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Consider the circuit below. Let \$V_i\$ be switched on at t = 0. We want to find an expression that related the output voltage \$V_o\$ to the input \$V_i\$, in terms of R1, R2, C1, C2.

Here is my attempt. The current through all these elements is the same since the op-amp draws no current:

$$i_{C_1}=i_{C_2} \therefore C_1 \frac{dV_{C_1}}{dt} = C_2 \frac{dV_{C_2}}{dt}$$ $$V_i = V_{C_1} + V_{R_1} \quad , \quad -V_o = V_{C_2} + V_{R_2}$$ $$V_{R_1} = iR_1 \quad, \quad V_{R_2} = iR_2$$

Now if I plug the value of Vc1 back to the equation above, I will have a derivative in terms of V1 (which is what I want) but also in terms of Vr1 (which I don't know). Same for Vc2. So this approach is not correct.

Am I missing something?

enter image description here

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    \$\begingroup\$ As a side note, you don't have a DC path for the non-inverting bias current, so if you were to build this circuit you would probably have a problem with the bias current charging the caps. \$\endgroup\$
    – John D
    Commented Mar 17 at 0:58
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    \$\begingroup\$ Did you mean the inverting input?...@JohnD \$\endgroup\$
    – user319836
    Commented Mar 17 at 2:52
  • \$\begingroup\$ If you want Vo in terms of Vi, R1, R2, C1, C2 then you're asking for a frequency domain answer. But you've posed your question as a time domain problem as if you care only about the step response. Is that what you're really interested in? \$\endgroup\$
    – td127
    Commented Mar 17 at 17:33

1 Answer 1

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Note as allready pointed that there is no DC path ...

Here is what is happening. You see the "shift" of the output ...

enter image description here

Why don't you use "s" variable (or "operator"), without "shift" ...

Here is what this give. Made with Maple sheet.

A low frequency, it give \$Vo,0=C1/C2\$.
At high frequency, it give \$Vo,inf=R2/R1\$.

enter image description here

Here, you can see the effect of limited bandwidth. LF13741 has only a 1 MHz GBW.

enter image description here

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  • \$\begingroup\$ Can you clarify the DC path point? \$\endgroup\$
    – student1
    Commented Mar 17 at 18:29
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    \$\begingroup\$ Both opamp inputs require a small amount of DC current. The non-inverting input in your circuit gets that from ground. But your inverting inputs has capacitors in both paths. DC doesn't flow through capacitors but rather builds up a DC voltage on its plates. This will eventually cause the opamp to be unable to maintain feedback. Solution is to add another resistor across one (or both) of your caps. This will be the current path for DC and low frequencies. At higher frequencies the capacitor conducts and becomes the dominant impedance element. \$\endgroup\$
    – td127
    Commented Mar 17 at 18:39

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