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I don't quite understand how you are supposed to get the voltage going to the first op-amp. I do understand how I would get the answer from that point forward.

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  • \$\begingroup\$ Is the second opamp amplifying? \$\endgroup\$ Sep 21 at 19:09
  • \$\begingroup\$ Basically these are just ideal op amps from what I understand. I believe they are called voltage followers? \$\endgroup\$
    – John
    Sep 21 at 19:14
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    \$\begingroup\$ DC? Or are you looking for an AC analysis? \$\endgroup\$ Sep 21 at 19:19
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    \$\begingroup\$ Do you know KVL? You do that, but use the ideal op-amp assumption that the voltage at the inverting and non-inverting terminals since you can see there is negative feedback. You can even skip that in this case since you know the op-amps are both just buffers so Vin+ = Vout for each op-amp. Don't just replace the opamp with a piece of wire though because the currents don't flow through the opamp. Write all the KVL equations out mindlessly and see what comes out of them. \$\endgroup\$
    – DKNguyen
    Sep 21 at 19:23
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    \$\begingroup\$ Look at the derivation of a low pass sallen key filter and make simplifications due to the input op-amp. \$\endgroup\$
    – Andy aka
    Sep 21 at 19:36

3 Answers 3

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The op-amps are configured as voltage followers. Assuming that they're ideal, we have an equivalent circuit as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

The voltages at the input and output of each follower are of course the same, so we got only 3 nodes in the analysis: \$V_{IN},V_1,{\rm and\ }V_{OUT}\$.

The impedances are complex, namely:

$$ Z_1=Z_2=R,\\ Z_3=Z_4=-\frac{j}{\omega C}, $$

where \$\omega=2\pi f\$ is the angular frequency. The impedances do depend on the frequency: that's how we've got AC analysis here.

The circuit simplifies into two voltage dividers, with top and bottom driven from ideal voltage sources:

schematic

simulate this circuit

We assume that the input is an ideal voltage source - otherwise we could add its output impedance into \$Z_1\$. The voltage at node 1 is the voltage on a voltage divider sitting between \$V_{IN}\$ and \$V_{OUT}\$:

$$ V_1=V_{OUT} + (V_{IN}-V_{OUT}) \cdot a, $$ where \$a=\frac{Z_4}{Z_1+Z_4}\$.

The voltage at the output node also follows from a voltage divider:

$$ V_{OUT}=V_{1}\cdot b, $$ where \$b=\frac{Z_3}{Z_2+Z_3}\$.

Substituting \$V_1\$ into the formula for \$V_{OUT}\$, we get

$$\begin{aligned} V_{OUT} &= \left[V_{OUT} + (V_{IN}-V_{OUT}) \cdot a \right] \cdot b\\ V_{OUT} &= V_{OUT}\cdot b + V_{IN}\cdot ab - V_{OUT}\cdot ab\\ V_{OUT}(1-b+ab) &= V_{IN}\cdot ab \\ V_{OUT} &= V_{IN} \cdot \frac {ab}{1+ab-b}. \end{aligned}$$

All that's left is substituting the impedances \$Z_1\ldots Z_4\$ into \$a\$ and \$b\$, and substituting that into the formula for \$V_{OUT}\$. I'll let you do that :)

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In your place, I would proceed in 3 steps :

  1. Assume the voltage for the + input of the first op-amp is a "known" function V1+(t), and use this to compute Vout(t) (it will depend on V1+(t)) 2). Assume Vout(t) known, and use it to compute V1+(t) (hint: V1+(t) depends on Vin, Vout, the resistor on the left (R), and the capacitor on the top (C)).
  2. Now you have 2 equations (Vout(t) = f(Vin(t),V1+(t)) and V1+(t) = g(Vin(t),Vout(t)), with 2 unknowns (Vout(t) and Vin(t)): you should be able to solve for Vout(t).

NB : either you work with differential equations directly, or you can work directly assuming complex voltages in the form A·ejωt. It's up to you (if you learned the second way, it should be easier, it not, then the first one will also work).

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  • \$\begingroup\$ Hi, I understand all these steps and I have actually tried to so it this way, but I don't know how to compute V1+(t). That's where I'm stuck. How should this be done? \$\endgroup\$
    – John
    Sep 21 at 19:40
  • \$\begingroup\$ To compute V1+(t), as there is no current entering the + pin of the op-amp, it is just a complex "voltage divider" between Vin and Vout, with impedences R and 1/(jCw) \$\endgroup\$
    – Sandro
    Sep 22 at 5:32
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I don't quite understand how you are supposed to get the voltage going to the first op-amp.

The voltage ( I will call \$V_{a}\$) at the non-inverting input of the first operational amplifier is the superposition of contributions \$V_{a}\$ from the input and \$V_{a}\$ from the output.

schematic

simulate this circuit – Schematic created using CircuitLab

Using Laplace domain and superposition:

$$V_{a}(s)=V_{in}(s)\frac{Z_{C}}{Z_{R}+Z_{C}}+V_{out}(s)\frac{Z_{R}}{Z_{R}+Z_{C}}\tag{1}$$

Then, through the forward path:

$$V_{out}(s)=V_{a}(s)\frac{Z_{C}}{Z_{R}+Z_{C}}\tag{2}$$

At this point there is a choice:

  1. Substitute for Va and solve for Vout.
  2. Substitute for Vout and solve for Va.

For the second choice:

$$V_{a}(s)=V_{in}(s)\frac{RCs+1}{\left(RCs\right)^{2}+RCs+1}; Z_{C}=\frac{1}{Cs}, Z_{R}=R\tag{3}$$

To find \$V_{out}\$, substitute Equation 2 into equation 3 and solve for \$V_{out}\$.

Transform to sinusoidal domain by substituting \$s=j\omega \$.

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