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Opamp RC-circuit -- Capacitor time constant calculation

  1. Jan 8, 2017 #1
    1. The problem statement, all variables and given/known data
    for the opamp circuit give in the figure, find Vo(t) for t > 0
    the figure is attached.

    2. Relevant equations
    I know that tau = R * C
    that is the time constant of a capacitor discharging is equal to the product of the resistor that it discharges through and its capacitance.


    3. The attempt at a solution
    The solution is in the file attached, I understand the whole thing. However, when I calculate the time constant my intuition tells me that the capactior discharges through the 100KΩ, the 20KΩ and the 10KΩ resistors. But the solution tells us that the 10KΩ resistor isn't included in the calculation of the time constant.
    My question is why???

    Thanks for the help
     

    Attached Files:

  2. jcsd
  3. Jan 8, 2017 #2

    gneill

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    Staff: Mentor

    Consider the properties of an ideal op-amp. In your circuit, what's the potential at the - input of the op-amp?
     
  4. Jan 8, 2017 #3
    Yes, since it's an ideal op-amp the voltages of the - and + inputs of the opamps would be the same.
    Therefore, the answer to your question would be: the voltage at the negative input would be 0V since the positive input is grounded and they must be equal.
    But where does that lead me...?
    Thank you sir, I really appreciate it
     
  5. Jan 8, 2017 #4

    gneill

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    Staff: Mentor

    The feedback circuit "sees" the "-" terminal as a virtual ground (fixed at 0 V). Essentially it can't "see" anything past that connection, as it's effectively "shorted out" by the virtual ground.
     
  6. Jan 8, 2017 #5
    But beyond that 0V there's a resistor of 10KΩ with 4V on the other end.
    Doesn't that count for anything? I mean because a voltage difference exists shouldn't that incur a current somehow?
     
  7. Jan 8, 2017 #6

    gneill

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    Staff: Mentor

    That fixed 0 V means that the source current cannot be influenced in any way by what happens to the capacitor voltage. The feedback circuit only sees the (virtual) ground connection. Similarly, the input voltage source only sees the same virtual ground (hence the input current is fixed by voltage source and input resistor only).

    Remember how when you want to determine the Thevenin resistance for a network you suppress fixed voltage sources by replacing them with a short circuit? The fixed input voltage at the - terminal becomes a short to ground.
     
  8. Jan 8, 2017 #7
    Alright cool, I somehow got it
    Thanks sir
     
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