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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!

source  enter image description here

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    \$\begingroup\$ 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? \$\endgroup\$
    – TimWescott
    Commented Sep 8, 2019 at 19:41
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    \$\begingroup\$ 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. \$\endgroup\$
    – TimWescott
    Commented Sep 8, 2019 at 19:42
  • \$\begingroup\$ thanks a lot! s \$\endgroup\$
    – tairit
    Commented Sep 8, 2019 at 19:43
  • \$\begingroup\$ 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. \$\endgroup\$
    – Huisman
    Commented Sep 8, 2019 at 19:51

1 Answer 1

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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: $$\omega_{\text{R}_2\to\infty}=\lim_{\text{R}_2\to\infty}\frac{1}{\text{CR}_2}\cdot\sqrt{\left(\frac{\text{R}_2}{\text{R}_1}\right)^2-1}=\frac{1}{\text{CR}_1}\tag7$$
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  • \$\begingroup\$ I was interested to find how to calculate 1/C*R1 \$\endgroup\$
    – tairit
    Commented Sep 8, 2019 at 23:04
  • \$\begingroup\$ @tairebit well according to my answer that is not the correct formula. \$\endgroup\$ Commented Sep 8, 2019 at 23:13
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    \$\begingroup\$ @tairebit in the calculation they did in your picture, they let \$\text{R}_2\to\infty\$. \$\endgroup\$ Commented Sep 9, 2019 at 18:01

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