequency is \$F_{upper} = \dfrac{1}{2\pi C R_1}\$

The easiest way to consider the upper frequency is the gain being 0 dB or unity. At this point the impedance of the capacitor equals R1 (assuming that R2 is much bigger than R1 of course. Now

\$R_1 = \dfrac{1}{2\pi F C}\$ and you just rearrange R1 and F to find F.

For the lower frequency a similar method is used but this time the impedance of the capacitor is equated to R2. This gives the "so-called" 3 dB point i.e. the frequency at which the gain R2/R1 drops by 3 dB. So, if R1 and R2 produced a DC gain of (say) 40 dB the 3 dB point would be: -

enter image description here

Picture stolen from here - a useful site for op-amp basics.

  • \$\begingroup\$ Thanks. I understand for inverting-amplifier, if R1=R2 without C, then the gain is unity. And here we assume R2 is much bigger than R1, but I still could not understand how you derive the upper frequencies. \$\endgroup\$
    – kevin
    Mar 26 '16 at 1:32
  • \$\begingroup\$ Upper frequency is when the magnitude of the gain is unity and this happens when the impedances of R1 and C are equal. The "0" on your graph is 0 dB or unity gain. \$\endgroup\$
    – Andy aka
    Mar 26 '16 at 10:36
  • \$\begingroup\$ Are you satisfied with this answer? If not please explain what you are having problems with. If you are satisifed, please feel free to formally accept it. \$\endgroup\$
    – Andy aka
    Dec 19 '16 at 13:18
  • \$\begingroup\$ From what I know, \$ Z_c = \frac{1}{j\omega C} \$ So I understand how \$ 2\pi\$ appears when you want to find the frequencies in Hertz, while the graph Kevin posted must be having the frequencies in radians. But I do not understand why you have not considered the "j", can we simply equate impedances like that? Won't the resistance be a real number and the capacitive impedance be imaginary? \$\endgroup\$
    – Aditya P
    Oct 31 '18 at 5:51
  • \$\begingroup\$ @Aditya I'm equating the magnitudes of the impedances and not their complex value. \$\endgroup\$
    – Andy aka
    Oct 31 '18 at 12:42

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