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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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    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..

    Klaus
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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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    Re: 2nd order opamp circuit

    Quote Originally Posted by frankrose View Post
    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.
    I don't think it is a typo. Because at t = 0+ v1(t) = vo(t).



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    Re: 2nd order opamp circuit

    Quote Originally Posted by paulmdrdo View Post
    I don't think it is a typo. Because at t = 0+ v1(t) = vo(t).
    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.

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    Then the values of the components are substituted.

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    And vo, the output voltage is calculated from the system of 3 equations.

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    vo is evaluated as amp goes to infinity

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    Finally, the inverse Laplace is found.

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    Here is a plot of the function.

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    As you can see, the exponential term eventually causes the voltage to go to zero.

    Now, can you calculate the voltage at v1?

    Ratch
    Hopelessly Pedantic


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