# 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) ???

## 1 Answer

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. Commented Mar 26, 2016 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. Commented Mar 26, 2016 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. Commented Dec 19, 2016 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? Commented Oct 31, 2018 at 5:51
• @Aditya I'm equating the magnitudes of the impedances and not their complex value. Commented Oct 31, 2018 at 12:42