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enter image description here

This is my solution so far:

  • Vth would simply be the output of the inverting op amp which is Vth = -(Rf/Ri)Vi
  • Rth would simply be R2 || R3.

schematic

simulate this circuit – Schematic created using CircuitLab

And now we have a simple RC circuit with a DC source. So at t = inf, its right to assume that the capacitor would just charge to the V of its source. Am I right?

And then the step response would be:

  • Using the total response formula of an RC circuit (wouldn't go to all the derivation)

V(t) = (V(0) - V(steady-state))e^(1/RC) + V(steady-state)

  • Tau would just be C x (R2 || R3)

So the step response would look like

V(t) = (0 - Vth)e^(1/(R2||R3)C) + Vth

Did I get this right? I'm having doubts because it's an op-amp and i'm not sure if i'm oversimplifying things.

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Well, we are trying to analyze the following circuit (assuming an ideal opamp):

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} 0=\text{I}_1+\text{I}_2\\ \\ \text{I}_6=\text{I}_2+\text{I}_3\\ \\ \text{I}_3=\text{I}_4+\text{I}_5\\ \\ \text{I}_7=\text{I}_4+\text{I}_5\\ \\ \text{I}_7=\text{I}_1+\text{I}_6 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2-\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2-\text{V}_3}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_3}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_3}{\text{R}_5} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} 0=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_2-\text{V}_1}{\text{R}_2}\\ \\ \text{I}_6=\frac{\text{V}_2-\text{V}_1}{\text{R}_2}+\frac{\text{V}_2-\text{V}_3}{\text{R}_3}\\ \\ \frac{\text{V}_2-\text{V}_3}{\text{R}_3}=\frac{\text{V}_3}{\text{R}_4}+\frac{\text{V}_3}{\text{R}_5}\\ \\ \text{I}_7=\frac{\text{V}_3}{\text{R}_4}+\frac{\text{V}_3}{\text{R}_5}\\ \\ \text{I}_7=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\text{I}_6 \end{cases}\tag3 $$

Now, using an ideal opamp, we know that:

$$\text{V}_+=\text{V}_-=\text{V}_1=0$$

So we can rewrite equation \$(3)\$ as follows:

$$ \begin{cases} 0=\frac{\text{V}_\text{i}}{\text{R}_1}+\frac{\text{V}_2}{\text{R}_2}\\ \\ \text{I}_6=\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_2-\text{V}_3}{\text{R}_3}\\ \\ \frac{\text{V}_2-\text{V}_3}{\text{R}_3}=\frac{\text{V}_3}{\text{R}_4}+\frac{\text{V}_3}{\text{R}_5}\\ \\ \text{I}_7=\frac{\text{V}_3}{\text{R}_4}+\frac{\text{V}_3}{\text{R}_5}\\ \\ \text{I}_7=\frac{\text{V}_\text{i}}{\text{R}_1}+\text{I}_6 \end{cases}\tag4 $$

Now, we can solve for:

  • $$\text{V}_3=-\frac{\text{R}_2\text{R}_4\text{R}_5\text{V}_\text{i}}{\text{R}_1\text{R}_5\left(\text{R}_3+\text{R}_4\right)+\text{R}_1\text{R}_3\text{R}_4}\tag5$$
  • $$\text{I}_5=-\frac{\text{R}_2\text{R}_4\text{V}_\text{i}}{\text{R}_1\text{R}_5\left(\text{R}_3+\text{R}_4\right)+\text{R}_1\text{R}_3\text{R}_4}\tag6$$

Where I used the following Mathematica-code:

In[1]:=V1 = 0;
FullSimplify[
 Solve[{0 == I1 + I2, I6 == I2 + I3, I3 == I4 + I5, I7 == I4 + I5, 
   I7 == I1 + I6, I1 == (Vi - V1)/R1, I2 == (V2 - V1)/R2, 
   I3 == (V2 - V3)/R3, I4 == V3/R4, I5 == V3/R5}, {I1, I2, I3, I4, I5,
    I6, I7, V2, V3}]]

Out[1]={{I1 -> Vi/R1, I2 -> -(Vi/R1), 
  I3 -> -((R2 (R4 + R5) Vi)/(R1 R3 R4 + R1 (R3 + R4) R5)), 
  I4 -> -((R2 R5 Vi)/(R1 R3 R4 + R1 (R3 + R4) R5)), 
  I5 -> -((R2 R4 Vi)/(R1 R3 R4 + R1 (R3 + R4) R5)), 
  I6 -> -((((R2 + R3) R4 + (R2 + R3 + R4) R5) Vi)/(
    R1 R3 R4 + R1 (R3 + R4) R5)), 
  I7 -> -((R2 (R4 + R5) Vi)/(R1 R3 R4 + R1 (R3 + R4) R5)), 
  V2 -> -((R2 Vi)/R1), 
  V3 -> -((R2 R4 R5 Vi)/(R1 R3 R4 + R1 (R3 + R4) R5))}}

In order to calculate the equivalent Thevenin circuit seen by the resistor \$\text{R}_5\$, we can write:

  • For the Thevenin voltage \$\text{V}_\text{th}\$: $$\text{V}_\text{th}=\lim_{\text{R}_5\to\infty}\text{V}_3=-\frac{\text{R}_2\text{R}_4\text{V}_\text{i}}{\text{R}_1\left(\text{R}_3+\text{R}_4\right)}\tag7$$
  • For the Thevenin current \$\text{I}_\text{th}\$: $$\text{I}_\text{th}=\lim_{\text{R}_5\to0}\text{I}_5=-\frac{\text{R}_2\text{V}_\text{i}}{\text{R}_1\text{R}_3}\tag8$$
  • For the Thevenin resistance \$\text{R}_\text{th}\$: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_3\text{R}_4}{\text{R}_3+\text{R}_4}\tag9$$

Where I used Mathematica-code, again:

In[2]:=Limit[-((R2 R4 R5 Vi)/(R1 R3 R4 + R1 (R3 + R4) R5)), R5 -> Infinity]

Out[2]=-((R2 R4 Vi)/(R1 R3 + R1 R4))

In[3]:=Limit[-((R2 R4 Vi)/(R1 R3 R4 + R1 (R3 + R4) R5)), R5 -> 0]

Out[3]=-((R2 Vi)/(R1 R3))

In[4]:=FullSimplify[(-((R2 R4 Vi)/(R1 R3 + R1 R4)))/(-((R2 Vi)/(R1 R3)))]

Out[4]=(R3 R4)/(R3 + R4)

When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

  • $$\text{R}_5=\frac{1}{\text{sC}}\tag{10}$$
  • The input voltage, \$\text{V}_\text{i}\$, is a Heaviside theta function (because of the step response): $$\text{v}_\text{i}\left(\text{s}\right)=\mathcal{L}_t\left[\theta\left(t\right)\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\tag{11}$$

So, we get for \$\text{V}_3\$ (using \$(5)\$):

$$\text{v}_3\left(\text{s}\right)=-\frac{\text{R}_2\text{R}_4\cdot\frac{1}{\text{sC}}\cdot\frac{1}{\text{s}}}{\text{R}_1\cdot\frac{1}{\text{sC}}\cdot\left(\text{R}_3+\text{R}_4\right)+\text{R}_1\text{R}_3\text{R}_4}\tag{12}$$

Using inverse Laplace transform we get:

$$\text{V}_3\left(t\right)=\frac{\text{R}_2\text{R}_4\left(\exp\left(-\frac{\left(\text{R}_3+\text{R}_4\right)t}{\text{C}\text{R}_3\text{R}_4}\right)-1\right)}{\text{R}_1\left(\text{R}_3+\text{R}_4\right)}\tag{13}$$

Using your values we end up with:

$$\text{V}_3\left(t\right)=\frac{5\left(\exp\left(-75 t\right)-1\right)}{3}=\frac{5}{3}\cdot\left(e^{-75 t}-1\right)\tag{14}$$

Where I used Mathematica-code, again:

In[5]:=Vi = 1/s;
R5 = 1/(s*c);
FullSimplify[
 InverseLaplaceTransform[-((R2 R4 R5 Vi)/(
   R1 R3 R4 + R1 (R3 + R4) R5)), s, t]]

Out[5]=((-1 + E^(-(((R3 + R4) t)/(c R3 R4)))) R2 R4)/(R1 (R3 + R4))

In[6]:=R1 = 20000;
R2 = 50000;
R3 = 10000;
R4 = 20000;
c = 2*10^(-6);
FullSimplify[((-1 + E^(-(((R3 + R4) t)/(c R3 R4)))) R2 R4)/(
 R1 (R3 + R4))]

Out[6]=5/3 (-1 + E^(-75 t))
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When analyzing transfer functions, a good way is to break down the system into blocks. The problem is it looks like the analysis for the opamp was done incorrectly The s transfer functions look like this: enter image description here

Then the total equation would be

K1 is orange
K2 is green
\$ V_{out} = K_1K_2V_{in}\$

As s goes to infinity (as c charges) c ends up with the same voltage As r3

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  • \$\begingroup\$ And the voltage at R3 will be Vout, right? Also, what would the relationship be of K1 and K2? And how would they affect the output? \$\endgroup\$
    – user263783
    Sep 27 '20 at 3:32
  • \$\begingroup\$ Yes, the voltage at R3 is Vout. K1 and K2 are just gains (from the pic) \$\endgroup\$
    – Voltage Spike
    Sep 27 '20 at 4:03
  • \$\begingroup\$ Thank you! I understand it now. \$\endgroup\$
    – user263783
    Sep 27 '20 at 4:53

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