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Am I right about this Transfer Function:

Opamp schematic

My work:

$$\text{I}_{\text{in}}(t)+\text{I}_{\text{p}}(t)=0\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=-\text{I}_{\text{in}}(t)\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}$$


$$\text{I}_{\text{p}}(t)=\text{I}_{\text{C}}(t)+\text{I}_{\text{R}_2}(t)\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}$$


$$-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}=\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}\Longleftrightarrow$$ $$\mathcal{L}_t\left[-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}\right]_{(s)}=\mathcal{L}_t\left[\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}\right]_{(s)}\Longleftrightarrow$$ $$-\frac{\text{V}_{\text{in}}(s)}{\text{R}_1}=\text{C}s\text{V}_{\text{out}}(s)+\frac{\text{V}_{\text{out}}(s)}{\text{R}_2}\Longleftrightarrow$$ $$\color{red}{\frac{\text{V}_{\text{out}}(s)}{\text{V}_{\text{in}}(s)}=-\frac{\text{R}_2}{\text{R}_1\left(1+\text{C}\text{R}_2s\right)}}$$

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    \$\begingroup\$ Yes - looks good. However, there is a simpler (and quicker) way for arriving at the transfer function: Calculate from the beginning in the s-domain (without the necessity to perform the Laplace transformation). \$\endgroup\$
    – LvW
    Feb 20, 2016 at 16:34
  • \$\begingroup\$ @LvW Thanks for your response, can you show me that method? \$\endgroup\$ Feb 20, 2016 at 17:10
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    \$\begingroup\$ @JanEerland Here's a brief introduction s-domain circuit analysis. Basically, you can transform R, L, and C into complex impedances in terms of s and then all the math becomes much easier. \$\endgroup\$
    – uint128_t
    Feb 20, 2016 at 23:00

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The solution you have arrived at is correct. The circuit is a practical integrator. The resistor in parallel with capacitor limits low frequency gain and minimizes variations in output. Here is a simpler and quicker solution:

Since the opamp is in inverting configuration, the transfer function is:

$$ Av = \frac{-Z2(s)}{Z1(s)} $$ Note that all impedances are in s-domain. Z2(s) happens to be the parallel combination of R2 and 1/sC $$ Z2(s) =\frac{R2\cdot\frac{1}{sC}}{R2+\frac{1}{sC}} $$ $$ Z1(s) =R1 $$ $$ \frac{vo(s)}{vin(s)} = -\frac{\frac{R2\cdot\frac{1}{sC}}{R2+\frac{1}{sC}}}{R1} $$

Which on simplification reduces to: $$ \frac{vo(s)}{vin(s)} = -\frac{R2}{R1\cdot(1+R2Cs)} = -\frac{\frac{R2}{R1}}{1+R2Cs} $$

So the gain at low frequencies is -R2/R1 which without R2 would have quickly rolled off.

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    \$\begingroup\$ Typo in 4th line: read "inverting" instead of "non-inverting". \$\endgroup\$
    – LvW
    Feb 23, 2016 at 14:02
  • \$\begingroup\$ @LvW Corrected. :) \$\endgroup\$
    – JGalt
    Feb 23, 2016 at 14:07

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