Opamp output voltage

Discussion in 'Homework Help' started by liviu93, Nov 15, 2015.

  1. liviu93

    Thread Starter New Member

    Nov 15, 2015
    1
    0
    Hello to all,
    I have the following circuit
    StaticShot_15-11-2015_15-41-40.png
    and I need to find [​IMG].
    Because this is a non-inverting opamp [​IMG] = [​IMG](R2/R1 + 1).
    https://en.wikipedia.org/wiki/RC_circuit#Parallel_circuit says :
    "the output voltage [​IMG] is equal to the input voltage [​IMG]" .
    So, Uin = Uin+.
    is that correct?
    Thanks in advance .
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Yes and No, it all depend on input signal frequency. If input signal is a DC voltage (F = 0Hz) than yes Uin = Uin+
    But for example if input signal has a frequency equal to F = 1/(2 * pi *R3 * C2) ≈ 0.16/(R3 *C2) = 0.16/(1kΩ*1μF) = 160Hz then Uin+ = 0.707*Uin Because this RC circuit at the input act just like a low pass filter - voltage divider whose attenuation (gain) depends on the input signal frequency.
    https://en.wikipedia.org/wiki/RC_circuit#Series_circuit
     
    liviu93 likes this.
  3. hsazerty2

    New Member

    Sep 25, 2015
    22
    1
    This is a non-inverting amplifier, so that's right, we have:

    Uout=(Uin+)*(1+R2/R1)

    Next is to find a relation between Uin+ and V2; It's an RC circuit, with the capacitor charging from 0v (if we assume zero initial condition like i guess) to 1V. The equation is then:

    Uin+=(V2)*(1-Exp(-t/(R*C)))

    So finally:

    Uout=(1+R2/R1)*(V2)*(1-Exp(-t/(R*C)))

    I guess that's it.
     
  4. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,347
    Hello,

    @hsazerty2
    In this home work forum, you do not give complete solutions, but hints to get to the solution.
    The topic starter will not learn much from complete solutions.
    When they have to do some work on their own, the learned stuff will be better understanded.

    Bertus
     
  5. hsazerty2

    New Member

    Sep 25, 2015
    22
    1
    Ooops, sorry.
     
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