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Homework Help: Op-amp circuit peak current and swtiched circuit transient

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data
    I have uploaded the questions in the attachment. I am having trouble with 2. e,f and 3.c
    untitled.png is the parts that I am having trouble with. untitled1.png and untitled2.png is the entire question.

    2. Relevant equations
    y(t)= steady state+transient

    3. The attempt at a solution
    For part E I have tried to change x(t) into a phasor and so according to the question x(t) becomes A. Then letting X=A, and then subbing in ω=1/RC I got Y/X=2, then finally Y=2A...
    The problem I have is I don't know how to get from that part to calculating the peak current. There was an earlier part of the question which asked me to derive the transfer function, and I did so by writing out the KCL equations for V1,V2,V3 using a common voltage V and assuming V+=V- and that the op-amps are ideal. I wonder if the approach to this question is using the KCL equations for the V1/V2/V3 nodes and subbing in X=A for them? But then I don't know what to do with the unknown R and C

    for part f I am confused because I worked out the corner frequency earlier to be ω=1/RC and so using the transfer function, where Y/X=2 when you sub in ω=1/RC and on the right I sub in ω=2∏1000. I attempted to separate imaginary and real parts so that I can find the real parts to be the R value and imaginary the capacitance, but then in the end the terms with RC cancel out and I'm left with something that doesn't make sense.

    Then for part C I earlier worked out the steady states for both open and closed. The steady state for open was y(t)=2.8cosωt+0.89sinωt and for closed it was 4.49cosωt+2.8sinωt. When the transient amplitude is equal to zero, that means when the the y(t0) is equal to the steady state when the circuit is closed right? So I now have y(t0)=4.49cosωt0+2.8sinωt0+Ae^(-t0/τ) where τ is the time constant and A here is the transient amplitude, which is equal to zero. But since I don't know what t0 is how can I find out what y(t0) is?


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    Last edited: Sep 1, 2012
  2. jcsd
  3. Sep 2, 2012 #2

    rude man

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    2e: you need to compute the remaining unknown node voltage W. Then you can write the KVL equations & solve for op ap output currents.

    2f: obviously, you need to do 2e before doing 2f. You know RC but what R and C are separately is determined by your op amp output capability.

    3c: toughie! First, you know what Y is for t=0-. A constant in other words. So now you have an applied voltage of sin{w(t-t0)}, t0 unknown. So your problem is to use this input, suitably transformed, with a known initial condition on Y, and compute the output y(t). For some (minimum) value of t0 you will see that there is no exponential decay term, or rather that the coefficient before the exponential decay term is zero.

    This is an exercise in understanding Laplace transforms and including initial conditions in the reactive components (in this case a capacitor). I don't believe this problem can be solved using combinations of steady-state solutions as you have attempted. I'll give it some more thought though.

    BTW 3d is a similar problem.
  4. Sep 3, 2012 #3
    Thank you for your help. I think I have also approached 3B wrongly as well, since I basically ignored the part of the question where it said "the switch has been open for a long time and is closed at t=0" since I thought it was irrelevant since it asked for y(t) for t>=0. So is the right way to use the steady state waveform for the opened switch and sub in 0 and let it equal to the steady state when the switch is closed+transient? so it becomes A=2.8-4.49
  5. Sep 3, 2012 #4

    rude man

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    3b is solved by laplace transforms. You have an input vin(t) = Asin(wt) and a transfer function H(s) = 1/(Ts+1) so in the absence of an initial condition,
    Vout(s) = Vin(s)H(s).

    But you also have an initial condition = vout(0+) due to the phase shift between vin and vout before the switch was switched on.

    Here is how I like to handle initial conditions like this one:

    1. Inverse-transform your H(s) back to the time domain: in this case,
    Vout(s)/Vin(s) = 1/(Ts+1) so (Ts+1)Vout(s) = Vin(s) and the corresponding diff. eq. is
    Td(vout)/dt + vout = vin.

    2. Laplace-transform that equation per the derivative rule: L{df/dt} = sF(s) - f(0+) where F(s) = L{f(t)}. Now you have a trasform equation with the initial condition on vout included and you can inverse-transform each term of Vout(s) to get the time response at the output.

    I don't want to do the detailed steps for you, that's not allowed, but if you give me your answer to 3b I will tell you if it's mine also.

    BTW notice I use lower-case for time variables and upper-case for transformed variables.
  6. Sep 3, 2012 #5
    but I didn't learn laplace transforms yet...anyway the answer I got to 3B was y(t)= 4.49cosωt+2.8sinωt-1.6e^(-t/τ)
  7. Sep 3, 2012 #6

    rude man

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    OK, gotta go right now but will check your answer and also try to figure out how that's doable without Laplace! If your course hasn't covered Laplace yet then there must be a way ...

    Oh, of course it can also be done by classical diff. eq. solving - did your course cover that way maybe?
  8. Sep 4, 2012 #7

    rude man

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    OK, mine was similar in form but different numbers.

    According to your answer your initial condition on C (just before the switch is turned ON) is 4.49 -1.6 = +2.89V. However, the i.c. has to be negative since the output phase is lagging the input in the steady-state, and the input is of course zero at t=0.

    I got -2.85V fot the i.c. which is {10/√(1 + w2T2)}sin{tan-1(wT)} with w = 2000 and T = 1.6e-3. I computed -1.27 rad for the lagging phase angle.

    I also used Excel and did some double-checking, including verifying that vout(0)= i.c. as it must from physical reasoning (when the switch is switched ON there is no vin so the only voltage at the output should be the i.c.

    Make sense, any of this?

    I am looking into how this could be solved without the Laplace transform & will let you know.
  9. Sep 4, 2012 #8

    rude man

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    Update: yes you CAN do this without Laplace - and it's quite straight-forward. Funny how we tend to do things the obvious if messy way without stopping to think of easier ones.

    So: 1. Find Vout for case 1 at t=0 (just BEFORE the switch is closed). Let that Vout = v1.

    2. Compute Vout for the case where the switch has been CLOSED a long time. Find Vout(t=0) for this case also = v2.

    3. Then the answer is the same as what it would be had the switch been closed a long time , except there is an additional term vdelta*e-t/τ where vdelta = v1 - v2 and τ is the post-closing time constant, i.e. 10K*80nF.

    I have checked both ways & got the same answer. The rationale for this solution is superposition: any imbalance in the initial condition acts independently of the driving function, decaying as e-t/τ.
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