# AC Op-amp Integrator with DC Gain Control

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

• 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. – kevin Mar 26 '16 at 1:32
• 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. – Andy aka Mar 26 '16 at 10:36
• 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. – Andy aka Dec 19 '16 at 13:18
• 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? – Aditya P Oct 31 '18 at 5:51
• @Aditya I'm equating the magnitudes of the impedances and not their complex value. – Andy aka Oct 31 '18 at 12:42