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I'm studying for my circuit analysis exam and I stumbled upon a problem I can't understand. Here it is:

enter image description here

The voltage source is 0V for t<0s and 10V for t>0s, and I have to find the Vout voltage formula for t>0s.

I can't understand what happens as soon as the voltage source gets turned on. I calculated the final Vout voltage using KVL (Vout = -10V, assuming that the voltage in all the capacitors is the same and no current flows through the resistor after a while), so my result would be:

$$ Vout = 10(e^{\frac{-t}{τ}}-1) [V] $$

I know the result is correct but can anyone explain the logic behind this circuit? Also how do I calculate the time constant τ ?

Thanks!

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    \$\begingroup\$ Start by assuming the input RC is tied to ground. What is the input current? (hint, it's an inverse exponential, just like the capacitor voltage - and for the same reason). Once you know the current into the - input, you can calculate the output voltage by integration (which is what the feedback cap will provide). \$\endgroup\$ – WhatRoughBeast Jan 4 '17 at 2:00
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    \$\begingroup\$ It's not a real life circuit - there are vital components missing that would make it a practical circuit. \$\endgroup\$ – Andy aka Jan 4 '17 at 8:32
  • \$\begingroup\$ Do you need to solve it in the time domain, or can you use Laplace transform techniques? \$\endgroup\$ – Matt L. Jan 4 '17 at 11:16
  • \$\begingroup\$ I need to solve it in the time domain, we haven't learned Laplace techniques yet. I tried with @WhatRoughBeast method but I get a wrong result :( \$\endgroup\$ – squeck Jan 4 '17 at 12:29
  • \$\begingroup\$ So, what exactly did you get for the current, and what exactly did you get for the output voltage? "I get a wrong result" is not remotely informative. Edit your answer to show what you did and why. \$\endgroup\$ – WhatRoughBeast Jan 4 '17 at 13:35
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Note that you have an ideal op-amp with zero current into the inputs and with zero voltage between the inputs. Assuming a current \$i(t)\$ through the feedback capacitor (from right to left), we get

$$i(t)=C\frac{dv_o(t)}{dt}\quad\text{and}\quad v_o(t)=\frac{1}{C}\int i(t)dt\tag{1}$$

Since no current flows in or out of the inverting input of the op-amp, the same current also flows through the series connection of \$R\$ and \$C\$ at the input:

$$v_i(t)+Ri(t)+\frac{1}{C}\int i(t)dt=0\tag{2}$$

Plugging \$(1)\$ into \$(2)\$ we get

$$v_i(t)+RC\frac{dv_o(t)}{dt}+v_o(t)=0\tag{3}$$

Using the time constant \$\tau=RC\$, \$(3)\$ can be rewritten as

$$\frac{dv_o(t)}{dt}+\frac{1}{\tau}v_o(t)=-\frac{1}{\tau}v_i(t)\tag{4}$$

The homogeneous solution of \$(4)\$ is

$$v_{o,h}(t)=Ae^{-t/\tau},\quad t>0\tag{5}$$

(with some constant \$A\$ to be determined) and, since \$v_i(t)\$ is constant for \$t>0\$ (equal to \$v_i(0^+)=10V\$), a particular solution is

$$v_{o,p}(t)=-v_i(0^+),\quad t>0\tag{6}$$

The total solution is the sum of the homogeneous and the particular solution:

$$v_o(t)=Ae^{-t/\tau}-v_i(0^+),\quad t>0\tag{7}$$

The constant \$A\$ must be determined from the initial condition \$v_o(0^+)=0\$:

$$v_o(0^+)=A-v_i(0^+)=0\quad\Longrightarrow\quad A=v_i(0^+)\tag{8}$$

which finally gives us the complete solution

$$v_o(t)=v_i(0^+)(e^{-t/\tau}-1),\quad t>0\tag{9}$$

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  • \$\begingroup\$ Thanks a lot! This is very helpful :) However, there's a thing I don't understand. Since the inverting input of the op-amp is 0V, I would assume the voltage drop across the resistor and the input capacitor to be opposite in sign than Vout and the voltage drop across the feedback capacitor. I would have written formula (2) like this $$v_i(t)-Ri(t)-\frac{1}{C}\int i(t)dt=0\tag{2}$$ but this looks strange because the voltage drops don't follow the sign of the i(t) current. Drawing here: i.imgur.com/SdcxO9i.jpg Can you explain what's going on? Thank you so much. \$\endgroup\$ – squeck Jan 4 '17 at 17:12
  • \$\begingroup\$ @squeck: I defined the current to flow from right to left (just like in your drawing), but then the voltages are in the same direction as the input voltage. Your + and - signs for the voltages across the resistor and the capacitor at the input are the wrong way around. \$\endgroup\$ – Matt L. Jan 4 '17 at 20:29

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