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I have to find the differential equation which describes the relation between input and output voltage for the circuit below. Per how the RC components are arranged, I simply can not find a differential equation only including \$V_\text{in}\$. All I got so far is the differential equation including the current \$i\$:

$$\frac{dV_\text{out}}{dt} = \frac{dV_\text{in}}{dt} - \frac{di}{dt}(R_1+R_2) - i\left(\frac{1}{C_1} + \frac{1}{C_2}\right) $$

I also couldn't find any similar examples in literature or on the internet. So my question is, am I overseeing something that lets me replace the current terms by \$V_\text{in}\$ terms or is it simply not possible to get such an equation for this circuit (that's what I'm starting to believe right now).

enter image description here

Null
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M. Matthias
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  • What is \$i\$ ? – Hearth Mar 21 '19 at 22:47
  • Oops, didn't realize it wasn't drawn in this circuit. i1 would be the current through R1,C1, i2 through R2,C2. Since i1=i2 I just named them i. – M. Matthias Mar 21 '19 at 22:55
  • Using R1 and R2 ONLY what is the gain? (-R2/R1) now add both caps in series using cap impedance Zc(ω)=1/(jωC) Now what do you get? – Tony Stewart EE75 Mar 21 '19 at 23:00
  • @SunnyskyguyEE75 That would lead me to the transfer function and not the differential equation. – M. Matthias Mar 21 '19 at 23:01
  • Use integrals., since the i*R = Vr( = integral of cap voltage) and current sum=0 at Vin- then convert to derivative – Tony Stewart EE75 Mar 21 '19 at 23:13
  • Because it's an ideal op-amp, you can split the problem into two pieces. First, because it's an ideal op-amp, \$V_- = V_+\$. So \$V_- = 0\$. Second, because it's an ideal op-amp, there is no current flowing into the negative node. So find \$i_1\$ as a function of \$V_{in}\$, and \$i_2\$ as a function of \$V_{out}\$, then equate them. You have two caps, so you should end up with a 2nd-order differential equation. – TimWescott Mar 21 '19 at 23:26
  • @TimWescott Slight correction: Because it's an ideal op-amp _with negative feedback_. If it had no feedback or positive feedback that would not be the case. – Hearth Mar 21 '19 at 23:43
  • @Hearth True. But then, if it had no or positive feedback the universe in which it existed would implode because of the infinite power being created by the infinite voltage at its output. (Well, unless *your* ideal op-amp has rails, but then it wouldn't be *really* ideal, now would it? :) – TimWescott Mar 22 '19 at 00:01
  • Here's what I tried: $$-V_{in}+v_{R1}+v_{c1}=0$$, analogously for Vout R2,C2 $$i_1=\frac{V_{in}-v_{C1}}{R_1}$$ $$i_2=\frac{-V_{out}-v_{C2}}{R_2}$$ which eliminates the i*R terms. Then from $$\frac{V_{in}-1/C_1\int{i dt}}{R_1}=\frac{-V_{out}-1/C_2\int{i dt}}{R_2}$$ I was able to get $$i=a*(\frac{dV_{in}}{dt R_1}+\frac{dV_{out}}{dt R_2})$$ (with $$a=\frac{R_1C_1R_2C_2}{R_1C_1+R_2C_2}$$). Putting this back into the KVL at the output I finally reach $$V_{out}(1+\frac{a}{R_2C_2})+a\frac{dV_{out}}{dt}=-\frac{a}{R_1C_2}V_{in}-a\frac{R2}{R1}\frac{dV_{in}}{dt}$$ – M. Matthias Mar 22 '19 at 00:21
  • It's pretty late here so I'll head to bed now and try verifying this result tomorrow. In case anyone derived it on their own, I'd be happy to hear about it. Anyways, thanks to everyone for your help. – M. Matthias Mar 22 '19 at 00:25
  • correction for a $$a=\frac{R_1C_1R_2C_2}{R_2C_2-R_1C_1}$$ – M. Matthias Mar 22 '19 at 00:29
  • @TimWescott perhaps a 1st order derivative equations for both in and out because the integrated currents are equal but opposite – Tony Stewart EE75 Mar 22 '19 at 01:55

2 Answers2

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As pointed out in the comments, the voltages at the op amp's inputs are zero and no current flows into them. So you have your current \$i(t)\$ going into the input node and flowing out of the output node. This zero voltage between the RC blocks allows you to separate the analysis into two circuits.

The input: $$v_{in}(t) = R_1 i(t) + \frac{1}{C_1}\int i(\tau)d\tau $$ The output: $$v_{out}(t) = -R_2 i(t) - \frac{1}{C_2}\int i(\tau)d\tau $$ Which gives a bit more information than what you wrote: $$i(t)=\frac{v_{in}(t)-\frac{1}{C_1}\int i(\tau)d\tau}{R_1}$$ and $$-\int i(\tau)d\tau = C_2\left(v_{out}(t)+R_2i(t)\right).$$ Then $$i(t)=\frac{v_{in}(t)+\frac{C_2}{C_1}(v_{out}(t)+R_2i(t))}{R_1} $$ So $$i(t)\left(1-\frac{C_2R_2}{C_1R_1}\right)=\frac{v_{in}(t)+\frac{C_2}{C_1}(v_{out}(t))}{R_1} $$ which allows you to write \$i(t)\$ as a function of \$v_{in}(t)\$ and \$v_{out}(t)\$.

xuva
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  • We don't give full answers to homework. – Chu Mar 22 '19 at 01:40
  • @Chu should I delete this? it’s not the full answer, but it delivers the main point i guess – xuva Mar 22 '19 at 01:43
  • @xuva People have differing opinions about completed answers and what quaifies as one. You get to make your own choices. I feel that while a completed answer may hurt the questioner (who is 'successfully cheating' because of it), it can help hundreds more who work such problems in private. Over time, you may find a balancing point where you provide the more difficult aspects that leaves an almost-complete answer. But where the OP (and others who may later encounter your help) still needs to take a few further steps they should be able to achieve. – jonk Mar 22 '19 at 02:32
  • @xuva I don't believe there is a hard, bright line anyone can point to about how much or how little you write. Just write well and accurately and with consideration for others. Listen, but think for yourself and remember that you alone get to make your own choices as you see best fits your own perspective and what and how you want to contribute. – jonk Mar 22 '19 at 02:35
  • @jonk Thanks for the comments. I will keep the answer as I think I’m helping the OP who did significant work in the comments of his post and I still left some work to be done – xuva Mar 22 '19 at 04:20
  • @xuva This place is a community. Not everyone sees things the same way. That's how communities are. Different opinions. So long as we have each others' backs and help push each other towards being better people and learning while we are at it, the remaining differences aren't that important to worry about. You might get dinged (down-voted) from time to time. That is what it is. Someone disagreeing with you. No more. No less. If they choose to comment about why, it's golden. Because then you learn, or they learn, something and we all improve from that. Don't worry too much. Just be yourself. – jonk Mar 22 '19 at 05:36
  • You answer came just when I was typing out my result. However, since I think it'll help people out pretty well if anyone ever finds this thread again, you'll get the accepted answer :) – M. Matthias Mar 26 '19 at 11:12
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your result

\$V_{out}(1+\frac{a}{R_2C_2})+a\frac{dV_{out}}{dt}=-\frac{a}{R_1C_2}V_{in}-a\frac{R2}{R1}\frac{dV_{in}}{dt} ; ~~~~~ a=\frac{R_1C_1R_2C_2}{R_2C_2-R_1C_1}\$

s domain
\$V_{out}/V_{in}=-\dfrac{R_2+\dfrac{1}{sC_2}}{R_1+\dfrac{1}{sC_1}}=\dfrac{sR_2C_2+1}{sR_1C_1+1}\cdot \dfrac{C_1}{C_2} \$ .........(1)

for \$let~V_{out}(t)=y,~~~ V_{in}(t)=x, ~~dy/dt = y'\$

Inverse Laplace of (1) ?? I forget

\$R_2C_1C_1~y'+C_1~y = R_1C_1C_2~x'+C_2~x\$

Tony Stewart EE75
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