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14th April 2019, 09:30 #1
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2nd order opamp circuit
I was trying to understand the way this problem was solved and I got confused with the latter part of the solution. I encircled the part that confused me.
They seem to contradict each other. If dv(0+)/dt = 0 why is it dv(0+)/dt = 1 in the other one? Please explain. TIA!

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14th April 2019, 09:52 #2
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Re: 2nd order opamp circuit
Hi,
I did go in detail through the formulas...
But I think it's because of phase shift by 90° caused by C2.
Right side of C2 is varying signal, but right side is virtual ground.
To ensure virtual ground the Opamp has to regulate..
Since dV/dt on the left side of C2 causes proportional current through C2...the same current will cause a 90° phase shifted voltage at the right side of R2 (the output).
But I'm not sure..
KlausPlease don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.

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14th April 2019, 10:07 #3
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Re: 2nd order opamp circuit
I think in the 1st encircled part there is a typo, it should be dv1(0+)/dt, not dvo(0+)/dt. Then the 2nd encircled part makes sense, it explains why C2 is missing from the expression.
"Try SCE to AUX." /John Aaron/

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14th April 2019, 18:09 #4
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16th April 2019, 15:28 #5
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Re: 2nd order opamp circuit
I will solve this problem using Laplace transforms and taking into consideration the amplification of the opamp. First I write the equations of the circuit. "amp" is the amplification of the opamp which can be several thousand times.
Then the values of the components are substituted.
And vo, the output voltage is calculated from the system of 3 equations.
vo is evaluated as amp goes to infinity
Finally, the inverse Laplace is found.
Here is a plot of the function.
As you can see, the exponential term eventually causes the voltage to go to zero.
Now, can you calculate the voltage at v1?
RatchHopelessly Pedantic
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