# Integrator with DC control gain

I am analyzing the integrator circuit and I can't get from which equation can I get the frequency where there is the intersection with x axis - 1/C*R1 ?

Thanks!

• If $2\pi f C R_2$ is way larger than 1, you can take the 1 out of the denominator of the AC voltage gain expression. 0dB happens when gain is 1, so you end up (remembering that you're concerned with absolute values) with $\frac{R_2}{R_1} \frac{1}{2 \pi f C R_2} = 1$. Does this help? Commented Sep 8, 2019 at 19:41
• Just a comment -- there are a number of inaccuracies in that. It looks like it's oversimplified, but poorly. You may want to find a different book to work from. Commented Sep 8, 2019 at 19:42
• thanks a lot! s Commented Sep 8, 2019 at 19:43
• This seems to be homework, but I think I understand the confusion. It depends on the type of frequency being used (angular frequency or regular frequency). The graph uses different type than the equations. Commented Sep 8, 2019 at 19:51

Well, first of all we can look at the wiki-page of 'Operational amplifier applications' and look under 'Inverting amplifier', to find that your TF is given by:

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=-\frac{1}{\text{R}_1}\cdot\frac{\text{R}_2\cdot\frac{1}{\text{Cs}}}{\text{R}_2+\frac{1}{\text{Cs}}}=-\frac{\text{R}_2}{\text{R}_1}\cdot\frac{1}{1+\text{R}_2\text{Cs}}\tag1$$

In the complex analysis of this circuit we can write:

$$\text{s}=\text{j}\omega\tag2$$

So, we get for the amplitude:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\left|-\frac{\text{R}_2}{\text{R}_1}\cdot\frac{1}{1+\text{R}_2\text{C}\omega\text{j}}\right|=\frac{\text{R}_2}{\text{R}_1}\cdot\frac{1}{\sqrt{1+\left(\text{CR}_2\omega\right)^2}}\tag3$$

In dB's we get:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|_\text{dB}=20\log_{10}\left(\frac{\text{R}_2}{\text{R}_1}\cdot\frac{1}{\sqrt{1+\left(\text{CR}_2\omega\right)^2}}\right)=$$ $$20\log_{10}\left(\frac{\text{R}_2}{\text{R}_1}\right)-10\log_{10}\left(1+\left(\text{CR}_2\omega\right)^2\right)\tag4$$

Now, solve:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|_\text{dB}=0\space\Longrightarrow\space\omega=\frac{1}{\text{CR}_2}\cdot\sqrt{\left(\frac{\text{R}_2}{\text{R}_1}\right)^2-1}\tag5$$

EDIT:

In order to provide feedback also at very low frequencies, they put $$\\text{R}_2\$$ in parallel over the capacitor. But they did not include it in the formula's, what they did:

• For the amplitude: $$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|_{\text{dB}\space\text{&}\space\text{R}_2\to\infty}=-20\log_{10}\left(\omega\text{CR}_1\right)\tag6$$
• For the zero-crossing: $$\omega_{\text{R}_2\to\infty}=\lim_{\text{R}_2\to\infty}\frac{1}{\text{CR}_2}\cdot\sqrt{\left(\frac{\text{R}_2}{\text{R}_1}\right)^2-1}=\frac{1}{\text{CR}_1}\tag7$$
• I was interested to find how to calculate 1/C*R1 Commented Sep 8, 2019 at 23:04
• @tairebit well according to my answer that is not the correct formula. Commented Sep 8, 2019 at 23:13
• @tairebit in the calculation they did in your picture, they let $\text{R}_2\to\infty$. Commented Sep 9, 2019 at 18:01