# 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? – TimWescott Sep 8 '19 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. – TimWescott Sep 8 '19 at 19:42
• thanks a lot! s – tairit Sep 8 '19 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. – Huisman Sep 8 '19 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 – tairit Sep 8 '19 at 23:04
• @tairebit well according to my answer that is not the correct formula. – Jan Sep 8 '19 at 23:13
• @tairebit in the calculation they did in your picture, they let $\text{R}_2\to\infty$. – Jan Sep 9 '19 at 18:01