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I am studying the function of an Op-Amp integrator with offset voltage that is in the picture below: Op-Amp integrator with offset voltage

I have derived the output voltage formula myself, but I am not sure if it is correct. The formula is given below: $$V_o = {V_{off}\over{RC}} - {1\over{RC}}\int_0^tV_{in}dt$$

Can anybody please tell me if the formula is correct or not? I have searched the literature and the internet, but I couldn't find the answer. Thank you for your time.

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3 Answers 3

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Well, if we assume an ideal opamp we know that the current trough \$\text{R}\$ must be the same current trough \$\text{C}\$. So we can write:

$$\text{i}_\text{R}\left(\text{s}\right)=\text{i}_\text{C}\left(\text{s}\right)\space\Longleftrightarrow\space\frac{\text{v}_\text{in}\left(\text{s}\right)-\text{v}_-\left(\text{s}\right)}{\text{R}}=\frac{\text{v}_-\left(\text{s}\right)-\text{v}_\text{o}\left(\text{s}\right)}{\frac{1}{\text{sC}}}\tag1$$

We also know that the ideal opamp wants to establish \$\text{V}_+=\text{V}_-\$. And in your circuit we know that \$\text{V}_+=\text{V}_\text{off}\$. So using this, we can rewrite \$(1)\$ as follows:

$$\frac{\text{v}_\text{in}\left(\text{s}\right)-\text{v}_\text{off}\left(\text{s}\right)}{\text{R}}=\frac{\text{v}_\text{off}\left(\text{s}\right)-\text{v}_\text{o}\left(\text{s}\right)}{\frac{1}{\text{sC}}}\space\Longleftrightarrow\space\text{v}_\text{o}\left(\text{s}\right)=\text{v}_\text{off}\left(\text{s}\right)-\frac{\text{v}_\text{in}\left(\text{s}\right)-\text{v}_\text{off}\left(\text{s}\right)}{\text{sCR}}\tag2$$

Which is not hard to transform back to the time domain. I let you do that.

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  • \$\begingroup\$ We also know that the ideal opamp wants to establish \$V_+=V_ −\$. This is only the case for a negative feedback configuration. If there was positive feedback instead this would not hold. \$\endgroup\$
    – Carl
    Commented Dec 31, 2021 at 9:12
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No, the integral should contain (VIN-Voff)

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The node equation for \$V_-\$ is $$\frac{V_--V_\text{in}}{R}+C\frac{\text{d}(V_--V_\text{o})}{\text{d}t} =0$$ Due to the negative feedback of the op-amp configuration, the principle of a virtual short can be applied, meaning that \$V_+=V_- \Rightarrow V_-=V_\text{off}\$

Substituting gives you $$\frac{V_\text{off}-V_\text{in}}{R}+C\frac{\text{d}(V_\text{off}-V_\text{o})}{\text{d}t} =0 $$ Can you do the rest?

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  • \$\begingroup\$ Thank you for your time and help jp314 and Carl. I managed to do the rest. The correct formula is given below (posted in case somebody runs into the same problem): $$Vo = V_off+{1\over{RC}}\int_0^t(V_{off}-Vin)dt$$ \$\endgroup\$ Commented Dec 30, 2021 at 17:50
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    \$\begingroup\$ @NikolaRistic Good work! You could (and should, for the record) accept an answer to show others that the question-and-answer of this post has been achieved. Whether you write your own answer and accept that, or accept mine or jp314 is up to you. \$\endgroup\$
    – Carl
    Commented Dec 30, 2021 at 20:08

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